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.
[tex]x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150[/tex]
Multiplying both sides by 100, we get
[tex]7x+16x+27x=15000[/tex]
[tex]50x=15000[/tex]
Dividing both sides by 50, we get
[tex]x=\dfrac{15000}{50}[/tex]
[tex]x=300[/tex]
So, the amount invested at 7% is [tex]300[/tex].
The amount invested at 8% is
[tex]2(300)=600[/tex]
The amount invested at 9% is
[tex]3(300)=900[/tex]
Hence, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.
