SOLUTION: The annual interest on a $10,000 investment exceeds the interest earned on an $8000 investment by $100. The $10,000 is invested at a 0.5% higher rate of interest than the $8000.
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Question 995264: The annual interest on a $10,000 investment exceeds the interest earned on an $8000 investment by $100. The $10,000 is invested at a 0.5% higher rate of interest than the $8000. What is the interest rate of each investment?
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x=interest on $10000
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(x+0.005)$10000=(x)$8000+$100
$10000x+$50=$8000x+$100
$2000x=$50
x=0.025
ANSWER 1: The rate on the $8000 investment is 2.5%.
x+0.005=0.03
ANSWER 2: The rate on the $10000 investment is 3%.
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CHECK:
(x+0.005)$10000=(x)$8000+$100
0.03($10000)=0.025($8000)+$100
$300=$200+$100
$300=$300
