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Question 1142477: im struggling with this:
algelica invested part of her 15000 dollar inheritance in an account earning 4 percent simple interest and the rest in an account earning 3 percent simple interest. How much did she invest in each account if she earned 520 dollars in total interest after one year
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! x = amount invested in 3% account.
y = amount invested in 4$ account.
x + y = 15000
.04 * x + .03 * y = 520
solve for y in the first equation to get y = 15000 - x.
replace y with 15000 - x in the second equation to get:
.04 * x + .03 * (15000 - x) = 520.
simplify to get .04 * x + .03 * 15000 - .03 * x = 520.
simplify further and combine like terms to get .01 * x + 450 = 520.
subtract 450 from both sides of the equation to get .01 * x = 70.
solve for x to get x = 70 / .01 = 7000.
that makes y = 15000 - 7000 = 8000.
confirm by replacing x with 7000 and y with 8000 to get .04 * 7000 + .03 * 8000 = 520, which becomes 280 + 240 = 520, confirming the solution is correct.
your solution is that she invested 7000 at 4% and 8000 at 3%.
Answer by greenestamps(13203)  (Show Source):
You can put this solution on YOUR website!
Here is an alternative to the standard algebraic method for solving mixture problems like this, as shown by the other tutor. If you understand how to use it, it will get you to the answer with much less effort than the algebraic method.
(1) Determine the amounts of interest if the whole $15,000 were invested at each rate.
$15,000 at 3% = $450
$15,000 at 4% = $600
(2) Picture the three interest amounts on a number line: 450, 520, and 600.
(3) Use mental arithmetic or any method you want to determine that 520 is 7/15 of the way from 450 to 600.
(4) That means 7/15 of the total was invested at the higher rate.
ANSWER: 7/15 of $15,000, or $7000, at 4%; the other $8000 at 3%.
CHECK: .04(7000)+.03(8000) = 280+240 = 520
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