Ssc_speed, Time And Distance

1

Speed, Time and Distance

Speed, Time and Distance Take another example, a train A moving with the speed of 60 km/hr will cross the another train B moving in same direction with the spee d of 20 km/hr with a re lative speed of (60 – 20 =) 40 km/hr which is less than the train A and greater than the train B.

1. Important Formulae (i)

Distance = Speed × Time Distance (ii) Time = Speed (iii) Speed =

Distance Time

3. Average Speed

5  (iv) x km/hr =  x   m/sec 18 

Total Distance Covered Total Time Taken For example, a person divides his total route of journey into three equal parts and decides to travel the three parts with speed of 40, 30 and 15 km/hr respectively. Find his average speed during the whole journey. Let the three equal parts of journey be x km. Time taken to travel first part of the journey  x  =   hour 40  Time taken to travel second part of the journey  x  =   hour 30  Time taken to travel third part of the journey  x =   hour 15 Average Speed =

18   (v) x m/sec =  x   km/hr 5

2. Relative Speed

K KUNDAN

(i) Wh en o bjects are m oving in op posite directions

The relative speed of one object with respect to the other will have magnitude greater than individual speed of each object. This is why, for example, a train A moving with speed 10 km/hr will cross another train B moving in opposite directions with speed 25 km/hr, with a relative speed of (10 + 25 =) 35 km/hr which is greater than the individual speed of either train. (ii) When objects are moving in same direction

x x   x    Total time taken =   hours 40 30 15  Total distance travelled = x + x + x = 3x km Total Distance Travelled Average Speed = Total Time Taken

The relative speed of one object with respect to the other will have magnitude either less than or greater than individual speed of each object. This is why, for example, a train A moving with the speed of 20 km/hr will cross the another train B moving in same direction with the speed of 15 km/hr, with a relative speed of (20 – 15 =) 5 km/hr which is less than the individual speed of either train.

=

x km/hr = 24 km/hr x x  x     40 30 15

Exercise 1.

6 of his usual speed, a 7 man becomes late by 25 minutes. What is his usual time taken for the journey? For a journey, walking

5 km/hr, she is late by 7 minutes. However, if she walks at the rate of 6 km/hr, she reaches the school 5 minutes earlier than the scheduled time. What is the distance of the school from her house?

[SSC Section Officers’ (Audit) Exam–2005]

2.

A motorist covers a distance from A to B at a speed of 20 km/hr and return journey from B to A at a speed of 30 km/hr. If he takes 5 hours for the whole journey, find the distance from A to B. [SSC UDC (Mains) Exam–2001]

3.

Shivangi starts from her house for her school at a certain fixed time. If she walks at the rate of

[SSC Section Officers’ (Audit) Exam–2005]

4.

If a scooterist drives at the rate of 24 km/hr, he reaches his destination 5 minutes too late. If he drives at the rate of 30 km/hr, he reaches his destination 4 minutes too soon. How far is his destination from the starting point? [SSC Tax Assistant Excam–2004]

2 5.

Walking at a speed of 6 km/hr, a student reaches his school 6 minutes early and walking at a speed of 4 km/hr, he reaches the school 4 minutes late. Find the distance of the school from the student’s house. [SSC UDC (Mains) Exam–2001]

6.

I will reach my destination 40 minutes late if I walk at the rate of 3 km/hr. However, I will reach 30 minutes before time if I walk at the rate of 4 km/hr. Find the distance of my destination from the starting point. [SSC UDC (Mains) Exam–2003; SSC Tax Assistant Excam–2008]

7.

A student travels to his school at a speed of 4 km/hr and reaches the school 15 minutes late. On travelling at a speed of 6 km/hr, he reaches the school 5 minutes early. At what speed must he travel to reach the school just in time? [SSC Section Officers’ (Audit) Exam–2001]

8.

A person travels a certain distance on a bicycle at a certain speed. Had he moved 3 km/hr faster, he would have taken 40 minutes less. Had he moved 2 km/hr slower, he would have taken 40 minutes more. Find the distance and original speed of the person.

