Mathematics Practice Book Class 9.pdf

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CLASS

9

Pearson IIT Foundation Series Mathematics Practice Book First Edition

Trishna Knowledge Systems

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Copyright © 2018 Pearson India Education Services Pvt. Ltd Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128, formerly known as TutorVista Global Pvt. Ltd, licensee of Pearson Education in South Asia. No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 978-93-528-6772-1 eISBN 978-93-530-6193-7 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector 16, Noida 201 301, Uttar Pradesh, India. Registered Office: 4th Floor, Software Block, Elnet Software City, TS 140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai - 600 113, Tamil Nadu, India. Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: [email protected]

Contents Prefacevii How to Use the Practice Book ix Series Chapter Flow x

Assessment Test III Assessment Test IV Answer Keys

5.7 5.10 5.12

CHAPTER 1 Coordinate Geometry and Trigonometry

CHAPTER 6 Number Systems

6.1

Assessment Test I Assessment Test II Assessment Test III Assessment Test IV Answer Keys

1.1 1.1 1.3 1.5 1.7 1.9

CHAPTER 2 Geometry-I and Geometry-II

2.1

Assessment Test I Assessment Test II Assessment Test III Assessment Test IV Answer Keys

2.1 2.4 2.7 2.11 2.14

CHAPTER 3 Linear Equations and Inequations

3.1

Assessment Test I Assessment Test II Assessment Test III Assessment Test IV Answer Keys CHAPTER 4  Logarithms

Assessment Test I Assessment Test II Answer Keys CHAPTER 5 Mensuration

Assessment Test I Assessment Test II

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3.1 3.3 3.5 3.7 3.9 4.1 4.1 4.3 4.4 5.1

Assessment Test I Assessment Test II Answer Keys CHAPTER 7 Number Systems and Logarithms

Assessment Test I Assessment Test II Answer Keys

6.1 6.4 6.5 7.1 7.1 7.3 7.4

CHAPTER 8 Polynomials and Square Roots of Algebraic Expressions 8.1

Assessment Test I Assessment Test II Assessment Test III Assessment Test IV Answer Keys

8.1 8.3 8.5 8.7 8.9

CHAPTER 9 Statistics and Probability; Applications of Percentages 9.1

Assessment Test I Assessment Test II Assessment Test III Assessment Test IV Answer Keys

9.1 9.4 9.6 9.9 9.11

Hints and Explanations 

A.1

5.1 5.4

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Preface Pearson IIT Foundation Practice Book Series is designed to accompany the course-books available in this series. Developed by a team of experienced faculties, this workbook series connects the subjective knowledge to its real world applications through various text and chapter level problems. Each chapter has a set of assessment tests which are mapped to chapters covered in the course-book. These worksheets will guide students step-by-step towards understanding the central concept of that particular chapter. These tests are recommended as after class material for further practice. Any suggestions for added or updated additional readings would also be welcome. Students can share their feedback at [email protected].

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How to Use the Practice Book Many times, students face significant challenges in answering application level questions in Physics, Chemistry and Mathematics. These Practice Books will enhance their problem-solving skill which will definitely lead to a strong subject foundation. The entire practice book series are recommended to be used alongside IIT Foundation course-books. Students can refer the following steps while using the practice books:

Study chapters from IIT Foundation Course-books Complete remaining Assessment Tests

Check self progress

Solve the chapter-end exercises

Solve one Assessment Test related to the chapter

While preparing for Foundation courses, students need to learn the fundamental concepts with utmost clarity. In order to successfully complete the IIT Foundation course, one must prepare profoundly. Consistent hard work, practice and perseverance are needed throughout the year. During any competitive examination, one must exercise clinical precision with speed since the average time available to respond to a question is hardly a minute. The aspirants should be conceptually excellent in the subject owing to the negative marking in the examination. A better practice to solve the paper would be to go for the easiest questions first and then gradually progress to the more complicated ones. Regular practice of MCQs will assist the aspirants in preparing for the examination. In a nutshell, hard work, conceptual clarity and self-assessment are the essential ingredients to achieve success in competitive examinations. IIT Foundation course-books play an important role in understanding the concepts. Student need to read-up on all concepts/theories in a regular and systematic manner.

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Course-book Chapter Flow Class 7 Expressions and Special Products

Indices

1

5

3 2

4 Geometry

Ratio and Its Applications

Number Systems

Equations and their Applications

Statistics 10

8

6 7

9 Set Theory

Formulae

Mensuration

Class 9

Linear Equations and Inequations

Logarithms 3

1

5

2

4 Polynomials and Square Roots of Algebraic Expressions

Number Systems

Probability

Geometry

Quadratic Expressions and Equations

11

9

7

10

12 Banking and Computing

6

8 Statistics

Matrices Percentages, Profit and Loss, Discount, and Partnership

Locus

Mensuration

17

15

13

Trigonometry

Coordinate Geometry Time and Work 23

Sales Tax and Cost of Living Index

Ratio, Proportion and Variation 21

22

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18

16

14

Time and Distance

Sets and Relations

Significant Figures

19 20

Shares and Dividends

Simple Interest and Compound Interest

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xi

Course-book Chapter Flow

Class 8 Squares and Square Roots and Cubes and Cube Roots

Polynomials, LCM and HCF of Polynomials

1

5

3 2

Real Numbers and LCM and HCF

4 Formulae

Indices Simple Interest and Compound Interest

Percentages 8

10

6

9

7 Profit and Loss, Discount and Partnership

Time and Work, Pipes and Cisterns Linear Equations and Inequations

Ratio, Proportion and Variation Geometry

Statistics 13

11

17

15

12

16

14 Sets

Time and Distance

Mensuration

Matrices

Class 10 Quadratic Equations and Inequalities

Polynomials and Rational Expressions

Sets, Relations and Functions

3

1

5 4

2

6

Linear Equations in Two Variables

Number Systems Mensuration

Statements

13

11

14

9

7 8

10

12 Geometry

Trigonometry

Matrices

Statistics

Remainder and Factor Theorems

Limits

Mathematical Induction and Binomial Theorem

Permutations and Combinations

Linear Programming 17

15

19

16

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Banking 23

25 24

26 Logarithms

Computing

Instalments

Partial Fractions 27

20

18 Modular Arithmetic

Coordinate Geometry

Progressions

Shares and Dividends

21 22

Taxation

Probability

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Coordinate Geometry and Trigonometry

1

Reference: Coursebook - IIT Foundation Maths Class 9; Chapters - Coordinate Geometry; Trigonometry; pp. 14.1−14.20; 16.1−16.13

Assessment Test I

Time: 30 min. Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following steps are involved in finding the area of a triangle formed by the line 6x − 5y − 30 = 0 with the coordinate axes. Arrange the steps in sequential order. 1 (A) Area of the triangle formed = product of the intercepts . 2 (B) 6x − 5y = 30 x y (C) Compare the equation with + = 1. a b (D) x-intercept is 5 and y-intercept is −6. y x (E) + =1 5 ( − 6) (a) ACBDE (c) BECDA

(b) BCDEA (d) BECAD

2. Find the distance between the points (a, 2a) and (3a, 4a) (in units). (a) 8a (b) 3a (c) 2 2 a (d) 2 3 a 3. Find the equation of a line parallel to 3x − 5y = 7 and passing through the point (5, 3). (a) 8x − 5y = 25 (b) 9x − 15y = 0 (c) 3x = 7y − 6 (d) 3x − 5y = 20 4. Find the equation of a line passing through the points (2012, 2014) and (2013, 2015). (a) x − y + 2 = 0 (b) x + y = 4026 (c) 2012x + 2015y = 4027 (d) 4025x = 4029y 5. A circle is passing through the origin O(0, 0) and intersects the x-axis and the y-axis at A(a, 0) and B(0, b), respectively, where a and b are integers. How many circles can we draw such that the diameter of the circle is 15 units? (a) Only one circle (b) Eight circles (c) Four circles (d) Infinite circles

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1.2

Chapter 1 Coordinate Geometry and Trigonometry

6. Find the area of a triangle formed by the lines x − 6 = 0, y + 8 = 0, and 4x − 3y = 24. (a) 24 sq. units (b) 48 sq. units (c) 12 sq. units (d) 10 sq. units

Space for rough work

7. Find the equation of the median AD of a triangle whose vertices are A(0, 4), B(0, 0), and C(8, 0). (a) x + y = 12 (b) x − y = 4 (c) 4x + 3y = 5 (d) x + y = 4 8. Find the points of intersection of the diagonals of a parallelogram ABCD formed by the vertices A(3, 5), B(3, k), C(m, − 4), and D(−2, 2).  1 1 (a) (1, 1) (b)  ,   2 2 1  (d)  , 1 3 

(c) (2, −2)

3 5 and cos B = , then find the value of sin (A + B). 5 13 63 33 (a) 1 (b) (c) 65 65 sin 30° × cos 40° × tan 50° × cot 60° 10. Find the value of . cosec 60° × sin 50° × cot 40° × sec 30° 9. If sin A =

(a)

3 2

(b)

3 4

(c)

3 6

(d) 0

(d)

3 8

11. If sin q + cosec q = 2, then find the value of sin2014q − cosec2013q. (a) 1 (b) 2012 (c) 2 (d) 0 Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A 12. sin

pc 4

13. sec 54° × sin 36° 14. tan

pc pc × cot 3 4

15. cos 100g

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Column B pc 4 pc (b) cos 4

(a) tan

(c) cot

pc 6

(d) cot

pc 2

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Assessment Test II

Assessment Test II

Time: 30 min.

1.3

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1.  The following steps are involved in finding the value of sin a, if sec a + tan a = 2. Arrange the steps in sequential order. 1 (A) sec a + tan a = 2 ⇒ sec a − tan a = 2 2 2 (B) Recall the formula sec q − tan q = 1. (C) (sec q + tan q ) and (sec q − tan q ) are reciprocals to each other. (D) (sec q − tan q ) (sec q + tan q ) = 1 (E) By adding the reciprocals, we get sec a and cos a and hence sin a. (a) ABCDE (b) BCDAE (c) BDCAE (d) BCADE

2.  What will be the radius of a circle if the end points of its diameter are A(2013, 2014) and B(2011, 2012) (in units). (a) 2012 (b) 2 (c) 1 (d) 2 2

3.  Which of the following is a line perpendicular to 2x + 3y = 7 and passing through (2000, 2013). (a) 2x − 3y = 1947 (b) 3x + 2y = 1974 (c) 3x − 2y = 1947 (d) 3x − 2y = 1974

4.  Find the equation of a line passing through the point (2012, 2013) and having slope 1. (a) x + y = 4025 (b) x − y = 1 (c) x + 2012 = 2014 (d) x + 1 = y





5.  A is a point on the x-axis, B is a point on the y-axis, and O is the origin. If A and B are equidistant from O, then find the number of circles that can be drawn such that the circle should pass through O, A, and B. (a) Only one circle (b) Four circles (c) Eight circles (d) Infinite circles

6.  Find the perimeter of a triangle formed by the lines x + y = 2, x − 5 = 0, and y − 3 = 0. (a) 18 units (b) 18 2 units (c) 6(2 + 2 ) units (d) 6(1 + 2 ) units

7.  Find the equation of the perpendicular bisector of the side AB of DABC whose two vertices are A(−2011, −2011) and B(2011, −2011). (a) x = 0 (b) y = 0 (c) y = x (d) y = −x



8.  Find the point of intersection of the diagonals of a rhombus PQRS, whose vertices are P(2000, 0), Q(x, y), R(−2012, 0), and S(a, b).

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1.4

Chapter 1 Coordinate Geometry and Trigonometry (a) (0, 0) (c) (−6, 0)

(b) (−2006, 0) (d) (6, 0)

4 8 and cos β = , then find the value of tan(a + β ). 5 17 − 84 − 32 81 (a) (b) (c) 17 13 13 tan 20° × cot 40° × sec 60° × co sec 80° 10. Find the value of . sec 10° × cosec 30° × tan 50° × cot 70°

Space for rough work

9. If cos a =

(a) 1 11. If tanq  + cotq  = (a) 32012

(b) 2 4 3

(c)

3 2

(d) 1

(d)

2 3

, then the value of tan2014q can be _______. (b) 31007

(c) 1

(d) 0

Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B (Value of q )

2p c 3

(a) 30°

13. q = 200g

(b) 45°

14. cos q = sin q

(c) 54°

p = cot q 5

(d) 90°

12. q =

15. tan

(e) 120° (f) 180°

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Assessment Test III

Assessment Test III

Time: 30 min.

1.5

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options.

1.  If sin (A − B) =

1 1 and cos (A + B) = , then find A and B. 2 2

The following steps are involved in solving the above problem. Arrange them in sequential order.

(A) 2A = 90 ⇒ A = 45 (B) A = 45 ⇒ B = 15 1 1 (C) sin (A − B) = ⇒ sin (A − B) = sin 30° and cos (A + B) = ⇒ cos (A + B) 2 2 = cos 60° (D) A − B = 30° and A + B = 60° (a) DCBA

(b) CDAB

(c) CADB

2.  If the point (−x, −y) ∈Q2, then (−x, y) belongs to _____. (a) Q1 (b) Q2 (c) Q3 3.  The value of 200g in circular measure is _______. 5p c 3p c (a) (b) (c) p c 2 4

(d) DCAB (d) Q4



(d) 2p c

4.  Find the coordinates of the point on the x-axis which is at equidistant from (2, 4) and (3, 2). 5   −5  (b) (a)  , 0  , 0 2 2 7   −7  , 0 (c) (d)  , 0   2 2



5.  The length of the minute hand of a wall clock is 30 cm. Find the distance covered by its tip in 42 minutes. (a) 66 cm (b) 130 cm (c) 132 cm (d) 264 cm 1   1   2 6.  The value of cot a  1 −  1+  is ______.  cosa   cosa 

(a) −1

(b) 0

(c) 1

(d) 2

7.  Find the equation of the line parallel to 3x − 2y + 2015 = 0 and passing through (1, 1). (a) 3x − 2y − 2 = 0 (b) 2x − 3y + 1 = 0 (c) 3x + 2y − 1 = 0 (d) 3x − 2y − 1 = 0

8.  If tanq  + cotq  = 2, then find the value of tan2015q + cot2015q. (a) 1 (b) 0 (c) 2

(d) −1

9.  Find the relation obtained by eliminating ‘q  ’ from the equation x = cosq + sinq and y = sinq − cosq.

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1.6

Chapter 1 Coordinate Geometry and Trigonometry (a) x2 + y2 = 2 (c) x2 − y2 = 2

(b) x2 + y2 = 22 (d) x2 − y2 = 22

Space for rough work

10. Find the area of the triangle formed by the line 3x + 4y − 24 = 0 and the coordinate axes (in sq. units). (a) 24 (b) 36 (c) 48 (d) 60 1 11. The slope of the line joining the points (2k + 1, 2k + 2) and (3k + 6, k + 10) is . 2 Find the value of k. 5 2

(a)

(b)

12 5

(c)

7 9

(d)

11 3

12. log sin30° + log sin45° + log sin60° = ______  3 (b) log    32  1  3 (d) log    12  4

(a) 0 1  3 log    32  2

(c)

13. Which of the following points is collinear with the points (4, −5) and (6, −7)? (a) (−8, 9) (b) (8, −9) (c) (7, −8) (d) (−8, 7) 14.

1 + tan q 1 − tan q + = ______ 1 − tan q 1 + tan q

(a) 2 cos2q

(b) 2 sin2q

(c) 2 tanq

(d) 2sec2q

15. Match the values of Column A with those of Column B. Column A

Column B

(i) sin 15°

(A) 2 − 3

(ii) cos 15°

(B) 2 + 3

(iii) tan 75°

(C)

3 −1 2 2

(iv) cot 75°

(D)

3 +1 2 2

(a) (b) (c) (d)

(i) → (A); (ii) → (B); (iii) → (C); (iv) → (D) (i) → (C); (ii) → (D); (iii) → (A); (iv) → (B) (i) → (B); (ii) → (A); (iii) → (D); (iv) → (C) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)

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Assessment Test IV

Assessment Test IV

Time: 30 min.

1.7

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1.  Find the third vertex of DABC, if two of its vertices are A(−2, 3), B(4, 5), and its centroid is G(1, 2). The following steps are involved in solving above problem. Arrange them in sequential order. y+8 x+2 (A)  = 1, = 2 ⇒ x = 1, y = −2 3 3 (B) Let C(x, y) be the third vertex, given centroid of DABC = G(1, 2). (C) The third vertex is (1, −2).

 −2 + 4 + x 3 + 5 + y  , (D)    = (1, 2) 3 3

(a) BADC

(b) ADCB

(c) BCAD

(d) BDAC

2.  If the point (x, y) ∈Q2, then (−x, y) belongs to _________. (a) Q1

(b) Q2

(c) Q3

(d) Q4

3.  The value of 75° in circular measure is ______. pc 5p c 3p c 7p c (a) (b) (c) (d) 12 12 12 12

4.  Find the coordinates of the point on the y-axis which is at equidistant from (2, 4) and (3, 5).  7  −7  (b) (a)  0 ,    0 , 2 2



(c) (0, 7)

(d) (0, −7)

5.  The length of the minute hand of a wall clock is 15 cm. Find the distance covered by its tip in 28 minutes. (a) 22 cm (b) 44 cm (c) 66 cm (d) 88 cm 1   6.  The value of  sin a −   sin a 

(a) −1

(b) 0

1  1     is ______.  cos a −   tan a + tan a  cos a  

(c) 1

(d)

−1 2

7.  Find the equation of the line perpendicular to 4x − 5y − 2016 = 0 and passing through (1, −1). (a) 5x + 4y + 1 = 0 (b) 4x − 5y − 1 = 0 (c) 5x + 4y − 1 = 0 (d) 5x − 4y − 1 = 0 4 4 4 8.  If sec q + tan q = , then the value of tanq ·secq is _____. 3 1 1 1 1 (b) ± (c) (d) ± (a) ± ± 3 3 8 6

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1.8

Chapter 1 Coordinate Geometry and Trigonometry

9. Find the relation obtained by eliminating ‘q  ’ from the equation x = 1 + a sinq, y = 1 + a cosq. (a) (x + 1)2 + (y + 1)2 = a2 (b) (x − 1)2 + (y − 1)2 = a2 (c) x2 + y2 = a2 (d) x2 − y2 = a2

Space for rough work

10. Find the area of the triangle formed by the line 4x − 5y − 20 = 0 with the coordinate axes (in sq. units). (a) 50 (b) 30 (c) 10 (d) 40 9 11. The slope of the line joining the points (2k + 6, 3k + 5) and (5k − 5, 6k − 4) is . 8 Find the value of k. (a) 11 (b) 17 (c) 9 (d) 5 12. log tan 30° + log cot 30° + log cosec 45° = ______ (a) log 2

(b) 2 log 2

(c)

3 log 2 2

(d)

1 log 2 2

13. Which of the following points is collinear with the points (1, 3) and (3, 7)? (a) (6, 5) (b) (3, 2) (c) (2, 4) (d) (2, 5) 14.

1 + tan q 1 − tan q − = ______ 1 − tan q 1 + tan q

(a) 2 tan2q

(b) 2 cot2q

(c) 2 cosec2q

(d) 2 cos2q

15. Math the values of Column A with those of Column B. Column A

Column B

(i) (4, −3), (2, −5)

(A) x + y − 7 = 0

(ii) (4, 3), (2, 5)

(B) 4x + y − 13 = 0

(iii) (4, −3), (2, 5)

(C) 4x − y − 13 = 0

(iv) (4, 3), (2, −5)

(D) x − y − 7 = 0

(a) (b) (c) (d)

(i) → (D); (ii) → (A); (iii) → (C); (iv) → (B) (i) → (B); (ii) → (D); (iii) → (A); (iv) → (C) (i) → (D); (ii) → (A); (iii) → (B); (iv) → (C) (i) → (B); (ii) → (C); (iii) → (D); (iv) → (A)

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Answer Keys

1.9

Answer Keys Assessment Test I 1. (c) 11. (d)

2. (c) 12. (b)

3. (b) 4. (a) 13. (a) 14. (c)

5. (b) 6. (a) 15. (d)

7. (d)

8. (b)

9. (b)

10. (d)

5. (d) 15. (c)

7. (a) 8. (c) 9. (b) 10. (a)

Assessment Test II 1. (c) 11. (b)

2. (b) 12. (e)

3. (d) 13. (f)

4. (d) 14. (b)

6. (c)

Assessment Test III 1. (b) 11. (d)

2. (c) 12. (c)

3. (c) 4. (d) 5. (c) 13. (b) 14. (d) 15. (d)

6. (c) 7. (d)

8. (c)

9. (a)

10. (a)

7. (c) 8. (d) 9. (b)

10. (c)

Assessment Test IV 1. (d) 11. (c)

2. (a) 12. (d)

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3. (b) 13. (d)

4. (c) 14. (a)

5. (b) 6. (c) 15. (c)

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Geometry-I and Geometry-II

2

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Geometry; pp. 12.1−12.39

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1. The following are the steps involved in proving a theorem. Arrange them in sequential order. (A) Proof (B) Given (C) Construction (D) Required to prove (a) ABCD (b) BCDA (c) BACD (d) BDCA 2. A chord subtends an angle 70° at the centre of a circle. Find the angle subtended by the chord in the minor segment of the circle. (a) 140° (b) 145° (c) 70° (d) 110° 3. In a DABC, D and E are the points on the sides AB and AC, and D is the midpoint of AB. DE is parallel to BC. F is the midpoint of AE and G is a point on AD such that GF is parallel to DE. If BC = 12 cm, then what is the length of GF in cm? (a) 2 (b) 3 (c) 4 (d) 6 4. ABCD is a trapezium inscribed in a circle and PQRS is the quadrilateral formed by joining the midpoints of AB, BC, CD, and DA in order. If AC = 15 cm, then find the perimeter of quadrilateral PQRS (in cm). (a) 20 (b) 25 (c) 30 (d) 45 5. I is the incentre of a right triangle ABC. If AB = BC, then find the measure of ∠AIC. (a) 120° (b) 135° (c) 140° (d) 115° 6. In a rhombus KLMN, the diagonal KM is equal to the side of the rhombus. Find the measure of ∠LMN. (a) 120° (b) 90° (c) 60° (d) 30° 7. VXYZ is a parallelogram and A is a point on VZ. If AYZ is an equilateral triangle and VZ = 2YZ, find ∠XAY. (a) 90° (b) 110° (c) 120° (d) 100°

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2.2

Chapter 2  Geometry-I and Geometry-II

8. I n the given figure, ‘O’ is the centre of the circle. DABC and DODE are equilateral triangles and AB is parallel to DE. Find ∠AED + ∠ABD + ∠ACD.

Space for rough work

C O B

A E

D



(a) 30°

(b) 60°

(c) 75°

(d) 45°

9. ABCDE is a pentagon, inscribed in a circle with centre ‘O’, CD = DE and AD = BD. ∠EAD = 30° and ∠BOC = 80°. Find ∠DAO + ∠BAO. D E

C O B

A



(a) 60°

(b) 70°

(c) 90°

(d) 110°

10. In the given figure, ABCD is a quadrilateral inscribed in a circle and BD is a diagonal passing through the centre of the circle ‘O’. Find ∠BOC + ∠AOD. C D

50°

20°

A



(a) 140°

O

(b) 100°

B

(c) 120°

(d) 180°

11. In the given figure, the two circles intersect at A and B. If AB is equal to the radius of the circle with centre C2 and it is equal to 6 cm and radius of the circle with centre C1 is 4 cm, then find the distance between the centres (in cm). A

C1



C2 B

(a) 7 + 3 3

(b) 7 3+3



(d) 7 + 3

M01_IIT-CH02.indd 2

(c) 3 10

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Assessment Test I Direction for questions 12 to 15: Match the statements of Column A with those of Column B. Column A

2.3

Space for rough work

Column B

12. An excircle and the circumcircle of a triangle.

(a) Incircle

13. A circle drawn by using the bisector of internal angles of a triangle.

(b) Circumcircle

14. A circle drawn by using the perpendicular bisectors of the sides of a triangle.

(c) Excircle

15. The incircle and the circum (d) Concentric circles circle of an equilateral triangle.

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2.4

Chapter 2  Geometry-I and Geometry-II

Assessment Test II 

Time: 30 min.

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following steps are followed to draw a circumcircle. Find the sequential order. (A) Draw the perpendicular bisectors of the sides of the triangle. (B) The distance between the circumcentre and the vertex is circumradius. (C) Draw the triangle with the given measurements. (D) Draw a circle which passes through all its vertices. (a) ABCD (b) CBAD (c) CABD (d) CDAB  ind the angle subtended by a chord, which makes 120° at the centre, in the 2. F minor segment.

(a) 100°

(b) 120°

(c) 140°

(d) 60°

3. I n the given figure, AE:EB = 2:3 and ABCD is a parallelogram. EG || BC and BC = 15 cm. Find the difference between EF and FG. D

A F E

G

C

B



(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

4. In a DPQR, M and N are the midpoints of PQ and PR, respectively. K and L are the midpoints of PM and PN, respectively. I and J are the midpoints of PK and PL, respectively. G and H are the midpoints of PI and PJ, respectively. If GH is 4 cm, then find QR . P

M

Q



(a) 48 cm

(b) 64 cm

N

R

(c) 32 cm

(d) 16 cm

5. I is the incentre of an isosceles triangle ABC. If ∠B = 100°, then find ∠AIC. (a) 120° (b) 135° (c) 140° (d) 115°

M01_IIT-CH02.indd 4

4/16/2018 12:40:10 PM

Assessment Test II 6. K  LMN is a rhombus. The diagonal LN is equal to the side of the rhombus. Find the measure of ∠LKM. (a) 30° (b) 45° (c) 60° (d) 120°

2.5

Space for rough work

7. VXYZ is a parallelogram and A is a point on VZ. If XBAV and ABYZ are rhombuses, and ∠ BYZ = 100°, then find ∠XAY. (a) 90° (b) 135° (c) 120° (d) 115° 8. In the given figure, ‘O’ is the centre of the circle. DABC and DODE are equilateral triangles. AB is parallel to DE . Find ∠AOE + ∠ACE. C O B

A E

D



(a) 100°

(b) 135°

(c) 120°

(d) 145°

9. ABCDE is a pentagon, inscribed in a circle with centre ‘O’. CD = DE and AD = BD. ∠EAD = 30°, ∠BOC = 80°. Find ∠ADE. D E

C O B

A



(a) 15°

(b) 30°

(c) 40°

(d) 50°

10. In the given figure, ABCD is a quadrilateral inscribed in a circle with centre ‘O’ and BD is a diagonal passing through the centre of the circle. Find ∠AEC. E D

C

50° O A



(a) 140°

(b) 100°

40°

B

(c) 120°

(d) 180°

11. I n the figure, the two circles intersect at A and B. If AB is equal to the radius of the circle with centre C2 and it is equal to 8 cm and radius of the circle with centre C1 is 6 cm, then find the distance between the centres (in cm).

M01_IIT-CH02.indd 5

4/16/2018 12:40:13 PM

2.6

Chapter 2  Geometry-I and Geometry-II A

Space for rough work C2

C1 B

(a) 5 + 4 3 (c) 3 10

(b) 7 3 +8 5 (d) 2 5+4 3

Direction for questions 12 to 15: Match the statements of Column A with those of Column B. Column A

Column B

12. The diagonals are equal but only one pair (a) Rhombus of opposite sides are equal. 13. The diagonals may not be equal but both the pairs of opposite sides are equal and the adjacent sides are not equal.

(b) Cyclic Trapezium

14. The diagonals are perpendicular and two pairs of adjacent sides are equal and opposite sides are not equal.

(c) Kite

15. The opposite sides are equal and diagonals are equal.

(d) Rectangle (e) Parallelogram (f) Square

M01_IIT-CH02.indd 6

4/16/2018 12:40:16 PM

Assessment Test III

Assessment Test III

Time: 30 min.

2.7

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. One angle of a decagon is 90° and all the remaining nine angles are equal. The following steps are involved in calculating the measure of each of the other angles. Arrange them is sequential order. (A) Let each of the other angles be equal to x°. (B) Given, one angle of the decagon = 90°. The number of sides (n) = 10. 1350° (C) x= = 150°. Each of the other angles = 150° 9 (D) 9x + 90° = (20 - 4) × 90° = 1440°

(a) ABDC

(b) BADC

(c) BACD

(d) ABCD

2. I n the given figure, ‘O’ is the centre of the circle and ∠ABC = 30°. Find the measure of reflex ∠AOC. A

C O

B



(a) 120°

(b) 270°

(c) 300°

(d) 330°

3. In the given figure (not to scale), AB||CD, ∠EAB = 20° and ∠EDC = 35°. Find the measure of ∠DEA. D

C

E

A



(a) 35°

(b) 45°

B

(c) 55°

4. The number of diagonals of a 21-sided polygon is ______. (a) 189 (b) 170 (c) 231

(d) 105°

(d) 190

5. In the given figure, ‘O’ is the centre of the circle and AB = BC = CD = DE. If ∠OAB = 80°, then what will be the measure of ∠OAE?

M01_IIT-CH02.indd 7

4/16/2018 12:40:19 PM

2.8

Chapter 2  Geometry-I and Geometry-II Space for rough work •E

O

•D • A



(a) 20°

• C

• B

(b) 50°

(c) 80°

(d) 160°

6. In the given figure, ‘O’ is the centre of the circle and B, C, and D are collinear. AC is the bisector of ∠BAD. If ∠ABC = 50°, then find ∠ADB. D C B O A



(a) 40°

(b) 50°

(c) 45°

(d) 55°

7. In the given figure, ABCD is a parallelogram and BDEC is a square and BD < AC. If the length of the shorter diagonal of the parallelogram is 5 cm, then the length of the longer diagonal of the parallelogram is ____ cm. E

C D K B

A



(a) 5 2

(b) 5 3

(c) 5 5

(d) 10 2

8. I n the given figure, DE||BC, AD = 5 cm, BC = 6 cm and DE = 2 cm. Find DB = ______. A

D

B



(a) 5 cm

(b) 10 cm

E

C

(c) 15 cm

(d) 7.5 cm

9. In the given figure, ABCD is a trapezium in which AB||CD . ‘E’ is a point on AB and AD = CE. If BCE is an equilateral triangle, then ∠DAE = ____.

M01_IIT-CH02.indd 8

4/16/2018 12:40:24 PM

Assessment Test III C

D

A



(a) 30°

2.9

Space for rough work

E

B

(b) 40°

(c) 50°

(d) 60°

10. In the given figure, ‘O’ is the centre of a circle. AB and CD are two equal and parallel chords of the circle. If the distance between them is 14 cm and AB = 48 cm, then the radius of the circle is ______. D

C O

B

A



(a) 12.5 cm

(b) 25 cm

(c) 50 cm

(d) 75 cm

11. ABC and PQR are similar triangles. The areas of DABC and DPQR are 100 cm2 and 225 cm2, respectively. If AD and PS are the corresponding altitudes of the triangles ABC and PQR, respectively, then the ratio of the lengths of AD and PS is _____. (a) 2:3 (b) 4:9 (c) 3:2 (d) 9:4 12. In the given figure, ‘O’ is the centre of the circle. AB and CD are diameters perpendicular to each other. Find the ratio of the perimeters of the semicircle ADB and triangle ADB. D

A

O

B

C

(a) (p + 2) ( 2 + 1):2 (c) p ( 2 - 1):2

(b) (p + 2) ( 2 − 1):2 (d) p ( 2 + 1):2

13. In the given figure, ABC is an isosceles triangle AB = AC. BD and CD are external bisectors of ∠B and ∠C, respectively. They meet at D. If ∠ABC = 70°, then ∠BDC = ______. A

B

C

D



M01_IIT-CH02.indd 9

(a) 40°

(b) 70°

(c) 80°

(d) 100°

4/16/2018 12:40:31 PM

2.10

Chapter 2  Geometry-I and Geometry-II

14. In the given figure, ABC is a triangle. BD and CD are internal and external bisectors of ∠B and ∠C, respectively. If ∠BAC = 50°, then ∠BDC = ____.

