Question

An investor invested a total of $14,000. She divided the money into three different accounts. At the end of the year, she had made $670 in interest. The annual yield on each of the three accounts was 4%, 5%, and 6%. If the amount of money invested in the 5% account was the same as the combined amount invested in the 4% and the 6% accounts, determine how much was invested in the account that paid 6% interest.

Answer 1

$2000

Explanation:

Given the following :

Total amount invested = 14000

Total interest earned = 670

Amount invested in 4% (0.04) account = a

Amount invested in 6% (0.06) account = b

Amount invested in 5% (0.04) account = a + b

a + b + (a + b) = 14000

2a + 2b = 14000

a + b = 7000 - - - (1)

Simple interest = principal * rate * time

(a * 0.04) + (b * 0.06) + [(a+b) * 0.05] = 670

0.04a + 0.06b + 0.05a + 0.05b = 670

0.09a + 0.11b = 670 - - - - (2)

from (1)

a = 7000 - b

0.09(7000 - b) + 0.11b = 670

630 - 0.09b + 0.11b = 670

= 630 + 0.02b = 670

0.02b = 670 - 630

0.02b = 40

b = 40/ 0.02

b = $2000

Amout invested in 6% account = $2000