[SSC UDC (Mains) Exam–2007]

14. An army bomb squad man set a fuse for blasting a rock to take place after one minute. He ran away from the site at the speed of 13 m/s. Sound travels at the speed of 325 m/s. Upto what distance could the army man run, before he heard the sound of blast? [SSC Graduate Level (Mains) Exam–2000]

15. On a particular day a person starts walking from a place X at 2 am and reaches place Y at 5 am. A second person starts walking from a place Y at 4 am and reaches place X at 9 am on the same day. At what time do they cross each other? [SSC LDC (Mains) Exam–1999]

16. A covers some distance in 50 days when he rests 9 hours a day. In how many days will he cover the double distance by resting twice as before? [SSC Assistant Grade (Mains) Exam–1995]

K KUNDAN [SSC UDC (Mains) Exam–2001]

9.

B at a distance of 12 km from Y. What is the speed of B?

Two places A and B are 80 km apart from each other on a highway. A car starts from A and anohter from B at the same time. If they move in the same direction, they meet each other in 8 hours. If they move in opposite directions towards each other, they meet in 1 hour 20 minutes. Determine the speeds of the cars.

[SSC Assistant Grade (Mains) Exam–2006]

10. A train covers a distance between stations A and B in 45 minutes. If the speed is reduced by 5 km/hr, it will cover the same distance in 48 minutes. What is the distance between the two stations A and B (in km)? Also, find the speed of the train.

11. A car covering half of a distance of 100 km develops some engine trouble and later travels at half of its original speed. As a result, it arrives 2 hours later than its normal time. What was the original speed of the car? [SSC UDC (Mains) Exam–2004]

12. A train cove rs a distance of 193

1 3

km in

1 hours with one stoppage of 10 minutes, two 4 of 5 minutes and one of 3 minutes on the way. Find the average speed of the train. 4

[SSC Section Officers’ (Audit) Exam–2001]

13. Distance between two places X and Y is 90 km. Two persons A and B start from X towards Y at the same time. Speed of B is 4 km/hr less than speed of A. A reaches Y, returns at once and meets

17. A man travelled a total distance of 3990 km, part of it by air, part by water and the rest by land. The time he spent in travelling by air, water and land was in ratio 1 : 16 : 2 respectively and the average speed of each mode of travel was in the ratio 20 : 1 : 3 respectively. If his overall average speed was 42 km/hr, find the distance covered by water. [SSC Assistant Grade (Mains) Exam–1997]

18. A goods train travelling from station A to station B meets with an accident one hour after starting. After stopping there for 30 minutes, it proceeds 4 at of its usual speed and arrives at B 2 hours 5 late. Had the train covered 80 km more before the accident, it would have been just one hour late. Determine the original speed of the train and the distance between A and B. [SSC Assistant Grade (Mains) Exam–2001; SSC UDC (Mains) Exam–2004]

19. A train, an hour after starting, meets with an accident which detains it for 30 minutes. After 3 this the train proceeds at th of its former speed 4 1 and arrives 3 hours late. Had the accident 2 happened 90 km farther along the line, it would have arrived only 3 hours late. Find the length of the journey. [SSC Assistant Grade (Mains) Exam–1997; SSC Assistant Grade (Mains) Exam–2000]

20. A train after travelling 50 km meets with an 3 accident and then proceeds at of its former 4 speed and arrives at its destination 25 minutes late. Had the accident occurred 24 km behind, it would have re ache d the de stination only

3

Speed, Time and Distance 35 minutes late. Find the speed of the train and the distance travelled by the train. 21. Ravi can walk a certain distance in 40 days, when he rests 9 hours a day. How long will he take to walk twice the distance, twice as fast and rest twice as long each day? [SSC UDC (Mains) Exam–1999

22. Two men set out at the same time to walk towards each other from two points A and B, 72 km apart. The first man walks at the rate of 4 km/hr. The

1 km 2 in the second hour, 3 km in the third hour and so on. Find the time after which the two men will meet.

leaves Q for P. Both trains meet at the end of the 6 hours. If one trains travels 8 km/hr faster than the other, find the speed of the other trains. [SSC LDC (Mains) Exam–2007]

26. On a 2-km road, a total number of 201 trees are plante d on e ach side of the road at e qual distances. How many such trees in all will be planted on both sides of a 50-km road such that the distance between two consecutive trees is the same as that of the consecutive trees on the 2-km road? [SSC UDC (Mains) Exam–2004]

second man walks 2 km in the first hour, 2

27. Two men A and B walk from P to Q a distance of 21 km, at 3 and 4 km an hour respectively. B reaches Q, returns immediately and meets A at R. Find the distance from P to R.

