In a certain Algebra 2 class of 28 students, 7 of them play basketball and 5 of them
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Given:
The rate of interest on three accounts are 7%, 8%, 9%.
She has twice as much money invested at 8% as she does in 7%.
She has three times as much at 9% as she has at 7%.
Total interest for the year is $150.
To find:
Amount invested on each rate.
Solution:
Suppose x be the amount invested at 7%. Then,
The amount invested at 8% = 2x
The amount invested at 9% = 3x
Total interest for the year is $150.
Multiplying both sides by 100, we get
Dividing both sides by 50, we get
So, the amount invested at 7% is .
The amount invested at 8% is
The amount invested at 9% is
Hence, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.