Manhattan GMAT Challenge Problem of the Week – 10 December 2012
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Operation F means “take the square root,” operation G means “multiply by constant c,” and operation H means “take the reciprocal.” For which value of c is the result of applying the three operations to any positive x the same for all of the possible orders in which the operations are applied?
The language of the question itself is convoluted. In simpler terms, it means this: imagine that you apply the 3 operations in some order to a positive number x. Say you do F first, then G, then H. You wind up with some other number—call it y. Now, if you apply the operations in a different order to x, you want to still get y—in fact, no matter which order you apply the operations, you want to get the same y. How do you get that outcome? By picking the right constant c, which affects what G does. You are asked for that special value of c.
There are 3 different operations, so they can appear in 3! = 6 different orders that need to be tested (FGH, FHG, etc.). That’s a lot of orders and operations to set up. What’s more efficient is to think about what operation G really is, think about what you want to happen, and scan the answer choices if necessary.
G means “multiply by constant c.” The answer choices may all seem “likely” possibilities (perhaps the ½ and –½ appear less likely, but the square root is the ½ power, so there’s a connection). You want the outcome to be the same when you apply the operations in any order. Well, simplify the question: what constant c would leave the outcome the same after multiplication? The answer is 1. In fact, this educated guess is the right answer to the problem! If G means “multiply by 1,” then it doesn’t change anything, so you can ignore it. Your only question then is to verify that doing F then H is the same as doing H then F. This is indeed the case: taking the square root, then the reciprocal, gives you 1/√x, or . Taking the reciprocal, then the square root, gives you √(1/x), which also equals . So the answer must be 1.
Taking an inspired guess saves you a bunch of time, but you can also write out all 6 cases if you really want to:
Case 1: F(G(H(x))) = √(c/x)
(By the way, be careful with how you interpret function notation. The case above applies H first, then G, then F. Always work your way out from the inside with “nested” functions.)
Case 2: F(H(G(x))) = √(1/cx)
Case 3: G(F(H(x))) = c √(1/x)
Case 4: G(H(F(x))) = c / √x
Case 5: H(F(G(x))) = 1/ √(cx)
Case 6: H(G(F(x))) = 1/ (c √x)
Setting all these cases equal, you can cancel out the that appears everywhere, leaving you with expressions in c that must all be the same:
√c = √(1/c) = c = 1/c
The only number for which all these expressions are the same is 1, as you can see by analyzing individual pairs. The equation c = 1/c is only true for 1 and for –1, while the equation √c = c is only true for 1 and 0.
The correct answer is E.
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