Space for rough work

A D

B



(a) 25°

C

(b) 50°

E

(c) 75°

(d) 100°

15. Match the statements of Column A with those of Column B. Column A

Column B

(i) If ∠A + ∠B = 180°, then ∠A and ∠B are (A) Equal



(ii) If ∠A + ∠B = 90°, then ∠A and ∠B are

(B) Supplementary

(iii) Vertically opposite angles are

(C) Complementary

(iv) In an isosceles triangle, the centroid, the orthocentre and the circumcentre are

(D) Collinear

(a) (i) → (C); (b) (i) → (D); (c) (i) → (B); (d) (i) → (D);

M01_IIT-CH02.indd 10

(ii) → (B); (iii) → (A); (ii) →(A); (iii) → (B); (ii) → (C); (iii) → (A); (ii) → (B); (iii) → (A);

(iv) → (D) (iv) → (C) (iv) → (D) (iv) → (C)

4/16/2018 12:40:32 PM

Assessment Test IV

Assessment Test IV

Time: 30 min.

2.11

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. In the given figure, the following steps are involved in finding the value of x. Arrange them in a sequential order. C

D x°

A

E 70°

B Y

(A) ∠CBE = 180° − 70° = 110° (B) In the given figure, ∠CBE is an exterior angle of the cyclic quadrilateral, x° = ∠CBE. (C) x° = ∠CBE = 110° (D) ∠CBE + ∠EBY = 180° (linear pair). (a) BADC (b) BACD (c) BDAC (d) BDCA 2. In the given figure, ‘C’ is the centre of the circle and ∠PQR = 55°. Find reflex ∠PCR. Q

C P



(a) 110°

R

(b) 250°

(c) 220°

(d) 305°

3. In the given figure (not to scale), SP||RQ. If ∠SPT = 40° and ∠RQT = 50°, then find ∠PTQ. S

P



(a) 90°

(b) 120°

T

R

Q

(c) 100°

4. The number of diagonals of a 17-sided polygon is ________. (a) 109 (b) 119 (c) 129

(d) 70°

(d) 150

5. In the given figure, ‘O’ is the centre of a circle and PQ = QR = RS = ST. If P, Q, R, S, and T are the points on the circle and ∠QPO = 70°, then what is the measure of ∠TPO?

M01_IIT-CH02.indd 11

4/16/2018 12:40:34 PM

2.12

Chapter 2  Geometry-I and Geometry-II Q

Space for rough work

R

P

S O



(a) 10°

T

(b) 40°

(c) 80°

(d) 160°

6. In the given figure, ‘O’ is the centre of the circle and A, E, and C are collinear. AB = BC If ∠ADE = 45°, then ∠ACB = ______. D O

A

B

E

C



(a) 40°

(b) 45°

(c) 55°

(d) 65°

7. In the given figure, ABCD is a rectangle and DECF is a square. If AB = 12 cm, then what is the length of each diagonal of the rectangle? F

C

D

A



E

(a) 4.5 5 cm (c) 8 5 cm

B

(b) 6 5 cm (d) 9 5 cm

8. I n the given figure, DE||AB. If EC = 12 cm, AB = 40 cm, and DE = 16 cm, then BE = _____. A

D

B



(a) 10 cm

M01_IIT-CH02.indd 12

(b) 15 cm

E

C

(c) 18 cm

(d) 20 cm

4/16/2018 12:40:42 PM

Assessment Test IV 9. I n the given figure, ABCD is an isosceles trapezium AB||CD . If ∠DAE = 30°, then ∠BCE = ______. D

(a) 60°

Space for rough work

C

B

E

A



2.13

(b) 20°

(c) 30°

(d) 40°

10. In the given figure, ‘O’ is the centre of the circle AB and CD are equal chords of lengths 80 cm each. If the distance between the chords is 18 cm, then the radius of the circle is ________ (in cm). A

C

O B



(a) 9

D

(b) 36

(c) 28

(d) 41

11. If the ratio of the corresponding medians of two similar triangles XYZ and MNP is 4:3, then the ratio of the areas of ∆XYZ and ∆MNP is _____. (a) 27:64 (b) 64:27 (c) 9:16 (d) 16:9 12. ‘O’ is the centre of a circle. ABCD is a square inscribed in the circle. Find the ratio of the perimeter of ABCD and perimeter of the semicircle ABC. (a) 2 2 :p (b) 2 2 :(p + 2) (c) 4 2 :p (d) 4 2 :(p + 2) 13. In the given figure, ABC is a triangle. BD and CD are external bisectors of ∠B and ∠C, respectively, and they meet at D. If ∠BDC = 40°, then ∠BAC = _____. A

C

B

D



(a) 40°

M01_IIT-CH02.indd 13

(b) 70°

(c) 80°

(d) 100°

4/16/2018 12:40:47 PM

2.14

Chapter 2  Geometry-I and Geometry-II

14. In the given figure, ABC is a triangle. BD and CD are internal and external bisectors of ∠B and ∠C, respectively. If ∠BAC = 60°, then ∠ACD - ∠CBD = _____.

Space for rough work

A

D

B



(a) 20°

C

(b) 30°

(c) 60°

(d) 40°

15. Match the statements of Column A with those of Column B. Column A



Column B

(i) Medians of a triangle

(A) Circumcentre

(ii) Altitudes of a triangle

(B) Centroid

(iii) Perpendicular bisectors of sides of a triangle

(C) Orthocentre

(iv) Angle bisectors of a triangle

(D) Incentre

(a) (i) → (B); (b) (i) → (D); (c) (i) → (B); (d) (i) → (A);

(ii) → (C); (ii) → (C); (ii) → (D); (ii) → (D);

(iii) → (A); (iii) → (B); (iii) → (C); (iii) → (B);

(iv) → (D) (iv) → (A) (iv) → (A) (iv) → (C)

Answer Keys Assessment Test I 1. (d) 11. (a)

2. (b) 12. (e)

3. (b) 13. (a)

4. (c) 14. (b)

5. (b) 15. (d)

6. (a)

7. (a)

8. (d)

9. (b)

10. (a)

4. (b) 14. (c)

5. (c) 15. (d)

6. (a)

7. (a)

8. (b)

9. (c)

10. (b)

4. (a) 14. (a)

5. (b) 15. (c)

6. (b)

7. (c)

8. (b)

9. (d)

10. (b)

4. (b) 14. (b)

5. (a) 15. (a)

6. (b)

7. (b)

8. (c)

9. (a)

10. (d)

Assessment Test II 1. (c) 11. (d)

2. (b) 12. (b)

3. (b) 13. (e)

Assessment Test III 1. (b) 11. (a)

2. (c) 12. (b)

3. (c) 13. (b)

Assessment Test IV 1. (c) 11. (d)

M01_IIT-CH02.indd 14

2. (b) 12. (d)

3. (a) 13. (d)

4/16/2018 12:40:48 PM

Linear Equations and Inequations

3

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Linear Equations and Inequations; pp. 4.1−4.25

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1. Find the sequential order of the steps in solving the two simultaneous linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 by substitution method.  −c1 − b1 y  (A) From one of the equations, find the value of x in terms of y  x = . a1   (B) Substitute the value of x in equation (2) in terms of y. (C) Mark the equations as (1) and (2). −c − b y (D) Substitute the value of y in x = 1 1 . a1 (E) Simplify and find the value of y. (a) ABCDE (b) CABDE (c) CADEB (d) CABED

2. If the common solution of the equations 3x + 2y = 5 and 9x - 2y = 3 is (p, q), then find the product of p and q. (a) 1 (b) 2 (c) 3 (d) 5 3. In a DXYZ if ∠X = 5 and ∠Y = 2 (∠Y + ∠Z), then what is the sum of ∠X and ∠Y? (a) 144° (b) 120° (c) 82° (d) 136° 4. Trains T1 and T2 are on parallel tracks. T1 starts from station P and T2 starts from station Q. T1 and T2 travel with uniform speeds of 50 kmph and x kmph, respectively. If they travel in opposite direction, then they will meet in 3 h, and if they travel in the same direction, then they will meet in 9 h. Find the value of x(x > 50) (in kmph). (a) 75 (b) 100 (c) 64 (d) 81 5. Find the values of a and b, if the linear equations 2x + 8y = 15 and ax + by = 45 have infinite solutions. (a) 6, 24 (b) 1, 5 (c) 1, 4 (d) 4, 6

M01_IIT-CH03.indd 1

4/16/2018 2:50:36 PM

3.2

Chapter 3  Linear Equations and Inequations

2 5 8 3 6. I f the common solution of + = 26 and − + 11 = 0 is (x, y), then find x y x y the value of 5xy. (a) 2 (b) 12 (c) 4 (d) 6

Space for rough work

7. I f ax + by = a - b and bx + ay = a + b have a common solution (x, y), then find the value of x. 2 ab 2 ab (a) 1 - 2 2 (b) 1 + 2 2 a −b a +b 2 ab 2 ab (c) 1 + 2 2 (d) 1 - 2 2 a −b a +b 8. A  person can row 20 km downstream in 2 h and 16 km upstream in 8 h. Find the speed of the person in still water (in kmph). (a) 12 (b) 8 (c) 6 (d) 4 9. In an examination consisting of 100 questions, three marks are given for every question attempted correctly and one mark is deducted for every question attempted wrongly. A student attempts all the questions and gets 100 marks. How many questions did he mark wrong? (a) 40 (b) 50 (c) 30 (d) 60 10. The sum of the digits of a three-digit number is 9. If 99 is added to the number, then the digits will be in the reverse order. The digit in the ten’s place is equal to 2/7 times the sum of the digits in the hundred’s place and unit’s place. Find the product of the digits in the number. (a) 24 (b) 36 (c) 48 (d) 64 11. The cost of 3 pencils, 5 rulers, and 7 erasers is `49. The cost of 5 pencils, 9 rulers, and 13 erasers is `78. Find the cost of 1 pencil, 1 ruler, and 1 eraser. (a) `15 (b) `20 (c) `25 (d) `30 Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A 3x −7 ≥ 5 2 5x 13. −7 = 3 6 8x 3 14. − <7 3 4 12.

15.

M01_IIT-CH03.indd 2

( x − 3)2 x−6

<x

Column B (a) 12 or 4.8 (b) [8, ∞)  −75 93  , (c)   32 32  (d) (- ∞, 6)

4/16/2018 2:50:40 PM

Assessment Test II

Assessment Test II

Time: 30 min.

3.3

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following are the steps involved in solving two simultaneous linear equations in two variables by elimination method. Arrange them in sequential order. (A) Observe the coefficients of variables in both the equations. (B) The given equations are indicated as Eq. (1) and Eq. (2). (C) Select the variable which is to be eliminated. (D) Add or subtract the modified equations according to the sign of the term. (E) Equate the coefficients of the variable by multiplying with suitable number.

(a) BACDE (c) ABCDE

(b) BACED (d) ABECD

2. I f the common solution of the equations 4x + 5y = 26 and 8x + 3y = 17 is (p, q), then find the value of p + q.

(a) 14/5

(b) 21/4

(c) 23/4

(d) 17/3

4 (∠P + ∠Q). Find the difference of the 5 largest angle and the smallest angle. (a) 55° (b) 65° (c) 40° (d) 30°

In a DPQR, ∠P = 3, ∠Q, and ∠R = 3.

3 5 12 1 4. If the common solution of + = 11 and − = 2 is (m, n), then find m + x y x y 2n. (a) 3 (b) 4 (c) 5 (d) 3/2 5. If px + qy = p2 + q2 and

(a)

p q x − y = q - p, then find the value of x. q p

q p2 p (b) (c) p q q

(d)

q2 p

6. T  wo cars C1 and C2 start from A with uniform speed and reach B. The total time taken by both the cars is 10 h. If C1 starts from A and C2 starts from B, then they meet at a point in between A and B in 2 h 24 min. If the distance between A and B is 360 km, then find the difference in speed of the cars (in kmph). (a) 30 (b) 10 (c) 15 (d) 20 7. If 7x + 8y = 10 and 9x - ky = 20 have a unique solution, then which of the following cannot be the value of k? −72 81 −67 64 (a) (b) (c) (d) 7 7 9 9

M01_IIT-CH03.indd 3

4/16/2018 2:50:44 PM

3.4

Chapter 3  Linear Equations and Inequations

8. S  um of the digits of a three-digit number is 12. If 99 is added to the number, then the digits will be in reverse order. The digit in the ten’s place is equal to 1/3 of the sum of the digits in the hundred’s place and unit’s place. Find the product of the digits in the number. (a) 60 (b) 40 (c) 36 (d) 70

Space for rough work

9. A person can row 30 km downstream in 3 h and 54 km upstream in 9 h. Find the speed of the man in still water. (a) 12 kmph (b) 8 kmph (c) 6 kmph (d) 4 kmph 10. An examination paper contains 100 questions. Two marks are awarded for every correct answer, 1 mark is deducted for every wrong answer. However, no mark is deducted for unattempted questions. Praveen gets 100 marks and he left as many questions as those he had answered wrongly. Find the number of questions he had answered correctly.

(a) 60

(b) 40

(c) 50

(d) 70

 he cost of 4 pens, 5 erasers, and 6 pencils is `37. The cost of 3 pens, 2 eras 11. T ers, and a pencil is `19. Find the cost of 3 pens, 3 erasers, and 3 pencils. (a) `16 (b) `20 (c) `18 (d) `24 Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A 12. 2 x −

5 ≥7 2

3x +4 =7 2 4 <7 14. 8 x − 3

13.

15.

M01_IIT-CH03.indd 4

( x − 2)2 4−x

< -x

Column B (a) 2 (b) 4.25 (c) 1 (d) -3

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Assessment Test III

Assessment Test III

Time: 30 min.

3.5

Space for rough work

Direction for questions 1 to 14: Select the correct answer from the given options. 1. If 2x + 1 + 2x - 1 = 160, then find the value of x. The following steps are involved in solving the above problem. Arrange them in sequential order. 1  (A) 2x = 26 ⇒ x = 6. (B) 2x ·  2 +  = 32 × 5  2

(C) 2x · 2 + 2x · 2−1 = 160

 5 (D) 2x   = 25 × 5  2



(a) ABCD

(c) CADB

(b) CBDA

If 3a + 5b + 6c = 33 and a + 3b + 4c = 19, then a + b + c = _____. 2. (a) 7 (b) 14 (c) 21

(d) CDBA

(d) 13

The number of solutions for x − 4 = 0, where x is a real number is ________. 3.

(a) 0

(b) 1

(c) 2

(d) 3

4. If a line 8x - 3y = 6 passes through a point (k - 1, k + 2), then k = ______. (a) 3 (b) 4 (c) 2 (d) 5 5. Solve for x: 2x + 3 ≥ 3x - 7. (a) x ≤ 10 (c) -x ≤ 10

(b) x ≥ 10 (d) -x ≥ 10

1 4 is added to a number, the sum is divided by 2 and is added to the 3 3 quotient, then the resultant sum is 19. Find the number. (a) 24 (b) 35 (c) 45 (d) 27

6. If

7. If 2x + 3y = 6 and kx + 6y = 15 has a unique solution, then find the set of values of k. (a) N (b) N - {4} (c) R (d) R - {4} 8. If a and b are two natural numbers and a + b = 5, then which of the following can be the value of ab? (a) 2 (b) 6 (c) 8 (d) 5 9. If a, b, c, and d are positive real numbers, then the minimum value of the  1 1 1 1 product (a + b + c + d)  + + +  is __________.  a b c d (a) 16 (b) 32 (c) 24 (d) 36

M01_IIT-CH03.indd 5

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3.6

Chapter 3  Linear Equations and Inequations

10. If

1 1 1 1 = = , then xy = ______. and 2x − 3 y 7 5 x − 6 y 11

13 (a) 3

(b) 13

(c)

Space for rough work

−13 (d) -13 3

11. I n an examination consisting of 100 questions, one mark is given for every correct answer and one-fourth mark is deducted for every wrong answer. A candidate attempts all the 100 questions and scores a total of 70 marks. Find the number of questions he marked correctly. (a) 66 (b) 76 (c) 86 (d) 82 12. Find the solution of |x + 1|< 5. (a) x < 6 (c) -5 ≤ x ≤ 6

(b) x > -5 (d) -6 < x < 4

13. The number of non-negative integer values of x that satisfy the inequation x - 4 < 0 is _______.

(a) 1

(b) 2

(c) 3

14. The maximum value of 3 -|2x - 1| is _________. (a) 3 (b) ∞ (c) -3

(d) 4

(d) 0

Direction for question 15: Match the values of Column A with those of Column B. 15.  Column A

Column B

(i) x - 2 ≥ 0 (ii)

−5 ≤0 x−2

(iii) x + 2 ≥ 0 (iv)

M01_IIT-CH03.indd 6

(A) x ≥ -2

3 ≥0 x+2

(a) (i) → (C); (b) (i) → (C); (c) (i) → (A); (d) (i) → (A);

( B ) x > 2 (C) x ≥ 2 (D) x > -2

(ii) → (B); (ii) → (A); (ii) → (D); (ii) → (B);

(iii) → (A); (iii) → (B); (iii) → (C); (iii) → (C);

(iv) → (D) (iv) → (D) (iv) → (B) (iv) → (D)

4/16/2018 2:50:50 PM

Assessment Test IV

Assessment Test IV

Time: 30 min.

3.7

Space for rough work

Direction for questions 1 to 14: Select the correct answer from the given options. 1. If (27)x - y = 81 and (9)x + y = 243, then the following steps are involved in finding the value of y. Arrange them in sequential order. (A) 3(x - y) = 4 and 2(x + y) = 5 (B) (33)x - y = 34 and 32(x + y) = 35 4 (C) x-y= (1) 3 and 5 x + y = (2) 2 4 5 (D) From Eqs (1) and (2), -2y = 3 2 7 (E) y= 12 (a) BACDE (b) BCADE (c) BADCE (d) BCADE 2. If 2a + 3b + 5c = 9 and 2a + b - c = 7, then what is the value of a + b + c? (a) 2 (b) 4 (c) 6 (d) 5 3. The number of solutions for x + 2 = 0, where x is a real number is ________.

(a) 0 (c) 2

(b) 1 (d) infinite

4. A line 13x + 7y = 10 passes through a point (k + 2, k + 2), then what is the value of k? −1 −3 −2 −1 (a) (b) (c) (d) 2 2 3 3 5. S  olve for x: x - 3 < 2x - 5. (a) x < 2 (b) x > 2

(c) -x < 2

(d) -x > 2

6. If the numerator and the denominator of a fraction are increased by 10 and 12, respectively, or decreased by 10 and 12, respectively, then the resulting fraction is the same. Find the fraction. 3 4 5 6 (a) (b) (c) (d) 4 5 6 7 7. I f 7x + 8y = 15 and 14x + py = 30 has infinite solutions, then find the value of p. (a) 8 (b) 16 (c) 24 (d) 32

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4/16/2018 2:50:55 PM

3.8

Chapter 3  Linear Equations and Inequations

8. I f x and y are two natural numbers and x + y = 7, then which of the following cannot be the value of yx? (a) 25 (b) 36 (c) 64 (d) 81

Space for rough work

9. If a, b, c, d, and e are positive real numbers such that abcde = 32, then the minimum value of (1 + a) (1 + b) (1 + c) (1 + d) (1 + e) is ______. (a) 300 (b) 98 (c) 147 (d) 243 10. If

1 1 4 1 1 1 = = , then x = _______. + and x y 3 x y 3

5 (a) 6 2 (c) 3

6 (b) 5 3 (d) 2

11. In an examination consisting of 75 questions, each correct answer gets 4 marks, and for each wrong answer, 2 marks are deducted. A student attempted all questions and he scored 132 marks. How many questions did he attempted wrongly? (a) 28 (b) 32 (c) 43 (d) 47 12. One of the solutions of |x - 2|> 10 is ______. (a) x = 7 (b) x = 10 (c) x = - 7 (d) x = -10 13. The number of positive integer values of x when y = 15 for the inequation y - x ≥ 10 is ______. (a) 5 (b) 10 (c) 25 (d) -25 14. What is the minimum value of 6 +|4 - 7x|? (a) - ∞ (b) 6 (c) -1

(d) 10

Direction for question 15: Match the values of Column A with those of Column B. 15.  Column A (i)

(A) x ≥ 3

(ii) x - 3 ≥ 0

(B) x > -3

(iii) x + 3 ≥ 0

(C) x < 3

(iv)

M01_IIT-CH03.indd 8

−3 >0 x−3

Column B

3 ≥0 x+3

(D) x ≥ -3

4/16/2018 2:50:59 PM

Answer Keys (a) (i) → (B); (b) (i) → (C); (c) (i) → (B); (d) (i) → (C);



(ii) → (D); (ii) → (A); (ii) → (D); (ii) → (A);

(iii) → (A); (iii) → (D); (iii) → (A); (iii) → (B);

(iv) → (C) (iv) → (B) (iv) → (C) (iv) → (D)

3.9

Space for rough work

Answer Keys Assessment Test I 1. (d) 11. (b)

2. (a) 12. (b)

3. (a) 13. (a)

4. (b) 14. (c)

5. (a) 15. (d)

6. (a)

7. (a)

8. (c)

9. (b)

10. (a)

4. (b) 14. (c)

5. (d) 15. (b)

6. (a)

7. (a)

8. (a)

9. (b)

10. (a)

4. (b) 14. (a)

5. (a) 15. (a)

6. (b)

7. (d)

8. (c)

9. (a)

10. (b)

4. (b) 14. (b)

5. (b) 15. (b)

6. (c)

7. (b)

8. (b)

9. (d)

10. (b)

Assessment Test II 1. (b) 11. (d)

2. (b) 12. (d)

3. (a) 13. (a)

Assessment Test III 1. (b) 11. (b)

2. (a) 12. (d)

3. (c) 13. (d)

Assessment Test IV 1. (a) 11. (a)

M01_IIT-CH03.indd 9

2. (b) 12. (d)

3. (a) 13. (a)

4/16/2018 2:50:59 PM

Thispageisintentionallyleftblank

4

Logarithms

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Logarithms; pp. 2.1−2.8

Assessment Test I

Time: 30 min.

Direction for questions 1 to 15: Select the correct answer from the given options.

Space for rough work

1. If log3 2 = x, then following steps are involved in finding the value of log24 72 in terms of x. Arrange them in sequential order. log24 3 (A) log3 72 (B) (C) (D) (a)

log3 9 + log3 8 2 + 3x 2 log3 3 + 3 log3 2 ABCD (b) BADC

(c) ABDC

(d) ADBC

(c) 12

(d) 13

2. 3log3 7 + 5log5 6 = ________ (a) 10 3. log2 (a) 4. log

2

(b) 11 32 = ________

4 3

(b)

5 3

(c)

2 3

(d)

3 2

(c)

15 8

(d)

8 15

k k k k = ________

k

15 (a) 32

(b)

32 15

5. If log 2 = 0.3010, then the number of digits in 21000 is _______. (a) 300 6. If log

256

(b) 301 k2 =

(a) 0

(c) 302

3  k , then what is the value of log   ?  8 4 (b) 1 (c) 8

(d) 303

(d) 32

7. 1 + log2 3 = __________ (a) log2 3

M01_IIT-CH04.indd 1

(b) log2 6

(c) log3 6

(d) log3 2

4/12/2018 2:30:30 PM

4.2

Chapter 4  Logarithms

8. I f ax = by = cz and y2 = xz, then what is the value of

(a) 1

(logb a) (logc b)

Space for rough work

?

(b) logy x/logz y (c) xz/y (d) xyz

x+y = ______. xy (a) 1 (c) 6

9. If 3x = 2y = 6, then

(b) 2 (d) 30

10. log2 7 log3 5 log7 2 log5 3 = ______ (a) log3 1 (b) log3 2 (c) log2 2 11. I f x = log5 10 , then x +

(d) 2

1 + 2 = _______. x

2 ⋅ log 50 (a) log 5 ⋅ log 10

(b)

5 log 10 (c) 5

log5 10 + log10 5 (d)

(log 50)2 log 5 ⋅ log 10

12. 10 log10 3 + 6log6 5 = _____

(a) 16

(b) 15

(c) 8

(d) log15

13. If log10 25 = 1.3979, then log10 5 = ___. (a) 0.4659 (b) 0.6020

(c) 0.6989

(d) 0.3979

14. If log10 3 = 0.4771, then log10 300 = ___. (a) 1.4771 (b) 2.4771

(c) 3.4771

(d) 1 .4771

15. Match the values of Column A with those of Column B. Column A



M01_IIT-CH04.indd 2

Column B

(i) logb a × loga b

(A) a

(ii) log b

(B)

a

b

1 a

(iii) log a b b

(C) b

(iv) log b × log b a a b

(D) 1

(a) (i) → (D); (b) (i) → (D); (c) (i) → (B); (d) (i) → (B);

(ii) → (B); (ii) → (A); (ii) → (D); (ii) → (C);

(iii) → (A); (iii) → (B); (iii) → (C); (iii) → (D);

(iv) → (C) (iv) → (C) (iv) → (A) (iv) → (A)

4/12/2018 2:30:34 PM

Assessment Test II

Assessment Test II

Time: 30 min.

4.3

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. The following steps are involved in finding the value of x when log3 log9 (x2 4x + 4) = 0. Arrange them in a sequential order. (A) (x2 - 4x + 4) = 91 (B) x = -1 or x = 5 2 (C) x - 4x - 5 = 0 (D) log9 (x2 - 4x + 4) = 30 (E) (x + 1) (x - 5) = 0 (a) DACEB (b) DAECB (c) DCAEB (d) DCABE 2log3 7 - 7log3 2 = ________ 2. (a) log3 7 (b) log3 2 (c) log7 2 (d) log7 1 3. log3

3

729 = ________

(a) 2

4. log

y

(b) 3

(c) 4

(d) 6

y y y = _________

4 7 14 7 (a) (b) (c) (d) 7 4 8 16 5. I f log 5 = 0.6989, then number of digits in 510 is _______. (a) 5 (b) 6 (c) 7 6. If log81 x5 = 1

(a) 0

7. 1 +

1 , then find the value of log3 x2 . 4 (b) 1 (c) 2

(d) 8

(d) 3

1 = _________ log5 3

(a) log3 10 (c) log3 15

(b) log10 3 (d) log15 3

log a log b log c = = , then b² = _________. 5 6 7 (a) ac (b) a² (c) bc (d) ab

8. If

9. I f 7x = 8y = 3136, then

1 1 + = _________. x y



(a) 2



(c) 1

M01_IIT-CH04.indd 3

1 2 1 (d) 3

(b)

4/12/2018 2:30:38 PM

4.4

Chapter 4  Logarithms

10. Solve for: log3 10 log2 25 log10 3 log5 2 .

Space for rough work

(a) log5 5 (b) log5 25 (c) log2 9 (d) log5 1 1 + 1 = _________. x+1 (b) log3 28

11. If x = log3 2, then

(a) log18 6

12. 5log5 7+ 13log13 11 = _________ (a) 18 (b) 19

(c) log6 18

(d) log3 4

(c) 17

(d) 13

13. If log10 2 = 0.3010, then log10 8 = _________.

(a) 1.3010

(b) 0.0961

(c) 0.903

14. If log10 3 = 0.4771, then what is the value of log10 0.03? (a) 1.4771 (b) 2 .4771 (c) 2.4771

(d) 0.69

(d) 1.4771

15. Match the values of Column A with those of Column B. Column A

(i) logx = 42.3168

(A) 41

(ii) logx = 41.8671

(B) 43

(iii) logx = 39.1568

(C) 42

(iv) logx = 40.2929

(D) 40

(a) (i) → (A); (b) (i) → (B); (c) (i) → (B); (d) (i) → (C);



Column B (Number of Digits in Integral Part of x)

(ii) → (B); (ii) → (C); (ii) → (A); (ii) → (B);

(iii) → (C); (iii) → (D); (iii) → (D); (iii) → (A);

(iv) → (D) (iv) → (A) (iv) → (C) (iv) → (D)

Answer Keys Assessment Test I 1. (c) 11. (b)

2. (d) 12. (c)

3. (b) 13. (c)

4. (c) 14. (b)

5. (c) 15. (a)

6. (a)

7. (b)

8. (a)

9. (a)

10. (c)

4. (b) 14. (b)

5. (c) 15. (b)

6. (c)

7. (c)

8. (a)

9. (b)

10. (b)

Assessment Test II 1. (a) 11. (c)

M01_IIT-CH04.indd 4

2. (d) 12. (a)

3. (a) 13. (c)

4/12/2018 2:30:40 PM

Mensuration

5

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Mensuration; pp. 13.1−13.23

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1. The following steps are involved in finding the area of a circle whose radius is equal to the diagonal of a square and area of the square is AS. Arrange them in sequential order. (A) Recall the formula for area of a circle. (B) Recall the formula for area of a square in terms of diagonal. (C) d = 2As (D) Ac = 6.28 As (a) ABCD (c) BACD

(b) BCAD (d) ADBC

2. The diagonal of a rectangle ABCD is the side of a rhombus AEFC. What is the ratio of the areas of the rectangle and the rhombus? (a) 1:4 (b) 2:3 (c) 3:4 (d) 1:2 3. What is the area of the triangle (in sq. cm), when base of an isosceles triangle is 8 cm and its perimeter is 20 cm? (a) 9 5 (b) 8 5 (c) 6 5 (d) 5 5 4. In the above figure, a circle is inscribed in a hexagon of side 6 cm and the hexagon is circumscribed by a circle. What will be the difference between the areas of the regions A and B (in sq. cm)? A

B

A B

B

A

A B

B A

M01_IIT-CH05.indd 1

B

A

(a) 3(11 - 6 3 )

(b) 11 - 6 3

(c) 2(11 - 6 3 )

(d) 5(11 - 6 3 )

4/16/2018 2:36:19 PM

5.2

Chapter 5  Mensuration

5. T  he edge of a cube is a cm. A cuboid is constructed with the dimensions as the longest diagonal of the cube, face diagonal of the cube, and the edge of the cube as length, breadth, and height, respectively. Find the ratio of the volumes of the cube and the cuboid. (a) 1:9 (b) 6 :5 (c) 1: 6 (d) 3: 5

Space for rough work

6. A  student made a model of a prism with three identical rectangular and two identical triangular cardboard pieces of negligible thickness. If the dimensions of a rectangular piece are 10 3 cm × 10 cm, then what can be the minimum volume of the air enclosed in the prism (in cu cm)? (a) 750 (b) 250 (c) 700 3 (d) 250 3



7. A  saint constructed a square pyramid with 99 m as the side of the base and height as well. It is believed that the person who performs “dhyana” under the pyramid for 6 h will get divine powers. If approximately 2700 l of air is required for a person for 6 h, then find the number of persons, who could be allowed inside the pyramid to perform “dhyana” for 6 h. (a) 1,19,790 (b) 1,90,900 (c) 99,990 (d) 1,01,010







8. O  n 13 November 2012, the people of Green Gold Apartment burst the crackers. A student felt bad about the pollution created. He wanted to study the situation and collected all the used flowers pots of same size (cone shaped) in the apartment. He found that the inner measures of the flowerpot are of radius 3 cm and height 7 cm. If the number of flower pots used by the people of Green Gold Apartments is 200, then calculate approximately, how much magnesium powder is burnt in the form of flower pots? (a) 13,200 cm3 (b) 23,600 cm3 (c) 14,300 cm3 (d) 12,400 cm3 9. A  capsule, made of gelatine is in the form of cylinder with hemi-spherical ends. It is packed with 97 cu mm of medicinal powder, as shown in the figure: 7 mm

4.2 mm

If a pharmaceutical company produces 1,00,000 capsules in a year, then what will be the quantity of gelatine required for one year (approximately to nearest thousands)?

