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Angelo invested $9500 in two accounts: one that pays 2.9% simple interest and one that pays 1.9% simple interest. At the end of the first year, his total interest earned was $260.5. How much did he deposit in the account that pays 2.9% simple interest?

1 Answer

5 votes

Ykay, this is the solution:

Let x to represent the amount invested in the account that pays 2.9% simple interest, therefore 9,500 - x would be the amount invested in the account that pays 1.9% simple interest.

The time of the investment is 1 year or 12 months.

Now, let's recall the formula of the simple interest:

A = P * (1 + rt), where:

A = Final amount

P = Initial principal balance

r = Annual interest rate

t = Time in years

For our case, A would be:

9,500 + 260.50

9,760.50

rate 1 = 2.9% = 0.029

rate 2 = 1.9 = 0.019

In consequence, our equation to solve for x is:

x (1 + 0.029 * 1) + (9,500 - x) (1 + 0.019 * 1) = 9,760.50

1.029x + (9,500 - x) (1.019) = 9,760.50

1.029x + 9,680.50 - 1.019x = 9,760.50

0.01x = 9,760.50 - 9,680.50

0.01x = 80

Multiplying by 100 at both sides:

0.01x * 100 = 80 * 100

x = 8,000

9,500 - x = 9,500 - 8,000 = 1,500

Thus, Angelo invested:

• $ 8,000 in the account that pays 2.9% simple interest ($ 232 of interest)

,

• $ 1,500 in the account that pays 1.9% simple interest ($ 28.50 of interest)

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