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Question 435031: $2000 total deposit with 2 savings accounts. One pays interest rate of 6%the other at rate of 8% if a total earned interest is $144 how much is each deposit
Found 4 solutions by mananth, ikleyn, josgarithmetic, greenestamps:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! Part I 6.00% per annum
Part II 8.00% per annum
x+y=2000 ------------------------1
6.00%x+ 8.00%y= $144.00
Multiply by 100
6x + 8 y = $14,400.00 --------2
Multiply (1) by -5
we get
-5x-5y= -10000.00
Add this to (2)
3y= $4,400.00
divide by 3
y=$1,466.67 investment at 8.00%
Balance $533.33 investment at 6.00%
Answer by ikleyn(53419)  (Show Source):
You can put this solution on YOUR website! .
$2000 total deposit with 2 savings accounts. One pays interest rate of 6%,
the other at rate of 8% if a total earned interest is $144 how much is each deposit
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The solution and the answer in the post by @mananth both are incorrect.
His solution has arithmetic errors and does not withstand the standard check.
So, I came to bring a correct solution.
I will solve the problem using one single unknown, since it provides a shorter way.
Let x be the amount invested at 8%, in dollars.
Then the amount invested at 6% is (2000-x) dollars.
Write the total annual interest equation
0.08*(x + 0.06(2000-x) = 144 dollars.
Simplify and find x
0.08x + 120 - 0.06x = 144,
0.08x - 0.06x = 144 - 120,
0.02x = 24
x = 24/0.02 = 2400/2 = 1200.
ANSWER. $1200 invested at 8%; $the rest, $2000 - $1200 = $800 invested at 6%.
CHECK. 0.08*1200 + 0.06*800 = 144 dollars. PRECISELY as it is given in the problem.
Solved correctly.
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This problem can be solved algebraically using one equation in one unknown or two equations in two unknowns.
It also can be solved mentally.
Since this site is to teach algebra, I place here algebraic solution.
Since the solution with one equation is simpler and shorter, I prefer this form with one unknown, which I present here.
What is really important, is to check the final solution for validity.
If you do not check, nothing will prevent you from mistakes.
Answer by josgarithmetic(39701)  (Show Source):
Answer by greenestamps(13258)  (Show Source):
You can put this solution on YOUR website!
This is essentially a mixture problem -- we are mixing money invested with a return of 6% and money invested at 8%.
Here is a solution using a method that can be used in any 2-part mixture problem like this.
All $2000 invested at 6% would yield a return of $120; all invested at 8% would yield a return of $160; the actual return is $144.
The ratio in which the money is invested in the two places is exactly determined by where the actual return of $144 lies between $120 and $160.
Use a number line if it helps to determine that $144 is three-fifths of the way from $120 to $160.
(The difference between $160 and $120 is $40; the difference between $144 and $120 is $24. $24 is 24/40 = 6/10 = 3/5 of $40)
That means 3/5 of the total $2000 was invested at the higher rate.
3/5 of $2000 is $1200, so $1200 of the total $2000 was invested at 8%.
ANSWERS: $1200 was invested at 8%; $800 at 6%.
CHECK: .08(1200)+.06(800) = 96+48 = 144
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