M01_IIT-CH05.indd 2

(a) 3000 cm3 (c) 6000 cm3

(b) 4000 cm3 (d) 5000 cm3

4/16/2018 2:36:23 PM

Assessment Test I 10. T  he surface area of a regular octahedron is 216 3 sq. cm. Find the volume of the octahedron (in cu cm). (a) 648 6 (b) 516 6 (c) 343 6 (d) 216 6

5.3

Space for rough work

11. T  he lateral surface area of a regular tetrahedron is 450 3 sq. cm. What is the volume of the regular tetrahedron (in cu cm)? (a) 100 3 (b) 900 3 (c) 1000 3 (d) 1350 3 Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B

(Total Surface Area (Name of the of A Solid in Sq. cm) Solid) 12. 4p2 R r

(a) Cone

13. 2 p r(h + r)

(b) Sphere

14. p [R(R + l) + r(r + l)]

(c) Hemisphere

15. p r(l + r)

(d) Cone frustum (e) Torus (f) Cylinder

M01_IIT-CH05.indd 3

4/16/2018 2:36:27 PM

5.4

Chapter 5  Mensuration

Assessment Test II

Time: 30 min.

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options.



1. A  rectangle is inscribed in a circle. The following steps are involved in finding the area of the circle, if the area of the rectangle is AR and one of its sides is r. Arrange them in sequential order. (A) Find the diagonal of the rectangle. A (B) The other side of the rectangle is R . r (C) Recall the formula for area of a rectangle. (D) Recall the formula for area of circle when diameter is given. (E) Diagonal of the rectangle inscribed is equal to the diameter of the circle. (a) ABCDE (c) CABED

2. The two adjacent sides of a rectangle PQRS are, the halves of the diagonals of the rhombus PMNR, respectively. If PQ = 4 cm and QR = 3 cm, then find the ratio of the perimeter of the rectangle and the rhombus.



(b) BCDEA (d) CBAED

(a) 1:2

(b) 9:7

(c) 7:10

3. One of the equal sides of an isosceles triangle is 16 cm and its perimeter is 62 cm. What is the area of the triangle (in sq. cm)?



(a) 15 31

(b) 32 31



(c) 16 30

(d) 15 30



(d) 3:4

4. I n the given figure, a circle is inscribed in a square and the square is circumscribed by another circle. If the diagonal of the square is 14 cm, then find the difference between the areas of the region A and B (in sq. cm). A B

B A

A B A



(a) 8

3 4

(b) 10

B

(c) 9

1 2

(d) 11

2 3

5. T  here is a cube of edge 6 cm. What is the volume of the cuboid whose length, breadth, and height are the longest diagonal of the cube, face diagonal of the cube, and the edge of the cube, respectively (in cu cm)? (a) 36 (b) 24 6 (c) 216 3 (d) 30 2

M01_IIT-CH05.indd 4

4/16/2018 2:36:31 PM

Assessment Test II 6. A  sweet shop wanted to make sweets in the form of triangular prisms. The base of the model prism used to make sweets has sides 2.5 cm, 2.5 cm and 4 cm, and height is 6 cm. If the shopkeeper has 1800 cm3 of sweet mix, then find how many pieces of sweets can be made out of it? (a) 180 (b) 100 (c) 120 (d) 90



5.5

Space for rough work

7. T  he top of a building is in the shape of a square pyramid roofed with glass material. Each face of the pyramid is an equilateral triangle of edge 12 m and has a window of dimensions 1 m × 0.8 m. Find the number of glass plates of area 1 sq. m for the roof (take 3 = 1.732). (a) 246 (b) 247 (c) 250 (d) 248

8. ‘Glow well Fireworks Company’ is manufacturing flower pots 100 times the number of that year. If the company wanted to produce cone-shaped flowerpots with inner measures of radius 3 cm and slant height 5 cm, then find the volume of the cracker mixture required for the year 2016 (in cm3). (a) 96,22,600 (b) 76,03,200 (c) 51,63,800 (d) 45,63,800





9. A  goods train has 49 identical wagons in the shape of cylinder with hemispherical ends. The inner dimensions of the each tank are shown in the figure. What is the quantity of crude oil that can be transferred by one train (in litres)? 8m



(a) 3.465 × 106 (c) 2.345 × 106

3m

(b) 3.456 × 106 (d) 2.354 × 106

10. V  olume of a regular octahedron is 36 cu cm. Calculate the surface area of the octahedron (in sq. cm).

(a) 36 3

(b) 36 2



(c) 36

(d) 36 6

11. V  olume of a regular tetrahedron is 72 cu cm. What will be the surface area of the regular tetrahedron (in sq. cm)?

(a) 72

(b) 72 2



(c) 72 3

(d) 72 6

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4/16/2018 2:36:34 PM

5.6

Chapter 5  Mensuration

Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B

(Volume of a Solid)

(Surface Area of the Solid)

12. 2 p r3 3

(a) 2 p r2

13. 1 p r2 h 3

(b) 3 p r2

14. p R2 h - p r2 h

(c) 4 p r2

15.

2 2 p R3 - π r3 3 3

Space for rough work

(d) 3 p R2 + p r2 (e) 3 p R h2 - 2 p r2 yh (f) p r2 + p r l (g) 2p h(R + r) + 2p (R2 - r2)

M01_IIT-CH05.indd 6

4/16/2018 2:36:36 PM

Assessment Test III

Assessment Test III

Time: 30 min.

5.7

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options.

1. T  he following steps are involved in finding the length of the longest diagonal of a cuboid of dimensions 4 cm × 3 cm × 2 cm. Arrange them in a sequential order.



(A) Longest diagonal of a cuboid = l2 + b2 + h2 (B) Let l = 4, b = 3, and h = 2 (C) Required diagonal length = 29 cm



(D) Diagonal of the cuboid =



(a) ABCD



42 + 32 + 22

(b) BDAC

(c) BADC

(d) ABDC

 he area of a triangle whose sides are 17 cm, 8 cm, and 15 cm is _________. 2. T (a) 30 cm2 (b) 40 cm2 (c) 50 cm2 (d) 60 cm2

3. A  circle of radius 6 cm is inscribed in an equilateral triangle. What is the area of the equilateral triangle (in sq. cm)? (a) 24 3 (b) 48 3 (c) 108 3 (d) 96 3

4. If the edge of a cube increases by 5%, then calculate the percentage increase in its volume. (a) 5% (b) 10.625% (c) 15.7625% (d) 20.625%

5. The number of vertices of a pyramid whose base is a pentagon is _________. (a) 4 (b) 6 (c) 10 (d) 5 6. If the radius of a hemisphere is 30 cm, then what is the total surface area of the hemisphere (in sq. cm)? (Take p = 3.14) (a) 8378 (b) 8478 (c) 8578 (d) 8678



7. In the given figure, AB = 15 cm, BC = 12 cm, and CD = AF = 2 cm. Find the area of the polygon AFEDCA (in cm2). F

A

15 12 B E



M01_IIT-CH05.indd 7

(a) 168

(b) 158

C D

(c) 148

(d) 128

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5.8

Chapter 5  Mensuration

8. A  sphere fits exactly in a hollow cube. What will be the ratio of the volumes of the cube and the sphere? (a) 12:p (b) 3: p (c) 2:p (d) 6:p



Space for rough work

9. T  he diameters of the circular ends of a bucket are 14 cm and 7 cm. If the height of the bucket is 12 cm, then find the total surface area of the bucket. (a) 41.5 cm2 (b) 588.5 cm2 (c) 605 cm2 (d) 718.5 cm2

10. A  cow is tied by a rope of length 14 m to a pole at a corner of a grassy field of length 100 m and breadth 80 m. What is the area of the field it can graze?

(a) 16 p m2 (c) 36 p m2

(b) 98 p m2 (d) 49 p m2

11. T  he base of a right pyramid is a regular hexagon. The perimeter of the base is 36 cm. If the vertical height of the pyramid is 12 cm, then find the volume of the right pyramid. (a) 216 3 cm3 (b) 432 3 cm3

(c) 636 3 cm3

(d) 648 3 cm3

12. T  he sum of the radius and height of a solid cylinder is 35 cm. If the total surface area of the solid is 1540 cm2, then what is the volume of the solid (in cm3)? (a) 1078 (b) 2156 (c) 4312 (d) 8624 13. T  he base radius of a conical tent is 150 cm and its slant height is 200 cm. The area of the canvas required to make N such tents is 9,42,000 sq. cm. Find the value of N. (a) 10 (b) 20 (c) 5 (d) 40 14. A  n edge of a regular tetrahedron is 5 6 cm. What will be the vertical height of the tetrahedron (in cm)? (a) 5 2 (b) 5 3 (c) 10 (d) 15 15. Match the statements of Column A with the values of Column B. Column A

Column B

(Given a Cube of Edge 10 cm)

M01_IIT-CH05.indd 8

(i) LSA of the cube (in cm2)

(A) 10 3

(ii) Volume of the cube (in cm3)

(B) 400

(iii) Longest diagonal of the cube (in cm)

(C) 600

(iv) TSA of the cube (in cm2)

(D) 1000

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Assessment Test III

M01_IIT-CH05.indd 9

(a) (i) → (C); (b) (i) → (A); (c) (i) → (B); (d) (i) → (B);

(ii) → (D); (ii) → (B); (ii) → (D); (ii) → (D);

(iii) → (A); (iii) → (D); (iii) → (A); (iii) → (C);

(iv) → (B) (iv) → (C) (iv) → (C) (iv) → (A)

5.9

Space for rough work

4/16/2018 2:36:42 PM

5.10

Chapter 5  Mensuration

Assessment Test IV

Time: 30 min.

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. T  he following steps are involved in finding the edge of the cube, if the length of the diagonal of a cube is 5 3 cm. Arrange them in sequential order. (A) Edge of the cube = 5 cm (B) Diagonal of the cube = 3 a (C) 3 a=5 3 =a=5 (D) Let a be the edge of the cube. (a) BACD (b) DBCA (c) DCBA (d) CDBA



 he area of a triangle whose sides are 40 cm, 9 cm, and 41 cm is _________. 2. T (a) 160 cm2 (b) 170 cm2 (c) 180 cm2 (d) 200 cm2

3. A  circle of radius 2 cm is inscribed in an equilateral triangle. What can be the area of the circle circumscribing the triangle? (a) 8p (b) 24p (c) 16p (d) 32p

4. I f the length and the breadth of a rectangle are increased by 20% and 10%, respectively, then find the percentage increase in its area. (a) 30% (b) 32% (c) 35% (d) 31%

5. The number of edges of a tetrahedron is _________. (a) 6 (b) 9 (c) 12

(d) 24

6. I f the surface area of a sphere is 5024 cm2, then find the radius of the sphere (take π = 3.14). (a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm 7. I n the given figure, ABCD is a rectangle. DE = EC, CF = FB, EH = HC, and CG = GF. If the area of ABCD is 96 cm2, then what is area of ∆HCG? E

D

H

C G F

A



(a) 3

M01_IIT-CH05.indd 10

cm2

(b) 6

B

cm2

(c) 12 cm2

(d) 24 cm2

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Assessment Test IV

5.11

8. A  cube fits exactly in a sphere. Find the ratio of the total surface area of the Space for rough work sphere and that of the cube. (a) p :2 (b) p :3 (c) p :6 (d) p :2





9. T  he diameters of the circular ends of a bucket are 18 cm and 50 cm. If the total surface area of the bucket is 1862 p  cm2, then what is the height of the bucket? (a) 24 cm (b) 25 cm (c) 30 cm (d) 34 cm

10. A horse is tied by a rope of length 12 m to a pole at a corner of the squareshaped grassy field of length 50 m. What is the area of the field it can graze? (a) 16 p m2 (b) 25 p m2 2 (c) 36 p m (d) 49 p m2 11. A  regular hexagonal pyramid is 10 cm high and side of the base is 8 cm. Find the slant height of the pyramid (in cm). (a) 2 37 (b) 2 39 (c) 2 41 (d) 2 43 12. T  he height of a solid cylinder exceeds its radius by 7 cm. If the lateral surface area of the solid is 616 cm2, then what is the volume of the solid (in cm3)? (a) 1078 (b) 2156 (c) 4312 (d) 8624 13. The base radius of a conical tent is 160 cm and its height is 300 cm. Find the area of canvas required to make 10 such tents (in sq. m) (take p  = 3.14). (a) 75.36 (b) 150.72 (c) 170.816 (d) 231.104 14. An edge of a regular tetrahedron is 4 3 cm. Find the slant height of the tetrahedron (in cm).

(a) 8 3

(b) 6

(c) 8

(d) 6 3

15. Match the values of Column A with those of Column B. Column A

Column B

(CSA of the Solid)

(TSA of the Solid)

(i) p  r

(

r2 + h2

(A) p R2 + p r2 + p (R + r)

(ii) 2p r2

(B) 3p r2

(iii) 2p (R2+ r2)

(C) 3p R2 + p r2

(iv) p  (R + r)

M01_IIT-CH05.indd 11

)

(R − r )2 + h2

(D) p  r (r +

(R − r )2 + h2

r2 + h2 )

4/16/2018 2:36:48 PM

5.12

Chapter 5  Mensuration (a) (i) → (A); (b) (i) → (A); (c) (i) → (D); (d) (i) → (D);

(ii) → (B); (ii) → (C); (ii) → (C); (ii) → (B);

(iii) → (C); (iii) → (B); (iii) → (B); (iii) → (C);

(iv) → (D) (iv) → (D) (iv) → (A) (iv) → (A)

Space for rough work

Answer Keys Assessment Test I 1. (b) 11. (c)

2. (d) 12. (e)

3. (b) 13. (f)

4. (a) 14. (d)

5. (c) 15. (a)

6. (a)

7. (a)

8. (a)

9. (b)

10. (d)

4. (a) 14. (g)

5. (a) 15. (d)

6. (b)

7. (b)

8. (b)

9. (a)

10. (a)

4. (c) 14. (c)

5. (b) 15. (c)

6. (b)

7. (c)

8. (d)

9. (c)

10. (d)

4. (b) 14. (b)

5. (a) 15. (d)

6. (c)

7. (a)

8. (a)

9. (c)

10. (c)

Assessment Test II 1. (d) 11. (c)

2. (c) 12. (b)

3. (a) 13. (f)

Assessment Test III 1. (c) 11. (a)

2. (d) 12. (c)

3. (c) 13. (a)

Assessment Test IV 1. (b) 11. (a)

M01_IIT-CH05.indd 12

2. (c) 12. (b)

3. (c) 13. (c)

4/16/2018 2:36:48 PM

6

Number Systems

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Number Systems; pp. 1.1−1.19

Assessment Test I

Time: 30 min.

Direction for questions 1 to 15: Select the correct answer from the given options. 1. The following steps are involved in rationalizing the numerator of

1− 1+ x

Arrange them in sequential order. (A) (C)

(

x

( x − 1) 1 + 1 + x 1 − (1 + x )

(

( 1+ x) (1 − x ) (1 + 1 + x ) 12 −

(D)

(b) BDCA

(c) BDAC

2

(d) BADC

5 - 2 and ab is a rational number, then what can be the value of b?

2. If a = (a)

)

.

1− 1+ x 1+ 1+ x × 1− x 1+ 1+ x

)

(1 − x ) 1 + 1 + x

(a) BCDA

(B)

1− x

Space for rough work

5+2

(b) 2 ( 5 + 2)

(c) 3 ( 5 + 2)

(d) All of these

3. 3 12 - 3 27 + 2 48 = _________ (a) 3 3

(b) 4 3

(c) 5 3

(d) 6 3

4. If a = 2 , b = 3 3 , and c = 4 4 , then _________. (a) a < b < c (c) a < c < b

(b) a = c < b (d) b = a < c

5. If x = 2 6 + 5, then solve for x2 - 10x - 1. (a) 5 - 2 6 6.

(

)(

324 + 2 323 −

(a) 2 (c) 36

M01_IIT-CH06.indd 1

(b) 2 6 - 5

(c) 10

(d) -2

)

324 − 2 323 = _________ (b) 1 (d) 2 323

4/16/2018 12:43:51 PM

6.2

Chapter 6  Number Systems 49 50

87 60

Space for rough work

− 73

93

7.  = _________ 07 07 08 (63)3 + (81)5 + ( 49)6 (a) 1 (b) 2

(c) 3

(d) 4

8. If 3x = 25, then find the value of 32x + 1. (a) 625 (b) 875

(c) 1225

(d) 1875

9. The conjugate of (3 2 + 6) - (9 + 2 2 ) is _________. (a) 2 + 3 (c) - 2 + 3

(b) - 2 - 3 (d) 2 - 3

10. 12. 04 = _________ 1084 271 1084 542 (a) (b) (c) (d) 99 225 999 45 11. 3 4 3 = _________ 4 4 4 4 (a) 9 (b) 27 (c) 243 (d) 729

12. If 3x = 9y = 27z = 81, then what is the value of x - y + z?

(a) 0

13. 

6

3− 3 (a) 3 +

(b)

22 10 14 (c) (d) 3 3 3

= _________ (b) 6 (3 + 3 )

3

3+ 3 (c) 3 14. 

3 3 +1

+

3 5+ 3

+

(a) 23

3 7+ 5

(d) +…+

3 25 + 23

(b) 3 23

3− 3 3

= _________ (c) 3

(d) 6

15. Match the values of Column A with those of Column B and select an option which is true about the given data. Column A (i)

3

9,

3

Column B 24

(ii)

12 + 3 , 12 − 3

(B) Conjugate surds

(iii)

3 + 2, 2 − 3

(C) Mixed surds

(iv) 3 2, 2 2

M01_IIT-CH06.indd 2

(A) Similar surds

(D) Rationalizing factors

4/16/2018 12:44:03 PM

Assessment Test I

M01_IIT-CH06.indd 3

(a) (i) → (A); (b) (i) → (A); (c) (i) → (D); (d) (i) → (D);

(ii) → (C); (ii) → (B); (ii) → (B); (ii) → (C);

(iii) → (B); (iii) → (D); (iii) → (C); (iii) → (B);

(iv) → (D) (iv) → (C) (iv) → (A) (iv) → (A)

6.3

Space for rough work

4/16/2018 12:44:03 PM

6.4

Chapter 6  Number Systems

Assessment Test II

Time: 30 min.

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. The following steps are involved in finding the value of x +

1 , if x = 5 + 2 6 . x

Arrange them in sequential order. 1 (A) = 5 − 2 6 x 1 5−2 6 = (C) x (5)2 − (2 6 )2

(a) DCAB

2. If x = 7 +

(B) x+

1 = 5+2 6 +5−2 6 x 1

1 (D) = x 5+2 6

(b) ADCB

(c) CDAB

(d) BDCA

3 and xy is a rational number, then what can be the value of y?



(a) 2 (7 + 3 )

(b) 3 ( 7 − 3)



(c) 7 -

(d) All of these

3

3. Simplify: 2 125 - 5 5 + 3 3125 .

(a) 75 5

4. I f a = 7 , b =

(b) 85 5 3

12 , and c =

4

2 -

3 , then find the value of x +

(a) - 2 3 (b) 2 3 6. F  ind the value of

(d) 70 5

25 , then _________.

(a) c > b > a (c) c < b < a 5. If x =

(c) 80 5

(b) a > c > b (d) a
16 + 2 55 + 16 − 2 55 .

(a) 2 5 (b) 10 + 5 + 5 (c) 1 + 2 5 + 15 (d) 2 11 7 04

23

7. (81)

9 08

24

− (19)

(a) 6100

= _________ (b) 6200

(c) 6300

8. If 6x = 30, then what will be the value of 6x - 1 + 6x + 1? (a) 30 (b) 150 (c) 185 9. The conjugate of

(d) 900

5 + 4 is _________.

(a) 5 -4 (c) - 5 - 4

M01_IIT-CH06.indd 4

(d) 6400

(b) 5 + 4 (d) - 5 + 4

4/16/2018 12:44:17 PM

Answer Keys 10. 9. 324 = _________ 3077 3077 3077 3077 (a) (b) (c) (d) 333 300 330 33

6.5

Space for rough work

11. 2 3 4 = _________ 3 3 3 3 (a) 8 (b) 16 (c) 32 (d) 108

12. If 2x = 3y = 4z, for some integer values of x, y, and z, then what is the value of x + y + z? 1 14 (a) 0 (b) 1 (c) (d) 3 3 11 13.  = _________ 5+ 3 5+ 3 5− 3 5− 3 (c) (d) (a) 5− 3 (b) 3 2 22 1 1 1 1 14. + + + ... + = _________ 2 +1 3+ 2 4+ 3 9+ 8

(a) 1

(b) 2

(c)

2

(d) 2 2

15. Match the values of Column A with those of Column B and select an option which is true about the given data. Column A (i) 0.1 6 (ii) 0. 16 (iii) 0.16 (iv) 1.6 (a) (i) → (B); (b) (i) → (B); (c) (i) → (C); (d) (i) → (C);



Column B 5 3 16 (B) 99 1 (C) 6 (A)

(D)

4 25

(ii) → (C); (ii) → (D); (ii) → (D); (ii) → (B);

(iii) → (D); (iii) → (C); (iii) → (A); (iii) → (D);

(iv) → (A) (iv) → (A) (iv) → (B) (iv) → (A)

Answer Keys Assessment Test I 1. (b) 11. (c)

2. (d) 12. (c)

3. (c) 13. (a)

4. (b) 14. (d)

5. (d) 15. (d)

6. (a)

7. (b)

8. (d)

9. (b)

10. (d)

4. (c) 14. (b)

5. (a) 15. (d)

6. (d)

7. (b)

8. (c)

9. (d)

10. (c)

Assessment Test II 1. (a) 11. (c)

M01_IIT-CH06.indd 5

2. (c) 12. (a)

3. (c) 13. (c)

4/16/2018 12:44:26 PM

Thispageisintentionallyleftblank

Number Systems and Logarithms

7

Reference: Coursebook - IIT Foundation Maths Class 9; Chapters - Number Systems; Logarithms; pp. 1.1−1.19; 2.1−2.8

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1 . x The following steps are involved in solving the above problem. Arrange them in sequential order. 1 x −1 (A) = 2 − 3 (B) = 2+ 3 −2+ 3 x x 1 2− 3 1 1 = (C) = (D) x 4−3 x 2+ 3

1. If x = 2 +

3 , then find the value of x -

(a) DCAB

(b) ADCB

(c) CDAB

2. Solve for x: log(x² - 4) - log(x + 2) = log3. (a) 1 (b) 5 (c) 8

(d) BDCA

(d) 10

3. Simplify: log( 343 )4 (2401)3 . (a) 2/3

(b) 3/4

4. Solve for x:

( 81)

2x + 4

(a) 1 5.

(

729

3 +1

+

3 5+ 3

23

+

3 7+ 5

(c) 4

(d) 6

2x + 2

+ ... +

(b) 3 23

(d) 0

)

(b) 3

3 (a)

=

(c) 1

3

= _________.

25 + 23

(c) 3

(d) 6

6. Which of the following is the largest in value? (a) 61/2 (b) 71/3 (c) 81/4

(d) 91/5

7. Find the positive square root of 18 + 308 . (a) 7 + 11 (c)

M01_IIT-CH07.indd 1

7 + 11

(b)

15 + 3

(d) 7 − 11

4/16/2018 1:02:55 PM

7.2

Chapter 7  Number Systems and Logarithms

8. I f log3 = 0.4771, log5 = 0.6990, then find the value of log180. (a) 2.3553 (b) 2.2552 (c) 3.2553 (d) 3.3552

Space for rough work

9. If log5 = 0.6990, log3 = 0.4771, then what is the number of digits in (150)250. (a) 544 (b) 545 (c) 543 (d) 540  ab   bc   ca  10. log   + log   + log   = _________  c  a  b

(a) 1 (c) log ax log bx log c

11. Compare the surds a =

(b) 0 (d) log abc

5 + 31 and b = 11 + 21 .

(a) a > b (c) a = b

(b) a
Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B

12. logb a × logb c × loga b2

(a) logb c2

13. aloga b × blogb c

(b) loga abc

( )

14. log alogc

(c) loga 1

 b  c 15. log   − log    a  a

(d) loga - logb

1

M01_IIT-CH07.indd 2

(e) logc - logb

4/16/2018 1:02:59 PM

Assessment Test II

Assessment Test II

Time: 30 min.

7.3

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following steps are involved in finding the value of x, if log3 2 = x. Arrange them in sequential order.

log24 3

in terms of

(B) log3 9 + log3 8

(A) log72 3

log24 72

(C) 2 + 3x (a) ABCD (b) BADC

(D) 2 log3 3 + 3 log3 2 (c) ABDC (d) ADBC

 ind the value of x, if log(x + 1) + log(x - 1) = log15. 2. F (a) 16 (b) 4 (c) -4 (d) Both (b) and (c)  ind the value of log525 (3125)3125 . 3. F

(a) 625

(b) 125

4. What is the value of x, if

(a) 3

5. 

1 2 +1

(a) 1

3

(c) 15625

52 x + 1 =

6

254 x + 5 ? (c) -2 (d) -3

(b) 2 +

1 3+ 2

+

1 4+ 3

(d) 3125

+ ... +

1 9+ 8

= _________

(b) 2

6. F  ind the largest of the surds

(c)

2

(d) 2 2

3 , 3 9 , 9 3 , and 6 18 .

3 3 6 (a) 3 (b) 9 (c) 3 (d) 18

7. F  ind the positive square root of 14 − 6 5 . (a) 5 + 3

(b) 3− 5

(c) 3 + 5

(d) 5 − 3

If log 3 = 0.4771 and log 2 = 0.3010, then what is the value of log 405? 8. (a) 2.6074 (b) 3.9084 (c) 2.6084 (d) 2.9134 9. If log 3 = 0.4771, log 2 = 0.3010, then find the number of digits in 6060. (a) 107 (b) 106 (c) 66 (d) 67

M01_IIT-CH07.indd 3

4/16/2018 1:03:05 PM

7.4

Chapter 7  Number Systems and Logarithms

10. logb a × logc b × loga clogc = _________ (a) 0 (b) 1

(c) log c

Space for rough work

(d) logca

11. Compare the following pairs of surds and relate them: a = 7 + 21 and b = 6 + 22 (a) a > b (c) a = b

(b) a
Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B

12. log3 2 × log5 3 × log4 25

(a) 1/2

log 5 log 2 + log 100 log 10

(b) 1

14. 3 log3 3 + log625 25

(c) 2

1 15.   log2 8 + 2 log32 128  15 

(d) 3

13.

(e) 5 2

Answer Keys Assessment Test I 1. (a) 11. (b)

2. (b) 12. (a)

3. (c) 13. (b)

4. (a) 14. (c)

5. (d) 15. (e)

6. (a)

7. (c)

8. (b)

9. (b)

10. (d)

4. (c) 14. (c)

5. (b) 15. (d)

6. (b)

7. (b)

8. (a)

9. (a)

10. (c)

Assessment Test II 1. (c) 11. (a)

M01_IIT-CH07.indd 4

2. (b) 12. (b)

3. (a) 13. (a)

4/16/2018 1:03:08 PM

Polynomials and Square Roots of Algebraic Expressions

8

Reference: Coursebook - IIT Foundation Maths Class 9; Chapter - Polynomials and Square Roots of Algebraic Expressions; pp. 3.1−3.23

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1. The following steps are involved in showing that if a + b + c = 0, then a3 + b3 + c3 = 3abc. Arrange them in sequential order. (A) a3 + b3 + 3ab(-c) = -c3 (B) (a + b)3 = (-c)3 (C) a + b + c = 0 ⇒ a + b = - c (D) a3 + b3 + 3ab(a + b) = -c3 (E) a3 + b3 + c3 = 3abc (a) ABDCE (b) BCDAE (c) CBDAE (d) CADBE 2. Factorize: 4x4 + 16. (a) 4(x2 + 2) (x2 - 2) (c) 4(x2 + 2x + 2) (x2 - 2x + 2) 3. Find the square root of (a) 4x2 - 12x + 9 (c) 2x2 - 6x + 9 4.

(b) 2(x2 + 4) (x2 - 2) (d) 4(x2 - 2x - 2) (x2 + 2x - 2)

16 x4 − 96 x3 + 216 x2 − 216 x + 81 . (b) 4x2 + 12x - 9 (d) 2x2 - 2x + 3

( x4 − 3 x2 + 3 x − 1)( x3 + x2 − 2 x + 1)( x − 1) = _________ (a) x4 - 3x2 + 2x -1 (c) x4 - 2x + 1

(b) x4 - 2x - 2 (d) x4 - 3x2 + 3x - 1

5. If the LCM of two polynomials f(x) and g(x) is (x2 - 4) (x + 3), then what can be the HCF of f(x) and g(x)? (a) x2 + 5x + 6 (b) x2 + x - 6 2 (c) x + 5x - 6 (d) None of these 6. Find the HCF of the polynomials f(x) = (x - 1)2 (x2 - 1)2 (x + 2)2 and g(x) = (x2 - 2x + 1) (x + 2)3 (x + 1)3. (a) (x2 - 2x + 1) (x2 + 4x + 4) (b) (x - 1)2 (x + 1)2 (c) (x - 1)2 (x + 1)2 (x + 2)2 (d) (x - 1)5 (x + 1)3 (x + 2)2

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8.2

Chapter 8  Polynomials and Square Roots of Algebraic Expressions

7. W  hat will be the value of k + l, if x4 - 14x3 + 71x2 + kx + l is a perfect square? (a) -33 (b) 154 (c) -121 (d) 0

Space for rough work

8. Find the square root of (8x2 + 10x - 25) (2x2 + 15x + 25) (4x2 + 15x - 25). (a) (2x + 5) (4x - 5) (x + 5) (b) (2x - 5) (4x + 5) (x - 5) (c) (2x + 5) (4x + 5) (x - 5) (d) (2x + 5) (4x - 5) (x - 5) 9. I f ax4 + bx3y + cx2y2 + dxy3 + ey4 is a complete homogenous and symmetric expression of degree 4 in two variables x and y, then which of the following necessarily be true? (a) a = e (b) a=b=c=d=e (c) b = d (d) Both (a) and (c) 10. Find the square root of (4x2 - 12x + 9) (4x2 + 12x + 9). (a) 4x2 + 9 (b) 2x2 + 3 (c) 4x2 - 9

(d) 2x2 - 3

11. F  ind the square root of (3x + 3y + xy)2 - 12xy (x + 3). (a) 3x - 3y - xy (b) 3x + 3y - xy (c) 3x - 3y + xy (d) 3x + 3y + xy Direction for questions 12 to 15: Match the values of Column A with those of Column B. Column A

Column B

12. (3x - 2y)2

(a) Homogeneous

13. (x + y + z)2

(b) Symmetric

14. 3x2 - 2xy + 5x + 5y + 3y2

(c) Homogeneous and cyclic

15. 3x3 - 3y2 - 3x + 3y

(d) Non-homogenous and non-symmetric

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Assessment Test II

Assessment Test II

Time: 30 min.

