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x and y are positive integers, where (x + y)/3 = 36. What is [#permalink]

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06 May 2006, 10:43

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x and y are positive integers, where (x + y)/3 = 36. What is x?
(1) x > y
(2) | xy | > 32

The answer is explained below. My question is simple: How the heck do you manipulate (x + y)/3 = 36, to (x + y) = 12

(x+y)/3 = 36 ---> multiply both sides by 3 and x+y = 108.

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Principle - Missing the Constraint
In this problem, we are given a value for (x + y) where x and y are positive integers: (x + y)/3 = 36, which we manipulate to:(x+y) = 12

As we know that x and y are positive integers, there are a limited number of value for each of x and y: 1 + 11 where x = 1 and y = 11, 2 + 10 where x = 2 and y = 10 . . . up to 11 + 1 where x = 11 and y = 1.

(1) INSUFFICIENT. Statement 1 indicates that x is greater than y. This means that x could be 7, 8, 9, 10, or 11. Eliminate A and D.
(2) INSUFFICIENT. Statement 2 tells us that | xy | > 32.
This information restricts the values for x substantially, as seen in the following:
1 x 11 = 11
2 x 10 = 20
3 x 9 = 27
4 x 8 = 32
5 x 7 = 35
6 x 6 = 36
7 x 5 = 35
8 x 4 = 32, etc.
In order for statement 2 to be true, x could only be 5, 6, or 7. Eliminate B.

(1) + (2) SUFFICIENT. The only value for x in which x > y and | xy | > 32 is 7. This question utilized a hidden constraint on the number of values for x based on the information that x and y are positive integers and that x + y = 12.