SOLUTION: Last year, Chau had $30,000 to invest. He invested some of it in an account that paid 6% simple interest per year, and he invested the rest in an account that paid 8% simple intere
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-> SOLUTION: Last year, Chau had $30,000 to invest. He invested some of it in an account that paid 6% simple interest per year, and he invested the rest in an account that paid 8% simple intere
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Question 1125192: Last year, Chau had $30,000 to invest. He invested some of it in an account that paid 6% simple interest per year, and he invested the rest in an account that paid 8% simple interest per year. After one year, he received a total of $2260 in interest. How much did he invest in each account? Answer by greenestamps(13200) (Show Source):
Let x be the amount invested at 6%; then (30000-x) is the amount invested at 8%.
The interest earned is then 6% of x plus 8% of (30000-x). Since the earned interest was 2260,
Perhaps a bit awkward to solve because of the decimals; still basic algebra that you should be able to solve in a minute or two.
But here is a method which, if you understand it, will get you to the answer much faster and with much less work.
(1) The interest earned if all $30000 were invested at 6% is $1800; the interest if all were invested at 8% is $2400. Where the actual interest of $2260 lies between $1800 and $2400 determines the ratio in which the money was split between the two investments.
(2) Think of it on a number line if it helps. 2260-1800=460; 2400-1800=600. So the actual amount of interest is "460/600 of the way from 1800 to 2400". That means 460/600 of the total investment was at the higher rate.
(3) 460/600 = 230/300 = 23000/30000. So $23000 was invested at the higher 8% rate; the other $7000 was invested at 6%.