8.3

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following steps are involved in dividing a polynomial by another polynomial. Arrange them in sequential order. (A) The first term of the quotient is obtained by dividing the first term of the dividend by the first term of the divisor. (B) Multiply all the terms of the divisor by the first term of the quotient and subtract the result form the dividend. (C) Arrange the terms of the dividend and the divisor in the descending order of their degree. (D) Repeat this process till we obtain a remainder ‘0’ or a polynomial whose degree is less than that of the divisor. (E) Consider the remainder as new dividend and proceed as before.

(a) CAEBF (c) BACED

(b) ABDCE (d) CABED

 hich of the following is a factor of x4 + x2 + 1? 2. W (a) x2 + x - 1 (b) x2 - x - 1 2 (c) x - x + 1 (d) -x2 - x - 1 3. Find the square root of

81x4 − 108 x3 + 18 x2 + 12 x + 1 .

(a) 9x2 - 3x - 1 (c) 9x2 - 6x + 1

(

)(

(b) 3x2 + 6x - 1 (d) 9x2 - 6x - 1

)

4. Solve for x4 + x2 − 2 x3 − x2 + 2 x − 2 ( x + 1) . (a) x2 + x - 2 (c) x2 - x + 2

(b) x4 + x2 - 2 (d) x4 + x3 - 2

5. If the LCM of two polynomials f(x) and g(x) is (x - 1) (x - 2)2 (x - 3)3, then f(x)·g(x) can be _________. (a) (x - 1)3 (x - 2) (x - 3)3 (b) (x - 1) (x - 2)2 (x - 3)3 3 2 (c) (x - 1) (x - 2) (x - 3) (d) (x - 1)2 (x - 2)2 (x - 3)3 6. F  ind the LCM of the polynomials f(x) = (x + 1)2 (x - 1)3 (x + 2)4 and g(x) = (x - 2)4 (x + 1)3 (x + 2) (x - 1)2. (a) (x + 2)4 (x - 2)4 (x2 - 1)2 (b) (x + 2) (x - 2)4 (x - 1)2 (x + 1)2 (c) (x2 - 2)2 (x2 - 1)3 (d) (x2 - 4)4 (x2 - 1)3

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8.4

Chapter 8  Polynomials and Square Roots of Algebraic Expressions

7. I f 9x4 - 12x3 + 22x2 - kx + m is a perfect square, then what will be the value of k + m? (a) 6 (b) 9 (c) 15 (d) 21

Space for rough work

8. Find the square root of (x2 + 8x + 15) (x2 + x - 6) (x2 + 3x - 10). (a) (x + 3) (x - 5) (x + 2) (b) (x - 3) (x + 5) (x - 2) (c) (x + 3) (x + 5) (x - 2) (d) (x - 3) (x - 5) (x + 2) 9. I f f(x) and g(x) are two complete homogenous and symmetric expressions in x and y, then which of the following necessarily be a complete homogeneous and symmetric expression? (a) f(x) + g(x) (b) f(x) - g(x) (c) f(x).g(x)

(d) f(x)/g(x)

10. Find the square root of (x2 - 10x + 25) (x2 + 10x + 25). (a) x2 + 25 (b) x2 + 5 (c) x2 - 5

(d) x2 - 25

11. F  ind the square root of (2x + y)2 - 4 (x + 2y) (x - y). (a) 3x - y (b) 3x + y (c) x - 3y

(d) 3y

Direction for questions 12 to 15: Match the statements of Column A with the values of Column B. Column A

Column B

12. Homogenous but not complete

(a) x + y

13. Symmetric but not homogenous

(b) ax2 + 2bxy + by2

14. Complete homogenous but not symmetric

(c) 3x2 - 5xy + 3y2 - 7x - 7y + 6

15. A complete homogeneous symmetric

(d) 8x2 - 3xy + y (e) 5x2 - 8xy

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Assessment Test III

Assessment Test III

Time: 30 min.

8.5

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. The following steps are involved in factorizing x4 + x2 + 1. Arrange them in sequential order. (A) (x2 + 1)2 - x2 (B) x4 + x2 + 1 = x4 + 2x2 + 1 - x2 (C) (x2 + 1 + x) (x2 + 1 - x) (D) (x2)2 + 2(x2) (1) + 12 - x2 (a) DBCA (b) DBAC (c) BDAC (d) BDCA 2. Solve for (2x + 3y)2 + (2x - 3y)2.

(a) 8x2 + 18y2

(b) 4x2 + 9y2

(c) 12xy

(d) 24xy

3. If the degrees of two polynomials f(x) and [ f(x)·g(x)] are 9 and 27, respectively, then what is the degree of g(x)? (a) 3 (b) 9 (c) 18 (d) 27 4. The remainder obtained when x4 + 3x2 - 2x - 10 is divided by x + 2 is _________. (a) 16 (b) 18 (c) 20 (d) 22

(

)(

)(

)

5. Solve for  x2 + 7 x + 12 x2 + 8 x + 15 x2 + 9 x + 20 .

(a) (x + 2) (x + 6) (x + 5) (c) (x + 3) (x + 4) (x + 2)

(b) (x + 3) (x + 4) (x + 5) (d) (x + 4) (x + 5) (x + 2)

6. If f(x, y) = 5x5 - 3y5 + 2x3ya - 7xby is a homogeneous expression, then a - b = _________. (a) -4 (b) -2 (c) 2 (d) 4 7. I f f(x + 3) = x2 - 2x + 8, then the remainder, when f(x) is divided by (x - 2) is _________. (a) 9 (b) 11 (c) 13 (d) 15 8. Solve for 8x3 + y3. (a) (2x - y) (4x2 + y2 - 2xy) (c) (2x - y) (4x2 + y2 + 2xy)

(b) (2x + y) (4x2 + y2 - 2xy) (d) (2x + y) (4x2 + y2 + 2xy)

9. Simplify: (x + y) (x2 + y2) (x4 + y4) (x8 + y8). (a) x16 + y16 (b) x15 + y15 16 16 x −y x15 − y15 (d) (c) x−y x−y

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8.6

Chapter 8  Polynomials and Square Roots of Algebraic Expressions

10. If the LCM of x9 ya and xby6 is x10y10, then find the value of a + b. (a) 10 (b) 20 (c) 15 (d) 25 11. T  he square root of x2 + 2x + 3 + 1 x + + 3 (a) x 1 (c) x+ + 1 x

Space for rough work

2 1 + 2 is _________. x x 1 (b) x+ -1 x 1 (d) x + - 2 x

12. If (x + 1) (x + 3) (x + 5) (x + 7) + p is a perfect square, then p = _________. (a) 16 (b) 39

(c) 64

(d) 36

13. Which of the following is a factor of x (y2 - z2) + y(z2 - x2) + z(x2 - y2)? (a) (x + y + z) (b) (x + y - z) (c) (x + y) (d) (x - y) 14. If x + 1 is a factor of x2 + 8x + a, then the value of a is _________. (a) 5 (b) 6 (c) 7 (d) 8 15. Match the values of Column A with the statements of Column B. Column A

Column B

(i) x(y - z) + y(z - x) + z(x - y) (A) Only homogenous



M01_IIT-CH08.indd 6

(ii) 3x2 - 4xy + 2y2

(B) Only homogenous and symmetric

(iii) x3 + y3 + z3 + 3xyz

(C) Only homogenous and cyclic

(iv) 5x3 + 3xy2 + 3x2y + 5y3

(D) Homogeneous, symmetric and cyclic

(a) (i) → (C); (b) (i) → (D); (c) (i) → (D); (d) (i) → (C);

(ii) → (A); (ii) → (C); (ii) → (C); (ii) → (A);

(iii) → (B); (iii) → (A); (iii) → (B); (iii) → (D);

(iv) → (D) (iv) → (B) (iv) → (A) (iv) → (B)

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Assessment Test IV

Assessment Test IV

Time: 30 min.

8.7

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 2 1 1. The following steps are involved in factorizing x2 + 2x - + 2 - 1. Arrange x x them in sequential order. 2

1 1   2 (A)  x −  + 2  x −  + 1 x x

1 1    (B)  x2 + 2 − 2 + 2  x −  + 1    x x 2

1   (C)  x − + 1 x (D) x2 + 2x

2 1 + 2 -1= x x

 2 1  x + 2  + 2 x

(a) BDAC (c) DABC

1   x −  -1 x (b) BADC (d) DBAC

2. Solve for (5x + 4y)2 - (5x - 4y)2. (a) 50x2 + 32y2 (c) 40xy

(b) 25x2 + 16y2 (d) 80xy

3. If the degree of two polynomials f(x) and g(x) are 15 and 10, respectively, then the degree of [ f(x)·g(x)] is _________. (a) 15 (b) 25 (c) 150 (d) 10 4. The remainder obtained when 2x5 - 3x3 + x2 - 18 is divided by x - 2 is _________. (a) 26 (b) 27 (c) 28 (d) 29

(

)(

)(

)

5. Solve for x2 − 3 x + 2 x2 − 7 x + 6 x2 − 8 x + 12 .

(a) (x - 2) (x - 6) (x - 3) (c) (x - 1) (x - 2) (x - 6)

(b) (x - 3) (x - 6) (x - 4) (d) (x - 4) (x - 2) (x - 1)

If f(x, y) = 3x6 + 4y6 + 8xayb is a homogenous expression, then which of the 6. following is true? (a) a=b (b) a>b (c) a
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8.8

Chapter 8  Polynomials and Square Roots of Algebraic Expressions

7. I f f(x - 2) = 2x2 + 4x + 9, then the remainder when f(x) is divided by (x + 4) is _________. (a) 7 (b) 9 (c) 11 (d) 13 8. x2 - y2 + 2xz + z2 = _________ (a) (x + y + z) (x - y + z) (c) (x + y - z) (x + y - z)

Space for rough work

(b) (x - y - z) (x + y + z) (d) (x + y + z) (y + z - x)

9. Solve for (a4 + b4) (a2 + b2) (a - b). (a) a8 - b8

a8 − b8 (b) a+b

(c) a7 - b7

a7 − b7 (d) a+b

10. If the HCF of x3y2 and xayb is x3y2, then what is the minimum value of a + b. (a) 5 (b) 6 (c) 7 (d) 4 11. The square root of x4 + 6x2 + 4x + 9 + 2 (a) x2 + + 3 x 3 (c) x2 + + 3 x

12 4 + 2 is _________. x x 1 (b) x2 + +3 x 1 (d) x2 + -3 x

12. If (2a + 1) (2a + 2) (2a + 3) (2a + 4) + q is a perfect square, then q = _________. (a) 24 (b) 12 (c) 1 (d) - 8 13. Which of the following is a factor of x(y3 - z3) + y(z3 - x3) + z(x3 - y3)? (a) z + x (b) z-x (c) x - y + z (d) x+y-z 14. If x - 1 is a factor of x2 - bx + 8, then the value of b is _________. (a) 9 (b) 8 (c) 7 (d) 6 15. Match the values of Column A with those of Column B. Column A

M01_IIT-CH08.indd 8

Column B

(i) HCF of (x + 1)3 (x - 1) and (x - 1)2 (x + 2)2

(A) (x - 1)2 (x + 1)2

(ii) HCF of (x + 1)2 (x + 2)2 and (x + 1)2 (x + 3)2

(B) (x - 1)3 (x + 1)3

(iii) LCM of (x - 10) (x + 1)2 and (x - 1)2 (x + 1)

(C) (x - 1)

(iv) LCM of (x - 1)2 (x + 1)3 and (x - 1)3 (x + 1)

(D) (x + 1)2

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Answer Keys (a) (i) → (C); (b) (i) → (D); (c) (i) → (C); (d) (i) → (D);



(ii) → (D); (ii) → (C); (ii) → (D); (ii) → (C);

(iii) → (A); (iii) → (A); (iii) → (B); (iii) → (B);

(iv) → (B) (iv) → (B) (iv) → (A) (iv) → (A)

8.9

Space for rough work

Answer Keys Assessment Test I 1. (c) 11. (a)

2. (c) 12. (a)

3. (a) 13. (c)

4. (d) 14. (b)

5. (d) 15. (d)

6. (c)

7. (a)

8. (a)

9. (d)

10. (c)

4. (b) 14. (b)

5. (b) 15. (a)

6. (d)

7. (d)

8. (c)

9. (c)

10. (d)

4. (d) 14. (c)

5. (b) 15. (d

6. (b)

7. (b)

8. (b)

9. (c)

10. (b)

4. (a) 14. (a)

5. (c) 15. (a)

6. (d)

7. (b)

8. (a)

9. (b)

10. (a)

Assessment Test II 1. (d) 11. (d)

2. (c) 12. (e)

3. (d) 13. (c)

Assessment Test III 1. (c) 11. (c)

2. (a) 12. (a)

3. (c) 13. (d)

Assessment Test IV 1. (d) 11. (a)

M01_IIT-CH08.indd 9

2. (d) 12. (c)

3. (b) 13. (b)

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Thispageisintentionallyleftblank

Statistics and Probability; Applications of Percentages

9

Reference: Coursebook - IIT Foundation Maths Class 9; Chapters - Statistics; Probability; Percentages, Profit and Loss; Simple Interest and Compound Interest; pp. 9.1−9.26; 10.1−10.6; 17.1−17.14; 19.1−19.9

Assessment Test I

Time: 30 min.

Direction for questions 1 to 11: Select the correct answer from the given options.

Space for rough work

1. The following steps are involved in constructing a histogram for a raw data of 100 entries from 0 to 50. Arrange them in sequential order. (A) Divide the data into 5 classes and construct a frequency table. (B) Determine the scale of the graph on the x-axis and y-axis. (C) Find the height of each bar according to the scale and frequency of each class. (D) Take the size of the class on the x-axis and frequency on the y-axis. Draw bars for corresponding frequencies. (a) ABCD (b) ACBD (c) ABDC (d) BCAD 2. 40% of the employees of an office are women. If 70% of the employees are postgraduates, then what is the least percentage of the women who are postgraduates? (a) 10 (b) 15 (c) 20 (d) 25 3. Find the average of P, Q and R, if P is the mean of first 25 natural numbers, Q is the mean of the first 20 odd natural numbers and R is the mean of the squares of the first 13 natural numbers. (a) 32 (b) 23 (c) 63 (d) 21 4. If the difference between the mean and median of a data is x, then what will be the difference between the mean and mode of the data? x x (d) (a) 2x (b) 3x (c) 2 3 5. Find the median of a grouped data, if the lower boundary of the median class is 20, the cumulative frequency of the median class is 24, frequency of the median class is 8, the total frequency of the data is 40 and the length of each class interval is 5. (a) 22 (b) 23 (c) 21.5 (d) 22.5

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9.2

Chapter 9  Statistics and Probability; Applications of Percentages

6. C  alculate the mean of a grouped data for which the assumed mean is 35, total frequency is 40, size of the class is 10 and the sum of the product of the frequencies and their deviation is -8. (a) 38 (b) 37 (c) 33 (d) 30

Space for rough work

7. A certain data is divided into 5 classes. The frequencies of the classes 20-40, 40-60, and 60-80 are 6, 10 and 9, respectively. And there are at least 2 entries in each of the remaining classes. The total frequency of the data is 35. Find the mode of the data. (a) 56 (b) 45 (c) 42 (d) 47 8. A retailer sold an article at a 10% loss. If he had sold it at `14 more, then he would have gained 10%. What will be the cost price of the article (in `)?

(a) 70

(b) 65

(c) 72

(d) 84

 die is rolled 100 times and recorded each outcome. If 1 occurred ten times, 9. A 2 twenty times, 3 fifteen times, 4 fifteen times and 5 twenty times, then what is the probability of getting a 6? 1 (a) 6

1 (b) 5

1 (c) 4

(d)

1 3

10. On a page, there are hundred words of equal number of letters. Of them, 60 start with a consonant, 10 words start with ‘a’, 12 words start with ‘e’, 8 words start with ‘i’ and 4 words start with ‘o’. Without looking at the page, if a word is selected, then what is the probability that the word starts with ‘u’? 3 (a) 50

4 (b) 70

5 (c) 60

3 (d) 40

11. In an office, 40% of employees are women. All employees are classified as Group I, Group II, Group III and Group IV. The number of employees who are men in Group I, II, III and IV are 10%, 20%, 50% and 20%, respectively, of the total number of men employees. The company wanted to select the luckiest person and give a car. Hence, the names of all the employees were written on distinct slips and are put in a box. If a slip is picked randomly, then what is the probability of a man being selected from Group II? 3 (a) 50

7 (b) 60

4 (c) 25

3 (d) 25

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Assessment Test I Direction for questions 12 to 15: Match the statements of Column A with those of Column B. Column A

Column B

(Situation)

(Suitable Measure)

12. To know the deviation of the data

(a) Mean

13. To know the popular brand of an article

(b) Median

14. When extremes are not important

(c) Mode

15. When all observations are important

(d) Range

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9.3

Space for rough work

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9.4

Chapter 9  Statistics and Probability; Applications of Percentages

Assessment Test II

Time: 30 min.

Space for rough work

Direction for questions 1 to 11: Select the correct answer from the given options. 1. The following steps are involved in constructing a frequency polygon for the data divided into 5 groups with classes 5-10, 11-16, … Arrange them in sequential order. (A) Find the mid-value of each class. (B) Convert the class into continuous classes. (C) Mark the points on the graph for the mid-values and their corresponding frequencies.

(D) Join these points with the scale. (E) Take the mid-values on the x-axis and frequencies on the y-axis. (a) BAECD (b) BACDE (c) ABDCE (d) BACED

 he students of a class appeared for a test. Of them, 40% of the total number 2. T of students of the class are boys. 25% of the number of girls secured at least 70% marks in the test. Find the percentage of the number of girls who scored less than 70% in the class.

(a) 50%

(b) 45%

(c) 30%

(d) 60%

 ind the median of A, B, C and D, where A is the mean of first 21 natural 3. F numbers. B is the mean of the squares of first 7 natural numbers. C is the sum of the first 10 odd natural numbers and D is the sum of the first 10 even natural numbers. (a) 50 (b) 60 (c) 40 (d) 30 4. I f the difference between the median and the mean of an ungrouped data is 6, then what will be the difference between the mode and the median of the data? (a) 18 (b) 10 (c) 11 (d) 12 5. For a grouped data, the median class is 30-40. The frequency and the cumulative frequency of the median class are 10 and 25, respectively. The total frequency of the data is 40. Find the median of the grouped data. (a) 34 (b) 35 (c) 32 (d) 36 6. Calculate the mean of a grouped data in which the sum of the frequencies is 50 and sum of the products of the frequency and mid-values of each class is 1400. (a) 20 (b) 25 (c) 26 (d) 28 7. In a secondary school, due to a fire accident, data pertaining to marks was partially destroyed. The IX standard had 50 students in this school.

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Assessment Test II The marks scored by them in Biology were subjected to analysis. The statistics teacher could gather that, the data was divided in classes 0-10, 10-20, 20-30 and 90-100. Furthermore, the frequencies of the classes 30-40, 40-50, and 50-60 are 9, 13 and 7, respectively, and there are at least 2 entries in each of the remaining classes. The teacher wanted to find the mode of the data. Is it possible to find the mode? If yes, what is the mode of the data? (a) 46 (b) 44 (c) 48 (d) 45



9.5

Space for rough work

8. A man sold an article at 20% loss. If he had sold it for `40 more, then he would have gained 30%. Find the cost price of the article (in `). (a) 90 (b) 60 (c) 100 (d) 80 9. If a die is rolled, which of the following events has the least probability? (a) Getting an odd number (b) Getting a prime number (c) Getting a composite number (d) Getting an even number 10. There are 60 cell phone numbers in a page. Of them, 20 start with 99, 15 start with 84, 12 start with 98 and the remaining numbers start with 90. Without looking into the page, if you select a number, what is the probability that the number should start with 84? 1 1 1 1 (a) (b) (c) (d) 2 3 4 5 11. A company EMIT produces playing articles for children. The production of the articles in the year 2012 is as follows: Grade I

Grade II

Grade III

Grade IV

Total

70%

20%

x%

4%

100%

If an article is selected from the production of the year 2012 randomly, then what is the probability that it is a Grade III article? 3 2 1 3 (c) (d) (a) (b) 50 35 40 40 Direction for questions 12 to 15: Match the statements of Column A with those of Column B. Column A

Column B

(Statistical Measure)

(One of the Required Information)

12. Mean

(a) Cumulative frequency

13. Median

(b) Mid-values of the classes

14. Mode

(c) Maximum value

15. Range

(d) Maximum frequency

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9.6

Chapter 9  Statistics and Probability; Applications of Percentages

Assessment Test III

Time: 30 min.

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. The following steps are involved in finding profit (or) loss percentage for a shopkeeper, who bought an article for `1000 and sold it for `1500. Arrange them in sequential order. (A) Gain = S.P. - C.P. = `1500 - `1000 = `500 (B) As S.P. > C.P., there is a gain. Gain (C) Gain percentage = × 100% C.P. 500 (D) Gain percentage = × 100% = 50% 1000

(a) BACD (c) BCAD

(b) ABCD (d) BADC

2. T  he range of the scores 30, 24, 37, 52, 28 and x, where x < 24 is 31. Find the value of x. (a) 28 (b) 31 (c) 24 (d) 21 3. W  hen two coins are tossed, find the number of outcomes for getting at least one head. (a) 1 (b) 2 (c) 3 (d) 4 4. I f the difference of the median and the mean of a data is 16, then the difference of the mode and the mean is _________. (a) 16 (b) 32 (c) 48 (d) 64 1 1 + ? x y 1 4 2 3 (a) (b) (c) (d) 5 5 5 5 5. I f (2x% of y) + (x% of 3y) = 12.5% of (x + y), then what is the value of

6. T  hree biased coins were tossed 1000 times simultaneously. The outcomes are given in the table below: Outcome

No tail

One tail

Two tails

Frequency

160

330

x

The occurrence of two tails was twice that of all tails. Find the value of x.

(a) 270

(b) 540

(c) 340

7. Find the quartile deviation of 2, 4, 9, 13, 23, 29 and 32. (a) 10.5 (b) 11.5 (c) 12.5

M01_IIT-CH09.indd 6

(d) 510

(d) 13.5

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Assessment Test III 8. A  sum becomes four times itself in 5 years at simple interest. In how many years will the same sum become sixteen times itself at the same rate? (a) 18 (b) 25 (c) 22 (d) 14

9.7

Space for rough work

9. The value of an old article decreases every year at the rate of 16% over that of the previous year. If its value at the end of two years will be `194,481 then find its present value. (a) `200,625 (b) `225,525 (c) `257,825 (d) `275,625 10. The mean deviation of 24 and 32 is _________. (a) 4 (b) 14 (c) 7

(d) 28

11. The median of a data consisting of odd number of observations is 30. If two observations 10 and 40 are removed from the data, then the median of the new data is _________. (a) 30 (b) 25 (c) 35 (d) 20 12. The mode of the following data is 36.25. Find the value of x.



Class

0-10

10-20

20-30

30-40

40-50

50-60

Frequency

2

7

x

23

20

10

(a) 16

(b) 18

(c) 20

(d) 21

13. If the cost price of an article is increased by 60%, then by what percentage should it be decreased to bring it back to its original price? 1 1 4 (c) 28 % (d) 37 % (a) 60% (b) 28 % 7 2 2 14. Mr Reddy wants to borrow `9000 at 10% simple interest and lend the same amount at compound interest at same rate of interest for two years. What would be his income in the above transaction? (a) `80 (b) `90 (c) `100 (d) `110 15. Match the statements of Column A with the values of Column B. Column A

M01_IIT-CH09.indd 7

Column B

(i) Mean of first 101 natural numbers

(A) 50

(ii) Mean of the squares of first 101 natural numbers

(B) 101

(iii) Mean of first 101 odd natural numbers

(C) 51

(iv) Mean of first 101 whole numbers

(D) 3451

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9.8

M01_IIT-CH09.indd 8

Chapter 9  Statistics and Probability; Applications of Percentages (a) (i) → (C); (b) (i) → (A); (c) (i) → (C); (d) (i) → (A);

(ii) → (D); (ii) → (D); (ii) → (D); (ii) → (D);

(iii) → (A); (iii) → (C); (iii) → (B); (iii) → (C);

(iv) → (B) (iv) → (B) (iv) → (A) (iv) → (B)

Space for rough work

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Assessment Test IV

Assessment Test IV

Time: 30 min.

9.9

Space for rough work

Direction for questions 1 to 15: Select the correct answer from the given options. 1. A shopkeeper gains 20%, by selling an article for `48. The following steps are involved in calculating the gain at 50%. Arrange them in sequential order.  40 × 150   100 × 48  = `60 = `240 (B) S.P. =  (A) C.P. =   100   100 + 20  (C) C.P. = `40, Gain = 50% (D) S.P. = `48, Gain = 20% (a) ADBC (b) ABCD (c) CDAB (d) DACB 2. The range of the scores 26, 19, 32, 23 and x is 10, where x < 19. What will be the value of x? (a) 19 (b) 14 (c) 12 (d) 16 3. When three coins are tossed, find the number of outcomes for getting at least 2 heads. (a) 1 (b) 2 (c) 3 (d) 4 4. If the difference of the median and the mean of a data is 20, then what will be the difference of the mode and the median? (a) 20 (b) 40 (c) 60 (d) 80 5. If (x% of y) + (3% of xy) = 12% (x + y), then what is the value of

1 1 + ? x y

1 2 1 3 (a) (b) (c) (d) 3 3 4 4 6. T  he outcomes are given in the table below, when three biased coins were tossed 600 times simultaneously: Outcome

No head

One head

Two heads

Frequency

x

250

50

The occurrence of no heads was two-third that of all the three heads. Find the value of x.

(a) 300

(b) 180

(c) 220

(d) 120

7. F  ind the quartile deviation of the following data 6, 10, 12, 13, 14, 18, 22, 26. (a) 4 (b) 7 (c) 9 (d) 5

M01_IIT-CH09.indd 9

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9.10

Chapter 9  Statistics and Probability; Applications of Percentages

8. A  sum becomes twice itself at compound interest in six years. In how many years will the same become sixteen times itself? (a) 18 (b) 24 (c) 9 (d) 27

Space for rough work

9. The value of a machine depreciates at the rate of 7.5% p.a. every year. What was the value of the machine two years ago, if its present value is `1711.25? (a) `2500 (b) `2000 (c) `1950 (d) `2100 10. The mean deviation of 2, 6, 5, 10, 14, 8 and 17 about the median is _________. (a) 4 (b) 5 (c) 6 (d) 7  he median of a data consisting of odd number of observations is 45. If two 11. T observations 34 and 48 are included to the data, then what is the median of the new data? (All observations are integers.)

(a) 34 (c) 45

(b) 44 (d) 48

12. The mean of the following data is 29. Find the value of m. Class

1-9 10-18

Frequency 9

(a) 27

11

19-27 28-36 37-45 46-54 8

(b) 81

m

3

11

(c) 65

(d) 54

13. If A gets 50% more than B, then by what percentage does B get less than A?

(a) 50%

(b) 33

1 % 3

(c) 16

2 % 3

(d) 150%

14. The difference between simple interest and compound interest on `40,000 at the end of two years is `400. The rate of interest is same in both the cases. What is the rate of interest? (a) 40% (b) 20% (c) 10% (d) 50% 15. Match the values of Column A with those of Column B. Column A (i) 40 cm = ?% of 6 m

(A) 125

(ii) 87.5% of 125 + 12.5% of 125 = ?% of 100

(B) 6

5 times C.P., then loss percentage is 6 6 (iv) If S.P. is times C.P., then profit percentage is 5

(iii) If S.P. is

M01_IIT-CH09.indd 10

Column B

2 3

(C) 16

2 3

(D) 20

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Answer Keys

(a) (i) → (C); (b) (i) → (A); (c) (i) → (B); (d) (i) → (B);

(ii) → (A); (ii) → (B); (ii) → (A); (ii) → (A);

(iii) → (A); (iii) → (D); (iii) → (D); (iii) → (C);

9.11

(iv) → (D) (iv) → (C) (iv) → (C) (iv) → (D)

Answer Keys Assessment Test I 1. (a) 11. (d)

2. (d) 12. (d)

3. (a) 13. (c)

4. (b) 14. (b)

5. (d) 15. (a)

6. (c)

7. (a)

8. (a)

9. (b)

10. (a)

4. (d) 14. (d)

5. (b) 15. (c)

6. (d)

7. (b)

8. (d)

9. (c)

10. (c)

4. (c) 14. (b)

5. (c) 15. (c)

6. (c)

7. (c)

8. (b)

9. (d)

10. (a)

4. (b) 14. (c)

5. (a) 15. (d)

6. (d)

7. (a)

8. (b)

9. (b)

10. (a)

Assessment Test II 1. (a) 11. (a)

2. (b) 12. (b)

3. (b) 13. (a)

Assessment Test III 1. (a) 11. (a)

2. (d) 12. (b)

3. (c) 13. (d)

Assessment Test IV 1. (d) 11. (c)

M01_IIT-CH09.indd 11

2. (d) 12. (d)

3. (d) 13. (b)

4/16/2018 12:46:01 PM

Thispageisintentionallyleftblank

Hints and Explanation CHAPTER 1 Coordinate Geometry and Trigonometry Assessment Test I

(1) A(9, 0), B(0, 12) (2) A(12, 0), B(0, 9) (3) A(-9, 0), B(0, 12) (4) A(-12, 0), B(0, 9) (5) A(-9, 0), B(0, -12) (6) A(-12, 0), B(0, -9) (7) A(+9, 0), B(0, -12) (8) A(12, 0), B(0, -9) Eight circles can be drawn. Hence, the correct option is (b).

1. BECDA is the required sequential order. Hence, the correct option is (c). (3 a − a)2 + ( 4 a − 2 a)2

2. =

4 a2 + 4 a2

= 2 2 a units Hence, the correct option is (c). 3. Let the required equation be 3x - 5y = k It is passing through (5, 3) ⇒ 3(5) - 5(3) = k ⇒ k = 0 \ 3x - 5y = 0 or 9x - 15y = 0 Hence, the correct option is (b).

6.

x=6 0 4x − 3y = 24

4. Two points form of a line is y −y y - y1 = 2 1 (x - x1) x2 − x1 y - 2014 =

6 units (0, −8)

2015 − 2014 ( x − 2012) 2013 − 2012

5. 7.

B1 O

1 A A

OAB is a right triangle OA = 9, OB = 12, and AB = 15.

M01_IIT-Hints and Solutions.indd 1

(6, 0)

X

8 units (6,−8)

x-6=0⇒x=6 y + 8 = 0 ⇒ y = -8 4x - 3y = 24 From the figure, 1 Area of the triangle = × 6 × 8 = 24 sq. units 2 Hence, the correct option is (a).

y - 2014 = x - 2012 x - y - 2012 + 2014 = 0 ⇒x-y+2=0 Hence, the correct option is (a). B

Y

Y A(0, 4)

B(0, 0)

D(4, 0)

C(8, 0)

X

4/16/2018 11:31:28 AM

A.2

AD intersects the x-axis at (4, 0) and the y-axis at (0, 4). \ Equation of median AD is x + y = 4. Hence, the correct option is (d).

