McGraw-Hill Education GRE 2019

Step 4: Solve for the unknown. At this point, you have two algebraic

relationships:

l + s = 70

l = s + 20

Since the question asked for the length of the shorter piece, the final step is to

use substitution to solve for s. Substitute (s + 20) for l in the first equation: (s

+ 20) + s = 70. Solve for s:

2 s + 20 = 70

2 s = 50

s = 25

Price and Quantity Relationships

Some word problems will require you to recognize the following relationship:

(price/unit) × (number of units) = total price

Though it might be intimidating, this is a translation that most people use every

day. To illustrate this, look at the following example:

If Bob purchased 15 $30 shirts, how much money did he pay for all the

shirts?

SOLUTION: Plug the values into the given formula. The price per shirt is $30,

and the number of shirts is 15. Thus the total price is 30 × 15 = $450.

Now look at a more GRE-like example:

Bob spends a total of $140 at a certain shop, where he purchases a total of 20

shirts and ties. If the price of each shirt is $10, and the price of each tie is $5,

how many shirts does he buy?

Step 1: Assign variables. Since you are not given a value for the number of

shirts or the number of ties, let s = the number of shirts and t = the number

of ties.

Step 2: Identify relationships. There are two relationships in the question:

Relationship 1: Bob purchases a total of 20 shirts and ties. Thus t + s = 20.

Relationship 2: The total amount Bob pays for the shirts and ties is

$14 0. This amount will be the sum of the amount he paid for the shirts

and the amount he paid for the ties. If each shirt costs $10, then he paid

10 s for all the shirts. If each tie costs $5, then he paid 5t for all the ties.

The algebraic relationship will be 10s + 5t = 14 0.

CHAPTER 12 ■ FROM WORDS TO ALGEBRA 315

04-GRE-Test-2018_313-462.indd 315 12/05/17 12:03 pm