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The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

Get concrete right away. Say n = 5. Then Z = {1, 3, 5, 7, 9}. Since n > k, pick a k that’s less than n—say, k = 3.

x = max value of the sum of k distinct members of Z. Thus, you must pick the 3 largest values (5, 7, and 9). Their sum is 21, so x = 21.

y = min value of the same sort of sum. So you must pick the 3 smallest values (1, 3, and 5). Their sum is 9, so y = 9.

x + y = 21 + 9 = 30. Now, given that the target number is 30, while n = 5 and k = 3, check the answer choices.

(A) kn = 3 × 5 = 15
(B) kn + k2 = 3 × 5 + 32 = 15 + 9 = 24
(C) kn + 2k2 = 3 × 5 + 2 × 32 = 15 + 18 = 33
(D) 2knk2 = 2 × 3 × 5 – 32 = 30 – 9 = 21
(E) 2kn = 2 × 3 × 5 = 30

Only (E) fits.

The proof that (E) always works is substantially harder on a conceptual level. The sum of the smallest k positive odds equals k2, while the sum of the k largest odds in this set (where the largest odd is 2n – 1) winds up being 2knk2. The k2 terms cancel, leaving 2kn.

However, you should recognize that the plug-numbers approach is faster and easier.