Question:

# help needed in math question?

by Guest6320  |  11 years, 3 month(s) ago

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William opened two investment accounts for his grandsonÃ¢Â€Â™s college fund. The first year, these investments, which totaled \$18,000, yielded \$831 in simple interest. Part of the money was invested at 5.5% and the rest at 4%. How much was invested at each rate?

Tags: Help, math, QUESTION

1. Guest7654
x+y=18000

.04x + .055y = 831

.04x + .055(18000-x) = 831

.04x + 990 - .055x = 831

.015x = 159

x = 10600

y = 18000 - 10600 = 7400

2. Guest3973
\$18,000 was invested, so split that into the two parts. \$x amount was invested at 5.5% and (\$18,000-x) was invested at 4%

.055x + (18,000 - x).04 = 831

.055x + 720 - .04x = 831

.015x = 111

x = 7,400

\$7,400 was invested at 5.5%

\$10,600 was invested at 4.0%

Piece of cake.
3. Guest2404
interest = p * 18000 * 0.055 + (1-p) * 18000 * 0.04

= 831

=&gt; 270 p + 720 = 831

=&gt; 270 p = 111

=&gt; p = 0.4111111....

So 41.111 % of the money was invested at 5.5 %

or 7400 \$. The rest (10600 \$) was invested at 4 %.
4. Guest6863
let x and y be the two parts

x + y = 18000

x(0.055) + y(0.04) = 831

x = 18000 - y

(18000 - y)(0.055) + y(0.04) = 831

990 - 0.055y + 0.04y = 831

0.015y = 159

y = 10600

x = 7400

Check

7400(0.055) + 10600(0.04) = 831

407 + 424 = 831

.

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