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ludmilkaskok [199]
3 years ago
12

I won $60,000 I invested part in a CD at 2% interest the rest at 3%interest in a cd at the end of the year I earned $1600 in int

erest. How much did I invest in each cd
Mathematics
2 answers:
zaharov [31]3 years ago
5 0
Let x = amount invested in 2% CD and y = amount invested in 3% CD
x + y = 60000
0.02x + 0.03y = 1600
SOLVE THE 1st EQUATION FOR x AND SUBSTITUTE RESULT IN 2nd
0.02(60000 - y) + 0.03y = 1600  
1200 - 0.02y + 0.03y = 1600
0.03y = 400
y = 13333.34
x = 46666.66
11Alexandr11 [23.1K]3 years ago
3 0
At 2% interest:  i2 = x * 0.02 is the amount of interest earned
At 3% int:  i3 = (60000-x) * 0.03 is the amt of interest earned.

Note that i2 + i3 = $1600, and that x + y = 60000.
   x * 0.02 + (60000-x) * 0.03 = $1600  can be solved for x, the amt invested at 2%.

0.02x + 1800 - 0.03x = 1600 simplifies to $200 = 0.01x.  Thus, x = $20000 at 2% and y = 60000-20000 = $40000 at 3%.
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