[SSC UDC (Mains) Exam–1999]

23. Two trains start out towards each other from points 650 km apart. If they start out at the same time, they will meet in 10 hours, but if one of them starts out 4 hours and 20 minutes after the other, they will pass each other 8 hours following the departure of the latter. Determine the average speed of each other.

[SSC UDC (Mains) Exam–2000; [SSC LDC (Mains) Exam–2000]

28. A train met with an accide nt 3 hours after starting, which detains it for one hour, after which it proceed at 75% of its original speed. It arrives at the destination 4 hours late. Had the accident taken place 150 km farther along the railway line,

K KUNDAN 1 hours late. 2 Find the length of the trip and the original speed of the train. the train would have arrived only 3

[SSC UDC (Mains) Exam–1999]

24. Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 am and 9.51 am respectively at the speed of 25 km/hr and 20 km/hr respectively for journey towards Y. A train R leaves station Y at 11.30 am at a speed of 30 km/hr for journey towards X. When and where will P be at equal distance from Q and R. [SSC UDC (Mains) Exam–1999]

25. Two places P and Q are 336 km apart. A train leaves P for Q and at the same time another train

[SSC UDC (Mains) Exam–1999]

29. A man covers a certain distance on scooter. Had he moved 3 km/hr faster, he would have taken 40 minutes less. If he had moved 2 km/hr slower, he would have taken 40 minutes more. Find the distance (in km) and original speed. [SSC Assistant Grade (Mains) Exam–2007]

Answers and explanations

1.

Let the usual speed of the person be x km/hr and the distance of his journey be D km.

D His usual time to cover the distance =   hour x  Now, according to the question, Speed =

6  6 of his usual speed =  x  km/hr 7 7 

Time taken to cover the distance D km

Again,

   D  7D  =  hour  6x  6x    7 

7D D  25     6x x  60  

D7 5    1  x 6  12

D  5  5   6  hours x  12  2 1 hours.  Usual time = 2 2 Let the distance from A and B is x km.  Time taken to cover the distance from A to B at 20 x km/hr = hours. 20 and time taken to cover the distance from B to A at x 30 km/hr = hours. 30  Total time taken = 5 hours. (Given) x x  5  20 30 

2.

3x  2x 5 60  5x = 60 × 5 = 300  x =

300 = 60 km 5

4 3.

Let the required distance be x km.

x x 7   3 4 6

x Time taken to walk at 5 km/hr = hour 5 x  =   60  minutes = 12x minutes 5 

x Time taken to walk at 6 km/hr = hour 6 x  =   60  minutes = 10x minutes 6   Since the difference between the two times taken is (7 + 5 =) 12 minutes  12x – 10x = 12  2x = 12

x hours 4 15 1 hours  60 4

K KUNDAN x 1 be    hours 4 4

Time taken in second case =

early.  Actual time for reaching the school in time should

1  x  hours be    6 12 

From the above, we have

1  x 1 x      4 4 6 12    

x 1 7 5  7 12 1      30 12 60 60 60 5

30 = 6 5 Hence the required distance = 6 km Solve as Q.No. 3. Try yourself. [Ans : 18 km] Solve as Q.No. 3. Try yourself. [Ans : 2 km] Let the required distance be x km. Difference of times taken at different speeds = (40 + 30) minutes = 70 minutes 70 7 hours = hours 60 6

Times taken at 3 km/hr =

x hours 3



x  1 2x  1  4 12

 12x  12  8x  8

 4x  20  x = 5 km  The distance of the school be 5 km and

8.

actual

(x  1) time to reach school in time = = 1 hour 4  The required speed is 5 km/hr. Let the original speed and distance be V km/hr and D km respectively. Time taken to complete the whole journey D = hours. V When the person moves 3 km/hr faster, then D D 40   V  3 V 60

x Times taken at 3 km/hr = hours 4

According to the question,

x hours 6

 5  1  But this time is 5 minutes early or  hours  60  12

x x  7   1    t    t   5 6  60   12 

=

Solve as Q.No. Let the distance of the school be x km.

late  Actual time for reaching the school in time should

x 1 t  ....(ii) 6 12 Subtracting equation (ii) from equation (i), we get

6.

7  12 = 14 6

But this time is 15 minutes late or

x 5 t 6 60

5.

x 7  12 6

Time taken in first case =

x 7 t ....(i) 5 60 When Shivangi walks at 6 km/hr, then

4.



