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?
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
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)
Since we have the X value, we can find the Y value as well.
Y=18355 – 11242
Therefore the amount invested at a 13% interest rate would be X = 11242. The amount invested at 7% interest rate would be Y= 7113 .
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
18355=18355.Therefore, the solutions we arrived work for the Equation1
Let us cross-check with the equation2.
96355=96355. Therefore, the solutions we arrived work for the Equation2 as well.
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