Bunuel wrote:
How much greater is the square of the sum of three different positive integers than the sum of their squares?
(1) The sum of the products of all possible pairs of two different integers out of the original set of three is 61.
(2) The largest of the three integers, 7, is equal to the sum of the two smaller integers.
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:Translate this word problem into algebra. First, name the three integers x, y, and z (making x the smallest and z the largest integer, since they’re all different and it might be handy to keep track of their relative sizes).
The question refers to the square of the sum of the integers, or (x + y + z)^2. The question also refers to the sum of the squares of these integers, or x^2+y^2+z^2. Finally, you are asked “how much greater” the first quantity is than the second quantity. In other words, what is their difference:
(x + y + z)^2 – (x^2+y^2+z^2) = ?
Before going further, try to simplify this expression. Expand (x + y + z)^2:
(x + y + z)^2
= (x + y + z)(x + y + z)
= x^2+ xy + xz + yx + y^2 + yz + zx + zy + z^2
= x^2+ y^2+z^2+ 2xy + 2xz + 2yz
So the difference we’re looking for is just 2xy + 2xz + 2yz, or 2(xy + xz + yz). Ultimately, we can simplify the question to this:
xy + xz + yz = ?
Statement (1): SUFFICIENT. This statement tells you that the sum of every possible pair of different integers out of the set of x, y, and z is 61. In algebraic terms, you are told that xy + xz + yz = 61.
Statement (2): INSUFFICIENT: You know that z = 7 = x + y, so you can get rid of two variables in the target expression, as shown:
xy + xz + yz = x(7 – x) + 7x + 7(7 – x) = 7x – x^2 + 7x + 49 – 7x = 7x – x^2 + 49. You do know that x can be no smaller than 1 and no larger than 5 (to allow y to be 6, different from either x or z and between them both). However, these constraints are not enough to determine a single value for the difference you’re looking for.
The correct answer is A.
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