1. ## 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

2. Hello, helloying!

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

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

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

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