8. In a parallelogram, diagonals bisect each other. ⇒ Midpoint of the diagonal AC is the mipoint of the diagonal BD  3 + m 5 − 4  3 − 2 k + 2  1 1 , , ⇒  =  = ,   2 2   2 2   2 2 Hence, the correct option is (b). 9. 3, 4, 5 and 5, 12, 13 are Pythagorean triplets. 3 4 sin A = ⇒ cos A = 5 5 5 12 cos B = ⇒ sin B = 13 13 sin (A + B) = sin A cos B + cos A sin B =

15 + 48 63 3 5 4 12 × + × = = 5 13 5 13 65 65

Hence, the correct option is (b). sin 30° × cos 40° × tan 50° × cot 60° 10.  cosec 60° × sin 50° × cot 40° × sec 30° 1 1 × sin 50° × cot 40° × 2 3 = 2 2 × sin 50° × cot 40° × 3 3 1 1 3 3 3 = × × × = 2 2 2 8 3 Hence, the correct option is (d). 1 =2 sinq ⇒ sin2 q + 1 = 2 sin q ⇒ sin2 q - 2 sin q + 1 = 0

11. sin q + cosec q = 2 ⇒ sin q + 

⇒ (sin q - 1)2 = 0 ⇒ sin q = 1 ⇒ cosec q = 1 \ sin2014q - cosec2013q = 1 - 1 = 0 Hence, the correct option is (d). pc pc 12. Option (b): sin = sin 45° = cos 4 4

M01_IIT-Hints and Solutions.indd 2

13. Option (a): sec 54° × sin 36°   = sec 54° × cos 54°   ( 36° + 54° = 90°) ∴



Chapter 1  Coordinate Geometry and Trigonometry

  = 1 = tan 14. Option (c): tan

pc 4

pc pc × cot 3 4

  = tan 60° × cot 45° = cot 30° = cot 15. Option (d): cos

100g

  = cot

3 ×1 =

3 =

pc 6

= cos 90° = 0

pc 2

Assessment Test II 1. B  DCAE is the required sequential order of steps. Hence, the correct option is (c). 2. Diameter of the circle =

( x2 − x1 )2 + ( y2 − y1 ) 2

= (2011 − 2013)2 + (2012 − 2014)2 = 4 + 4 = 2 2 \ Radius of the circle =

2 2 = 2

2 units

Hence, the correct option is (b). 3. Equation of a line perpendicular to 2x + 3y = 7 is 3x - 2y = k. It is passing through (2000, 2013). ⇒ 3(2000) - 2(2013) = k \ k = 6000 - 4026 ⇒ k = 1974 \ The required equation in 3x - 2y = 1974. Hence, the correct option is (d). 4. Point slope form of a line: y - y1 = m(x - x1) y - 2013 = 1(x - 2012) x - y - 2012 + 2013 = 0 ⇒x-y+1=0 Hence, the correct option is (d). 5. I nfinite circles are possible with the given conditions. Hence, the correct option is (d).

4/16/2018 11:31:37 AM

Hints and Explanation 6.  (−1, 3)

y−3=0

A

(5, −3) B

x1 1 2 3 4 5 x+y=2

C

x=5=0 (5, −3)

y1

From the figure, A B = (1 + 5) = 6 units, B C = 3 + 3 = 6 units, and A C = (5 + 1)2 + ( −3 − 3)2 = 6 2 units \ Perimeter of the triangle ABC = 6 2 + 6 + 6 = 6(2 + 2) units Hence, the correct option is (c). 7. A(-2011, -2011), B(2011, -2011) ⇒ AB is parallel to x-axis since y-coordinates of A and B are same. Also, A are B are equidistant from y-axis. Therefore, the midpoint of AB is on y-axis and the perpendicular bisector of AB is y-axis. \ Its equation is x = 0. Hence, the correct option is (a). 8. The diagonal PR is the part of the x-axis. The  2000 − 2012 0 + 0  , . midpoint of PR is   2  2 = (-6, 0) Hence, the correct option is (c). 4 3 9. cos a = ⇒ tan a = 5 4 8 15 cos b = ⇒ tan b = 17 8 tan a + tan b tan(a + b ) = 1 − tan a tan b 3 15 + − 84 21 32 = = 4 8 × = 3 15 8 32 − 45 13 1− × 4 8 Hence, the correct option is (b). 10. 

tan 20° × cot 40° × sec 60° × cosec 80° sec 10° × cosec 30° × tan 50° × cot 70°

M01_IIT-Hints and Solutions.indd 3

=

A.3

tan 20° × cot 40° × 2 × cosec 80° =1 cosec80° × 2 × cot 40° × tan 20°

Hence, the correct option is (a). 4 11. t an q + cot q = 3 tan2 q + 1 4 1 4 tan q + = ⇒ = tanq tan q 3 3 3 tan2 q + 3 = 4 tan q 3 tan2q - 4 tan q + 3 = 0 3 tan2q - 3 tan q - tan q + 3 = 0 3 tan q (tan q - 3 ) - 1 (tan q - 3 ) = 0 (tan q - 3 ) ( 3 tan q - 1) = 0 ⇒ tan q - 3 = 0 or 3 tan q - 1 = 0 1 ⇒ tan q = 3 or tan q = 3 2014 2014 1007 tan q = ( 3) =3 Hence, the correct option is (b). 2p c 180° 12. Option (e):  = 2 × = 120° 3 3 9° = 180° 13. Option (f): 200g = 200 × 10 14. Option (b): cos q = sin q ⇒ q can be 45° pc 180° 15. Option (c): tan = tan = tan 36° = cot 54° 5 5   (since 36° + 54° = 90°)

Assessment Test III 1. T  he required sequential order is CDAB. Hence, the correct option is (b). 2. Given (−x, −y) ∈Q2 ⇒ x > 0, y < 0 \ (−x, y) ∈ Q3 Hence, the correct option is (c). 3. Given, G = 200g pc We have = 100g ⇒ p c = 200g 2 Hence, the correct option is (c). 4. Let C(x, 0) be the point on x-axis. Let A(2, 4) and B(3, 2) Given, AC = BC AC2 = BC2 (x − 2)2 + (0 − 4)2 = (x −3)2 + (2 − 0)2 x2 + 4 − 4x + 16 = x2 + 9 − 6x + 4

4/16/2018 11:31:51 AM

A.4

Chapter 1  Coordinate Geometry and Trigonometry 2x = −7 −7 x= 2

 −7  , 0 . \ Required point = (x, 0) =   2  Hence, the correct option is (d).  he length of the minute hand of a wall lock (r) 5. T = 30 cm We know that 1 min = 6° 42 min = 6° × 42 = 252° q Required distance = × 2p  r 360° 252° 22 ×2× × 30 360° 7 7 22 = ×2× × 30 10 7 = 132 cm Hence, the correct option is (c). 1   1   −1 1+ 6. Given, cot2a    cosa   cosa   1  − 1 = cot2a   cos2 a   1 − cos2 a  = cot2a   cos2 a   sin2 a  = cot2a   cos2 a  = cot2a (tan2 a  ) = 1 Hence, the correct option is (c). 7. Given, 3x − 2y + 2015 = 0  (1) Let 3x − 2y = k be a line which is parallel to 3x − 2y + 2015 = 0 and passing through (1, 1). ⇒ 3(1) − 2(1) = k ⇒3−2=k ⇒k=1 \ Required line is 3x − 2y − 1 = 0. Hence, the correct option is (d). Given, tanq + cotq = 2 8. 1 tanq  +  = 2 tanq tan2 q + 1 =2 ⇒ tan q ⇒ tan2q + 1 = 2 tan q

M01_IIT-Hints and Solutions.indd 4

⇒ tan2q + 1 − 2 tanq = 0 ⇒ (tanq − 1)2 = 0 ⇒ tanq = 1 ⇒ cot q = 1 \ tan2015q + cot2015q = (1)2015 + (1)2015 =1+1=2 Hence, the correct option is (c). 9. Given, x = cosq + sinq , y = sinq − cosq x2 = cos2q + sin2q + 2 sinq  cosq  (1) 2 2 2 y = sin q + cos q − 2 sin q cosq  (2) 2 2 2 2 From Eqs (1) + (2), ⇒ x + y = 2(sin q + cos q ) x2 + y2 = 2 Hence, the correct option is (a). 10. Given, line 3x + 4y − 24 = 0 ⇒ 3x + 4y = 24 3x 4 y + ⇒ =1 24 24 x y =1 ⇒ + 8 6 1 Required area = |8 × 6| 2 1 = × 48 = 24 sq. units 2 Hence, the correct option is (a). 11. Given points (2k + 1, 2k + 2), (3k + 6, k + 10) y − y1 Slope = 2 x2 − x1 1 (k + 10) − (2k + 2) = 2 (3 k + 6) − (2k + 1) 1 k + 10 − 2k − 2 = 2 3 k + 6 − 2k − 1 1 −k + 8 ⇒ = 2 k+5 ⇒ k + 5 = −2k + 16 ⇒ 3k = 11 11 ⇒k= 3 Hence, the correct option is (d). 12. log sin 30° + log sin 45° + log sin 60° = log (sin 30° × sin 45° × sin 60°) 1 1  3  3 × = log  = log  ×   2  2 2  4 2 1

 3 2 = log    32 

4/16/2018 11:32:02 AM

Hints and Explanation =

1  3 log   .  32  2

(ii) cos  15° = cos(45° − 30°) =

Hence, the correct option is (c). 13. Given points A(4, −5), B(6, −7) From option (b), let C(8, −9) ( −9 + 5)2

AC =  4 2 units

+ (8 − 4)2

=  16 + 16 =  32 =

AB =  ( −7 + 5)2 + (6 − 4)2 =  4 + 4  = 8 = 2 2 units BC =  ( −9 + 7 )2 + (8 − 6)2 =  4 + 4 = 8 = 2 2 \ AB + BC = AC. Alternative method: Given (4, −5), (6, −7) and (x, y) are collinear. y+5 −5+7 = = −1 x−4 4−6 From option (b), −9 + 5 = −1 8−4 Slope = −1 Hence, the correct option is (b). 14. Given =

1 + tan q 1 − tan q + 1 − tan q 1 + tan q

(1 + tan q )2 + (1 − tan q )2 (1 − tan q )(1 + tan q )

1 + tan2 q + 2 tan q + 1 + tan2 q − 2 tan q 1 − tan2 q 2 2(1 + tan q ) = 1 − tan2 q =

=

2(sec2 q )  cos2 q − sin2 q    cos2 q

2 cos 2q = 2 sec2q Hence, the correct option is (d).

=

15.   (i) sin  15° = sin (45° − 30°) = sin 45°cos 30° − 3 −1 1 3 1 1 − = sin 30°cos 45° = 2 2 2 2 2 2

M01_IIT-Hints and Solutions.indd 5

A.5

3 +1 2 2

(iii)  tan 75° = tan(45° + 30°) 1 3 = = 1 1− 3 1+

(iv)  cot 75° =

3 +1 3 −1

=2+ 3

1 1 = = 2− 3 tan 75° 2 + 3

(i) → (C); (ii) → (D); (iii) → (B); (iv) → (A) Hence, the correct option is (d).

Assessment Test IV 1. T  he required sequential order BDAC. Hence, the correct option is (d). 2. Given (x, y) ∈ Q2 ⇒ x < 0, y > 0 \ (−x, y) ∈ Q1 Hence, the correct option is (a). 3. Given, R = 75° c p We have 90° = 2 pc 1° = 180 pc 75° = × 75 180 5p c = 12 Hence, the correct option is (b). 4. Let C(0, y) be the point on y-axis. Let A(2, 4) and B(3, 5) Given AC = BC ⇒ AC2 = BC2 (0 − 2)2 + (y − 4)2 = (0 − 3)2 + (y − 5)2 4 + y2 + 16 − 8xy = 9 + y2 + 25 − 10y 2y = 14 y=7 \ Required point (0, y) = (0, 7). Hence, the correct option is (c). 5. G  iven, the length of the minute hand of a wall clock (r) = 15 cm.

4/16/2018 11:32:13 AM

A.6

Chapter 1  Coordinate Geometry and Trigonometry We know that 1 min = 6°. 28 min = 28 × 6° q 28 × 6° Required distance= × 2p r = ×2 360 360° 22 × × 15 7

= 44 cm Hence, the correct option is (b). 1   1   6. G  iven,  sin a −   cos a −   sin a   cos a  1    tan a +  tan a  − 1 sin a  

 sin2 a

=  

− 1  cos a  

 cos2 a

+ 1 tan a 

tan2 a

2  − cos2 a   − sin2 a   sec a  =     sin a   cos a   tan a 

 1   cos a  = (cos a  ) (sin a  )   cos2 a   sin a  =1 Hence, the correct option is (c). 7. Given, 4x − 5y − 2016 = 0 Let 5x + 4y = k be a line which is perpendicular to 4x − 5y − 2016 = 0 and passing through (1, −1) ⇒ 5(1) + 4(−1) = k ⇒5−4=k ⇒1=k \ Required line is 5x + 4y = 1. \ Required line is 5x + 4y − 1 = 0. Hence, the correct option is (c). 8. Given, sec4q + tan4q = (sec2q )2 + (tan2q )2 =

4 3

4 3

4 (sec2q )2 + (tan2q )2 − 2sec2q  .tan2q  = − 2 sec2q  . 3 tan2q  4 (sec2q − tan2q )2 = − 2 sec2q  tan2q  3 4 (1)2 = − 2sec2q  tan2q  3 4 ⇒ −2sec2q  tan2q  = 1 − 3

M01_IIT-Hints and Solutions.indd 6

⇒ −2sec2q  tan2q = ⇒ 2sec2q   tan2q = ⇒ sec2q tan2q = ⇒ secq  tanq = ±

1 3

−1 3

1 6 1

. 6 Alternative method: We know, sec2q + tan2q = 1 ⇒ sec4q + tan4q − 2sec2q .tan2q = 1 4 1 ( 2 secq ·tanq )2 = − 1 = ⇒secq ·tanq = 3 3 1 ± 6 Hence, the correct option is (d). 9. G  iven, x = 1 + a sin q, y = 1 + a cosq x − 1 = a sinq, y − 1 = a cosq (x − 1)2 = a2sin2q 

(1) (2)

(y − 1)2 = a2cos2q  From Eqs (1) and (2) ⇒ (x −1)2 + (y − 1)2. = a2sin2q + a2cos2q (x − 1)2 + (y − 1)2 = a2 Hence, the correct option is (b). 10. Given line 4x − 5y − 20 = 0 ⇒ 4x − 5y = 20 4x 5y − ⇒ =1 20 20 x y =1 ⇒ − 5 4 1 Required area = |5x (−4)| 2 1 = × 20 2 = 10 sq. units Hence, the correct option is (c). 11. Given points (2k + 6, 3k + 5), (5k − 5, 6k − 4) y − y1 Slope = 2 x2 − x1 9 6 k − 4 − (3 k + 5) = 8 5k − 5 − ( 2k + 6)

4/16/2018 11:32:26 AM

A.7

Hints and Explanation 9 6k − 4 − 3k − 5 = 8 5k − 5 − 2k − 6

AC =

9(3k − 11) = 8(3k − 9) 27k − 99 = 24k − 72 3k = 27 k=9 Hence, the correct option is (c).

= =

1 log 2 2 Hence, the correct option is (d).

)

1 − tan2 q

cos2 q  sin q  × = 4   cosq  cos 2q (sin q cosq ) =4 cos 2q 2(2 sin q cosq ) = cos 2q (sin 2q ) =2 cos 2q

13. Given, A(1, 3) and B (3, 7) Let the other point C be (x, y). y−3 7−3 4 \ = = =2 x −1 3 −1 2 5−3 From option (d), = 2 ⇒ 2 = 2. 2−1

M01_IIT-Hints and Solutions.indd 7

(

1 + tan2 q + 2 tan q − 1 + tan2 q − 2 tan q

4 tan q sin2 q 1− cos2 q 4 tan q = 2  cos q − sin2 q   (cos2 q ) 

= log 2 =

4+1 =

(1 + tan q )2 − (1 − tan q )2 (1 − tan q )(1 + tan q )

=

 1  = log  × 3 × 2  3 

(5 − 7 )2 + (2 − 3)2 =

5 units

1 + tan q 1 − tan q 14.  − 1 − tan q 1 + tan q

12. log tan 30° + log cot 30° + log cosec 45° 1 + log 3 + log 2 = log 3

BC =

4+1 =

\ AB + BC = AC We can check for slope from the options. Hence, the correct option is (d).

9 3k − 9 = 8 3 k − 11

Alternative method: Given, points A(1, 3) and B(3, 7) From option (d), Let C(2, 5). AB = (7 − 3)2 + (3 − 1)2 = 66 + 4 = 2 5 units

(5 − 3)2 + (2 − 1)2 =

20 = 5 units

= 2tan2q. Hence, the correct option is (a). 15. (i) → (D); (ii) → (A); (iii) → (B); (iv) → (C) Hence, the correct option is (c).

4/16/2018 11:32:38 AM

A.8

Chapter 2  Geometry-I and Geometry-II

CHAPTER 2 Geometry-I and Geometry-II Assessment Test I

B

1. B  DCA is the required sequential order. Hence, the correct option is (d).

I

2. 

C

A

∠B ∠AIC = 90° + = 90° + 45° = 135° 2 Hence, the correct option is (b).

C 70° A

B

D

6. KLMN is a rhombus. KL = LM = MN = NK KM = LM = LK. = MN = NK (given) ⇒ ∠KLM = 60° ⇒ ∠ LMN = 120° Hence, the correct option is (a).

∠ACB = 70° Reflex angle ACB = 290° 290° \ ∠ADB = = 145° 2 Hence, the correct option is (b).

7. 

V

A

3. 

120°

G

E

60° X

B

12 cm

C

R

S A

C Q

P

B

In DABC, PQ = (1/2) AC, and in DADC, SR = (1/2) AC. \ PQ + SR = AC In an isosceles trapezium, the diagonals are equal. \ SP + RQ = AC \ PQ + QR + RS + SP = 2AC = 30 cm Hence, the correct option is (c). 5. In D ABC, AB = BC. ⇒ ∠B = 90°

M01_IIT-Hints and Solutions.indd 8

60° Y

B

VXYZ is a parallelogram and AYZ is an equilateral triangle. VZ = 2YZ = 2AZ \ A is the midpoint of VZ. Join AB such that AB || VX . Now VABX and ABYZ are two congruent rhombuses. ∠VAB = 60°⇒ ∠XAB = 30° ∠BAZ = 120°⇒ ∠BAY = 60° \ ∠XAY= 30° + 60° = 90° Hence, the correct option is (a).

BC 12 DE = = = 6 cm 2 2 DE 6 GF = = = 3 cm 2 2 Hence, the correct option is (b). D

Z 60° 60°

30° 60°

F

D

4. 

A 30°

C

8.

O B

A D

E

Since AB || DE , AD = BE. Since ∠ACB = 60°, ∠AOB = 120°, and ∠DOE = 60°,

4/16/2018 11:32:48 AM

Hints and Explanation ∠AOD + ∠EOB = 60° Since AD = BE, ∠AOD = ∠EOB = 30°. 30° \ ∠ACD = ∠AED = ∠ABD = = 15° 2 \ ∠AED + ∠ABD + ∠ACD = 3 × 15 = 45° Hence, the correct option is (d). 9. S  ince AD = BD and CD = DE, AE = BC. ⇒ ∠BOC = ∠AOE = 80° D

E 30° A

\ ∠BOA = 360° - 2(140°) = 80° 180° − 80° ⇒ ∠BAO = = 50° 2 \ ∠DAO + ∠BAO = 20° + 50° = 70° Hence, the correct option is (b). C

10.

B

A

\ ∠BOC + ∠AOD = 2 ∠BDC + 2 ∠ABD = 100° + 40° = 140° Hence, the correct option is (a). A 6

3 O

C2

B

Let AB and C1C2 intersect at O. 16 − 9 =

7

OC2 = 62 − 32 = 36 − 9 = 27 \ OC1 + OC2 = 7 + 3 3 Hence, the correct option is (a).

M01_IIT-Hints and Solutions.indd 9

Assessment Test II

C 120° B

A D

 1 ∠ADB =   × (Reflex ∠ACB)  2  1 =   × 240° = 120°  2 Hence, the correct option is (b). 3. 

D

A F

B

O

42 − 32 =

15. Option (d): In an equilateral triangle, circumcircle and incircle are concentric circles.

E

D

OC1 =

14. Option (b): Circumcircle

B

 180° − 80°  ⇒ ∠EAO =   = 50°  2 ⇒ ∠DAO = 50° - 30° = 20° ⇒ ∠AOD = 180° - 2 × 20° = 140° ⇒ ∠BOD = 140°

C1

13. Option (a): Incircle

 2.

O

4

12. Option (e): An excircle and the circumcircle of a triangle are always intersecting circles.

1. C  ABD is the required sequential order. Hence, the correct option is (c).

C

11.

A.9

G C

AE 2 = EB 3 DAEF and DABC are similar triangles \ EF:BC = 2:(2 + 3) = 2:5 BC = 15 cm EF 2 = ⇒ EF = 6 cm 15 5 DCAD ~ DCFG CG:CD = 3:5 AD = 15 cm GF 3 GF 3 = = ⇒ AD 5 15 5 GF = 9 cm \ GF - EF = (9 - 6) cm = 3 cm Hence, the correct option is (b).

4/16/2018 11:35:49 AM

A.10

Chapter 2  Geometry-I and Geometry-II

4. I J = 2 × GH = 2 × 4 = 8 cm

8.

C

KL = 2 × IJ = 2 × 8 = 16 cm O

MN = 2 × KL = 2 × 16 = 32 cm QR = 2 × MN = 2 × 32 = 64 cm

B

A

Hence, the correct option is (b).

E

D

5.

B

Since AB || DE , AD = BE Since ∠ACB = 60°, ∠AOB = 120°, and ∠DOE = 60°, ∠AOD + ∠EOB = 60° ⇒ ∠AOD = ∠EOB = 30° ( AD = BE)

I C

∴

A

∠B = 90° + 50° = 140° 2 Hence, the correct option is (c). ∠AIC = 90° +

6.

∠AOE = 30° + 60° = 90° 90° = 45° 2 \ ∠AOE + ∠ACE = 90° + 45° = 135° Hence, the correct option is (b). ∠ACE =

K

L

N

9. Since AD = BD and CD = DE, AE = BC ⇒ ∠BOC = ∠AOE = 80°  1  1 ⇒ ∠ADE =   ∠AOE =   80° = 40°  2  2 Hence, the correct option is (c).

M

KLMN is a rhombus. KL = LM = MN = NK

10. ∠BDC = 50° = ∠BEC ∠ABD = 40° ⇒ ∠AOB = 180° - 2 × 40° = 100°  1 ⇒ ∠AEB =   ∠AOB  2 \ ∠AEC = ∠AEB + ∠BEC

LN = KL LM = MN = NK (given) ⇒ ∠LKN = 60° ⇒ 60° ∠LKM = = 30° 2 Hence, the correct option is (a). 7.

V

A 40° 40°

Z 50°

50° X

B

= 50° + 50° = 100° Hence, the correct option is (b).

C1 Y

AV = VX = XB = BA = AZ = ZY = YB ABYZ

are

congruent

∠BYZ = 100° = ∠BAZ⇒∠BAY = 50°, ∠VAB = 80°, ∠XAB = 40° \ ∠XAY= 40° + 50° = 90° Hence, the correct option is (a).

M01_IIT-Hints and Solutions.indd 10

8

4 O

C2

B

Let AB and C1C2 intersect at O. OC1 =

\ A is the midpoint of VZ. and

6

50°

VXYZ is a parallelogram. ABYZ is a rhombus.

Now, VABX rhombuses.

A

11.

62 − 42 =

20 = 2 5

OC2 = = 64 − 16 = 48 = 4 3 \ OC1 + OC2 = 4 3 + 2 5 Hence, the correct option is (d). 82

− 42

12. Option (b): Isosceles trapezium is a cyclic quadrilateral.

4/16/2018 11:36:29 AM

Hints and Explanation

4. G  iven, number of sides (n) = 21. Number of diagonals of an n-sided polygon = n (n − 3) , where n = 21. 2

13. Option (e): Parallelogram 14. Option (c): Kite 15. Option (d): Rectangle

Assessment Test III

2. Given, ∠ABC = 30°. An angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc at any point on the remaining part of the circle. ∠AOC = 2 ∠ABC = 60° \ Reflex ∠AOC = 360° - 60° = 300°

O θ 30°

O

A

80° B

E C

D

Given, ∠OAB = 80° Join OB, OC, and OD. Now, OAB is an isosceles triangle ( OA = OB = radii). \ ∠AOB = 180° - 2(80°) = 20° Similarly, ∠BOC = ∠COD = ∠DOE = 20° ( AB = BC = CD = DE) \ ∠AOE = ∠AOB + ∠BOC + ∠COD + ∠DOE = 20° + 20° + 20° + 20° ∠AOE = 80° Now, OAE is an isosceles triangle. ∠OAE = ∠OEA In ∆OAE, ∠AOE + ∠OAE + ∠OEA = 180°. 80° + ∠OAE + ∠OAE = 180° 2 ∠OAE = 100° ∠OAE = 50°. Hence, the correct option is (b). ∴

B

Hence, the correct option is (c). D

C 35°

P

a E

ℓ

20° B

Given AB || CD Draw a line (l) passing through E, parallel to AB. Let us say l meets AD at P. \ AB || l ⇒CD || l ∠AEP = ∠EDC = 35° ( Alternate angles) ∠PEA = ∠EAB = 20° ( Alternate angles) \ ∠DEA = ∠DEP + ∠PEA = 35° + 20° ∠DEA = 55° Hence, the correct option is (c). ∴ ∴

M01_IIT-Hints and Solutions.indd 11

5. 

∴

C

A

A

21 ( 21 − 3 ) 21 (18 )  =  = 189 2 2 Hence, the correct option is (a). =

1. T  he required sequential order is BADC. Hence, the correct option is (b).

3. 

A.11

6. 

D 50°

C

40° 40°

50°

B

O

A

Given, ∠ABC = 50°

4/16/2018 11:37:02 AM

A.12

Chapter 2  Geometry-I and Geometry-II ∴

7. 

E

C

D K B

A

Given, ABCD is a parallelogram and BDEC is a square. Given, BD = 5 cm. ( BD is the shorter diagonal of the parallelogram.) BC = CE = ED = DB = 5 cm ( BDEC is a square.) AC and BD bisect each other at K (say) ( ABCD is a parallelogram). 5 \ BK = cm 2 We know that each angle in a square is 90°. \∠CBD = 90° In ∆BKC, CK2 = BK2 + BC2.

\ Required diagonal length = AC = AK + CK = 2 CK ( AK = CK) ∴

∠BCA = 90° ( An angle in a semicircle is a right angle.) In ∆ABC, ∠CAB + ∠ABC + ∠BCA = 180° ∠CAB + 50° + 90° = 180° ∠CAB = 40° \ ∠CAD = ∠CAB = 40° In ∆ACD, ∠CAD + ADC + ∠DCA = 180° 40° + ∠ADC + 90° = 180° ∠ADC = 50° \ ∠ADB = ∠ADC = 50°. Hence, the correct option is (b).

 5 5 =2   = 5 5 cm  2  Hence, the correct option is (c). 8. D  E || BC A 5 D

B

E

2 6

C

\ ∆ADE ~ ∆ABC AD AB = DE BC 5 AD + DB = 2 6 30 = 5 + DB ⇒ DB = 15 - 5 2 ⇒ DB = 10 cm Hence, the correct option is (b). D

9. 

C

∴

∴

CK2 =

25 + 25 4

CK2 =

125 4

CK =

5 5 2

M01_IIT-Hints and Solutions.indd 12

E

B

Given, ∆BEC is an equilateral triangle and AB || CD. \ ∠BCE = ∠CEB = ∠EBC = 60° \ ∠DAE = ∠CEB = 60° ( ∠DAE and ∠CEB are corresponding angles.) Hence, the correct option is (d). ∴

∴

2

 5 CK2 =   + (5)2  2

A

10. 

48 cm

C

D 14 cm

O A

48 cm

B

Given, AB = CD = 48 cm and AC = BD = 14 cm ∆ABD is a right-angled triangle. AD2 = AB2 + DB2 AD2 = 2304 + 196

4/16/2018 11:37:08 AM

Hints and Explanation AD2 = 2500 ⇒ AD = 50 AD 50 \ required radius = = = 25 cm 2 2 Hence, the correct option is (b).

13. 

A.13

A 40° B

70°

70° C

11. Given, area of ∆ABC = 100 cm2 Area of ∆PQR = 225 cm2 We know that Area of ∆ABC AD2 = Area of ∆PQR PS2 100 AD2 AD = = ⇒ 225 PS2 PS ⇒

D

100 225

⇒

In ∆ABC, AB = AC. \ ∠ABC = ∠ACB and ∠ACB = ∠ABC = 70° (given) In ∆BCD, ∠BDC = 180° - (∠DBC + ∠DCB)

AD 10 = PS 15

AD 2 = \ required ratio = 2:3 PS 3

 180° − ∠ABC 180° − ∠ACB  + = 180° -    2 2

Hence, the correct option is (a). 12. 

D

A

∠ABC + ∠ACB = 70° 2 Hence, the correct option is (b).

=  B

O

14. 

A

Given, AB and CD are diameters of the circle. Let r be the radius of the circle. Clearly, the quadrilateral ACBD is a square ( OA = OB = OC = OD). Let a be side of the square. \ In ∆BOD, BD2 = OD2 + OB2. (a)2 = r2 + r2 (a)2 = 2r2 (a) = 2 r Perimeter of ∆ADB = 2r + 2a = 2r + 2 2r = 2r 1 + 2 Perimeter of the semicircle ADB = p r + 2r = (p + 2)r \ required ratio = (p + 2)r:2r + 2a = (p + 2)r:2r 1 + 2 = (p + 2): 2 2 + 1 × ∴

(

(

(

)

2 −1 2 −1

(

)

)

)

= (p + 2) 2 − 1 :2 Hence, the correct option is (b).

M01_IIT-Hints and Solutions.indd 13

D

50°

C

(

)

x B

x

180 − 2y

y y C

E

Given, ∠BAC = 50° Let ∠ABC = 2x and ∠ACE = 2y. Therefore ∠DBC = x, ∠ACD = y and ∠ACB = 180° - 2y° In ∆ABC, ∠ABC + ∠BCA + ∠CAB = 180° 2x + (180° - 2y) + 50° = 180° 2x - 2y = - 50 ⇒ y - x = 25 In ∆BCD, ∠BDC + ∠DCB + ∠CBD = 180° ∠BDC + (180 - y) + x = 180° ∠BDC = y - x ∠BDC = 25° Hence, the correct option is (a).

4/16/2018 11:37:16 AM

A.14

Chapter 2  Geometry-I and Geometry-II

15.   (i) If ∠A + ∠B = 180°, then ∠A and ∠B are supplementary angles. (ii) If ∠A + ∠B = 90°, then ∠A and ∠B are complementary angles. (iii) Vertically opposite angles are equal. (iv) In a triangle centroid, orthocentre and circumcentre are collinear. Hence, the correct option is (c).

4. G  iven, the number of sides (n) = 17 Number of diagonals of a 17-sided polygon =

n (n − 3) 2

17 (17 − 3 ) = 119 2 Hence, the correct option is (b). =

5. 

Q

Assessment Test IV

S

1. T  he required sequential order is BDAC. Hence, the correct option is (c).

T

Given, ∠QPO = 70° ∠POQ = 40° (

∆OPQ is an isosceles triangle.)

Similarly, ∠QOR = ∠ROS = ∠SOT = 40° (

PQ = QR = RS = ST)

∴

Now, ∠POT = ∠POQ + ∠QOR + ∠ROS + ∠SOT

Q

∠POT = 40° +40° +40° + 40°

θ

∠POT = 160°

C

In ∆ POT, ∠POT + ∠TPO + ∠PTO = 180° 160° + ∠TPO + ∠TPO = 180°

R

Hence, the correct option is (b). 3.  T

R

40° 50° Q

A

Given, SP || RQ Draw a line l passing through ‘T’ and parallel to RQ. Let us say l meets PQ at A. \ RQ || TA ⇒ SP || TA ∠PTA = ∠SPT = 40° ( Alternate angles) ∠QTA = ∠RQT = 50° ( Alternate angles) ∠PTQ = ∠PTA + ∠QTA = 40° + 50° = 90° Hence, the correct option is (a).

