# Money problem

• Nov 1st 2008, 07:24 PM
helloying
Money problem
Tom borrowed a sum of money at 10% per annum compound interest. If he repaid the sum only after 2 years, he would have to repay $4356. I already found the sum of money that he had borrowed to be$3600.

The question is if he repaid the sum by two equal annual instalments, at the end of the first year and the second year, how much was each instalment?

I didn't quite understand the question .pls teach me thanks
• Nov 1st 2008, 07:56 PM
Soroban
Hello, helloying!

Quote:

Tom borrowed a sum of money at 10% per annum compound interest.
If he repaid the sum only after 2 years, he would have to repay $4356. I already found the sum of money that he had borrowed to be$3600. Yes!

If he repaid the sum by two equal annual installments,
at the end of the first year and the second year,
how much was each installment?

This is a tricky problem . . .

It is an Amortization problem (time payments).
There is a formula for this situaton, but we can derive it ourselves.

Tom owes $3600. .The bank charges 10% interest per year. . . Let $x$ = amount of his two equal payments. During the first year, the bank charges 10% interest: . $0.10 \times 3600 \:=\:360$ . . At the end of year one, Tom owes: . $3600 + 360 \:=\:3960$ dollars. Then Tom pays $x$ dollars. . . So he owes a balance of $3960 - x$ dollars. During the second year, the bank charges 10% interest: . $0.10(3960-x)$ . . At the end of year two, Tom owes: . $(3960-x) + 0.10(3960-x)$ dollars Then Tom pays $x$ dollars and he owes$0.

Hence: . $(3960 -x) + 0.10(3960-x) \:=\:x$
. .
[His last balance is equal to his final payment.]

Solve for $x\!:\;\;3960 - x + 396 - 0.1x \:=\:x\quad\Rightarrow\quad 4356 \:=\:2.1x$

. . $x \:=\:\frac{4356}{2.1} \:=\:2074.285714... \quad\Rightarrow\quad \boxed{x \:=\:\2074.29}$