Question

Francisco invests $8,900 in two different accounts. The first account paid 14%, the second account paid 12% in interest. At the end of the first year, he had earned $1,170 in interest. How much was in each account?

A. $5,000 in the first account and $3,900 in the second account
B. $6,000 in the first account and $2,900 in the second account
C. $5,500 in the first account and $3,400 in the second account
D. $4,000 in the first account and $4,900 in the second account

Answer 1

None of the given options accurately reflect the correct amounts in each account. After setting up a system of equations and solving it, Francisco invested $5,100 in the first account at 14% interest and $3,800 in the second account at 12% interest to earn a total of $1,170 in one year.

Step-by-step explanation:

To determine how much was invested in each account, we can set up a system of equations. We know that the total investment is $8,900 and the total interest earned is $1,170 at the end of one year. We can let x be the amount invested in the first account at 14% interest, and y be the amount in the second account at 12% interest.

The system of equations based on the information given would be:

• x + y = 8,900 (Total investment)
• 0.14x + 0.12y = 1,170 (Total interest)

Now, we can solve this system using substitution or elimination. Let's use substitution:

From the first equation, we can express y in terms of x: y = 8,900 - x. Now substitute y in the second equation:

0.14x + 0.12(8,900 - x) = 1,170

Simplifying this, we get:

0.14x + 1,068 - 0.12x = 1,170

Now, combine like terms and solve for x:

0.02x = 102

x = 5,100

Now, we can find y using the expression y = 8,900 - x:

y = 8,900 - 5,100

y = 3,800

Therefore, the correct answer is $5,100 in the first account and $3,800 in the second account, which means that none of the given options (A, B, C, D) are correct.