2 ∠TPO = 180° - 160° ( triangle.)

∴

P

P

O

∴

2. Given, ∠PQR = 55°. An angle subtended by an arc at the centre of the circle is double the angle subtended by the same arc at any point on the remaining part of the circle. ∠PCR = 2(∠PQR) = 110° \ Reflex ∠PCR = 360° - 110° = 250°

S

R

P

∆POT is an isosceles

2 ∠TPO = 20° ∠TPO = 10° Hence, the correct option is (a). 6. 

D A

O

45° B

E

∴

∴

M01_IIT-Hints and Solutions.indd 14

C

Given, ∠ADE = 45° Join BE ∠ABE = ∠ADE = 45°

4/16/2018 11:37:19 AM

Hints and Explanation ( Angles in the same segment of a circle are equal.) ∠AEB = 90° ( An angle in a semicircle is a right angle.) In ∆ ABE, ∠ABE + ∠BEA + ∠EAB = 180° 45° + 90° + ∠EAB = 180° ∠EAB = 45° ⇒ ∠CAB = 45° \ (A = BC) ∠ACB = ∠CAB = 45° Hence, the correct option is (b).

8. 

∴

A

∴

7. 

F

K

D

A

E

6

C

6

B

Given, ABCD is a rectangle and DECF is a square. Given, AB = 12 cm. AE = =

DE2 − AD2

EC2 − BC2 = EB AE = EB = 6 cm.

∴

Join EF, CD, and EF bisect each other at K (say) ( DECF is a square).

A.15

D 40 16 12

E

B

C

DE || AB ∠ABC = ∠DEC = 90° \ ∆ABC ~ ∆DEC AB DE = BC EC 40 16 = BE + 12 12 40 × 12 BE + 12 = 16 BE + 12 = 30 BE = 18 cm Hence, the correct option is (c). 9. G  iven, ABCD is an isosceles trapezium and AB||CD . Given ∠DAE = 30° ⇒ ∠ABC = 30° In DEBC, ∠BCE = 180° - ∠ABC - ∠BEC. = 180° - 30° - 90° = 60° Hence, the correct option is (a). 10. In DCEB, ∠CEB = 90° ∠EBC = 30°. ⇒ ∠ECB = 60° A

C

∴

EF = CD and CD = AB = 12 cm 1 EF) 2 ⇒ AD = BC = EK = 6 cm \ EK = 6 cm (

EK =

∴

In ∆ABC, AC2 = AB2 + BC2. AC2 = (12)2 + (6)2 = 144 + 36 AC2 = 180 AC =

180

AC = 6 5 cm. Each diagonal of the rectangle is 6 5 cm. Hence, the correct option is (b).

M01_IIT-Hints and Solutions.indd 15

80 cm

O

⇒ EF = 12 cm

B

18 cm

D

Given, AB = CD = 80 cm Let BD = 18 cm. In ∆CBD, CB2 = BD2 + DC2 CB2 = 182 + 802 CB2 = 324 + 6400 CB2 = 6724 CB = 82 cm

4/16/2018 11:37:25 AM

A.16

Chapter 2  Geometry-I and Geometry-II Given, ∠BDC = 40° y x Let ∠DBC = and ∠DCB = . 2 2 \ ∠ABC = 180° - x and ∠BCA = 180° - y In ∆BDC, ∠BDC + ∠DCB + ∠CBD = 180° y x 40° + + = 180° 2 2 x+y = 140° 2 x + y = 280° In ∆ABC, ∠BAC + ∠BCA + ∠ABC = 180° ∠BAC + (180°- y) + (180°- x) = 180° ∠BAC = 180° - 180° + x + y - 180° ∠BAC = 280° - 180° ∠BAC = 100° Hence, the correct option is (d).

CB 82 = = 41 cm 2 2 Hence, the correct option is (d).

\ required radius =

11. G  iven, A median of the ∆XYZ 4 = The corresponding median of the ∆MNP 3 We know that, =

Area of the ∆XYZ Area ofthe ∆MNP

( A median of the ∆xyz )2

(The corresponding median of the ∆MNP)2

Area of the ∆XYZ ( 4 ) 16 = = 2 Area of the ∆MNP ( 3 ) 9 2

Hence, the correct option is (d). 12. C

D r O

r A

r

a

A

B

a

Given, ‘O’ is the centre of the circle. ABCD is a square. Let a be the side of the square. Clearly, ∆ABC is an isosceles right triangle. \a= 2r Perimeter of the square = 4a Perimeter of the semicircle = p r + 2r = (p + 2)r \ required ratio = 4a:(p +2)r = 4 2r :(p + 2)r = 4 2 :(p + 2) Hence, the correct option is (d).

( )

13.

14.

A

D

60°

x B

x 180 − 2y

y

y

C

E

Given, ∠BAC = 60° Let ∠CBD = x and ∠ACD = y. In ∆ABC, ∠ABC + ∠BCA + ∠CAB = 180° 2x + 180 - 2y + 60 = 180° 2(x - y) = -60 y - x = 30° \ ∠ACD - ∠CBD = 30° Hence, the correct option is (b). 15. (i) → (B); (ii) → (C); (iii) → (A); (iv) → (D) Hence, the correct option is (a).

B x 2

y 2

x 2

C y 2

40° D

M01_IIT-Hints and Solutions.indd 16

4/16/2018 11:37:32 AM

Hints and Explanation

A.17

CHAPTER 3 Linear Equations and Inequations Assessment Test I 1. C  ABED is the sequential order. Hence, the correct option is (d). 2. 3x + 2y = 5  9x - 2y = 3 

(1) (2)

Eq. (1) + Eq. (2) ⇒ 12x = 8 ⇒ x =

2 . 3

2 + 2y = 5 from Eq. (1) 3 2y = 3

3×

y=

3 2

 2 3 \ (p, q) =  ,   3 2 2 3 pq = × = 1 3 2 Hence, the correct option is (a). 3. X  + Y + Z = 180° (angles of X = 5Y = 2(Y + Z) is given. X = 5Y, X = 2Y + 2Z X - 2Y = 2Z 5Y - 2Y = 2Z ⇒ 3Y = 2Z

Dle)

3Y = 180° 2 ⇒ X + Y + Z = 180° Z=

15Y = 180°⇒ Y = 24° 2 ⇒ X = 5Y = 5 × 24° \ X = 120° ⇒ X + Y = 144° Hence, the correct option is (a). 4. L  et the distance between P and Q be d km. d = 3 h (opposite direction) x + 50 d ⇒ d = 3 (x + 50) and = 9 h (same x − 50 direction) ⇒ d = 9 (x - 50)

M01_IIT-Hints and Solutions.indd 17

\ 3 (x + 50) = 9 (x - 50) 3x + 150 = 9x - 450 600 = 6x ⇒ x = 100 kmph Hence, the correct option is (b). 5. 2 x + 8y = 15 and ax + by = 45 have infinite solu2 8 15 . tions⇒ = = a b 45 2 8 1 = = a b 3 2 1 8 1 ⇒ = , = a 3 b 3 ⇒ a = 6 and b = 24 Hence, the correct option is (a). 2 5 6.  + = 26  x y 8 3 − + 11 = 0  x y 1 1 = b. Let = a and x y Eq. (1) ⇒ 2a + 5b = 26  Eq. (2) ⇒ 8a - 3b = -11  Eq. (3) × 4 - Eq. (4) 8a + 20b = 104 8a - 3b = -11 (-) (+) (+) _____________ 23b = 115 b=5 Substitute b = 5 in Eq. (3) 2a + 5(5) = 26. 2a = 26 - 25

(1) (2)

(3) (4)

1 2 1 1 \ =a= ⇒x=2 x 2 1 1 =b=5⇒y= 5 y 1 5xy = 5 × 2 × = 2 5 Hence, the correct option is (a).

a=

4/16/2018 11:37:40 AM

A.18

Chapter 3  Linear Equations and Inequations

7. a x + by = a - b  (1) bx + ay = a + b  (2) Eq. (1) a - Eq. (2) b ⇒ x (a2 - b2) = a2 - 2ab - b2 a2 − b2 2 ab x= 2 2 − 2 2 a −b a −b 2 ab \ x = 1− 2 2 a −b Hence, the correct option is (a). 8. Let the speed of the man in still water be x kmph and speed of the stream be y kmph. 20 16 =2, =8 x+y x−y ⇒ x + y = 10  (1) x - y = 2  (2) Eq. (1) + Eq. (2) ⇒ 2x = 12 ⇒ x = 6 Hence, the correct option is (c).  et the questions the student marked wrong be 9. L x. Then, he marked (100 - x) questions as correct. Then, (100 − x) 3 + x (-1) = 100 ⇒ x = 50 Hence, the correct option is (b). 10. Let x, y, and z be the 100s, 10s, and units digits of a three-digit numbers. x + y + z = 9  (1) 100x + 10y + z + 99 = 100z + 10y + x ⇒ 99z - 99x = 99 z - x = 1 

(2)

Also y = 2/7 (x + z) 7y = 2x +2z 

(3)

From Eq. (1), x + z = 9 - y ⇒ 2x + 2z = 18 - 2y 9y = 18 ⇒ y = 2 From Eq. (3), 2x + 2z = 7 (2) From Eq. (2) and Eq. (4), z - x + x + z = 1 + 7 2z = 8 z=4 ⇒x=3

M01_IIT-Hints and Solutions.indd 18

Hence, the correct option is (a).  et the cost of each pencil, ruler, and eraser be 11. L p, r, and e, respectively. 3p + 5r + 7e = 49  (1) 5p + 9r + 13e = 78  (2) 2 x Eq. (1) - Eq. (2) ⇒ _________________________ p__________________________ + r + e = 49 × 2 - 78 = 20 \The cost of 1 pencil, 1 ruler, and 1 eraser is `20. Hence, the correct option is (b). 3x −7 ≥ 5 2 3x 3x - 7 ≤ - 5 or -7≥5 ⇒ 2 2 3x 3x ≤ 2 or ≥ 12 2 2 4 x≤ or x ≥ 8 3

12. Option (b):

x∈ ( −∞ , 4 3  or x ∈ [8, ∞ ) 13. Option (a):

5x −7 = 3 6

5x -7=±3 6 5x = 10 or 4 6 5x = 60 or 24 x = 12 or 4.80 14. Option (c):

8x 3 − <7 3 4

8x 3 - <7 3 4 3 8x 3 <7+ -7+ < 4 3 4 −25 8 x 31 < < 4 3 4 −75 93 <x< 32 32 − 75 93   ,  ⇒ x ∈  32 32 

-7<

7y = 18 - 2y

x + z = 7 

\ The numbers 3, 2, and 4 product of the digits = 3 × 2 × 4 = 24

(4)

4/16/2018 11:37:50 AM

Hints and Explanation

15. Option (d):

( x − 3)2 x−6

3 5 4.  + = 11  x y 12 1 − = 2  x y 1 1 =b Let = a, x y 3a + 5b = 11  12a - b = 2  Eq. (3) + Eq. (4) × 5 ⇒3a + 5b = 11 60a - 5b = 10 _____________ 63a = 21

<x

x2 − 6 x + 9 -x<0 x−6 x2 − 6 x + 9 − x2 + 6 x <0 x−6 9 <0 x−6 ⇒ x - 6 < 0 ⇒ x < 6 ⇒

x∈(-∞,6)

Assessment Test II 1. B  ACED is correct sequential order. Hence, the correct option is (b). 2. 4x + 5y = 26  8x + 3y = 17  Eq. (1) x2 - Eq. (2) 8x + 10y = 52 8x + 3y = 17 (-) (-)   (-) ___________ 7y = 35 y=5 ⇒x=

1 4

1 (p, q) = ( , 5) 4 21 \p+q= 4 Hence, the correct option is (b). 3. ∠P + ∠Q + ∠R = 180°, ∠P = 3∠Q 4 3∠Q + ∠Q + (∠P + ∠Q) = 180° 5 4 4∠Q + (3∠Q + ∠Q) = 180° 5 16 ∠Q = 180° 4∠Q + 5 20∠Q + 16∠Q = 900° ∠Q = 25°⇒ ∠P = 3 × 25° = 75° 4 (75 + 25°) = 80° 5 The required difference = 80° - 25° = 55° Hence, the correct option is (a).

∠R =

M01_IIT-Hints and Solutions.indd 19

(1) (2)

A.19 (1) (2)

(3) (4)

1 3 1 ⇒ 3 × + 5b = 11 3 5b = 10, b = 2

⇒a=

1 1 1 1 = = 3, y = = a 13 b 2 1 \ (x, y) = (3, ) = (m, n) 2 1 \ m + 2n = 3 + 2 × =4 2 Hence, the correct option is (b). \x=

5. p x + qy = p2 + q2 (1) p q x − y = q - p (2) q p Eq. (2) p + Eq. (1) p2 x - qy = p(q - p) q px + qy = p2 + q2 __________________  p2  2  q + p x = pq + q q x = (p + q)q × p ( p + q) q2 x= p Hence, the correct option is (d). 6. Let the speeds of the cars C1 and C2 be C1 kmph and C2 kmph, respectively. 360 360 + = 10 h (1) C1 C2

4/16/2018 11:38:00 AM

A.20

Chapter 3  Linear Equations and Inequations 360 = 2 h 24 minutes C1 + C2 360 = 2.4 h C1 + C2 360 = C1 + C2 2.4 \ C1 + C2 = 150 (2)  C + C2  From Eq. (1) 360  1  = 10  C1 C2  C + C2 1 ⇒ 1 = C1 C2 36 150 1 = From Eq. (1) C1 C2 36 C1 C2 = 150 × 36. C1 - C2 =

(C1 + C2 )2 − 4C1C2

C1 - C2 =

(150)2 − 4 × 36 × 150

C1 - C2 =

150 (150 − 144 )

=

25 × 6 × 6 = 30

Hence, the correct option is (a). 7. 7x + 8y = 10 and 9x - ky = 20 have a common solution. 7 8 ≠ 9 −k 72 −72 ⇒ -k ≠ ⇒k≠ 7 7 Hence, the correct option is (a).

⇒

8. Let the digits in hundred’s place, Ten’s place and in the unit’s place be x, y, and z, respectively. x + y + z = 12 Given that 100x + 10y + z + 99 = 100z + 10y + x ⇒ 99x - 99z = -99 x - z = -1 1 Given that (x + z) × =y 3 x + z = 3y and x + y + z = 12 ⇒ x + z = 12 - y ⇒ 12 - y = 3y 12 = 4y \y=3 \x+z=3×3

M01_IIT-Hints and Solutions.indd 20

x+z=9 x - z = -1 ⇒ 2x = 8 ⇒ x = 4 ⇒ z = 5 \ Product of the digits = 4 × 3 × 5 = 60 Hence, the correct option is (a). 9. Let the speed of the man in still water be x kmph and speed of the stream be y kmph. 30 = 3 ⇒ x + y = 10 x+y 54 =9⇒x-y=6 x−y ⇒ (x + y) + (x - y) = 16 ⇒x=8 \ Speed of the man in still water = 8 kmph Hence, the correct option is (b). 10. Let the number of correct, wrong, and unattempted questions be c, w, and u, respectively. c + w + u = 100 2c - w = 100 and w = u By solving, we get c = 60. Hence, the correct option is (a).  et the pens, erasers, and pencils be denoted 11. L by x, y, and z, respectively. Given 4x + 5y + 6z = 37 3x + 2y + z = 19 ______________ 7x + 7y + 7z = 56 x+y+z=8 \ The cost of 3 pens, 3 erasers, and 3 pencils is = 3 × 8 = `24. Hence, the correct option is (d). 5 ≥7 2 5 5 2x - ≥ 7 or 2x - ≤ -7 2 2 4x - 5 ≥ 14 or 4x - 5 ≤ -14

12. Option (d): 2 x −

19 9 or x ≤ − 4 4 x ≥ 4.75 or x ≤ -2.25

x≥

13. Option (a):

3x +4 =7 2

4/16/2018 11:38:09 AM

Hints and Explanation ⇒

3. x − 4=0 ⇒ x = 4

3x +4=±7 2

3x = ±7 − 4 2 ⇒x = 2 or x =

⇒ x = ±4 \ The given equation has two solutions. Hence, the correct option is (c).

−22 3

14. Option (c): 8 x −

4. G  iven 8x - 3y = 6  (k - 1, k - 2) passes through the Eq. (1). 8(k - 1) -3(k + 2) = 6 8k - 8 - 3k - 6 = 6 5k = 20 k=4 Hence, the correct option is (b).

4 <7 3

24 x − 4 <7 3 -21 < 24x - 4 < 21 -17 < 24x < 25 -7 <

−17 25 <x< 24 24

( x − 2)2

6. L  et x be the number. 1 x+ 3 + 4 = 19 2 3 3x + 1 4 = 19 − 6 3 3 x + 1 57 − 4 = 6 3 3x + 1 = 53 2 3x + 1 = 106 ⇒ 3x = 105 x = 35 Hence, the correct option is (b).

Assessment Test III 1. C  BDA is the sequential order. (c) 2x 2 + 2x·2-1 = 160 1  (b) 2x  2 + 2  = 32 × 5

 5 (d) 2x   = 25 × 5  2 x (a) 2 = 26 ⇒ x = 6 Hence, the correct option is (b).

2. Given, 3a + 5b + 6c = 33  a + 3b + 4c = 19  Eq. (1) - Eq. (2) ⇒ 3a + 5b + 6c = 33 a + 3b + 4c = 19 -  -  -  _______________ 2a + 2b + 2c = 14 _______________ a+b+c=7 Hence, the correct option is (a).

M01_IIT-Hints and Solutions.indd 21

(1)

5. 2 x + 3 ≥ 3x - 7 2x - 3x ≥ - 7 - 3 -x ≥ - 10 x ≤ 10 Hence, the correct option is (a).

< -x 4−x x2 − 4 x + 4 − x2 + 4 x <0 4−x 4 <0 4−x ⇒4-x<0 -x < - 4 x>4

15. Option (b):

A.21

(1) (2)

7. Given, 2x + 3y = 6 kx + 6y = 15 The system has a unique solution, 2 3 i.e., ≠ ⇒ k ≠ 4. k 6 \ The value of k is any real number except 4. Hence, the correct option is (d). 8. Given, a + b = 5. Possibilities of (a, b) = {(1, 4), (2, 3), (3, 2), (4, 1)} Possibilities of ab are 1, 8, 9, and 4. Option (c) follows. Hence, the correct option is (c).

4/16/2018 11:44:15 AM

A.22

Chapter 3  Linear Equations and Inequations

 1 1 1 1 9. ( a + b + c + d)  + + +  = [1 + 1 + 1 + 1 +  a b c d  a b  a c   a d  b c   b d  +  +  +  +  +  +  +  +  +  b a c a d a c b d b 1    c d +  +  ≥  4 + 6 ( 2)  ... x + ≥ 2    d c x = 16 \ The minimum value is 16. Hence, the correct option is (a). 1 1 = ⇒ 2x - 3y = 7  (1) 2x − 3 y 7 1 1 = ⇒ 5x - 6y = 11  (2) 5 x − 6 y 11 From Eqs (1) and (2), we get Eq. (1) × 2 ⇒ 4x - 6y = 14    (2) ⇒ 5x - 6y = 11    - +     _______________

10. Given,

-x = 3   x = -3 Substituting x = -3 in Eq. (1), we get 5(-3) - 6y = 11 -15 - 6y = 11 -6y = 11 + 15 -6y = 26 −26 6 −13 y= 3

−13 3

−13 = 13 3 Hence, the correct option is (b). 11. Let c be the number of questions marked correctly. \ The number of questions marked wrongly = 100 - c 1 (100 - c) = 70 4 c c - 25 + = 70 4

M01_IIT-Hints and Solutions.indd 22

13. Given, x - 4 < 0 x < 4 and x ∈ W x ∈ {0, 1, 2, 3} Hence, the correct option is (d). 14. The maximum value of 3 -|2x - 1| is 3 as the minimum value of |2x - 1| is 0. Hence, the correct option is (a). 15.   (i)  x - 2 ≥ 0 ⇒ x ≥ 2 → (C) −5 (ii)  ≤ 0 ⇒ x - 2 > 0 ⇒ x > 2 → (B) x−2 (iii)  x + 2 ≥ 0 ⇒ x ≥ -2 → (A) 3 ≥ 0 ⇒ x + 2 > 0 ⇒ x > - 2 → (D) x+2 Hence, the correct option is (a). (iv) 

1. B  ACDE is the required sequential order. Hence, the correct option is (a).

\ xy = -3 ×

\ 1c -

12. |x + 1| < 5 ⇒ -5 < x + 1 < 5 ⇒ -6 < x < 4 Hence, the correct option is (d).

Assessment Test IV

y=

\ x = - 3 and y =

5c = 95 4 c = 76 \ The number of questions marked correctly = 76 Hence, the correct option is (b).

2. Given, 2a + 3b + 5c = 9  2a + b - c = 7  Eqs (1) + (2) ⇒ 2a + 3b + 5c = 9    2a + b - c = 7     4a + 4b + 4c = 16    _______________

(1) (2)

a+b+c=4    _______________ Hence, the correct option is (b). 3. x + 2 = 0 ⇒ x = − 2 \ x has no real value. Hence, the correct option is (a).

4/16/2018 11:44:23 AM

Hints and Explanation 4. G  iven, 13x + 7y = 10  (1) Point (k + 2, k + 2) passes through the Eq. (1). 13(k + 2) + 7(k + 2) = 10 13k + 26 + 7k + 14 = 10 20k + 40 = 10 20k = -30 −3 k= 2 Hence, the correct option is (b). 5. x - 3 < 2x - 5 x - 2x < -5 + 3 -x < -2 x>2 Hence, the correct option is (b). 6. Let the fraction be

x . y

x + 10 x − 10 = y + 12 y − 12 (x + 10) (y - 12) = (x - 10) (y + 12) xy - 12x + 10y - 120 = xy + 12x - 10y - 120 -12x - 12x = - 10y - 10y -24x = - 20y x −20 x 5 = = ⇒ y −24 y 6 5 6 Hence, the correct option is (c). \The required fraction =

7. Given, 7x + 8y = 15 14x + py = 30 The system has infinite solutions. 7 8 15 = = 14 p 30 8 1 = p 2 p = 16 Hence, the correct option is (b). 8. Given, x + y = 7. Possibilities of (x, y) are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1). Possibilities of yx are 6, 25, 64, 81, 32, and 1. Option (b) follows. Hence, the correct option is (b).

M01_IIT-Hints and Solutions.indd 23

A.23

9. For the minimum value of (1 + a) (1 + b) (1 + c) (1 + d) (1 + e), each of a, b, c, d, and e is minimum. Given, abcde = 32 ⇒ The minimum value of each of a, b, c, d, and e is 2, since abcde = (2)5. \ (1 + a) (1 + b) (1 + c) (1 + d) (1 + e) = (2)5 = 243 Hence, the correct option is (d). 10. Given, and

1 1 4 + =  x y 3

(1)

1 1 1 =  x y 3

(2)

Eqs (1) + (2) ⇒

1 1 4 = + x 3 y

1 1 1 =     x y 3    _______________ 2 5   = x 3 1 5 6 \ = ⇒x= x 6 5 Hence, the correct option is (b). 11. Let c be the number of questions attempted correctly by the student. \ (75 - c) is the number of questions wrongly attempted by the student. 4c - 2(75 - c) = 132 4c - 150 + 2c = 132 6c = 132 + 150 6c = 282 c = 47 \ Number of questions marked wrongly = 75 - 47 = 28. Hence, the correct option is (a). 12. |x - 2| > 10 x - 2 < -10 or x - 2 > 10 x < -8 or x > 12 Option (d) follows. Hence, the correct option is (d). 13. Given -x + y ≥ 10. -x + 15 ≥ 10 -x ≥ -5 x ≤ 5 and x ∈ N ⇒ x = {1, 2, 3, 4, 5}

4/16/2018 11:44:41 AM

A.24

Chapter 3  Linear Equations and Inequations \ The number of values of x is 5. Hence, the correct option is (a).

14. 6 + |4 - 7x| will have the minimum value when |4 - 7x| has the minimum value, which is 0. Hence, the minimum value of 6 + |4 - 7x| is 6. Hence, the correct option is (b).

M01_IIT-Hints and Solutions.indd 24

−3 > 0 ⇒ x - 3 < 0 ⇒ x < 3 → (C) x−3 (ii)  x - 3 ≥ 0 ⇒ x ≥ 3 → (A) (iii)  x + 3 ≥ 0 ⇒ x ≥ -3 → (D)

15.   (i) 

3 ≥ 0 ⇒ x + 3 > 0 ⇒ x > -3 → (B) x+3 Hence, the correct option is (b). (iv) 

4/16/2018 11:44:42 AM

Hints and Explanation

A.25

CHAPTER 4 Logarithms Assessment Test I

3

k2 = ( 44 )

log 72 = log372 = log39 + log38 1.  24 log24 3

k2 = 64

= 2 log33 + 3 log32 = 2 + 3x \ The required sequential order is ABDC. Hence, the correct option is (c).

(

2. 3log3 7 + 5log5 6 ... aloga k = k

)

32 = log

2

8

= log

k

15

k

1

2

k

8. x log a = y log b = z log c = k logb a (log a) (log c) = logc b (log b) (log b)

k

[(\ lognm = (log m)/(log n)]

 y  y 2  x   z  = y /xz = 1 Hence, the correct option is (a).

1

2

16

 15    16 logk k =  1   2 15 = 8 Hence, the correct option is (c). 5. log (2)1000 = 1000 (log2) = 1000 × (0.3010) (given) = 301.0 Number of digits = 301 + 1 = 302. Hence, the correct option is (c). 6. Given log256 k2 = 3

k2 = ( 256 )

M01_IIT-Hints and Solutions.indd 25

4

3 4

7. 1 + log2 3 = log2 2 + log2 3

Hence, the correct option is (b).

32

k k k k = log

log (k/8) = log (8/8) = log 1 = 0 Hence, the correct option is (a).

= log2 6.

= log 32 8 5 = log2 2 3 5 = 3 Hence, the correct option is (b). 4. Given, log

k = ± 8. But k > 0.

= log2 (2 × 3)

= 7 + 6 = 13 Hence, the correct option is (d). 3. log2

4

1 1 1 1 + + + k 2 4 8 16

9. 3x = 2y = 6 3x = 6

2y = 6

x = log3 6

y = log2 6

1 = log6 3 x

1 = log6 2 y

1 1 = log6 3 + log6 2 + x y x+y = log6 6 = 1 xy Hence, the correct option is (a). 10. log2 7 log3 5 log7 2 log5 3 log7 log5 log2 log3 = × × × log2 log3 log7 log5 = 1 = log2 2 Hence, the correct option is (c). 11. G  iven, x = log10 5 1 5 ⇒ = log10 x

4/16/2018 11:44:57 AM

A.26

Chapter 4  Logarithms 1 1  + 2 = (x + 1) +  + 1 x  x 10 5 5 = log5 + log5 + log10 + log10 10

log9 ( x2 − 4 x + 4) = 1

x+

(

) (

)

50 = log50 5 + log10

=

log50 log5

+

log50 log10

 1 1  + = log50    log5 log10   log5 + log10  = log50    log5 log10  2 log50 ) log50 ( = = log50 . log5 ⋅ log10 log5 ⋅ log10 Hence, the correct option is (b). 12. 10 log10 3 + 6log6 5 = 3 + 5 = 8 Hence, the correct option is (c). 13. log10 25 = 1.3979 log10 52 = 1.3979 2log10 5 = 1.3979 1.3979 = 0.6989 2 Hence, the correct option is (c). 14. Given, log10 3 = 0.4771 log10 300 = log10 3 + 2log10 10 = 0.4771 + 2 = 2.4771 Hence, the correct option is (b). log a log b × =1 log b log a 1 1 (ii) logb a b = logbb = a a

15.   (i) logb a loga b =

(iii) log a b b = a logbb = a log b log a ×b× =b log a log b

Option (a) follows. Hence, the correct option is (a).

Assessment Test II 1. G  iven, log3log9 (x2 - 4x + 4) = 0 log9 ( x2 − 4 x + 4) = 30

M01_IIT-Hints and Solutions.indd 26

log 7

2. 2

3

(

− 7log3 2 = 0 alogcb = blogc a

= log7 1

)

Hence, the correct option is (d). 3. log3 =

3

729 = log( 3 )3/2

(6 2) log (3 2)

3

36

3

=2 Hence, the correct option is (a). 4. log

log10 5 =

(iv) logab × log b b a =

(x2 - 4x + 4) = 91 ⇒ x2 - 4x - 5 = 0 ⇒ (x + 1) (x - 5) = 0 ⇒ x = -1 or x = 5 The required sequential order is DACEB. Hence, the correct option is (a).

y y y = log

y

= log

1 1 y2

1

1

y2 y 4 y8

7 1 y2

y8

 7   8 logy y =  1   2 7 4 Hence, the correct option is (b). =

5. log510 = 10(log5) = 10 × 0.6989 = 6.989 Number of digits = 6 + 1 = 7 Hence, the correct option is (c). 6. G  iven, log81 x5 = 1 log81 x5 = 5

1 4

5 4

( )4

x5 = 34 x5 = 35 x=3

4/16/2018 11:45:08 AM

Hints and Explanation =2 = log5 25

log3 x2 = log3 32 = 2 × 1 = 2 Hence, the correct option is (c). 7. 1 +

Hence, the correct option is (b).

1 = 1 + log3 5 log5 3

= log 3 3 + log3 5 = log3 15 Hence, the correct option is (c). log a log b log c = = =k 5 6 7 ⇒ log a = 5k ⇒ a = 105k log b = 6k ⇒ b = 106k log c = 7k ⇒ c = 107k b² = (106k)² = 1012k = 107k × 105k = ac Hence, the correct option is (a).

8. L  et

7x = 3136

8y = 3136

x = log7 3136

y = log8 3136

1 1 + = log3136 7+ log3136 8 x y 1 1 = log( 56 )2 56 = log56 56 = 2 2 Hence, the correct option is (b). 10. log3 10 log2 25 log10 3 log5 2 log10 log3

×

M01_IIT-Hints and Solutions.indd 27

2 log5 log2

×

log3 log10

11. x = log3 2 1 1 1 = ⇒ = log3 2 + 1 log3 2 + log3 3 x+1 1 = = log6 3 log3 6 1 + 1 = log6 3 + log6 6 = log6 18 x+1 Hence, the correct option is (c). 12. 5log5 7 + 13log13 11 = 7 + 11 = 18 Hence, the correct option is (a). 13. Given, log10 2 = 0.3010 Log10 8 = log10 23 = 3 log10 2 = 3(0.301) = 0.903 Hence, the correct option is (c). 14. Given, log10 3 = 0.4771 Log10 3 - log10 100 = 0.4771 - 2

9. Given, 7x = 8y = 3136

=

A.27

×

log2 log5

Log10 0.03 = 2.4771 Hence, the correct option is (b). 15.  Characteristics

Number of digits

(i) 42

42 + 1 = 43

(ii) 41

41 + 1 = 42

(iii) 39

39 + 1 = 40

(iv) 40

40 + 1 = 41

(i) → (B); (ii) → (C); (iii) → (D); (iv) → (A) Hence, the correct option is (b).

4/16/2018 11:45:16 AM

A.28

Chapter 5  Mensuration

CHAPTER 5 Mensuration Assessment Test I

a3: 3 a × 2 a × a = 1: 6 5.

