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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 $112 in interest. How much money was invested in each account? Which of the following equations could be used to solve the problem?

A) 0.02x + 0.04(800) = 112
B) 0.02x + 0.04(x + 800) = 112
C) 0.04x + 0.02(x + 800) = 112

Respuesta :

The correct answer is B
If you let x = amount invested at 2% (which is 0.02), then x + 800 = amount invested at 4% (which is 0.04).

varshamittal029

The money which was invested in each account was $1333.33 at 2% and $2133.33 at 4% and the equation will be 0.02x + 0.04(x+800) =  112.

What is interest?

Interest is the amount which we receive more than principal when we lent or borrow money at some rate for some time.

We have,

A certain amount of money at 2% interest,

And,

$800 more than that amount in another account at 4% interest.

And,

Interest at the end of one year = $112

Let,

Money invest at 2% = x

So,

Money invest at 4% = x + 800

Now,

Simple interest at 2%,

Simple interest = (Principal * Rate * Time) / 100

Simple interest = ( x * 2 * 1) / 100 = 0.02x

Now,

Simple interest at 4%,

Simple interest = ( (x+800) * 4 * 1) / 100 = 0.04(x+800)

Now,

The equation for one year interest will be,

0.02x + 0.04(x+800) =  112

So,

Option (C) s the correct answer.

NOw,

Solve for x to know money invested at different interest ,

0.02x + 0.04(x + 800) =  112

0.02x + 0.04x + 32 =  112

⇒

0.06x = 80

x = 1333.33

So,

Money invest at 2% = $1333.33

And,

Money invest at 4% = 1333.33 + 800 = $2133.33

Hence we can say that the money which was invested in each account was $1333.33 at 2% and $2133.33 at 4% and the equation will be 0.02x + 0.04(x+800) =  112.

Learn more about the interest here:

https://brainly.com/question/25845758

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