 Distance of the destination = 14 km 7.

12 = 6 2 Hence, the required distance is 6 km. Alternative Method: Let x km be the distance between her house and school and t hours be the time required to reach the school from her house. When Shivangi walks at 5 km/hr, then

 x=

4x  3x 7  12 6

x=

 x=







D D 2   V3 V 3

5

Speed, Time and Distance 

D D 2   V V3 3



DV  3D  DV 2  V V  3 3

3D 2  V V  3   3

2V V  3 ....(i) 9 When the person moves 2km/hr slower, then  D

D D 40   V  2 V 60 

D D 2   V2 V 3

4  80 3  x + y = 60 ...(ii) Solving equations (i) and (ii), we have x = 35 and y = 25  Speeds of the cars = 35 km/hr and 25 km/hr. 10. Suppose the distance is x km and the speed of the train is y km/hr. Thus we have two relationships: (x  y ) 

(1)

x 45 3 3   x  y y 60 4 4

(2)

x 48 4 4    x  (y  5) y  5 60 5 5

From (1) and (2), we have

3 4 y  (y  5 ) 4 5

D D 2    V2 V 3 

DV  DV  2D 2  V(V  2) 3

4 3 y    4 5 4

K KUNDAN

2D 2  V(V  2)  3

V(V  2) ....(ii) 3 Combining equations (i) and (ii), we get  D

2V(V  3) V(V  2)  9 3  2(V + 3) = 3(V – 2)  2V + 6 = 3V – 6  3V – 2V = 6 + 6  V = 12 km/hr Putting the value of V in equation (ii),

12  10 = 40 km. 3 Case I: W hen the cars are mov ing in the same direction. we get D =

9.



Let A and B be two places and C be the place of meeting. Let the speed of car starting from A be x km/hr and the car starting form B be y km/hr. Relative speed = (x – y) km/hr According to the question, (x – y) × 8 = 80  x – y = 10 ...(i) Case II: When the cars are moving in the opposite directions and they meet at point C.

Relative speed = (x + y) km/hr Time taken = 1 hour 20 minutes

1  4  = 1    hours 3  3  Again, according to the question,

4  20 = 80 km/hr 16  15 Therefore speed = 80 km/hr and distance  y =

3  80 = 60 km 4 11. Half of the original speed means double the normal time. It means that the car should have covered half of the distance of 100 km, ie 50 km, in 2 hours. Hence, the original speed of the car x =

50 = 25 km/hr 2 12. Distance covered by train =

1 580 km = km 3 3 Time taken by the train to cover this distance = 193

1 17 hours = hours 4 4 Total stoppage during the journey = 10 × 1 + 5 × 2 + 3 × 1 = 23 minutes = 4

23 hours 60 Actual time taken by the train to cover the above distance =

=

17 23  4 60

=

17  15  23 60

=

255  23 232 58 = hours  60 60 15

6



580 3 Average speed of the train = 58 15 =

km/hr

580  15 = 50 km/hr 58  3

If the person starting from X reaches the meeting point after t hours, person starting from Y will reach the meeting point after (t – 2) hours. Since the person starting from X starts moving at 2 am while the person starting from Y starts moving at 4 am. And the difference of time = (4 am – 2 am)= 2 hours  Distance (XP) travelled by the person starting from

D  X =   t  km 3 

13.

and the distance (YP) travelled by the person starting

D  from Y =  (t  2) km 5   Let A and B meet after t hours. Let the speed of B be x km/hr.  Speed of A = (x + 4) km/hr Distance covered by A in t hours = 60 + 12 = 72 km Distance covered by A in t hours = 60 – 12 = 48 km Now, according to the question, xt = 48 ....(i) (x + 4)t = 72 ....(ii) On dividing equation (ii) by equation (i), we have

780 = 2.5 sec 312 Now, during this time man would have travelled further. So, distance covered by man in 2.5 seconds = 2.5 × 13 = 32.5 m The total distance travelled by man = 780 + 32.5 = 812.5 metres. 15. X P Y l

D  D  =   t     (t  2)  D 3 5      t t 2 D  D  5  3

K KUNDAN

x  4 72 3   x 48 2  2x + 8 = 3x  x = 8  Speed of A = 8 km/hr 14. Time after which the bomb is set to explode = 1 minute = 60 seconds Speed of the man = 13 m/sec Distance covered by man in 60 sec = 13 × 60 = 780 metres So, distance to be travelled by sound before it catches up with army man = 780 metres Speed of the sound = 325 m/sec (given) Since the man and sound are travelling in the same direction, the relative speed of sound = 325 – 13 = 312 m/sec Time taken by sound to travel 780 metres =

