Eric deposits X into a savings account at time 0, which pays interest at a nominal rate of i, compounded semiannually. Mike deposits 2X into a different savings account at time 0, which pays simple interest at an annual rate of i. Eric and Mike earn the same amount of interest during the last 6 months of the 8th year. Calculate i.

i = 9.46%

Step-by-step explanation:

Money deposited by Eric is X.

Now formula for compound interest is;

A = P(1 + i)ⁿ

Where n = kt and k is the number of compounding per annum.

Since Eric compounded interest semi annually, we have;

A = X(1 + i/2)^(2t)

After 7.5 years, Eric will earn;

A(7.5) = X(1 + i/2)^(2 * 7.5)

A_7.5 = X(1 + i/2)^(15)

For the last half year, where n = 1, Eric earns; (A_7.5)(1 + i/2)¹

Thus, the interest eric earns = (A_7.5)(1 + i/2) - A_7.5

Interest = A_7.5 + (A_7.5)*(i/2) - A_7.5

Interest = (A_7.5)*(i/2)

We initially got A_7.5 = X(1 + i/2)^(15)

Thus, interest Eric earns is now ;

X(1 + i/2)^(15) * (i/2)

Which gives;

0.5Xi(1 + i/2)^(15)

Now to mike. He deposited 2X

Simple interest earned by mike will be;

A = 2X(1 + it)

During the last half year of any yeat, Mike earns; 2X * i/2 = Xi

Equating last half year interests of both Eric and Mike gives us;

0.5Xi(1 + i/2)^(15) = Xi

Divide both sides by 0.5Xi to give;

(1 + i/2)^(15) = 2

So,

1 + i/2 = 2^(1/15)

i/2 = 1.04729 - 1

i/2 = 0.04729

Multiply both sides by 2 to give;

i = 0.09458

Thus, interest is approximately 0.0946 or 9.46%

attachment?

step-by-step explanation:

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