McGraw-Hill Education GRE 2019

Plugging in Numbers with Fractions

In many fraction questions, you will be given unspecified amounts and asked to

solve for a relationship between these amounts. In these situations, the best strategy

is to choose values that satisfy the relationships in the question. When plugging in

numbers for fraction questions with no specified amounts, keep the following tips

in mind:

■ When choosing a value, choose a value for the whole. For example, if you

are told that^13 of the employees in a company are administrators and that^34 of

the administrators are men, then choose a value for the number of employees

(the whole), and not the administrators or men.

■ Choose a value for the whole that will be divisible by the denominators of

all the fractions in the question. For example, if you are told that^13 of the

employees in a company are administrators and that^34 of the administrators

are men, then the value you choose for the number of employees should be

divisible by 3 (the denominator of the first fraction) and 4 (the denominator

of the second fraction). Good numbers here would be multiples of 12. Why is

this strategy important? Because to the extent that it is possible, you want to

work with integers. If the number that you choose is not divisible by 3 or 4,

then the value for one of the parts will end up being a fraction, and the math

will almost certainly be messier.

Let’s look at an example:

Two-thirds of the cars in a lot are sedans, and the rest are trucks. If 109 of the

sedans are new and^38 of the trucks are new, what fraction of the cars in the

lot are used?

SOLUTION: Since the question does not provide any amounts, you should

choose numbers to determine values for the total number of cars and the

number of used cars. Based on the first tip provided earlier, you should

choose a value for the whole, which in this case is the total number of cars.

Based on the second tip, the value you choose should be divisible by 3, 8, and

10. The most obvious value here is 3 × 8 × 10 = 240. If there are 240 cars, then
    2
    3 (240) = 160 are sedans, and the other 80 are trucks. Now determine how
    many sedans and how many trucks are used. If 109 of the sedans are new, then
    the remaining 101 are used. 101 of 160 = 16, so there are 16 used sedans. If^38 of
    the trucks are new, then^58 are used.^58 of 80 = 50, so 50 of the trucks are used.
    In total, there are 50 + 16 = 66 used cars. The final fraction of used cars/total
    cars = 24066. Divide the numerator and denominator by 6 and arrive at^1140.

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