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Dmitrij [34]
3 years ago
5

Please help me i need it right now please

Mathematics
1 answer:
12345 [234]3 years ago
4 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:

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

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

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