algebra word problem

Question

A man paid the bank $4,080 at the end of 3 years. This was the amount borrowed plus 12% interest for 3 years. How much money had he borrowed?

Answer 1

For simple interest: A = p(1 +rt), where p = principal, r = rate, t = time

 

4080 = p(1 + 0.12(3))

4080 = p(1 + 0.36)

4080 = p(1.36)

3000 = p

 

For compound interest: A = p(1 + r/n)^tn, where n = number of calculations per year

 

4080 = p(1 + 0.12)³

4080 = p(1.12)³

4080 ≈ p(1.404928)

2904.06341 ≈ p

Answer 2

The amount he borrowed depends on how the interest was calculated.

If the interest is 12% per year but uncompounded - that is simple interest, then the original amount 

= 4080/1.36 = $3,000

If the interest charged is compounded annually

= 4080/(1.12)³ = $2,904.06

 

Thus if you borrow or lend money it is important to not just specify the interest but also to specify how the interest will charged or compounded.