McGraw-Hill Education GRE 2019

Step 3: Solve for the unknown: you are asked to solve for the amount

of money allocated for marketing, so you must solve for 5x. Using the

preceding equation:

10 x = 200,000

divide both sides by 2: ↓

5 x = 100,000

Situation 2: When you do not have quantities for either part of a ratio, but have

information about how the ratio changes.

In a certain school, the ratio of students to teachers is 20 to 1. If the school

adds 1,000 students and 100 teachers, the new ratio of students to teachers

will be 15 to 1. How many students are currently at the school?

Step 1: Set up the original ratio with x representing one slice of the pie

original number of students

original number of teachers =

20 x

1 x

Step 2: Use the quantities from Step 1 to create a proportion

new number of students/new number of teachers =^151 =^201 xx + 1,000 + 100.

Step 3: Cross-multiply and solve for 20x (the original number of students)

151 =^201 xx + 1,000 + 100

15(x + 100) = 20x + 1,000

15 x + 1,500 = 20x + 1,000

5 x = 500

x = 100

20 x = 2,000

Multiple Ratios with a Common Element

On some GRE questions, you will be given two ratios that share a common part

and be asked to solve for the ratio of the other parts. The most efficient way to

answer these questions is to manipulate the two ratios so that the common element

has the same value in both. How will you do this? By finding the least common

multiple of the common element.

At a certain pet store, there are 3 dogs for every 2 cats and 5 cats for every 7

birds. What is the ratio of the number of dogs to the number of birds?

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