GMAT Tips from Veritas Prep

Reduce, Reuse, Recycle—Part III

Reduce, Reuse, Recycle—Part III

Photograph by Ty Wright/Bloomberg


Before you start reading, make sure to read Part I and Part II of this series.

To recycle is to take a product and break it down and use the underlying essence (aluminum, paper, plastic, etc.) to create another product. Recycling on the GMAT means that when you are doing practice problems you always break them down to their essence so that if you see a similar (but not identical) problem on test day, you can apply the method or technique to the new problem.

Problem-Solving Recycling
Let’s look at an example of two problems, one that will help you to learn a technique and another where you will apply the technique.

Problem 1
If and x and y are positive integers and 260x = 390y, what is the minimum ratio of x:y?
A. 13:30
B. 2:3
C. 3:2
D. 3:1
E. 30:13

To solve this problem, you clearly need to find the smallest values of x and y that will balance the equation. To do this, reduce the coefficients in front of each variable to the lowest possible ratio. The variables will then equal the opposite coefficient. When reducing a ratio, it is often helpful to reduce numbers to their prime factors. In this case you have (2*2*5*13)x = (2*3*5*13)y. The greatest common factor of these numbers is 2*5*13 or 130. So the new equation is 2x = 3y. To balance the equation, x and y must be in the ratio of 3:2. So the answer is C.

Recycle the technique of balancing the equations using prime factors as you approach this problem from the Veritas Prep Arithmetic book:

Problem 2
If 375y = x² and x and y are positive integers, then which of the following must be an integer?
I   y/15
II   y/30
III   y²/ 25
A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III only

At first glance this problem may appear to be very different from problem 1. In this problem you have an exponent for one of the variables, and that variable does not have a coefficient (other than the implied coefficient of 1). The answer choices are in fractions and not in ratios. However, a closer look shows that the problems have the essential facts in common. They are both equations that need to be balanced and they both feature positive integers. Recycle the technique from problem 1 as you approach problem 2.

Begin by reducing 375 to its prime factors. The prime factors of 375 are 3*5*5*5. In other words 3*5³. So the equation becomes (3*5³)y = x². The problem with this coefficient for y is that 375 is not a perfect square and it cannot equal x-squared. Since x and y are integers, the right-hand side of the equation will be a perfect square. Therefore the variable y must be a number that when multiplied by 375 creates a perfect square on the left side as well. If y = 15 (or 3*5), then the left-hand side of the equation becomes 3254, which is a perfect square. Therefore, the smallest number that y can be is 15.

Evaluating the options above: Option I must be an integer since 15/15 is an integer. Option 2 is not necessarily an integer since 15/30 is not an integer and while y can be larger than 15 it is clear that this option is not one that must result in an integer. Option 3 must be an integer since y = 3*5 and y²= 3²*5² which means that y²/ 25 must result in an integer. The correct answer is D. You recycled the technique of using prime factors to help balance an equation!

You do still need the original 3 R’s on the GMAT—reading, writing, and ‘rithmetic—but you can definitely count on Reduce, Reuse, and Recycle to lessen your burden on test day!


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