Question

Belinda had $20,000 to invest. She invested part of it at 10% and the remainder at 12%. If
her income from the two investments was $2160, then how much did she invest at each rate?

Answer 1

Let x be the amount invested at 10% and y be the amount invested at 12%. We can
summarize all of the given information in a table:
We can write one equation about the amounts invested and another about the interest
from the investments:
x + y = 20,000 Total amount invested
0.10x + 0.12y = 2160 Total interest
Solve the first equation for x to get x = 20,000 - y. Substitute 20,000 - y for x in
the second equation:
0.10x + 0.12y = 2160
0.10(20,000 - y) = 0.12y  2160 Replace x by 20,000 - y.
2000 - 0.10y + 0.12y = 2160 Solve for y.
0.02y = 160
y = 8000
x = 12,000 Because x  20,000  y
To check this answer, find 10% of $12,000 and 12% of $8000:
0.10(12,000) = 1,200
0.12(8000) = 960
Because $1200 + $960 = $2160 and $8000 + $12,000 = $20,000, we can be certain
that Belinda invested $12,000 at 10% and $8000 at 12%.