## 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% by963.55. 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

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

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.

——————————–

——————————–

#### .

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

### 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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