Total distance travelled by both before meeting = Distance travelled by person from X + Distance travelled by person from Y

l

l

Let the speed of the person who starts from X be x km/hr and speed of the person who starts from Y be y km/hr. Time taken by the person who starts from X = 5 am – 2 am = 3 hours Time taken by the person who starts from Y = 9 am – 4 am = 5 hours Again, let the distance between X and Y be D km. Now, according to the question, x km/hr =

D 3

and

y km/hr =

D 5



t t 2  1 3 5

5t  3t  6 1 15  8t – 6 = 15  8t = 15 + 6 = 21 

21 5 2 hours 8 8 Converting this in hours, minutes and seconds, we get 2 hours 37 minutes and 30 seconds. t =

[2

5  5 hours = 2 hours +   60  minutes 8 8 

1  75    37 = 2 hours +  minutes 2  2 

= 2 hours + 37 minutes +

1 minutes 2

1  = 2 hour s + 37 minut es +   60   2  30 seconds = 2 hours 37 minutes 30 seconds] 16. Let the distance for A be x km Number of hours A walks daily = (24 – 9 =) 15 hours Number of days = 50 days

 Speed (in km/hr) =

x 50  15

..... (i)

In second situation Let the number of days be Y Distance = 2x Number of hours for which A walks daily = 6 hours  Speed in second case (in km/hr) =

Distance 2x  Time Y6

..... (ii)

7

Speed, Time and Distance In both the cases, the speed remains the same

   (x  80 )    30 d  (x  80 )    =  hours 4 x   x  60   5  

2x 2x  Y  6 50  15  Y × 6 = 50 × 15



50  15 = 125 days 6 17. Total distance travelled = 3990 km Ratio of time spent in travelling by air, water and land = 1 : 16 : 2 Ratio of respective speeds = 20 : 1 : 3 From the given fact, the ratio of respective distances will be 20 : 16 : 6 = 10 : 8 : 3 Sum of the ratios = 10 + 8 + 3 = 21 Distance travelled by steamer will be  Y =

8 = 1520 km 21 18. Let the distance between station A and station B be d km. Again, let the initial speed of the goods train be x km/hr. As the accident takes place after 1 hour  distance covered in 1 hour by the goods train = x km Remaining distance = (d – x) km Total time taken, if no accident happened = 3990 

 According to the question, x  80 30 d  (x  80 ) d    1 4x x 60 x 5  1

80 1 5[d  (x  80] d    1 x 2 4x x



80 1 5[d  (x  80 )] d    x 2 4x x



320  5d  5x  400 1 d   4x 2 x



5d  5x  80 1 d   4x 2 x

1 4d  5d  5x  80  2 4x  2x = 5x – d + 80 Putting the value of d from equation (i), we have 2x = 5x – 7x + 80  4x = 80

K KUNDAN

d  =   hours x 

Case I: Time taken by the goods train to cover the distance = 1

30 d  x  4x 60 5

 1 5(d  x )  = 1   hours 2 4x  

Now, according to the question,

1



80 = 20 4 Hence original speed of the train = 20 km/hr. Distance between the stations A and B = d = 7x (From i) = (7 × 20) = 140 km. 19. Solve as Q.No. 18. Try yourself. [Ans: Speed = 60 km/hr Length of the journey = 600 km] 2 0 . Let the distance be D km and speed be the x km/hr From the question, we have

 x =

50 (D  50) 4 D 25 D 5      x 3x x 60 x 12

1 5(d  x ) d   2 2 4x x



(d  x ) 5 d 1   4x x 2



150  4D  200 12D  5x  3x 12x



(d  x ) 5 d 1   4x x 2



4D  50 12D  5 x  3x 12x



5d  5x  4d 1  4x 2

d  5x 1  4x 2  2d – 10x = 4x  2d = 14x  d = 7x ....(i) Case II: If the goods train had covered 80 km more before the accident, then the distance of site of the accident = (x + 80) km Remaining distance = [d – (x + 80)] km Time taken to cover the whole of the distance