1. B  CAD is the required sequential order. Hence, the correct option is (b). 2. 

C

D

A

F

B

E

Area of ABC = Area of AEB = Area of ADC ⇒ Area of ABCD = Area of AEBC ⇒ Area of the rectangle ABCD = rhombus AEFC

1 Area of the 2

\ Required ratio is 1:2 Hence, the correct option is (d). 3. 

C 6

A

6

4

D

4

b = 8 cm and a = 6 cm Area =

b 4 a2 − b2 4

8 4 × 62 − 64 4

= 2 80 = 8 5 sq. cm Hence, the correct option is (b). 4. R  equired area = A - B

2   1 3 2  3 2 p 3 A - B = p × 62 ×6 -  ×6 −  × 6  6 4 6 2  4     9p 21p =6p-9 3 -9 3 + = − 18 3 2 2

= 33 - 18 3 = 3(11 - 6 3 ) sq. cm Hence, the correct option is (a).

M01_IIT-Hints and Solutions.indd 28

6. Since the rectangular pieces are identical, the triangular base should be an equilateral triangle of side 10 cm or 10 3 cm. \ Volume of the air inside the model of the prism = Ah 3 Case (i)  V = × (10)2 × 10 3 4 V = 750 cu cm 3 × (10 3 )2 × 10 4 = 750 3 cu cm Hence, the correct option is (a). Case (ii) V =

7. Volume of the pyramid =

1 × (99)2 × 99 3

= 32343 m3 Volume of the air in litres = 323433 × 100 × 100 × 100 1000 = 32,34,33,000 litres The number of persons can be allowed for 6 h 323433000 2700 = 119,790 Hence, the correct option is (a). =

B

Area of an isosceles triangle =

Hence, the correct option is (c).

8. The number of flower pots used × Volume of each flower pot = Volume of the magnesium powder 1 22 2 × × 3 × 7 = 13,200 cm3 3 7 Hence, the correct option is (a). 200 ×

9. Volume of the gelatine = Outer volume of the capsule - inner volume of the capsule =

4 22 22 × × 2.1 × 2.1 × 2.1 + × 2.1 × 2.1 × 7 − 97 3 7 7

= 88 × 0.441 + 97.02 - 97 = 38.8 ≈ 40 cu. mm \ Quantity required for a year = 40 × 1,00,000 cu mm

4/16/2018 11:45:58 AM

Hints and Explanation = 40,00,000 cu mm

2. 

S

4000000 = 4000 cm3 1000 Hence, the correct option is (b). =

10. Surface area of a regular octahedron is 216 3 sq. cm. 2 3 a2 = 216 3 a=

108 = 6 3 cm

Volume of the octahedron V=

2 3 2 a = (6 3 )3 3 3

2 × 216 × 3 3 3 = 216 6 cu cm =

Hence, the correct option is (d). 11. Lateral surface area of a regular tetrahedron = 3 3 2 a 4 3 3 2 a = 450 3 4 a2

= 600

a = 10 6 Volume of the regular tetrahedron V = a3 6 2

=

(10 6 )3 6 2

= 1000 3 cu cm Hence, the correct option is (c). 12. Option (e): 4 p 2 R r is the surface area of a torus. 13. Option (f): 2 p r(h + r) is the surface area of a cylinder. 14. Option (d): p [R(R + l) + r(r + l)] is the surface area of a frustum. 15. Option (a): p r(l + r) is the total surface area of a cone.

Assessment Test II 1. C  BAED is the required sequential order. Hence, the correct option is (d).

M01_IIT-Hints and Solutions.indd 29

P

A.29

R 4

3 Q

N

M

From the figure, P R =

42 + 32 = 5

Perimeter of the rectangle = 7 + 7= 14 cm Perimeter of the rhombus = 4 × 5 = 20 cm Ratio of their perimeters = 14:20 = 7:10 Hence, the correct option is (c). 3. Base of the triangle = 62 - 2 × 16 = 30 Length of equal sides a = 16 cm and b = 30 cm A=

b 4 a2 − b2 4

\ Area of the triangle =

30 4 × 256 − 900 4

30 124 = 15 31 sq. units 4 Hence, the correct option is (a). =

4. D  iagonal of the square = 14 cm 14 cm = 7 2 ⇒ Side of the square = 2 Diameter of the smaller circle = 7 2 cm Area of the smaller circle = p d2 1 22 = × ×7 2 ×7 2 4 4 7 = 77 cm2 1 Area of region B = [Area of the square - Area 4 of the smaller circle] 1 1 21 (7 2 )2 − 77  = [98 − 77 ] = cm2  4 4 4 Diagonal of the square = Diameter of the bigger circle Area of region A 1 = Area of the bigger circle 4 − Area of the square =

=

1  1 22 1 56  × × 14 × 14 − 98  = 154 − 98  = 4  4 7 4 4 

4/16/2018 11:46:09 AM

A.30

Chapter 5  Mensuration 56 21 35 \ The required difference = − = cm2 4 4 4 Hence, the correct option is (a).

5. l = b=

3a =

3× 6

2a =

2× 6

h=a=

6

= 3.465 × 106 Hence, the correct option is (a). 10. Volume of the octahedron 2 3 = a = 36 3 108 a3 = 2

Volume = l b h

a3 = 54 2

=

a3 = 27 × ( 2)3

3× 6× 2× 6× 6

= 36 cu cm Hence, the correct option is (a). 6. Number of pieces = =

1800 25 − 16 × 6

1800 4  4(2.5)2 − ( 4)2  6   4

= 100

Hence, the correct option is (b). 3 7. Required area = 4 × × (12)2 − 4 × 1 × 0.8 4 = 144 3 - 3.2 = 144(1.732 - 3.2) = 249.408 - 3.2 = 246.208 \ 247 glass plates are required Hence, the correct option is (b). 8. Required volume of the mixture 1 22 = 100 × 2016 × × × (3)2 × 52 − 32 3 7 1 22 = 100 × 2016 × × ×9×4 3 7 = 100 × 96 × 22 × 36 = 7603200 cu cm Hence, the correct option is (b). 4   9. Required volume = 49 ×  p r2 h + p r3    3 4   = 49 × p r2  h + × r   3  22 4 = 49 × × (1.5)2 [8 + × 1.5 ] 7 3 = 7 × 22 × 2.25 [8 + 2] = 154 × 22.5 = 3465 m3 = 34,65,000 litres

M01_IIT-Hints and Solutions.indd 30

a=3 2 Surface area = 2 3 a2 = 2 3 (3 2 )2 = 2 3 × 9 × 2 = 36 3 cm2 Hence, the correct option is (a). 11. Volume of the regular tetrahedron =

a3 6 2

= 72

⇒ a3 = 72 × 6 2 ⇒ a3 = 36 × 2 × 6 2 = (6 2)3 \a=6 2 Total surface area of the regular tetrahedron = 3 a2 =

3 × (6 2 )2 = 72 3 sq. cm

Hence, the correct option is (c). 2 12. Option (b):  p r3 → Volume of a solid 3 hemisphere 3 p r2 → Surface area of a hemisphere 1 13. Option (f):  p r2 h → Volume of solid cone 3 p r2 + p r l → Surface area of a solid cone 14. Option (g): p R2 h - p r2 h → Volume of a hollow cylinder 2p h(R + r) + 2p (R2 - r2) → Surface area of a hollow cylinder 2 2 15. Option (d):  p R3 - p r3 → Volume of the 3 3 hemispherical shell

4/16/2018 11:46:23 AM

Hints and Explanation 3 p R2 + p r2 → Surface area of hemispherical shell

Assessment Test III

2. 82 + 152 = 172 ∴Then the triangle is right-angled triangle. 1 Required area = × 15 × 8 = 60 cm2 2 Hence, the correct option is (d). 3. 

 105 105 105  × × − 1 × 100 =   100 100 100  =

1. B  ADC is the required sequential order. Hence, the correct option is (c).

A.31

(105 × 105 × 105 − 100 × 100 × 100)

× 100 100 × 100 × 100 1157625 − 1000000 157625 = = = 15.7625% 10000 10000 Hence, the correct option is (c). 5. n  +1=5+1=6 Hence, the correct option is (b). 6. L  et r be the radius of the hemisphere Given r = 30 cm Total surface area of the hemisphere = 3p r2 = 3 × (3.14) × (30) × 30 = 8478 cm2 Hence, the correct option is (b).

A

7. 

F

O

A

12 K

2

6 cm B

D

C

17

⇒ a = 12

15

15

Let a be the length of the side of the equilateral triangle. 1  3a  =6 ⇒In radius,  3  2 

12 C

B 2

3 cm 3 2 a 4

Area of the equilateral triangle =

3 3 × (12 3 )2 = × 144 × 3 = 108 4 4 Hence, the correct option is (c).

=

 105  ∴ The volume of the cube =  x  100 

D

14

3 cm2

Let FA and DC extended meet at K. Area of rectangle FEDK = 14 × 17 = 98 cm2 1 × 15 × 12 = 90 cm2 2 ∴ Required area = 238 - 90 = 148 cm2 Hence, the correct option is (c). Area of ∆ AKC =

4. Let x be the edge of the cube. ∴ Volume of the cube = x3 Given, the edge increases by 5%. i.e., New edge of the resultant cube =

2

E

105 x 100

3

8.  r r

2r

3

 105  x  − x3  100  ∴ required percentage = × 100 x3

M01_IIT-Hints and Solutions.indd 31

r

4/16/2018 11:52:15 AM

A.32

Chapter 5  Mensuration Let r be the radius of the sphere. Edge of the cube = 2r Required ratio = Volume of the cube:volume of the sphere 4 4 = (2r)3: p r3 = 8 r3: p r3 = 6:p 3 3 Hence, the correct option is (d).

9. 

7

Required area =

90° × p × 142 360°

1 × 196 = 49 p cm2 4 Hence, the correct option is (d). =

11. Given: Height (h) = 12 cm Perimeter of the base of the right pyramid (6a) = 36 cm ∴ Side of the base (a)= 6 cm Area of the base = 6 ×

12

Let R and r be the radii of the top and bottom circular ends of the bucket. Let l be the slant height of the bucket. 14 7 R= = 7 cm and r = = 3.5 cm 2 2 Given height (h) = 12 cm

=

(7 − 3.5)2 + 122

3 × 6 × 6 = 54 3 cm2 4 1 Volume of the pyramid = × area of the base 3 × height =6×

3.5

Slant height (l) =

=

1 × 54 3 × 12 3

= 216 3 cm3 Hence, the correct option is (a). 12. 

(R − r )2 + h2 = 12.5 cm

h

Total surface area of the bucket = p l (R + r) + p r2 + pR2 = p [(12.5) (7+3.5)+ 3.52 + 72] 22 [192.5] = 605 cm2 7 Hence, the correct option is (c).

=

10.  80 m

r

Let r and h be the radius and the height of the cylinder, respectively. TSA of the cylinder = 2p r (h + r) Given: l + h = 35 cm Total surface area = 2 p r (r + h) = 1540 cm2 22 × r (35) = 1540 7 1540 × 7 ⇒r= 35 × 22 × 2 ⇒ r = 7 cm ∴ Height (h) = 28 cm Volume of the cylinder = p r2 h ⇒2×

14 m

14 m

100 m

Given, the cow is tied by a rope of length 14 m to the pole at a corner of the field of length 100 m and breadth 80 m.

M01_IIT-Hints and Solutions.indd 32

3 × a2 4

22 × 72 × 28 = 4312 cm3 7 Hence, the correct option is (c).

=

4/16/2018 11:52:23 AM

Hints and Explanation

A.33

13. Given: radius of the conical tent (r) = 150 cm Slant height of the tent (l) = 200 cm Lateral surface area of the tent = p r l = 94200 sq. cm N (94200) = 942000 N = 10 Hence, the correct option is (a).

Given, inradius of an equilateral triangle = 2 cm ∴ Circumradius = 2(2 cm) = 4 cm Area of the circle circumscribing the triangle ABC = p(4)2 = 16p sq. cm Hence, the correct option is (c).

14. Let a be the edge of the regular tetrahedron.

4. L  et l and b be length and breadth of the rectangle, respectively. ∴ Area of the rectangle = l b Given, length and breadth are increased by 20% and 10%, respectively. ∴ Area of the resultant rectangle

Given: a = 5

6 cm

Vertical height of the tetrahedron = =

2 (5 6 ) 3

=

2 (5 2 )

2 a 3

= 10 cm Hence, the correct option is (c). 15. Given, a = 10 cm  (i) LSA = 4a2 = 400 cm2 (ii)  Volume = a3 = 1000 cm3 (iii)  Longest diagonal =

3a = 10 3 cm

(iv) TSA = = 600 (i) → (B); (ii) → (D); (iii) → (A); (iv) → (C) Hence, the correct option is (c). 6a2

cm2

 120 l   110  b =   1000   100 

  120 l   110   100   100 b − lb   Required percentage =  lb × 100  12 × 11 − 100  =   × 100 = (132 - 100) % = 32%  100 Hence, the correct option is (b). 5. Tetrahedron is a triangular pyramid. ∴Number of edges = 2x = 2(3) = 6 Hence, the correct option is (a). 6. Let r be the radius of the sphere. Surface area of the sphere = 4 p r2

Assessment Test IV 1. D  BCA is the required sequential order. Hence, the correct option is (b). 2. 402 + 92 = 412 ∴ The triangle is a right-angled triangle. 1 × 9 × 40 = 180 cm2 2 Hence, the correct option is (c). Required area =

4pr2 = 5024 5024 4 × 3.14 r2 = 400 r = 20 cm Hence, the correct option is (c). r2 =

7. 

D

E

H

C

A

3. 

G F O

B

M01_IIT-Hints and Solutions.indd 33

D

C

A

B

4/16/2018 11:52:28 AM

A.34

Chapter 5  Mensuration Given: ABCD is a rectangle DE = EC, CF = FB, EH = HC, and CG = GF HC =

DC CB and CG = 4 4

∴ Area of ∆ HCG

1 1  DC   BC  (HC) (CG)=    2 4  4  2

1 (DC)(BC) 32 1 = × 96 = 3 cm2 32 Hence, the correct option is (a).

8. 

Let l and h be slant height and height of the bucket, respectively. ∴ R = 25 cm and r = 9 cm Total surface area = p l (R + r) + p R2+ p r2=1862 p ⇒ p [l (25 +9) + (25)2+ (9)2] = 1862 p ⇒ l (34) + 625 +81 = 1862 ⇒ l (34) = 1156 ⇒ l = 34 cm

(R − r )2 + h2

Slant height (l) = 34 =

(25 − 9)2 + h2

156 = 256 + h2 ⇒ h2 = 900 ⇒ h = 30 cm Hence, the correct option is (c).

r a

10. 

r a

Let r be the radius of the sphere and a be the edge of the cube. Here, the diameter of the sphere = the diagonal of the cube i.e., 2r =

3a ⇒ a =

2 3

50 m 12 m 12 m

r 2

 2r  Required ratio = 4 p r2:6    3  4 = 4 p r2:6 × × r2 = p :2 3 Hence, the correct option is (a).

50 m

Required area =

90° × p 122 360°

1 × p × 144 = 36 p 4 Hence, the correct option is (c). =

11. 

A

9.  25

8 8 9

Let R and r be the radii of the top and bottom circular ends of the bucket, respectively.

M01_IIT-Hints and Solutions.indd 34

C

h 10 B 8

8 8

Given: height (h) = 10 cm Side of the base (a) = 8 cm Six equal equilateral triangles constitute the base of the regular hexagonal pyramid.

4/16/2018 11:52:35 AM

Hints and Explanation ∴ BC =

3 ×8=4 2

(l) =

3 cm

l=

In ∆ABC, AC2 = AB2 + BC2

=

AC2 = 102 + ( 4 3 )2

A.35

(160)2 + (300)2 25600 + 90000

115600

l = 340 cm Lateral surface area of the tent = p r l = 3.14 × 160 × 340 = 170816 sq. cm Required area = 170816 × 10 = 1708160 cm2 = 170.816 sq. m Hence, the correct option is (c).

AC2 = 100 + 48 AC2 = 148 AC = 2 37 cm Hence, the correct option is (a). 12. 

14. Let a be the edge of the regular tetrahedron. Given: a = 4 Let h and r be the height and the radius of the cylinder, respectively. Given: h - r = 7 cm ⇒h=r+7 Given: lateral surface area = 2 p r h = 616 cm2 22 × r (r +7) = 616 7 ⇒ r (r + 7) = 14 × 7 ⇒ r = 7 cm h = 14 cm Volume of the cylinder = p r2 h ⇒2×

22 × 72 × 14 7 = 2156 cm3 Hence, the correct option is (b).

=

13. Given: base radius of the conical tent (r) = 160 cm Height of the conical tent (h) = 300 cm ∴ Slant height (l) of the tent =

M01_IIT-Hints and Solutions.indd 35

3 cm

Slant height of the tetrahedron = 3 × 4 3 = 6 cm 2 Hence, the correct option is (b).

3 a 2

=

15.  Column A

Column B

(CSA of the Solid)

(TSA of the Solid)

(i) CSA of cone

(a) TSA of cone frustum

(ii) CSA of hemisphere

(b) TSA of hemisphere

(iii) CSA of hemispherical bowl

(c) TSA of hemispherical bowl

(iv) CSA of cone frustum

(d) TSA of cone

(i) → (D); (ii) → (B); (iii) → (C); (iv) → (A) Hence, the correct option is (d).

r2 + h2

4/16/2018 11:52:41 AM

A.36

Chapter 6  Number Systems

CHAPTER 6 Number Systems Assessment Test I

⇒ a = 21/2, b = 31/3, and c = 41/4

1− 1+ x 1− 1+ x 1+ 1+ x 1.  = × 1− x 1− x 1+ 1+ x 1 − (1 + x ) = (1 − x )(1 + 1 + x )

⇒ a = 26

= =

1

⇒ a = (64 )

( x − 1)(1 + 1 + x )

12

1

; b = (81)

2. Going by the options, 5+2

(

324 − 2 323 =

⇒

= 5 - 4 = 1 is a rational number.

∴ Required value =

5 - 2, b = 2 ( 5 + 2 )

87 60

93

1

ab = ( 5 − 2) × 3 ( 5 + 2) = 3(1) = 3 is a rational number. Hence, the correct option is (d). 3. 3 12 - 3 27 + 2 48 = 3 4 × 3 - 3 9 × 3 + 2 16 × 3 =5 3 Hence, the correct option is (c).

M01_IIT-Hints and Solutions.indd 36

)

2

323 − 1 =

323 − 1

323 + 1 − ( 323 − 1)

49 50

− 73 1

=

93 − 73 1 (63) + (81)1 + (79)1

=

93 − 73 9(7 ) + 92 + 72

=

(9 − 7 )(92 + 9(7 ) + 72 ) 92 + 9(7 ) + 72

=9-7=2 Hence, the correct option is (b).

=6 3 -9 3 +8 3

3 , and c =

(

323 + 1

7.  07 07 08 (63)3 + (81)5 + ( 49)6

5 - 2 and b = 3 ( 5 + 2)

3

2

Hence, the correct option is (a).

= 2 (1) = 2 is a rational number.

2,b=

12

= 323 + 1 − 323 + 1 = 2

⇒ ab = ( 5 − 2) 2( 5 + 2)

4. Given, a =

)

323 + 1 =

= ( 5 )2 - (2)2

(iii)  a =

1

; c = (64 )

⇒ (x - 5)2 = 24 ⇒ x2 - 10x + 25 - 24 = 0 ⇒ x2 - 10x + 1 = 0 ∴ x2 - 10x - 1= -2 Hence, the correct option is (d). 6.  324 + 2 323 =

 (i) ab = ( 5 − 2) ( 5 + 2)

(ii)  a =

12

⇒x-5=2 6

.

Hence, the correct option is (b). 5 - 2, b =

1

x = 2 6 + 5 5.

\ The required sequential order is BDCA.

a=

1

∴a=c
−x (1 − x )(1 + 1 + x ) x

1

( ) 12 ; b = (34 ) 12 ; c = ( 43 ) 12

4

4

8. Given, 3x = 25 32x + 1 = (32x) (31) = (3x)2 (3) = (25)2 (3)

4/16/2018 11:52:57 AM

Hints and Explanation = 1875 Hence, the correct option is (d).

15.  (i): 

2 -3- 2-3=-6

9 × 3 24 = 3 9 × 3 × 8 = 6

\ 3 9 and 3 24 are rationalising factors to each other.

9. (3 2 + 6) − (9 + 2 2 ) = 2 − 3 Option (b) follows, ( 2 − 3)( − 2 − 3) = - (2 - 9) = 7

3

A.37

(ii):  12 + 3 and 12 − 3 are mixed surds as well as RF to each other.

∴ - 2 - 3 is a conjugate of the given surd.

(iii):  3 + 2 and 2 − 3 are mixed surds as well as conjugates to each other.

Hence, the correct option is (b).

(iv): 3 2 and 2 2 are similar surds.

1204 − 120 102 − 101 1084 542 = = 90 45 Hence, the correct option is (d).

(i) → (D); (ii) → (C); (iii) → (B); (iv) → (A) Hence, the correct option is (d).

10. 12.04 =

Assessment Test II 1. x = 5 + 2 6 (given)

11. 3. 4 3 = 4 34 × 4 3 = 4 34 × 3

⇒

= 4 243 Hence, the correct option is (c).

1 1 = x 5+2 6

⇒

1 5−2 6 = x ( 5)2 − 2 6

= = = 81 3x = 81; 9y = 81; 27z = 81 x = 4; y = 2; 33z = 34

12. 3x

9y

⇒Z=

27z

4 3

4 10 = 3 3 Hence, the correct option is (c). ∴x-y+z=4-2+

13.

6 3− 3

=

6 3− 3

×

2

1 =5-2 6 x 1 ⇒x+ = 5 + 2 6 + 5 - 2 6 = 10 x ∴ The required sequential order is DCAB. Hence, the correct option is (a). ⇒

2. G  oing by the options, x = 7 + 3 , y = 7 - 3 xy = (7 + 3 ) (7 − 3 )

3+ 3

= 49 - 3 = 46 is a rational number. Hence, the correct option is (c).

3( 3 − 1) 3( 5 − 3 ) 3( 7 − 5 ) + + + ... 14.  2 2 2 3( 25 − 23 ) 2 3 = [ 3 − 1 + 5 − 3 + 7 − 5 + ... + 25 − 23 ] 2 3 3 = [−1 + 25 ] = [4] = 6 2 2 Hence, the correct option is (d).

M01_IIT-Hints and Solutions.indd 37

)

3+ 3

6(3 + 3 ) = = 3+ 3 9−3 Hence, the correct option is (a).

+

(

3. 2 125 - 5 5 + 3 3125 = 2 25 × 5 - 5 5 + 3 625 × 5 = 10 5 - 5 5 + 75 5 = 80 5 Hence, the correct option is (c). 4. Given, a = a=

1 72

,

3 1

7 ,b=

9,b

1 = 123

3

12 , and c =

, and c = 1

4

1 52 4

( )

25 1

= 52 1

( )6 , b = ((12)2 )6 , and c = (53 )6

a = 73

4/16/2018 11:53:15 AM

A.38

Chapter 6  Number Systems 1

1

∴ 5 + 4 and - 5 + 4 are conjugate to each other.

1

a = ( 343 )6 , b = (144)6 , and c = (125)6 ∴c
2 -

1 2− 3

×

Hence, the correct option is (d).

2+ 3

9231 990 3077 = 330 Hence, the correct option is (c). =

2+ 3

2+ 3 2−3

=- 2 -

3

(

1 x+ = 2 - 3 + − 2− 3 x = 2 - 3 - 2 - 3

3 11. 2. 3 4 = 23 × 3 4

)

= 3 23 × 4 = 3 32 Hence, the correct option is (c).

=-2 3 Hence, the correct option is (a). 6.  16 + 2 55 = 11 + 5 + 2 11 5 = Now,

12. Given, 2x = 3y = 4z It is possible when x = y = z = 0 \ x + y + z = 0 + 0 + 0 = 0. Hence, the correct option is (a).

11 + 5

16 − 2 55 = 11 − 5

\ 16 + 2 55 + 16 − 2 55 = 11 + 5 + 11 − 5 = 2 11 Hence, the correct option is (d). 47 30

89 40

30

40

1

8. Given, 6x = 30 6x 30 6x - 1 = = = 5; 6x + 1 =(6x)6 = (30)6 = 180 6 6 \6x - 1 + 6x + 1 = 5 +180 = 185 Hence, the correct option is (c). 9. Option (d) follows. ( 5 + 4)( − 5 + 4) = −(5 − 16) = 11 ( 5 + 4) + ( − 5 − 4) = 0

13. 

11 5+ 3

=

11 5+ 3

×

5− 3 5− 3

11(5 − 3 ) 5 − 3 . = 25 − 3 2 Hence, the correct option is (c). =

1

7. (81)2 − (19)2 = (81)2 − (19)2 = (81)2 − (19)2 = (81 +19) (81 - 19) = (100) (62) = 6200 Hence, the correct option is (b).

M01_IIT-Hints and Solutions.indd 38

9324 − 93 103 − 101

10. 9.324 =

3

14.  2 - 1 + = -1 +

3 - 2 + 4- 3 +…+ 9- 8 9 = -1 + 3 = 2

Hence, the correct option is (b). 16 − 1 15 1 = = 90 90 6 16 − 0 16 (ii)  0.16 = 2 = 10 − 1 99 16 4 (iii)  0.16 = = 100 25 16 − 1 15 5 (iv)  1. 6 = = = 9 9 3 (i) → (C); (ii) → (B); (iii) → (D); (iv) → (A) Hence, the correct option is (d).

15.    (i) 0.1 6 =

4/16/2018 11:54:15 AM

Hints and Explanation

A.39

CHAPTER 7 Number Systems and Logarithms Assessment Test I 1. x = 2 + 3 (given) 1 1 = x 2+ 3 1 2− 3 = x 4−3 1 =2- 3 x 1 ⇒x=2+ 3 -2+ 3 =2 3 x \ The required sequential order is DCAB. Hence, the correct option is (a).  ( x2 − 4 )  = log3 2. log   ( x + 2)  ⇒ log(x - 2) = log3 ⇒ (x - 2) = 3 ⇒ x = 5 Hence, the correct option is (b). 3. log( 343 )4 (2401)3 log

(73 )4

(74 )3

= log712 712 = 1 Hence, the correct option is (c). 4. 3 4(2x + 4)/2 = 36(2x + 2)/2 ⇒ 8x + 16 = 12x + 12 ⇒ 4x = 4 ⇒ x = 1 Hence, the correct option is (a). 3( 3 − 1) 3( 5 − 3 ) 3( 7 − 5 ) 5.  + + + ... 2 2 2 3( 25 − 23 ) 2 3 = [ 3 − 1 + 5 − 3 + 7 − 5 + ... + 25 − 23 ] 2 3 3 = [−1 + 25 ] = [4] = 6 2 2 Hence, the correct option is (d). +

6. 61/2, 71/3, 81/4, and 91/5 take the LCM of denominators of the powers of the numbers.

M01_IIT-Hints and Solutions.indd 39

LCM = 60 Raise the numbers with this LCM (61/2)60, (71/3)60, (81/4)60, (91/5)60 630, 720, 815, 912 Between 630 and 720 (63)10 and (72)10 21610 and 4910 21610 rel="nofollow"> 4910, i.e., 630 > 720 Between 630 and 815 (62)15 and 815 3615 > 815, i.e., 630 > 815 Between 630 and 912 (65)6 and (92)6 (7776)6 > 816 i.e. 630 > 912 \ 61/2 is the largest in value. Hence, the correct option is (a). 7. 18 +

308

= 18 + 2 77

( 7 ) + ( 11) = ( 7 + 11 ) =

2

2

+ 2 7 × 11

2

\ 18 + 308 = 7 + 11 Hence, the correct option is (c). 8. log180 = log10 + log32 + log2  10  = 1 + 2(0.4771) + log    5 = 1 + 0.9542 + 1 - 0.6990 = 2.2552 Hence, the correct option is (b). 9. Consider log150250 = 250[log10 + log5 + log3] = 250[1 + 0.699 + 0.4771] = 250[2.1761] = 5440.25 The characteristic is 544. \Number of digits is 545. Hence, the correct option is (b).

4/16/2018 11:54:23 AM

A.40

Chapter 7  Number Systems and Logarithms

 ab   bc   ca  10. log   + log   + log    c  a  b  ( abc)2  = log   abc  = log abc Hence, the correct option is (d). 11. a =

5 + 31

a2 = 5 + 31 + 2 155 = 36 + 2(12 to 13) = 60 to 62 b=

b2 = 32 + 2 231 = 32 + 2(15 to 16) = 62 to 64 a2 < b2 \a
13. Option (b): a

 3125  3. log525 (3125)3125 =  log555  25  3125 × 5 = log55 25 = 625 Hence, the correct option is (a). 5 4.

11 + 21

12. Option (a):

⇒ x² - 1 = 15⇒ x = ±4 As log of negative numbers does not exist, x is 4. Hence, the correct option is (b).

logb c

× b

= bc = loga abc 14. Option (c): logc1 = 0 \loga° = log1 15. Option (e): logc - loga - logb + loga = logc - logb

2 x+ 1 3

= [ (52)4x + 5]1/6…

2 x+ 1 3

4 x+ 5 3

2x + 1 4 x + 5 = 3 3 ⇒ 2x = - 4 ⇒ x = - 2 Hence, the correct option is (c). ⇒ 5

= 5

5.  2-1+

⇒

3 - 2+ 4 - 3 +…+ 9- 8

= -1 +

9 = -1 + 3 = 2

Hence, the correct option is (b). 1

3

6.  3 = 32 ; 6

2

9

1

3 = 39

1

18 = 186

3 , 3 9 , and 3

1

( )3 = 3 3

9 = 32

9

3 have the same base.

9 has the largest power among them.

\

3

9 is greater than

3 and

9

\ Largest of the surds = Largest of

( 9 )= 3

6

3 . 3

9 and

\

6

log 72 = log372 = log39 + log38 1.  24 log24 3

\

( 9) >(

\

3

9 > 6 18

\

3

9 is the largest of the given surds.

2. l og[(x + 1) (x - 1)] = log15 ⇒ log(x² - 1) = log15

M01_IIT-Hints and Solutions.indd 40

18

92 = 6 81

Assessment Test II

= 2 log33 + 3 log32 = 2 + 3x \The required sequential order is ABDC. Hence, the correct option is (c).

