GRE algebra practive question and solution

GRE Math Tutorial

GRE algebra practive question and solution


This is a problem used to prepare for the quantitative reasoning section of your GRE.

Sample Problem

Question source

Pat invested a total of $3000. Part of the money was invested in a money market account that paid 10 percent simple annual interest, and the remainder of the money was invested in a fund that paid 8 percent simple annual interest. If the interest earned at the end of the first year from these investments was $256, how much did Pat invest at 10 percent and how much at 8 percent?


Define Variables.
let x = money invested at 10%
therefore 3000 – x = money invested At 8%

Setting up Equation.
The profits for these investments were $256. Part of this would have come from the investment at 10% and the other part would have come from the investment at 8%.

So the total money earned would be the sum of the two amounts.

**Note: when using percentages in an algebraic expression you must convert it to its decimal form (e.g. 10% –> 0.1)

0.1*x + 0.08*(3000-x) = 256

here the 0.1*x represents the amount earned from the 10% interest portion and the 0.08*(3000-8) represents the amount earned from the 8% interest portion.

Solving the Equation.
first we should expand the portion in the brackets

0.1*x + 240 – 0.08*x = 256

next we will combine like terms by subtracting 240 from both sides of the equation

0.1*x + 240 – 240 – 0.08*x = 256 – 240
0.1*x – 0.08*x = 16
0.02*x = 16

lastly we will divide both sides by 0.02 to solve for x

0.02*x/0.02 = 16/0.02

now we have solved for x

x = 800

plug into our initial statement

3000 – 800 = 2200

The amount invested into each investment were:

$800 at 10%

$2200 at 8%

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