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Solved investment and Interest word problems

Problem

Ben invested $30,000, part of which at 5% annual interest rate, the rest at 9% annual interest rate. The interest earned from the investments was $2,100 at the end of one year. How much did he invest at each rate?

How to solve it:

The two unknowns
the amount invested at 5%.
the amount invested at 9%

The two relationships

Use the following relationships to set up an equation or a system of two equations

(1)

{the amount invested at 5%} + {the amount invested at 9%} = 30,000

(2)

{the interest earned at 5%} + {the interest earned at 9%} = 2,100

Depending on how the two relationships are used, the problem can be solved by setting up either:
a system of two equations, or
one equation in one variable.

Solution (by setting up an equation in one variable)

Step 1
Let x be the amount invested at 5%.

Step 2
Use relationship (1) to express the amount invested at 9% in terms of x.
So the amount invested at 9% = 30,000 - x

Step 3
Express the interest earned on each investment in terms of x.

	amount invested	annual rate	interest earned
5% investment	x	5%	x · 5%
9% investment	30,000 - x	9%	(30,000 - x) · 9%

the interest earned on the 5% investment = x · 5%
the interest earned on the 9% investment = (30,000 - x) · 9%

Step 4
Substitute the interest earned on each investment into relationship (2) to set up an equation.
x · 5% + (30,000 - x) · 9% = 2,100

Step 5
Solve the equation for x.
x · 5% + (30,000 - x) · 9% = 2,100
5x + 9(30,000 - x) = 210,000 (After multiplying both sides by 100)
5x + 270,000 - 9x = 210,000
-4x = - 60,000
x = 15,000

So Ben invests $15,000 at 5%, and (30,000 - x = 30,000 - 15,000 =) $15,000 at 9%.

Alternative Solution (by setting up a system of two equations)

Step 1
Let x represent the amount invested at 5%.
Let y represent the amount invested at 9%.

Step 2
Express the interest earned on each investment in terms of x or y.

	amount invested	annual rate	interest earned
5% investment	x	5%	x · 5%
9% investment	y	9%	y · 9%

the interest earned at 5% = x · 5%
the interest earned at 9% = y · 9%

Step 3
Use relationship (1) to set up the first equation.
x + y = 30,000
Substitute the interest earned on each investment into relationship (2) to set up the second equation.
x · 5% + y · 9% = 2,100

Step 4
Solve the simultaneous equations for x and y.
x + y = 30,000
x · 5% + y · 9% = 2,100

Multiplying the second equation by 100, we have
x + y = 30,000 ...........................(1)
6x + 9y = 210,000 ...................(2)

Solving equation (1) for y, we have
y = 30,000 - x ............................(3)

Substituting equation (3) into equation (2), we have
5x + 9(30,000 - x) = 210,000

Solving for x, we obtain
5x + 9(30,000 - x) = 210,000
5x + 270,000 - 9x = 210,000
-4x = - 60,000
x = 15,000

Substituting x = 15,000 into equation (3), we obtain
y = 30,000 - x = 30,000 - 15,000 = 15,000

So John invests $15,000 at 6%, and $15,000 at 9%.

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