Question: A builder borrows Rs. 2550 to be paid back with compound interest at the rate of 4% per annum by the end of 2 yr in two equal yearly instalments. How much will each instalment be?
Options:
A) Rs. 1352
B) Rs. 1377
C) Rs. 1275
D) Rs. 1283
Show Answer
Answer:
Correct Answer: A
Solution:
- A = Rs. 2550, R = Rs. 4% per annum, n = 2yr
Let each of the two equal instalments be Rs. x.
Present worth $ =\frac{Instalment}{{{( 1+\frac{r}{100} )}^{n}}} $
$ P _1=\frac{x}{{{( 1+\frac{4}{100} )}^{1}}}=\frac{x}{1+\frac{1}{25}}=\frac{x}{\frac{26}{25}} $
$ \Rightarrow $ $ P _1=( \frac{25}{26} )x $
Similarly, $ P _2={{( \frac{25}{26} )}^{2}}x=\frac{625}{676}x $
$ P _1+P _2=A $
$ \therefore $ $ \frac{25}{26}x+\frac{625}{676}x=2550 $
$ \Rightarrow $ $ \frac{(650+625)x}{676}=2550 $
$ \Rightarrow $ $ \frac{1275}{676}x=2550 $
$ \Rightarrow $ $ x=2550\times \frac{676}{1275} $
$ \Rightarrow $ $ x=Rs.,1352 $