18,355 is invested,part at 13% and the rest at 7%.If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 7% by $963.55

Question:18,355 is invested,part at 13% and the rest at 7%.If the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 7% by\$963.55. How much he invested at each rate?

Solution:

Given an investment of \$18,355. It is divided into two parts.

One part of the investment is at 13% interest. Let us say that the investment is  \$ X.

Another part of the investment is at 7%. Let us say that the investment is \$ Y.

Since we are having two unknowns X and Y, we need to frame two equations to solve for these variables X and Y.

Since the  total investment is given as \$18,355, the  first  equation can be

X+Y =18355—-> Equation1

Now we need to arrive at Equation 2.Let us find the interest amount for each of the investments made at different percentages.

The  interest  amount earned from \$ X  investment at 13% is interest rate  will be

X\cdot \dfrac {13}{100}=0.13X

SImilar way, the  interest  amount  earned from the \$ Y  investment at 7% will be

X\cdot \dfrac {7}{100}=0.07Y

The directions to the  problem say that the interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 7% by \$963.55

we can convert this sentence into an equation.

0.13X=0.07Y+963.55 .Multiply with 100 both sides

13X =7y+96355

13X-7Y =96355—->Equation2

Now we have two equations and two variables X and  Y to solve.

7X+7Y =7*18355=128485 (Multiplying the  equation1 with 7 both sides and adding with equation2)

13X-7Y =96355

——————————–

 20X=224840

——————————–

X=\dfrac {224840}{20}=11242 .

Since we have the X value, we can find the Y value as well.

X+Y =18355

11242+Y=18355

Y=18355 – 11242

Y=7113

Therefore the amount invested at a 13% interest rate would be  X = \$11242. The amount invested at 7% interest  rate would be Y=\$ 7113 .

Verification:

We can cross-check the answers we got for the X =11242 and  Y =7113 values.

When the values of X and Y are plugged back into the original equations, they must hold good.

Let us take up  the  equation1

x+y=18355

11242+7113=18355

18355=18355.Therefore, the solutions we  arrived work for the Equation1

Let us cross-check with the equation2.

13X-7Y =96355

13(11242)-7(7113) =96355

96355=96355. Therefore, the solutions we arrived work for the Equation2 as well.

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