Question

A sum of money,X was deposited in a saving account at 10 percent compounded daily on 25 july 1993. on 13 august 1993, rm600 was withdrawn and the balance as on 31 december 1993 was RM 8900.calculate the value of X using exact time and 360 day year.


(1+.10/360)^20 - 600)(1+.10/360)^108 = 8900
x = 9185.83

but the answer is 9112.37

Answer 1

What you wanted as an equation is ...

[ x( 1 + .1/360)^20 - 600 ](1+.1/360)^108 = 8900

x(1+.1/360)^20(1+.1/360)^108 - 600(1+.1/360)^108 = 8900
x((1.00557024)((1.03045024) - 618.2701442 = 8900
x(1.036190095 = 9518.270144
x = 9185.83

Answer 2

ooooohhh thanks

Answer 3

how to get the 20 and 108

Answer 4

This is the way how we can calculate it.

RM8,900 = [ X (1+0.1/360)^19 - RM600](1+0.1/360)^140
RM8,900/(1+0.1/360)^140 = X (1+0.1/360)^19 - RM600
RM8560.58 + RM600 = X (1+0.1/360)^19
RM9160.58 = X (1+0.1/360)^19
RM9160.58/(1+0.1/360)^19 = X
RM9112.36 = X

Therefore, X is RM9112.36#