Question

A person had $18,000 invested in two accounts, one paying 8% simple interest and one paying 9% simple interest. How much was invested in each account if the interest after 1 year is $1606?

a) $7,200 at 8%, $10,800 at 9%
b) $10,800 at 8%, $7,200 at 9%
c) $9,000 at 8%, $9,000 at 9%
d) $8,000 at 8%, $10,000 at 9%

Answer 1

After setting up a system of equations with x representing the amount at 8% simple interest and y for 9%, and solving using elimination, we find that $1,400 was invested at 8% and $16,600 at 9%. None of the options provided match this result, suggesting either a typo in the options or a miscalculation in the problem setup.

Step-by-step explanation:

To solve the problem of determining how much was invested in each account, we can set up a system of equations based on the information given:

• Let x be the amount at 8% simple interest.
• Let y be the amount at 9% simple interest.
• The total amount invested is x + y = $18,000.
• The total simple interest earned after one year from both accounts is x(0.08) + y(0.09) = $1,606.

We have two equations now:

1. x + y = 18,000
2. 0.08x + 0.09y = 1,606

Multiply the second equation by 100 to clear decimals, which gives us:

1. x + y = 18,000
2. 8x + 9y = 160,600

Now, we can solve these equations using the method of substitution or elimination. If we use elimination, we can multiply the first equation by 8 to get:

1. 8x + 8y = 144,000
2. 8x + 9y = 160,600

Subtracting the first new equation from the second, we have:

y = 160,600 - 144,000

y = 16,600

Since 1 y = $1, y = $16,600

Substituting y into the first equation, x + $16,600 = $18,000 resulting in x = $1,400.

Checking these values in the second original equation:

0.08($1,400) + 0.09($16,600) = 112 + 1,494 = $1,606 which is correct.

Therefore, the correct answer is $1,400 at 8% and $16,600 at 9%, which is not listed in the options provided and thus suggesting a miscalculation in the problem as presented.