 16D  200  12D  5x ... (i) and  4D – 5x = 200

50  24 (D  26 ) 4 D 35    x 3x x 60







26 4D  104 12D  7 x   x 3x 12x



78  4D  104 12D  7x  3x 12x

D 7 12D  7x   x 12 12x

8 4D  26 12D  7x  3x 12x



 4D  7x  104 .... (ii) Now, subtracting equation (ii) from equation (i), we have 2x = 96  x = 48 km/hr Put the value of x in equation (i) and find the distance (D)  4D – 5 × 48 = 200  4D = 200 + 240 = 440

440 = 110 km. (Ans) 4 21. Time for work per day in first condition = (24 – 9 =) 15 hours Time for work per day in second condition = (24 – 9 × 2 =) 6 hours Here we have four quantities : Speed, Distance, Work and Days. We have to calculate number of days. Hence, Days will be in t he last column. Her e following relationships exist. More speed, less days (Inverse) More distance, more days (Direct) Less hours of work, more days (Inverse)



7t + t2 + 16t = 288 t2 + 23 t - 288 = 0 t2 + 32 t – 9t - 288 = 0 t (t + 32) – 9 (t + 32) = 0 (t + 32) )(t – 9) = 0  t + 32 = 0  t = –32 (Not possible)  t – 9 = 0  t = 9 They meet after 9 hours. 23. Let the trains A and B travel at speed of x and y km/ hr respectively and meets 10 hours after departure.     

D =

From the figure it can be seen that AC = (x × 10) km BC = (y × 10) km  AC + BC = xX × 10 + y × 10 or, 650 = 10(x + x) or, x + y = 65 In the second situation when the other train starts after 4 hours and 20 minutes

K KUNDAN 4 hours and 20 minutes = 4

Distance covered by train A in

Hence,

2 : 1  1 : 2  :: 40 : x 6 : 15

1  2  15  40 = 100 2 1 6 Hence the required time = 100 days. 2 2 . Let A starts from point X, B starts from point Y and they meet after t hours. A X

13 hours 3

13 13x = 3 3 Both the trains meet 8 hours after train A leaves P. Now if they meet at C1 then PC1 = 8 × x = 8x km BC1 = 8 × y = 8y km According to the question, = AP = x ×

 2 × 1 × 6 : 1 × 2 × 15 :: 40 : x  2 × 1 × 6 × x = 1 × 2 × 15 × 40 Product of extremes = Product of means

 x =

20 1 13 4  hours 60 3 3

8x + 8y = 650 

 8x  y   650 

B

P  XP = 4 × t = 4t km YP = 2 + 2.5 + 3 + .... t terms This is an AP.

Y

n 2a  n  1d  2 where n = number of terms, a = first term and d = common difference Sum of an AP =

t  1 t  t 1  YP = 2 2  2  (t  1)  2   2 4  2  2      t2 t 7t t 2 7t  t 2     4 4 4 4 4 But it is given that XY = 72 or XP + PY = 72

 8  65  650 



13x 3

13x 3

13 x 3

13x  650  520  130 3

130  3 = 30 km/hr 13 Speed of train A = 30 km/hr Speed of train B = (65 – 30) km/hr = 35 km/hr 24. As given, speed of the train P = 25 km/hr Speed of the train Q = 20 km/hr Speed of the train R = 30 km/hr  x =

= 2t 

7t  t  4

Q

20 t

P

B

A

Q1

P1

R

1

R

2

+ 4t = 72

X 33 km 87.5 km

25 t

30 t Y

9

Speed, Time and Distance Distance travelled by train P between 8:00 to 11:30

1 7 175  25   87.5 km hours = 2 2 2 Distance travelled by train Q between 9 : 51 to 11 : 30 ie. in ie in 3

39 33  20   20 60 20 = 33 km Assume that trains P and Q are at A and B respectively at 11 : 30 am. Also assume that t minutes after 11 : 30 am, train P was equidistant from train Q and train R. At the equidistant position train P, Q and R were at P1, Q1 and R1.  XP1 = XA + AP1 = (87.5 + 25 t) km XQ1 = XB + BQ1 = (33 + 20 t) km P1 Q1 = XP1 – XQ1 = (87.5 + 25 t) - (33 + 20 t) = (54.5 + 5 t) km Distance RR1 = 30 t km P1R1 = Total distance - XP1 - RR1 = 220 - (87.5 + 25 t) - 30 t = (132.5 - 55 t) km  P1Q1 = P1R 1  5t + 54.5 = 132.5 - 55 t 1 hour 39 minutes = 1

