Question

A man invests $2,200 in three accounts that pay 6%, 8%, 9% in annual interest, respectively. he has three times as much invented at 9% as he does at 6%. if his total interest for year is $178, how much is invested at each rate?

Answer 1

Answer:

6%: $200

8%: $1400

9%: $600

Step-by-step explanation:

Let x represent amount invested at a rate of 6% and y represent amount invested at a rate of 8%.

We have been given that a man has three times as much invented at 9% as he does at 6%. So amount invested at a rate of 9% would be $3x$.

We are also told that a man invests $2,200 in three accounts. We can represent this information in an equation as:

$x+3x+y=2200...(1)$

$4x+y=2200...(1)$

$y=2200-4x...(1)$

We are also told that his total interest for year is $178. We can represent this information in an equation as:

$0.06x+0.09(3x)+0.08y=178...(2)$

$0.06x+0.27x+0.08y=178...(2)$

$0.33x+0.08y=178...(2)$

Upon substituting equation (1) in equation (2), we will get:

$0.33x+0.08(2200-4x)=178$

$0.33x+176-0.32x=178$

$0.01x+176=178$

$0.01x+176-176=178-176$

$0.01x=2$

$\frac{0.01x}{0.01}=\frac{2}{0.01}$

$x=200$

Therefore, an amount of $200 is invested at 6%.

Amount invested at a rate of 9% would be $3x\Rightarrow 3(200)=600$.

Therefore, an amount of $600 is invested at 9%.

Upon substituting $x=200$ in equation (1), we will get:

$y=2200-4(200)$

$y=2200-800$

$y=1400$

Therefore, an amount of $1400 is invested at 8%.