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Question

A shop sells two variants of chocolates - one that costs $3 and the other that costs $5. If the shop sold $108 chocolates on a given day, how many different combinations of (number of $3 sold, number of $5 sold) exist?

A. 6

B. 7

C. 15

D. 8

E. 12

Correct Answer : Choice D. 8

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Explanatory Answer

Let the shop sell x numbers of the $3variant and y numbers of the $5 variant.

So, 3x + 5y = 108.

The only constraint to keep in mind is that both x and y are non-negative integers.

We can rewrite the equation as x = (108 - 5y)/3

So, (108 - 5y) should be divisible by 3.

108 is divisible by 3. So, we need to find such values for y that will make 5y divisible by 3.

Or in other words y should be a multiple of 3.

The values that y can take such that x does not become negative are 0, 3, 6, 9, 12, 15, 18, and 21.

So, there are 8 different combinations (36, 0), (31, 3), (26, 6), (21, 9), (16, 12), (11, 15), (6, 18) and (1, 21).

Labels: GMAT Linear Equations, GMAT Number Properties, GMAT Number Theory, GMAT Numbers, GMAT Problem Solving, GMAT Problem Solving Practice