Now the rates of A and B are 3 : 4 and they have walked 42 km. Hence the distance PR travelled by A

3 of 42 km. = 18 km. 7 Alternative Method I: =

Let the required distance be x km. Now, according to the question, A and B both walk for the same distance  Distance travelled by B = (21 + 21 – x) = (42 – x) km

 42  x   hours Time taken by B =  4  Distance travelled by A = x km

 x Time taken by A =   hours 3

K KUNDAN

 60 t = 78 or t =

78  60 minutes 60

 t = 78 minutes So 78 minutes after 11 : 30 am ie at 12 : 48 pm train P will be equidistant from train Q and R.

78 XP1 = 87.5 + 25 t = 87.5 + 25 × = 87.5 + 32.5 60 1 XP = 120 km  At 120 km away from station X, trains would be at equal distances. 25.

Let R be the meeting point. Let the speed of train form P = x km/hr and that form Q = (x + 8) km/hr Both trains meet after 6 hours  (x × 6) + (x + 8) × 6 = 336  6x + 6x + 48 = 336  12x = 336 – 48 = 288

288 = 44 12  Speed of one train = 24 km/hr Speed of the other train = (24 + 8 =)32 km/hr 26. Distance between 2 trees on a 2-km road  x =

 2  1000   = 10 m =   201  1  Number of trees planted on both sides of a 50-km road

 50  1000     1 = 10002 = 2  10    27. When B meets A at R, B has walked the distance PQ + QR and A the distance PR. That is both of them have together walked twice the distance from P to Q, ie 42 km.

x 42  x  3 4  4x = 126 – 3x  7x = 126

126 = 18 7  required distance = 18 km Alternative Method II: A’s speed = 3 km B’s speed = 4 km Let us consider that A and B meets after t hours. Distance covered by A in t hours = 3t km Distance covered by B in t hours = 4t km Total distance covered by A and B = 3t + 4t = 7t km But the total distance covered by A and B is twice the distance between P and Q. So, 7t = 21 × 2  x =

21 7 t = 6 hours So, the distance between P and R = Distance travelled by A = 3 × 6 = 18 km. 28. Let the length of the trip be d km and the original speed of the train be x km/hr. As the accident takes place after 3 hours.  distance covered in 3 hours by the train = (3 × x) = 3x km Remaining distance = (d – 3x) km Total time taken by the train if no accident happens t = 2

d   hours x 

= 

Case I: Time taken by the train to cover the whole length of the trip

10 15x  50 29  x 2  30x – 100 = 29x  x = 100 Hence, speed = 100 km/hr and the length of the trip (d) = 12x = 12 × 100 = 1200 km

   (d  3x )  3 1 75  hours =    x 100  



4(d  3x )  = 4   hours 3x 

29. Suppose the distance is D km and the initial speed is x km/hr.

Now, according to the question,

4

4(d  3x ) d  4 3x x

Then, we have

4d  12x d  3x x  4d – 12x = 3d  d = 12x .... (i) Case II: If the train had covered 150 km more before the accident then the distance of the accident = (3x + 150) km Remaining distance = (d – (3x + 15)) km Time taken to cover the whole length of the trip

D D 40   and x  3 x 60 D D 40   x  2 x 60





D D 2   x x 3 3



3D 2  x (x  3) 3

.... (1)

K KUNDAN

3x  150 d  (3x  150) 1 75 x x 100

and



Now, according to the question,

2D 2  x (x  2) 3

3D 2D  x (x  3) x (x  2)

3(x  2)  2(x  3)



3x  150 4d  12x  600 d 7    1 x 3x x 2



12x 5 3x  150 4  12x  12x  600    x 3x x 2

 3x  6  2x  6

 d  12x 

9x  450  36x  600 29  3x 2



3x  150  12x  200 29  x 2



3x  150  12x  200 29  x 2

.... (2)

From (1) and (2), we have

3x  150 d  (3x  150) d 7 1   3x x x 2 4



D D 2   x 2 x 3

 x  12 km/hr Now, if we put this value in (1), we get D =

2 12  15  = 40 km. 3 3

Hence, the distance is 40 km and the original speed is 12 km/hr.

Speed, Time and Distance

11

K KUNDAN