SOLUTION: A woman invest $10 000;part at 9% annual interests, and the rest at 14%. In each case the interest is compounded annually. The total annual income is $1275. How much is invested in

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Question 1196477: A woman invest $10 000;part at 9% annual interests, and the rest at 14%. In each case the interest is compounded annually. The total annual income is $1275. How much is invested in each rate?
Found 2 solutions by greenestamps, josgarithmetic:
Answer by greenestamps(13195)   (Show Source): You can put this solution on YOUR website!


(1) by standard algebra....

x = amount invested at 9%
10000-x = amount invested at 14%

The total interest was $1275:






ANSWER: x = $2500 at 9%; 10000-x = $7500 at 14%

(2) using logical reasoning and mental arithmetic....

All $10,000 invested at 9% would have yielded $900 interest; all at 14% would have yielded $1400 interest.
Look at the three interest amounts $900, $1275, and $1400 (on a number line, if it helps) and observe/calculate that $1275 is 3/4 of the way from $900 to $1400 (1275-900 = 375; 1400-900 = 500; 375/500 = 3/4).
That means 3/4 of the $10,000 was invested at the higher rate.

ANSWER: $7500 at 14%, the other $2500 at 9%
Answer by josgarithmetic(39613)   (Show Source): You can put this solution on YOUR website!
How many years? Is this just one single year?
                     QUANTITY        RATE(decimal)     1-YEAR INTEREST

Part at 9%             10000-x      0.09                0.09(10000-x)

Part at 14%             x           0.14                0.14x

TOTAL                  10000                             1275

What to do with this should be clear and needs no discussion.
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