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.
![x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150](https://tex.z-dn.net/?f=x%5Ctimes%20%5Cdfrac%7B7%7D%7B100%7D%2B2x%5Ctimes%20%5Cdfrac%7B8%7D%7B100%7D%2B3x%5Ctimes%20%5Cdfrac%7B9%7D%7B100%7D%3D150)
Multiplying both sides by 100, we get
![7x+16x+27x=15000](https://tex.z-dn.net/?f=7x%2B16x%2B27x%3D15000)
![50x=15000](https://tex.z-dn.net/?f=50x%3D15000)
Dividing both sides by 50, we get
![x=\dfrac{15000}{50}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B15000%7D%7B50%7D)
![x=300](https://tex.z-dn.net/?f=x%3D300)
So, the amount invested at 7% is
.
The amount invested at 8% is
![2(300)=600](https://tex.z-dn.net/?f=2%28300%29%3D600)
The amount invested at 9% is
![3(300)=900](https://tex.z-dn.net/?f=3%28300%29%3D900)
Hence, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.