A college student earned $8900 during summer vacation working as a waiter in a popular

restaurant. The student invested part of the money at 10% and the rest at 9%. If the student
received a total of $831 in interest at the end of the year, how much was invested at 10%?

If x at 10%, the rest (8900-x) at 9%, then calculate the interest:

.10x + .09(8900-x) = 831
x = 3000

To determine how much was invested at 10%, we can set up a system of equations based on the given information.

Let's represent the amount invested at 10% as "x" and the amount invested at 9% as "y".

From the problem, we know that the total amount of money earned during summer vacation is $8900:
x + y = 8900 (equation 1)

We also know that the total interest earned at the end of the year is $831, with 10% interest on the amount invested at 10% and 9% interest on the amount invested at 9%:
0.1x + 0.09y = 831 (equation 2)

Now we can solve this system of equations using substitution or elimination.

Let's solve it using the substitution method:
1. Solve equation 1 for x: x = 8900 - y
2. Substitute this value of x into equation 2: 0.1(8900 - y) + 0.09y = 831
3. Simplify the equation: 890 - 0.1y + 0.09y = 831
4. Combine like terms: -0.01y = -59
5. Divide both sides by -0.01: y = 5900

Now that we know the value of y, we can substitute it back into equation 1 to find x:
x + 5900 = 8900
x = 8900 - 5900
x = 3000

Therefore, the student invested $3000 at 10% and $5900 at 9%.