Question

Please help me anyone i need this in 10 minutes

Answer 1

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.

$x\times \dfrac{7}{100}+2x\times \dfrac{8}{100}+3x\times \dfrac{9}{100}=150$

Multiply both sides by 100.

$7x+16x+27x=15000$

$50x=15000$

Divide both sides by 50.

$x=\dfrac{15000}{50}$

$x=300$

The amount invested at 7% is $300$.

The amount invested at 8% is

$2(300)=600$

The amount invested at 9% is

$3(300)=900$

Therefore, the stockbroker invested $300 at 7%, $600 at 8%, and $900 at 9%.