6

81 > 6 18 3

6

6

18

)

6

Hence, the correct option is (b). 7. 1 4 - 6 5 = 14 - 2 45

4/16/2018 11:54:41 AM

Hints and Explanation

( 2) + ( 5) − 2 9 × 5 = ( 9 − 5 ) = (3 − 5 ) 2

=

2

2

2

\ 14 − 6 5 = 3 − 5 Hence, the correct option is (b). 8. l og405 = 4 log3 + log5  10  = 4(0.4771) + log    2 = 1.9084 + log10 - log2 = 1.9084 + 1 - 0.3010 = 2.6074 Hence, the correct option is (a). 9. C  onsider log6060 = 60[log10 + log3 + log2] = 60[1 + 0.4771 + 0.3010] = 60[1.7781] = 106.686 Characteristic is 106. \The number of digits is 107. Hence, the correct option is (a).

b2 = 6 + 22 + 2 132 = 28 + 2(11 to 12) = 50 to 52 a2 > b2 \a>b Hence, the correct option is (a). 12. Option (b): log3 2 × log5 3 × log4 25 log 2 log 3 log 25 × × log 3 log 5 log 4 log 2 log 3 2 log 5 × × =1 = log 3 log 5 2 log 2

=

13. Option (a):

a2

7 + 21

= 7 + 21 + 2 147

= 28 + 2(12 to 13) = 52 to 54 b=

6 + 22

M01_IIT-Hints and Solutions.indd 41

log 5 log 2 + log 100 log 10 log 5 log 21/2 + = 2 log 10 log 10  1  1 =   log 5 +   log 2  2  2 1 1 = log 10  = 2 2

14. Option (c): 3 log3 3 + log625 25 3 1  3  2 =   log3 3 +   log5 5 = +  2  4 2 2

10. logb a × logc b × loga clogc log a log b log c = × × log c × = log c log b log c log a Hence, the correct option is (c). 11. a =

A.41

= 15. Option (d):

4 =2 2

1 log2 8 + 2 log32 128 15 3 2(7 ) = log2 2 + log2 2 15 5 =

1 14 15 + = =3 5 5 5

4/16/2018 11:55:37 AM

A.42

Chapter 8  Polynomials and Square Roots of Algebraic Expressions

CHAPTER 8 Polynomials and Square Roots of Algebraic Expressions Assessment Test I

6. f(x) = (x - 1)2 (x2 - 1) 2(x + 2)2 = (x - 1)4 (x + 1)2 (x + 2)2 g(x) = (x2 - 2x + 1) (x + 2)3 (x + 1)3 = (x - 1)2 (x + 2)3 (x + 1)3 \ HCF = (x - 1)2 (x + 2)2 (x + 1)2

1. C  BDAE is the required sequential order. Hence, the correct option is (c). 2. 4x4 + 16 = (2x2)2 + 42 = (2x2)2 + 2(2x2) 4 + 42 - 2(2x2) 4 = (2x2)2 + 16x2+ 42 - 16x2 = (2x2 + 4)2- (4x)2 = (2x2 + 4x + 4) (2x2 - 4x + 4) = 4(x2 + 2x + 2) (x2 - 2x + 2) Hence, the correct option is (c).

Hence, the correct option is (c). 7. x4 - 14x3 + 71x2 + kx + l is a perfect square. x2 - 7x + 11 x2

x4 2x2 - 7x

3. 

4x2 - 12x + 9

- 7x 2x2

16x4

⇒ k = - 154 and l = 121 k+ l = -154 + 121= -33

-  +   -

Hence, the correct option is (a).

0 \ Square root of the given expression is 12x + 9. Hence, the correct option is (a).

)(

-

)

4.  x4 − 3 x2 + 3 x − 1 x3 + x2 − 2 x + 1 ( x − 1)

(

)(

)

x 4 − 3 x2 + 3 x − 1 x 4 − 3 x2 + 3 x − 1

= + 3x - 1 Hence, the correct option is (d). x4

3x2

5. The HCF of f(x) and g(x) is a factor of the LCM of f(x) and g(x). \ Option (d) follows. Hence, the correct option is (d).

M01_IIT-Hints and Solutions.indd 42

22x2 - 154x + 121 0

72x2 - 216x + 81

4x2

22x2 + kx + l -   +    -

72x2 - 216x + 81



- 14x

-96x3 + 144x2

8x2 - 24x + 9

(

-14x3 + 49x2

+ 11

-96x3 + 216x2-216x + 81

-12x

-14x3 + 71x2 + kx +l +  -

4x2 16x4 - 96x3 + 216x2-216x + 81 8x2 - 12x

x4 - 14x3 + 71x2 + kx + l

8. ( 8x2 + 10x - 25) (2x2 + 15x + 25) (4x2 + 15x -25) =(8x2 + 20x - 10x -25) (2x2 + 10x + 5x + 25) (4x2 + 20x - 5x - 25) =(4x - 5) (2x + 5) (2x + 5) (x + 5) (4x - 5) (x + 5) (4x - 5)2 (2x + 5)2 (x + 5)2 \Square root of the given expression is (4x - 5) (2x + 5) (x + 5). Hence, the correct option is (a). 9. Given expression is a complete homogenous and symmetric expression in two variables x and y. \ a = e and b = d Hence, the correct option is (d).

4/16/2018 11:55:38 AM

Hints and Explanation

\ Square root of the given expression is 9x2 - 6x -1. Hence, the correct option is (d).

10. (4x2 - 12x + 9) (4x2 + 12x + 9) = (2x - 3)2 (2x + 3)2 \Square root of the given expression = (2x - 3) (2x + 3) = 4x2 - 9 Hence, the correct option is (c). 11. (3x + 3y + xy)2 -12xy (x + 3) = (3x + 3y + xy)2 -4(3x) (y) (x + 3) = (3x + 3y + xy)2 -4(3x) (xy + 3y) (A + B)2 - 4AB = (A - B)+2

∴

= [3x - (3y + xy)]2   = (3x - 3y - xy)2

\Square root of the given expression is 3x - 3y - xy Hence, the correct option is (a). 12. Option (a): (3x - 2y)2 = 9x2 -12xy + 4y2 is a homogeneous expression. 13. Option (c): (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx is a homogeneous cyclic expression. 14. Option (b): 3x2 - 2xy + 5x + 5y + 3y2 is a symmetric expression. 15. Option (d): 3x3 - 3y2 - 3x + 3y is a non-homogeneous and non-symmetric expression.

1. C  ABED is the required sequential order. Hence, the correct option is (d). 2. x4 + x2 + 1 = (x2 + 1)2 - x2 = (x2 + x + 1) (x2 - x + 1) Hence, the correct option is (c).

9x2

81x4

9x2

-81x4

- 6x - 12x - 1

6. f(x) = (x + 1)2 (x - 1)3 (x + 2)4 g(x) = (x - 2)4 (x + 1)3 (x + 2) (x - 1)2 LCM = (x + 1)3 (x - 1)3 (x + 2)4 (x - 2)4 = (x2 - 1)3 (x2 - 4)4 Hence, the correct option is (d). 7.

3x2 - 2x + 3.



3x2 9x4 - 12x3 + 22x2 - kx + m

-

6x2 - 2x - 2x

- 12x3 + 22x2 - kx + m - 12x3 + 4x2

6x2 - 4x

18x2 - kx + m

+3

18x2 - 12x + 9 (-)  (+)  (-)

108x3

+

18x2

+ 12x +

- 108x3 + 18x2 + 12x + 1 - 108x3 + 36x2 (+)  (-)

18x2

5. f(x) × g(x) = HCF × LCM When HCF = 1, f(x) × g(x) = LCM. \ Option (b) follows. Hence, the correct option is (b).

(+)    (-)

9x2 - 6x - 1

18x2 - 6x

4. (x4 + x2 - 2) (x3 - x2 + 2x - 2) (x + 1) (x4 + x2 - 2) (x4 - x3 + 2x2 - 2x + x3 - x2 + 2x - 2) (x4 + x2 - 2) (x4 + x2 - 2) (x4 + x2 - 2)2 \ Square root of the given expression is x4 + x2 - 2. Hence, the correct option is (b).

3x2 9x4

Assessment Test II

3. 

A.43

-18x2 + 12x + 1 -18x2

+ 12x + 1

(+)  (-)  (-)

0 ⇒ k = 12, m = 9 ⇒ k + m = 21 Hence, the correct option is (d). 8. ( x2 + 8x + 15) (x2 + x - 6) (x2 + 3x - 10) = (x + 3) (x + 5) (x +3) (x - 2) (x + 5) (x - 2) = (x + 3)2 (x + 5)2 (x - 2)2

0

M01_IIT-Hints and Solutions.indd 43

4/16/2018 11:55:39 AM

A.44

Chapter 8  Polynomials and Square Roots of Algebraic Expressions \ Square root of the given expression is (x + 5) (x + 3) (x -2). Hence, the correct option is (c).

9. f(x) and g(x) are two complete homogenous and symmetric expressions. f(x).g(x) is also a complete homogeneous expression. Hence, the correct option is (c). 10. (x2 - 10x + 25) (x2 + 10x + 25) = (x - 5)2 (x + 5)2 \ The required square root is ( x - 5) (x + 5). i.e., x2 - 25 Hence, the correct option is (d). 11. (2x + - 4 (x + 2y) (x - y) = (x + 2y + x - y)2 - 4 (x + 2y) (x - 2y) = [x + 2y - (x - y)]2 Q (a + b)2 - 4ab = (a - b)2 = 9y2 \ Square root of the given expression is 3y. Hence, the correct option is (d). y)2

12. Option (e): 5x2 - 8xy is homogenous but not complete. 13. Option (c): 3x2 - 5xy + 3y2 - 7x - 7y + 6 is symmetric but not homogenous. 14. Option (b): ax2 + 2bxy + by2 is complete homogenous but not symmetric. 15. Option (a): x + y is complete homogenous and symmetric.

Assessment Test III 1. T  he required sequential order of steps is BDAC. Hence, the correct option is (c). 2. We know that (a + b)2 + (a - b)2 = 2a2 + 2b2 \ (2x + 3y)2 + (2x - 3y)2 = 2(2x)2 + 2(3y)2 = 8x2 + 18y2 Hence, the correct option is (a). 3. Degree of f(x)·g(x) = Degree of f(x) + Degree of g(x)

M01_IIT-Hints and Solutions.indd 44

⇒ 9 + Degree g(x) = 27 ⇒ Degree of g(x) = 27 - 9 = 18 Hence, the correct option is (c). 4. Let f(x) = x4 + 3x2 - 2x - 10 and divisor = x + 2 Remainder = f(-2) = (-2)4 + 3(-2)2 - 2 (-2) - 10 = 16 + 12 + 4 - 10 = 22 Hence, the correct option is (d). 5. x2 + 7x + 12 = (x + 3) (x + 4) x2 + 8x + 15 = (x + 3) (x + 5) x2 + 9x + 20 = (x + 4) (x + 5)

( x2 + 7 x + 12) ( x2 + 8x + 15) ( x2 + 9x + 20) = ( x + 3)2 ( x + 4)2 ( x + 5)2 = (x + 3) (x + 4) (x + 5) Hence, the correct option is (b). 6. f(x, y) = 5x5 - 3y5 + 2x3ya - 7xby is a homogenous expression of degree 5. \3 + a = 5 and b + 1 = 5 ⇒ a = 2 and b = 4 ⇒ a - b = 2 - 4 = -2 Hence, the correct option is (b). 7. Given f(x + 3) = x2 - 2x + 8 (1) The remainder, when f(x) is divided by (x - 2) is f(2). Put x = -1 in Eq. (1) f(-1 + 3) = (-1)2 - 2 (-1) +8 ⇒ f(2) = 1 + 2 + 8 = 11 Hence, the correct option is (b). 8. 8x3 + y3 = (2x)3 + y3 = (2x + y) [(2x)2 + y2 - (2x) (y)] = (2x + y) (4x2 + y2 - 2xy) Hence, the correct option is (b). 9. (x + y) (x2 + y2) (x4 + y4) (x8 + y8) = =

(

)(

)(

( x − y )( x + y ) x2 + y2 x4 + y4 x8 + y8

)

x−y

( x2 − y2 ) ( x2 + y2 ) (x4 + y4 ) ( x8 + y8 ) x−y

4/16/2018 12:39:22 PM

Hints and Explanation

= =

( x4 − y4 ) ( x4 + y4 ) ( x8 + y8 ) (

x8

−

y8

)(

x−y x8 + y8

14. Let f(x) = x2 + 8x + a, and a factor is (x + 1). ⇒ f(-1) = 0 ⇒ (-1)2 + 8(-1) + a = 0 ⇒1-8+a=0 ⇒a=7 Hence, the correct option is (c).

)

x−y x16 − y16 = x−y Hence, the correct option is (c). 10. The LCM of x9 ya and xb y6 = x10y10 ⇒ a = 10 and b = 10 ⇒ a + b = 20 Hence, the correct option is (b). 2 1 11. x2 + 2x + 3 + + 2 x x 2  1  1  1 = (x)2 +   + 12 + 2(x)   + 2   (1) + 2(x)  x  x  x (1) 1   =  x + + 1  x  \

2

x2 + 2 x + 3 +

2 1 + x x2

1 +1 x Hence, the correct option is (c). 12. (x + 1) (x + 3) (x + 5) (x + 7) + p = (x2 + 8x + 7) (x2 + 8x + 15) + p = (a + 7) (a + 15) + p (Let us assume that x2 + 8x = a.) = a2 + 22a + 105 + p = a2 + 2 (a) (11) + (105 + p) is perfect square. ⇒ 105 + p = 112 p = 16 Hence, the correct option is (a). 13. Let f(x) = + + Put x = y ⇒ f(y) = y(y2 - z2) + y(z2 - y2) + z(y2 - y2) = y(y2 - z2) - y(y2 - z2) + 0 =0 \ x - y is a factor of f(x). Hence, the correct option is (d).

M01_IIT-Hints and Solutions.indd 45

z2)

y(z2

x2)

15. (i) → (C); (ii) → (a); (iii) → (d); (iv) → (b) Hence, the correct option is (d).

Assessment Test IV 1. T  he required sequential order of steps is DBAC. Hence, the correct option is (d). 2. We know that (a + b)2 - (a - b)2 = 4ab (5x + 4y)2 - (5x - 4y)2 = 4(5x)(4y) = 80xy Hence, the correct option is (d). 3. Degree of f(x)·g(x) = Degree of f(x) + Degree of g(x) = 15 + 10 = 25 Hence, the correct option is (b).

=x+

x(y2

A.45

z(x2

y2)

4. Let f(x) = 2x5 - 3x3 + x2 - 18 and divisor = x - 2 Remainder = f(2) = 2(2)5 - 3(2)3 + (2)2 - 18 = 64 - 24 + 4 - 18 = 68 - 42 = 26 Hence, the correct option is (a). 5. x 2 - 3x + 2 = (x - 1) (x - 2) x2 - 7x + 6 = (x - 1) (x - 6) x2 - 8x + 12 = (x - 2) (x - 6) \

( x2 − 3x + 2) ( x2 − 7 x + 6) ( x2 − 8x + 12)

=

( x − 1)2 ( x − 2)2 ( x − 6)2

= (x - 1) (x - 2) (x - 6). Hence, the correct option is (c). 6. f(x, y) = 3x6 + 4y6 + 8xayb is a homogeneous expression of degree 6. ⇒a+b=6

4/16/2018 11:55:46 AM

A.46

Chapter 8  Polynomials and Square Roots of Algebraic Expressions We cannot say anything about the values of a and b. Hence, the correct option is (d).

7. Given, f(x - 2) = 2x2 + 4x + 9 The remainder, when f(x) is divided by x + 4 is f(-4). Put x = -2 ⇒ f(-2 -2) = 2(-2)2 + 4(-2) + 9 ⇒ f(-4) = 8 - 8 + 9 =9 Hence, the correct option is (b). 8. x2 - y2 + 2xz + z2 = (x2 + 2xz + z2) - y2 = (x + z)2 - y2 = (x + z + y) (x + z - y) Hence, the correct option is (a). 9. (a4 + b4) (a2 + b2) (a - b) = = = =

( a4 + b4 ) ( a2 + b2 ) (a − b)(a + b) a+b

( a4 + b4 ) ( a2 + b2 ) ( a2 − b2 ) a+b

( a4 + b4 ) ( a4 − b4 ) a+b a8

− b8

a+b

Hence, the correct option is (b). 10. The HCF of x3y2 and xayb. = x3y2 ⇒ a = 3 and b = 2 The minimum value of a + b = 3 + 2 = 5 Hence, the correct option is (a). 11. x4 + 6x2 + 4x + 9 +

M01_IIT-Hints and Solutions.indd 46

12 4 + 2 x x

2

 2  2  2 = (x2)2 +   + (3)2 + 2(x2)   + 2   (3)  x  x  x 2 + 2(x ) (3) 2   =  x2 + + 3    x

2

\ x 4 + 6 x2 + 4 x + 9 +

12 4 + x x2

2 +3 x Hence, the correct option is (a). = x2 +

12. (2x + 1) (2x + 2) (2x + 3) (2x + 4) + q = (4x2 + 10x + 4) (4x2 + 10x + 6) + q = (a + 4) (a + 6) + q (Let us assume that 4x2 + 10x = a) = a2 + 10a + 24 + q = (a)2 + 2 (a) (5) + (24 + q) is a perfect square. ⇒ 24 + q = 52 ⇒q=1 Hence, the correct option is (c). 13. Let f(x) = x(y3 - z3) + y(z3 - x3) + z(x3 - y3) Put x = y ⇒ f(y) = y(y3 - z3) + y(z3 - y3) + z(y3 - y3) = y(y3 - z3) - y(y3 - z3) = 0 \ x - y is a factor of f(x). As f(x) is a cyclic expression, in x1 y and z. z - x is also a factor of f(x). Hence, the correct option is (b). 14. Let f(x) = x2 - bx + 8, and a factor is (x - 1). ⇒ f(1) = 0 ⇒ (1)2 - b(1) + 8 = 0 ⇒g-b=0 ⇒b=9 Hence, the correct option is (a). 15. (i) → (c); (ii) → (d); (iii) → (a); (iv) → (b) Hence, the correct option is (a).

4/16/2018 11:55:50 AM

Hints and Explanation

A.47

CHAPTER 9 Statistics and Probability; Applications of Percentages Assessment Test I 1. A  BCD is the required sequential order. Hence, the correct option is (a). 2. Let the total number of employees be 100. Number of female employees = 40 \ Number of male employees = 60 Total number of employees who are graduates = 70 To have the minimum percentage of female postgraduates, all the male employees must be postgraduates. In this case, number of postgraduate female employees must be 70 − 60 = 10. Minimum percentage of postgraduate fe10 × 100%  = 25%. male employees = 40 Hence, the correct option is (d). 25 + 1 3. M  ean of first 25 natural numbers = = 13 2 (20)2 Mean of first 20 odd natural numbers = 20 = 20 Mean of the squares of first 13 natural numbers =

13 × 14 × 27 = 63 13 × 6

13 + 20 + 63 96 = = 32 3 3 Hence, the correct option is (a). \ Required average =

4. M  ode = 3 median − 2 mean Mode - mean = 3 median - 3 mean Mode - Mean = 3(Median - mean) \ Mode - mean = 3x (Since median - mean = x) Hence, the correct option is (b). N  −F  2  5. M  edian = L +  ×C f   

M01_IIT-Hints and Solutions.indd 47

 40  − (24 − 8)  2  = 20 +   ×5 8    20 − 16  = 20 +  × 5 = 20 + 2.5  8  = 22.5 Hence, the correct option is (d).  = 35, n = 40, and c = 10 6. A S fi mi = −8 1 Mean = A + (Σ fi × mi ) × c n 1 ( −8)  × 10 = 35 + 40 = 33 Hence, the correct option is (c). L = 40, f = 10, f1 = 6, and f2 = 9 7. ( f − f1 ) ×c Mode = L + 2 f − ( f1 + f2 ) (10 − 6) 4 × 20 = 40 + = 40 + × 20 5 2 × 10 − (6 + 9) = 56 Hence, the correct option is (a). 8. L  et his cost price be `x. Selling price = `0.9x 0.9x + 14 = 1.1x 14 = 0.2x \ 70 = x Hence, the correct option is (a). 100 − (10 + 20 + 15 + 15 + 20) 100 100 − 80 = 100 20 1 = = 100 5 Hence, the correct option is (b).

9. P  (E) =

4/16/2018 11:55:57 AM

A.48

Chapter 9  Statistics and Probability; Applications of Percentages

10. P(E) =

40 − (10 + 12 + 8 + 4) 100

6 100 3 = 50 Hence, the correct option is (a). =

11. Let the total number of employees be 100. \ The number of men employees = 60 20% of 60 P(E) = 100 12 3 = = 100 25 Hence, the correct option is (d). 12. Option (d): To know the deviation of the data, Range is used. 13. Option (c): To know the popular brand of an article, Mode is used. 14. Option (b): Median is used, when extremes are not important. 15. Option (a): When all observations are important, Mean value is taken.

Assessment Test II 1. B  AECD is the required sequential order. Hence, the correct option is (a). 2. Let the total number of students be 100. \ Number of boys = 40 and the number of girls = 60 Number of girls who scored at least 70% = 25% of 60 = 15 Percentage of girls who scored less than 70% 45 × 100 = 45% 100 Hence, the correct option is (b). =

3. A  =

21 + 1 7 × 8 × 15 = 11, B = 6×7 2

= 20 C = 102 = 100

D = 10 × 11 = 110

M01_IIT-Hints and Solutions.indd 48

\ Median of the value 11, 20, 100, 121 is 100 + 20 = 60. 2 Hence, the correct option is (b). 4. Mode - Median = 2(Median - Mean) =2×6 =12 Hence, the correct option is (d). 5. L = 30, f = 10, F = 25 - 10 = 15 n = 40 n   − F 2 ×c \M=L+ f  40   − 15 2 = 30 + × 10 10 = 30 +

5 × 10 10

= 35 Hence, the correct option is (b). 6. S  fi xi = 1400 n = 50 Σ fi xi 1400 \M= = = 28 n 50 Hence, the correct option is (d). 7. Since each class has at least 2 entries, the class interval is 10. The other classes which are not mentioned in the problem cannot have more than 9. The data of the class 40 - 50 is \ f = 13, f1 = 9, f2 = 7, and L = 40, c = 10 ( f − f1 )

×c 2 f − ( f1 + f2 ) 13 − 9 = 40 + × 10 2(13) − (9 + 7 ) 4 = 40 + × 10 = 44 10 Hence, the correct option is (b). Mode = L +

8. L  et the cost price of the article be `x. \ Selling price = 0.8x New selling price =1.3x Given that 1.3x - 0.8x = `40 0.5x = `40

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Hints and Explanation \ x = `80 Hence, the correct option is (d). 9. The outcomes of a die are 1, 2, 3, 4, 5, and 6. Number of evens = 3 Number of odds = 3 Number of primes = 3 Number of composites = 2 \ Probability of getting a composite number is least. 2 1 = , which is the least. 6 3 Hence, the correct option is (c). 15 1 10. Required probability = = 60 4 Hence, the correct option is (c). i.e.,

11. The percentage of Grade III articles = 100 - (4 + 70 + 20) = 6 6% 3 \ Required probability is = 100% 50 Hence, the correct option is (a). 12. Option (b): Mean is the mid-values of the classes. 13. Option (a): Median is the cumulative frequency. 14. Option (d): Mode is the maximum frequency. 15. Option (c): Range deals with the maximum value.

Assessment Test III 1. B  ACD is the required sequential order. Hence, the correct option is (a). 2. Given, 30, 24, 37, 52, 28 and x Range = Maximum value - Minimum value 31 = 52 - x ⇒x = 52 - 31 ⇒ x = 21 Hence, the correct option is (d). 3. Total possible outcomes = {HH, HT, TH, TT} Favourable cases of getting at least one head are HH, HT, and TH.

M01_IIT-Hints and Solutions.indd 49

\ The number of favourable cases = 3 Hence, the correct option is (c). 4. Given, median - mean = 16 We know that mode = 3 median - 2 mean Mode - mean = 3 median - 2 mean - mean Mode - mean = 3 (median - mean) Mode - mean = 3 (16) \ Mode - mean = 48 Hence, the correct option is (c). 5. Given, (2x% of y) + (x% of 3y) = 12.5% of (x + y) 3 xy 125 2 xy + = (x + y) 100 100 1000 5 xy 125 = (x + y) 100 1000 x+y 1000 5 = × ⇒ 100 125 xy 1 1 2 ⇒ + = y x 5 Hence, the correct option is (c). 6. Given, two tails was twice that of all tails. x i.e., Frequency of three tails = 2 Outcome

No tail

One tail

Two tails

All tails (three tails)

Frequency

160

330

x

x 2

∴ Total frequency = 1000 160 + 330 + x + 490 +

x = 1000 2

3x = 1000 2

3x = 510 2 \ x = 340 Hence, the correct option is (c). 7. G  iven, 2, 4, 9, 13, 23, 29 and 32 N=7 th

th

 N + 1  7 + 1 Q3 = 3  term = 3  term  4   4  = 6th term = 29

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Chapter 9  Statistics and Probability; Applications of Percentages  N + 1 Q1 =   4 

Given a = 24 and b = 32

th

=

2nd

term = 4

Quartile deviation = (Q.D.) =

Q3 − Q1

29 − 4 25 = = = 12.5 2 2 Hence, the correct option is (c).

2

 iven, A = 4P 8. G We know that A = P + I 4P = P + I 3P = I PTR S.I. = 100 P×5×R 3P = 100 300 ⇒ =R 5 Given, the same sum becomes sixteen times itself. i.e., A = 16P A=P+I 16P = P + I ⇒ I = 15P P×T ×R 100 300 P×T × 5 15P = 100 \ T = 25 years Hence, the correct option is (b).

S.I. =

2

16   9. Given, P  1 − = `194,481  100  2 4  P  1 −  = `194,481  25   21   21  P     = `194,481  25   25  P=

` 194481 × 25 × 25 21 × 21

\ P = `275,625 Hence, the correct option is (d). 10. Mean deviation of two numbers a and b is | a − b| . 2

M01_IIT-Hints and Solutions.indd 50

|24 − 32| 8 = =4 2 2 Hence, the correct option is (a). \ Mean deviation =

11. Given, the median of the data is 30. From the data 10 and 40 are removed, those two observations are either side of 30. Therefore, the median never changes. Hence, the correct option is (a). 12.  Class

Frequency

0-10

2

10-20

7

20-30

x(f1)

30-40

23(f)

40-50

20(f2)

50-60

10

→    Modal class

Given, mode = 36.25. It lies between 30 and 40. ∴ 30-40 is the modal class. ⇒ f = 23, f1= x, and f2 = 20 L = 30 and C = 10 Mode = L + 36.25 = 30 + ⇒ =

( f − f1 ) × c 2 f − ( f1 + f2 )

(23 − x ) × 10 2 × 23 − ( x + 20 )

(23 − x ) × 10 46 − x − 20

(23 − x ) × 10 26 − x

= 36.25 − 30 = 6.25

⇒230 − 10x = 162.5 − 6.25x ⇒ 10x − 6.25x = 230 − 162.5 ⇒ 3.75x = 67.5 ⇒x=

67.5 3.75

⇒ x = 18 Hence, the correct option is (b).

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Hints and Explanation 13. Let cost price of an article be `100. \ The value of the article after increase of 60% = `160 160 − 100 Required percentage = × 100 160 =

60 300 1 × 100 = = 37 % 160 8 2

Hence, the correct option is (d). 14. Let P = `90,000 R = 10% T = 2 years 2

 R   10  = `90000  \ Difference = P  = `90  100   100  Hence, the correct option is (b).

(ii)

1 + 101 = 51 → (c) 2 101 (102) ( 203 ) 12 + 22 + 32 + . . . + 1012 = 6 × 101 101 = 17(202) = 3451 → (d)

(iii)

(101)2 101

= 101 → (b)

0 + 100 (iv) = 50 → (a) 2 (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A) Hence, the correct option is (c).

Assessment Test IV 1. D  ACB is the required sequential order. Hence, the correct option is (d). 2. Given, 26, 19, 32, 23, and x Range = Maximum value - Minimum value x = 32 - x ⇒ 2x = 32 ⇒x = 16 Hence, the correct option is (d). 3. Total possible outcomes = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} Favourable cases of getting at least two heads

M01_IIT-Hints and Solutions.indd 51

= {HHT, HTH, THH, HHH} \ Number of favourable cases = 4 Hence, the correct option is (d). 4. Given, Median - Mean = 20 We know that Mode = 3 Median - 2 Mean Mode - Median = 2 (Median - Mean) \ Mode - Median = 40 Hence, the correct option is (b). 5. Given, (x% of y) + (3% of xy) = 12% (x + y)

2

15.  (i)

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xy 3 xy 12 ( x + y ) + = 100 100 100 4 ( xy )

=

12 ( x + y )

100 100 xy = 3 (x +y) 1 1 1 + = x y 3 Hence, the correct option is (a). ⇒

6. Let y be the occurrence of all the three heads. Given, occurrence of no head was two-third that of all the three heads. i.e., x =

2 3x (y) ⇒ y = 3 2 No One head head

Outcome Frequency

x

250

Two head

Three heads

50

3x 2

∴ Total frequency = 600 x + 250 + 50 +

3x = 600 2

5x = 300 2 ∴ x = 120 Hence, the correct option is (d). 7. 6 , 10, 12, 13, 14, 18, 22, 26 N=8 th  N Q3 = 3   term = 6th term = 18  4  N Q1 =    4

th

term = 2nd term = 10

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Chapter 9  Statistics and Probability; Applications of Percentages Q − Q1 \ Quartile deviation (Q.D.) = 3 2 18 − 10 8 = = =4 2 2 Hence, the correct option is (a).

8. A  sum becomes doubled itself at compound interest in six years. Given, A = 2P n

Mean deviation |2 − 8|+ |5 − 8|+ |6 − 8|+ |8 − 8|+ |10 − 8|+ |14 − 8|+ |17 − 8| = 7 6 + 3 + 2 + 0 + 2 + 6 + 9 28 = =4 7 7 Hence, the correct option is (a). =

11. G  iven, the median of the data is 45. 34 and 48 added to the data, those two observations are either side of the median 45. ∴ The median of the data will not be changed. Hence, the correct option is (c).

r   P 1+ =A  100  6

r   P 1+ = 2P  100  6

r    1 +  =2 100 

12. 

64

 r  4   = (2)   1 + 100    r    1 +  100 

24

24

r   = 16 ⇒ P  1 + = 16P  100 

∴ In 24 years, the sum will become sixteen times itself. Hence, the correct option is (b). n

r   9. Given, P  1 − = 1711.25  100  2

7.5   P 1 − = 1711.25  100  2

171, 125 75   P 1 − =   1000  100

Class

(xi) 1-9

9

5

45

10-18

11

14

154

19-27

8

23

184

28-36

m

32

32m

37-45

3

41

123

46-54

11

50

550

171, 125  925   925  = P      1000   1000  100 P=

171, 125 × 1000 × 1000 925 × 925 × 100

\ P = `2000 Hence, the correct option is (b). 10. Given data, 2, 6, 5, 10, 14, 8, and 17. Arrange the data in ascending order: 2, 5, 6, 8, 10, 14, 17 ∴N=7 ⇒ Median = 8

M01_IIT-Hints and Solutions.indd 52

Frequency (fi)

Midvalues of Adjusted fi xi class

∑ Mean = 29 =

fi = 42 + m

∑ fi xi = 1056 + 32m

∑ fi xi ∑ fi

1056 + 32m 42 + m

⇒ 1056 + 32m = 29 (42 + m) ⇒ 1056 + 32m = 1218 + 29m ⇒ 3m = 162 ⇒ m = 54 Hence, the correct option is (d).

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Hints and Explanation 13.  Share of A

Share of B

150

100

 150 − 100  Required percentage =   × 100%  150  1 = 33 % 3 Hence, the correct option is (b).  R  14. C.I. - S.I. = P   100   R  400 = 40,000   100 

2

2

R2 = 100 ⇒ R = 10 Hence, the correct option is (c).

M01_IIT-Hints and Solutions.indd 53

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15.   (i)  40 cm = 9% of 600 cm 40 9= × 100 600 2 = 6 → (b) 3 (ii)  87.5% of 125 + 12.5% if 125 = 100% of 125 = 12.5% of 100 → (a)  1 (iii) Loss =   C.P.  6 1 2 Loss% = × 100% = 16 % → (c) 6 3  1 (iv) Profit =   C.P.  5 1 Profit% = × 100% = 20% → (d) 5 (i) → (B); (ii) → (A); (iii) → (C); (iv) → (D) Hence, the correct option is (d).

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