Please help me i need it right now please
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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:
Let 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]
Multiply both sides by 100.
[tex]7x+16x+27x=15000[/tex]
[tex]50x=15000[/tex]
Divide both sides by 50.
[tex]x=\dfrac{15000}{50}[/tex]
[tex]x=300[/tex]
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]
Thus, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.