GMAT Hard Maths Question | GMAT Algebra

This is a very interesting GMAT sample question in algebra - linear equations word problem - and number properties and is about finding number of possible values an unknown can take, given one linear equation in two variables. Uses elementary number properties concept along with concepts of solving a system of linear equations in two variables. Such questions are spotted every now and then in the GMAT problem solving section. A 700 level GMAT linear equations problem.

Question 6: A children's gift store sells gift certificates in denominations of $3 and $5. The store sold 'm' $3 certificates and 'n' $5 certificates worth $93 on a Saturday afternoon. If 'm' and 'n' are natural numbers, how many different values can 'm' take?

1. 5
2. 7
3. 6
4. 31
5. 18

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Video Explanation

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Explanatory Answer

Step 1 of solving this GMAT Linear Equation Question: Understand Key Data

1. Total value of all certificates sold = $93.
2. Certificates sold were in denominations of $3 and $5.
3. Both 'm' and 'n' are natural numbers.

Approach to solve this GMAT Algebra Word Problem

The value of all certificates sold, 93 is divisible by 3.
So, a maximum of 31 $3 certificates and no $5 certificates could have been sold.
However, the question states that both 'm' and 'n' are natural numbers.
Hence, at least 1 $5 certificate should have been sold.

Let us reduce the number of $3 certificates from theoretical maximum count of 31 by say 'x' and correspondingly increase $5 certificates by 'y'.
Evidently, 3x = 5y because the value of $3 certificates reduced should be the same as the value of $5 certificates increased.

It means that x has to be a multiple of 5 and y has to be a multiple of 3.
Or $3 certificates reduce in steps of 5 certificates.

Step 2 of solving this GMAT Algebra Question: List down possible values for 'm' and 'n'

The following combinations are possible.

1. m = 26, n = 3
2. m = 21, n = 6
3. m = 16, n = 9
4. m = 11, n = 12
5. m = 6, n = 15
6. m = 1, n = 18

The question is "How many different values can 'm' take?"
Hence, m can take 6 values.

Choice C is the correct answer.

An alternative way to think of the same concept - Replacing five $3 certificates with three $5 certificates leads to no change in the overall value of certficates sold and gives us a new combination each time. We need to see how many such combinations are possible.