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mixas84 [53]
2 years ago
9

This is due in 10 minutes please help me

Mathematics
1 answer:
valkas [14]2 years ago
7 0

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

Multiplying both sides by 100, we get

7x+16x+27x=15000

50x=15000

Dividing both sides by 50, we get

x=\dfrac{15000}{50}

x=300

So, 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

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

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