How do I do this algebra word problem?
5,000 $ were invested in two CDs. Once of the CDs paid 4% interest and the other one paid 5% interest. At the end of one year 228$ in simple interest was earned. How much did they invest at each interest rate?
5,000 $ were invested in two CDs. Once of the CDs paid 4% interest and the other one paid 5% interest. At the end of one year 228$ in simple interest was earned. How much did they invest at each interest rate?
Comments
4% CD—c:
0.04c + 0.05($5,000 - c) = $228
0.04c + $250 - 0.05c = $228
0.01c = $22
c = $2,200
5% CD:
= $5,000 - $2,200
= $2,800
Answer: 4% CD, $2,200; 5% CD, $2,800
The total investment was 5000.
Let x be the amount invested in the 1st CD, so 5000 - x is the amount in the 2nd. If the first pays 4% = .04 and the 2nd pays 5% = .05, and the total payoff is 228, then the equation is:
(x)(0.04) + (5000 - x)(0.05) = 228.
Solving for x should be relatively easy now, and do (5000 - x) for the amount invested in the 2nd one.
A = invested at 4%
B = invested at 5%
.04A + .05B = 228
A + B = 5000
.04A + .05(5000-A) = 228
(.04 - .05)A = 228 - .05(5000)
A = (250-228)/.01 = 2200
B = 5000 - 2200 = 2800
Check:
.04*2200 + .05*2800 = 228
228 = 228
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x+y = 5000, so y=5000-x
x(0.04) + (5000-x)(0.05) = 228
x = 2200, y=2800
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