Question

Phyllis invested $49,328, part at 9% simple interest and part at 6% simple interest for a period of 1 year(s). How much did she invest at each rate if each account earned the same interest?

a) $24,664 at 9%, $24,664 at 6%
b) $26,000 at 9%, $23,328 at 6%
c) $30,000 at 9%, $19,328 at 6%
d) $20,000 at 9%, $29,328 at 6%

Answer 1

By setting up a system of equations, we find that Phyllis invested $29,328 at 9% simple interest and $20,000 at 6% simple interest, since each investment yielded the same amount of interest over one year.

Step-by-step explanation:

To determine how much Phyllis invested at each rate, we will use a system of equations to solve the problem. Let's assume x is the amount invested at 9% simple interest, and y is the amount invested at 6% simple interest. We know the total amount invested is $49,328, which gives us the equation:

x + y = 49,328

We also know that she earned the same interest from both investments. The simple interest for one year is calculated as principal × rate × time. As time is 1 year for both investments, and interest earned is equal, this gives us the second equation:

0.09x = 0.06y

After simplifying the second equation by dividing both sides by 0.03, we get:

3x = 2y

Now we have our system of equations:

1. x + y = 49,328
2. 3x = 2y

We can solve for y by multiplying equation (1) by 2:

2x + 2y = 98,656

Now subtract equation (2) from the new equation:

2x + 2y - 3x = 98,656 - 49,328

-x = -29,328

So, x = 29,328. Substituting x into equation (1), we get:

29,328 + y = 49,328

y = 49,328 - 29,328

y = 20,000

Therefore, Phyllis invested $29,328 at 9% and $20,000 at 6% simple interest.