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Math Word Problems for the GRE™:

When Plugging Numbers into Formulas

Just Isn't Enough

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Two Sample Problems with No Numbers (All Variables)

For many students, the most intimidating word problems are those that use letters - or other variables - in place of actual numbers. In extreme cases, students become so confused that they cannot identify simple mathematical operations that would normally be easy for them.

And that's why we have included this chapter - to give you a few basic tips that will increase your likelihood of solving "all variable" problems on the GRE quickly and correctly.

Our first tip is to read the problem slowly and carefully. Try to determine the mathematical operation that is being described. If you can't, substitute a few random numbers for the variables. In most cases, the scenario will be significantly easier to understand.

As always, here are a few examples:

Example 1. Backwards Calculations

For her Halloween party, Janice went to a local candy store and bought X dozen candy bars at a price of Y per bar. When she left the store, Janice had Z cents left over. Assuming that she made no other purchases, how much money (in cents) did Janice have when she entered the candy store?

Solution: We can solve this problem by plugging in numbers or by doing a few simple "backwards" calculations. First, let's try Option 1, the plug- in approach.

Let's assume that Janice bought 10 dozen candy bars at a price of 50 cents per bar. Let's also assume that she had 20 cents left over. Hence, X = 10, Y = 50 and Z = 20.

Janice therefore spent (12)(10)(50 cents), which is 12XY. If she had 20 cents left over, then her original amount of money was 12XY + Z. Choice E is correct.

Option 2. If you don't want to plug in numbers, you can just reason the problem through. Janice bought X dozen candy bars, which = 12X. If they cost Y cents each, then she spent 12XY. Finally, she had Z cents left over, which we must add to her total amount of money. When we do, we get the same answer as we did with the plug-in method: 12XY + Z.

Example 2: Fun with Fractions

Carol's sugar bowl was one-third full. When she added K cc of granulated sugar, the bowl was nine-tenths full. How much sugar (in cc) would Carol's sugar bowl hold if it was completely full?

Solution: Most students overcomplicate this question and take too long to solve it. In this case, we can find the answer by doing a few quick calculations.

First, we must determine how much of the bowl was filled by K cc.
Mathematically, K cc = 9/10 - 1/3 = 27/30 - 10/30 = 17/30.

To be totally filled, the bowl would require 30/30 of sugar, which is equal to K times its inverse, or 30/17K. Choice E is correct.

Alternatively, we can solve the problem by substituting a number for K. For simplicity, let's assume that K = 100 cc.

Therefore, K = 100 cc = = 9/10 - 1/3 = 27/30 - 10/30 = 17/30. We can find the total capacity of the sugar bowl (X) by using a proportion:

100/X = 17/30
17X = (30)(100)
X = (30)(100)/17 = 30K/17. Once again, we confirm that Choice E is correct.

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