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3 years ago

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Suppose you invested $10,000 part at 6% annual interest and the rest at 9% annual interest. If you received $684 in interest aft

er one year, how much did you invest at each rate?

1 answer:

victus00 [196]3 years ago

5 0

Assuming simple interest (i.e. no compounding within first year), then
At 6%, interest = 10000*0.06=$600
At 9% interest = 10000*0.09 = $900

Two ways to find the ratio
method A. let x=proportion at 6%
then
600x+900(1-x)=684
Expand and solve
300x=900-684=216
x=216/300=0.72 or 72%
So 10000*0.72=7200 were invested at 6%
10000-7200=2800 were invested at 9%

method B: by proportions
Ratio of investments at 6% and 9%
= 900-684 : 684-600
=216 : 84
= 18 : 7
Amount invested at 6% = 18/(18+7) * 10000 = 0.72*10000 = 7200
Amount invested at 8% = 7/(18+7)*10000=0.28*10000=2800

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