GRE Math Challenge: what is the smallest number in terms of w and n ?
Q: The largest number in a series of consecutive even integers is w. If the number of integers is n, what is the smallest number in terms of w and n?
A) w – 2n
B) w – n + 1
C) w– 2(n– 1)
D) n – 6 + w
E) w- (n/2)
Explanation:
Correct Choice: (C)
This is a pretty difficult number properties question that asks us to represent an unknown number in terms of two variables. Solving this question conceptually is possible; since w is the largest even number in this sequence, each number must be two less than the previous. The next number would be w — 2, followed by w — 4, and so on. However, using the Picking Numbers strategy will be a much better option. Since the stem specifies a set of consecutive even numbers, let's choose {8, 10}. Therefore w = 10, n = 2, and our target number is 8:
Choice (A): w– 2n = 10 – 2(2) = 10 – 4 = 6 ≠ 8
Choice (B): w–n + 1 =10 – 2 +1 = 9 ≠ 8
Choice (C): w – 2(n– 1) = 10 – 2(2 – 1) = 10 –2 = 8 = 8
Choice (D): n– 6 + w = 2 – 6 + 10 = 6 ≠ 8
Choice (E): w - (n/2) = 10 - (2/2) = 10 - 1 = 9 ≠ 8
Only correct choice (C) matches our target number. Therefore, the correct answer is (C).
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