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1. ## Interest rate question

Hi, I got a problem about the interest rate. Hope someone can help.

A deposits 100 into a saving account at time 0, which pays nominal interest i, compounded semiannually.

B deposits 200 into a saving account at time 0, which pay simple interest at an annual rate of i

A and B earn the same amount of interest during the last 6 months of the 8th year, find i

my calculation is like this

A's interest
100(1 + i/2)^15 - 100

B's interest
100(7.5i)

so
100(1 + i/2)^15 - 100 =100(7.5i)

Thanks a lot.  Reply With Quote

2. In your solution you calculate the interest of the whole 7.5 years, but in the problem it means the interest during the last 6 months of the 8th year, yes, only 6 months!
So I think the solution may be:

A's interest
100(a(8)-a(7.5)) = 100((1+i/2)^16-(1+i/2)^15) or 100a(7.5)*i/2 = 100((1+i/2)^15)*i/2

B's interest
200(a(8)-a(7.5)) = 200(1+8i-1-7.5i) = 200*0.5i

So 100((1+i/2)^15)*i/2 = 200*0.5i and i = 9.46%  Reply With Quote

3. Thanks a lot.

I have misunderstood the question :>  Reply With Quote

4. Originally Posted by thtang Thanks a lot.

I have misunderstood the question :>
Most of the mistakes are caused by our carelessness, I hope our everyone will never do this in the real exam...  Reply With Quote

I don't understand how: 100((1+i/2)^15)*i/2=100(1+i/2)^16-100(1+i/2)^15

:confused-:  Reply With Quote

6. Originally Posted by Elk I don't understand how: 100((1+i/2)^15)*i/2=100(1+i/2)^16-100(1+i/2)^15

:confused-:
100(1+i/2)^16-100(1+i/2)^15
=100(1+i/2)^15 * (1+i/2)-100(1+i/2)^15
=100(1+i/2)^15 * ((1+i/2)-1)
=100(1+i/2)^15 * i/2  Reply With Quote