Question

An investor invested a total of $3100 in two mutual funds. One fund earned a 6% profit while the other earned a 3% profit. If the investor's total profit was $153, how much was invested in each mutual fund?

A. $1200 in the 6% fund, $1900 in the 3% fund
B. $1600 in the 6% fund, $1500 in the 3% fund
C. $1400 in the 6% fund, $1700 in the 3% fund
D. $1800 in the 6% fund, $1300 in the 3% fund

Answer 1

By setting up a system of equations and using the elimination method, we determined that the investor placed $2000 in the 6% fund and $1100 in the 3% fund, which does not match any of the answer choices provided.

Step-by-step explanation:

To solve the problem of how much was invested in each mutual fund, we can set up a system of equations based on the given information. We are told that the investor invested a total of $3,100 in two funds, and the combined profit from these funds was $153.

Let x be the amount invested in the 6% fund, and y be the amount invested in the 3% fund. We have the following two equations:

1. x + y = 3100 (the total amount invested)
2. 0.06x + 0.03y = 153 (the total profit made from both investments)

To solve the system, we multiply the second equation by 100 to get rid of the decimals:

6x + 3y = 15300

We can now solve this system using substitution or elimination. If we use elimination, we can multiply the first equation by 3 and subtract it from the second equation to eliminate y:

3x + 3y = 9300 (multiplied first equation by 3)

6x + 3y = 15300 (second equation)

Subtract the first from the second to get:

3x = 6000, and thus x = 2000

Substituting x back into the first equation:

2000 + y = 3100, so y = 1100

Therefore, the investor invested $2000 in the 6% fund and $1100 in the 3% fund, which is not listed in the multiple-choice options provided by the student. This indicates that there might be an error in the options given or a misunderstanding in the question.