Question

$16,281$⁢16,281 is invested, part at 15%15% and the rest at 13%13%. If the interest earned from the amount invested at 15%15% exceeds the interest earned from the amount invested at 13%13% by $1995.27$⁢1995.27, how much is invested at each rate? (Round to two decimal places if necessary.)

Answer 1

Answer:

Ans. The amount invested at 13% was $1,595.97 and $14,685.03 were invested at 15%

Explanation:

Hi, you can solve this by using 2 equations, so let X be the portion of the money invested at 15% and Y be the amount invested at 13%. So the equation for the whole amount is:

$X+Y=16,281$

Now, the problem says that the money that you earn by investing at 15% exceeds the money received as interest in your investment of 13% by $1,995.27, this leads us to the second equation.

$0.15X=0.13Y+1995.27$

Now, to make it a little more friendly, we just have to go ahead and divide everything by 0.15, so we get.

$X=0.8667Y+13,301.8$

Now, in our first equation, we substitute X fo 0.8867(Y)+13,301.8 and we will see this.

$0.8667Y+13,301.8+Y=16,281$

Now, we solve for Y

$1.8667Y=16,281-13,301.8$

$Y=\frac{2,979.2}{1.8667} =1,595.97$

So the money invested at 13% was $1,595.97 therefore, the money invested at 15% was $16,281 - $1,595.97 = $14,685.03

And we can check this results like this. The money invested at 15% will return an amount of:

$14,685.03*0.15=2,202.75$

And the money invested at 13% will return

$1,595.97*0.13=207.48$

Substracting, we would found that the difference is:

$2,202.75-207.48=1,995.27$

Best of luck.