Question 1141993: A bank loaned out $15,000, part of it at the rate of 4% annual interest, and the rest at 10% annual interest. The total interest earned for both loans was $1,170.00. How much was loaned at each rate?
Found 2 solutions by josmiceli, greenestamps:
Answer by josmiceli(19441)   (Show Source):
Answer by greenestamps(13195)  (Show Source):
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Here is an alternative to the standard algebraic method for solving this and many other kinds of "mixture" problems. If you understand it and have moderately good mental math skills, it will get you to the answers to problems like this much faster, and with less effort, than the algebraic method.
The total $15,000 at 4% interest would earn $600 interest; the total at 10% would earn $1500 interest. The difference between $600 and $1500 is $900.
The actual interest was $1,170. The difference between $1,170 and $600 is $570.
Picture those three interest amounts on a number line: $600, $1170, $1500.
The actual interest amount is 570/900 of the way from $600 to $1500.
That means 570/900 of the total was invested at the higher rate.
Now simplify that fraction. 570/900 = 57/90 = 19/30.
1/30 of the total $15,000 is $500; 19/30 is 19*$500 = $9500.
ANSWER: $9500 at 10%; the rest, $5500, at 4%.
CHECK: .10(9500)+.04(5500) = 950+220 = 1170
All the words of explanation make the process look long. Here without the words are the few calculations required to solve the problem.
.04(15000) = 600
.10(15000) = 1500
1500-600 = 900; 1170-600 = 570
570/900 = 57/90 = 19/30
19/30(15000) = 19*500 = 9500
15000-9500 = 5500
ANSWER: 9500 at 10%; 5500 at 4%
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