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jason invested $6000. he invested part of the money at 6.5% and the remainder at 7.5% the money invested at 6.5% earns $177.30 more interest than the second amount earned. how much did he invest at each rate?

User Erik Duymelinck
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1 Answer

8 votes
8 votes

Answer:

Amount invested at 6.5% us $4,480.70 and

Amount invested at 7.5% = $1,519.30

Explanation:

This is a problem involving algebraic equations in two variables. We start by defining variables to represent the two unknowns.

Let x represent the money invested at 6.5% and y represent the money invested at 7.5%

Total Money Invested = $6000
This means x + y = 6000 (1)

We first convert all interest percentages to decimals for computational purposes

6.5% = 6.5/100 = 0.065
7.5% = 7.5/100 = 0.075

Interest earned on $y at 6.5% = 0.065y
Interest earned on $x at 7.5% = 0.075x

The difference = 0.065x - 0.075y and we are given that this difference is $177.30

Therefore we have a second equation:
0.065x - 0.075y = 177.30 (2)

In order to solve for x and y from equations (1) and (2) we have to eliminate one of the variables by making the coefficient of the other variable equal in both equations and either adding or subtracting

Let's multiply equation (1) by 0.075

This gives us
0.075x + 0.075y = 0.075 x 6000

Simplifying we get
0.075x + 0.075y = 450 (3)

Therefore we have two equations where the coefficients of y are the same but have opposite signs

0.065x - 0.075y = 177.30 (2)
0.075x + 0.075y = 450 (3)

Seeing that the coefficients of y are the same but with opposite signs we can add both the equations to cancel out the y terms

0.065x - 0.075x + 0.075x - 0.075y = 177.30 + 450

Simplify
0.14x =627.30

x = 627.30 / 0.14

x = 4,480.71

From equation (1)

x + y = 6000

4,480.71 + y = 6000

y = 6000 - 4480.71

y = 1,519.29

So amount invested at 6.5% us $4,480.70 and amount invested at 7.5% = $1,519.30

User Joeeee
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