Question

Cody invests $4500 in two different accounts. The first account paid 9%, the second account paid 8% in interest. At the end of the first year, he had earned $373 in interest. How much was in each account?

A. $2700 at 9%, $1800 at 8%
B. $2000 at 9%, $2500 at 8%
C. $3000 at 9%, $1500 at 8%
D. $3500 at 9%, $1000 at 8%

Answer 1

To solve this problem, we can set up a system of equations and solve using the substitution method. The amounts of money in the first and second accounts are $2700 at 9% and $1800 at 8%, respectively.

Step-by-step explanation:

To solve this problem, we can set up a system of equations. Let's represent the amount of money in the first account as 'x' and the amount of money in the second account as 'y'. We know that the total amount of money invested is $4500, so we have the equation x + y = 4500. We also know that at the end of the first year, Cody earned $373 in interest. Therefore, we have the equation 0.09x + 0.08y = 373.

To solve these equations, we can use the substitution method. Rearranging the first equation, we get x = 4500 - y. Substituting this into the second equation, we get 0.09(4500 - y) + 0.08y = 373. Solving this equation will give us the values of x and y, which represent the amounts of money in each account.

By simplifying and solving the equation, we find that y = 1800 and x = 2700. Therefore, the amounts of money in the first and second accounts are $2700 at 9% and $1800 at 8%, respectively. So, the correct answer is A. $2700 at 9%, $1800 at 8%.