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A man invests a certain amount of money at 2% interest and $800 more than that amount in another account at 4% interest. At the end of one year, he earned $92 in interest. How much money was invested in each account?

User Shunan
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2 Answers

3 votes
x*2/100 + (x+800)*4/100 = 92
2x+4x+3200 = 9200
6x = 9200-3200
6x = 6000
x = 1000
So the man invested $ 1000 at 2% interest and $ 1800 at 4% interest.
User Mtraut
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3 votes

Answer:

He invested
\$1000 in first account and
\$1800 in the second account.

Step-by-step explanation:

Let the man invest
\$\:x in the first account.

It is given that he gets a 2% interest on the first amount.

Let
I_(1) represents the interest received after 1 year.

We know that
I=(P* R* T)/(100)

Here,
P=x,\,R=2, T=1

So,
I_(1)=(x* 2* 1)/(100)


I_(1)=(2x)/(100)

It is given that he invest
\$800 more at 4% interest on the second amount.

Let
I_(2) represents the interest received after 1 year.

We know that
I=(P* R* T)/(100)

In that case,
P=x+800,\,R=2, T=1

So,
I_(2)=((x+800)* 4* 1)/(100)

I_{2}=\frac{4x+3200}{100}

Also, it is given that the total interest he earned is
\$92

So, we have
I_(1) +I_(2) =92


(2x)/(100)+(4x+3200)/(100)=92


(2x+4x+3200)/(100)=92


6x+3200=9200


6x=9200-3200


6x=6000


x=1000

Hence, he invested
\$1000 in first account and
\$1000+\$800=\$1800 in the second account.

User Radu Luncasu
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7.0k points