Sa#2

A high end Computer set was purchased under these terms: P20,000 down and P2,400 each quarter for 3 years and 3 months. If money is worth 15% compounded quarterly, find the cash price of the computer.

Down payment = 20,000 R = 2,400 j = 15% m = 4  i = 15/4 = 3.75 or 0.375 t = 3.25 n = 3.25x4 = 13

CP = 20,000 + A = 2,400 (1-(1+0.03575)^-13/0.0375 CP = 20,000 + A = 2,400 (0.380338335)/0.0375 CP = 20,000 + A = 2,400 (10.1423556) CP = 20,000 + A = 24,341.65344 CP = 20,000 + 24,341.65344 CP = 44,341.65

Romy owes the following obligations: a. P350,000 due in 4 years, and b. P600,000 due in 6 years with the accumulated interest from today at 14% compounded quarterly.

If the man would want to replace these obligations by a payment of P400,000 on the second year and another payment at the end of 5 ½ years, how much is the second payment if money is worth 16% compounded semi-annually?

Given: A.  P = 350,000 t=4 j = 16% m=2 i = 16/2 = 0.08 n = 4x2 = 8

B  F = 600,000 t = 0.5 j = 14% m=4 i = 14/4 = 0.035 n = 0.5x4 = 2

C =  x = 400,000  t=2

j = 12% m=2 i = 12/2 = 0.08 n = 2x2 = 4

D =  t = 5.5 j = 12% m=2 i = 16/2 = 0.08 n = 5.5 x 2 = 11

Solution: x(1+i)^n + x(1+i)^n = P (1+i)^n + F(1+i)^-n 400,000(1+0.08)^4 + x(1+0.08^11 = 350,000 (1+0.08)^8 + 600,000(1+0.035)^-2 544195.584 + 2.331638997x = 647825.5736 + 560106.4202 544195.584 + 2.331638997X = 663736.4098 2.331638997x = 663736.4098 - 544195.584 2.331638997 = 199540.8258 x= 199540.8258/2.331638997 x = 51,269.01118

Mrs. Sara Samonte owes PBN Bank P3,050,000 due in 3 years and 6 months and P2,340,000 due in 9 years, if she plans to settle these obligations by a single payment on the 6th year, how much should she pay if money is worth 14%, m=4?

Given: A.  P = 2,340,000 t=9 j = 14% m=4 i = 14/4 = 0.035 n = 9x4 = 36

B =  x = 3,050,000 t = 3.5 j = 14% m=4 i = 14/4 = 0.035 n = 3.5x4 = 14

C =  t = 5.5 j = 12%

m=2 i = 12/2 = 0.06 n = 5.5 x 2 = 11

Solution: x(1+i)^n + x(1+i)^n = P (1+i)^n + F(1+i)^-n x(1+0.035)^n + 3050000(1+0.035)^14 = 2340000 (1+0.035)^8 + 2340000(1+0.035)^-12 2.283328487 = 4937018.294 + 1548572.918 2..283328487 = 6485591.212 6485591.212/2.283328487 2,840,410.939

Margarita owes the following obligations: a. P5,000,000 due in 4 years, and b. P600,000 due in 6 years with the accumulated interest from today at 14% compounded quarterly. If the man would want to replace these obligations by a payment of P500,000 on the second year and another payment at the end of 5 ½ years, how much is the second payment if money is worth 12% compounded semi-annually?

Given: A.  P = 5,000,000 t=4 j = 12% m=2

i = 12/2 = 0.06 n = 4x2 = 8

B  F = 600,000 t = 0.5 j = 14% m=4 i = 14/4 = 0.035 n = 0.5x4 = 2

C =  x = 500,000  t=2 j = 12% m=2 i = 12/2 = 0.06 n = 2x2 = 4

D =  t = 5.5 j = 12% m=2

i = 12/2 = 0.06 n = 5.5 x 2 = 11

Solution: x(1+i)^n + x(1+i)^n = P (1+i)^n + F(1+i)^-n 500,000(1+0.06)^4 + x(1+0.06)^11 = 5,000,000 (1+0.06)^8 + 600,000 (1+0.035)^-2 631,238.48 + 1.898298558x = 7,969,240.375 + 560,106.42 631,238.48 + 1.898298558x = 8,529,346.795 1.898298558x = 8,529,346.795 - 631,238.48 1.898298558x = 7,898,108.315 7,898,108.315/1.898298558x  x= 4,160,624.935

A dining set was purchased under these terms: P5,000 down and P1,500 each quarter for 3 years and 3 months. If money is worth 11.5% compounded quarterly, find the cash price of the computer.

Down payment = 5,000 R = 1,500 j = 11.5% OR 0.115 t = 3.25 m=4 i = 11.5/4 = 2.875 Or 0.02875 n = 3.25x4 = 13

CP = 5,000 + A = 1,500 (1-(1+0.02875)^-13                                               0.02875

CP = 5,000 + A = 1,500 (1-0.691786325)                                              0.02875

CP = 5,000 + A = 1,500 (0.308213675)                                              0.02875

CP = 5,000 + A = 1,500 (10.72047565) CP = 5,000 + A = 16,080.71348 CP = 5000 + 16,080.71348 CP = 21,080.71348

Julie owes BN Bank P350,000 due in 3 years and P850,000 due in 8 years, if she plans to settle these obligations by a single payment on the 5th year, how much should she pay if money is worth 20%, m=4?

 a. Accumulate (F) 350,000 by 3 years, , t= 3,  j=20%, m=4, i = 20%/4 = 0.05, n = 3x4= 12      b. Discount (P) 850,000 by 8 years, t=8, j= 20%, m =4, i = 20%/4 = 0.05, n = 3x4=12 Payment      t= 5 j = 20% m=4  i = 20%/4 = 0.05, n = 5x4 = 20

  x(1+ 0.05)20 = 350,000(1 + 0.05)12 + 850,000(1 + 0.05)-12 x(2.653297705)= 350,000 (1.795856326) + 850,000(0.556837418) 2.653297705x = 628,549.7141 + 473311.0053 2.653297705x = 1,101,860.719 1,101,860.719/2.653297705x X= 415,279.7166