Question

1000 is borrowed at a rate of 16% interest per annul. Find the amount due at the end of two years if the interest is compounded annually?

Answer 1

I tried it, but my answer is wrong. I don't know what Im doing wrong tho.

Answer 2

1000 x 0.16 = 160
1160 x 0.16 = 185.6
so 1160 + 185.6
answer = 1345.6

Answer 3

What did you do? What were the results of your trying?

Answer 4

I have the formula A(t)= P(1+i/n)^nt were P is the initial amount, I is the interest rate, t is time, and n is the compound years.

Answer 5

for simple interest like this just do what i do :) for more difficult use the formula

Answer 6

A(2)=1000(1+1.16)^(0.16)(2)

Answer 7

I have to use the formula because my teacher wants to see it, I have a test tomorrow on this and I don't think I can learn a new way now.

Answer 8

A(t)= P(1+i/n)^nt

That's fine,
n = 1
t =2
P = 1000

Answer 9

yes

Answer 10

first year is 1000 x 0.16 ( interest) giving 160
add 160 to 1000 to give end of year 1 amount
1160x 0.16 ( second year interest) = 185.6
add total year 1 with interest of year 2
1160+185.6 is = 1345.6

Answer 11

Yes, you can do it a year at a time if you like. This is very impractical for longer periods and shorter compounding periods. Let the algebra help you.

\(1000(1 + 0.16)^{2} = 1345.6\)

Answer 12

why is it to the power of two?

Answer 13

instead of 0.32

Answer 14

I = 0.16
n = 1
t = 2

There is no "I" in the exponent.

Answer 15

I know what I was doing wrong, I always mix up the i and n. Thank you

Answer 16

That will, indeed, produce an incorrect calculation. Good work!