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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:

(1) SUFFICIENT: This statement yields three possibilities for A, B, and C, in some order: {1, 3, 5}, {3, 5, 7}, {5, 7, 9}; it makes no difference which of A, B, and C is which digit, because they are all added together.
• If A, B, and C are 1, 3, and 5 (in whatever order), then the digit E must be 1. This means the digits aren’t distinct, so it’s impossible.
• If A, B, and C are 3, 5, and 7 (in whatever order), then E = 1. That means D is one of the leftover digits 0, 2, 4, 6, 8, 9. If these values are plugged into D one at a time, the only one that yields all distinct digits is D = 4 (34 + 54 + 74 = 162).
• If A, B, and C are 5, 7, and 9 (in whatever order), then E = 2. That means D is one of the leftover digits 0, 1, 3, 4, 6, 8. If these values are plugged into D one at a time, the only one that yields all distinct digits is D = 8 (58 + 78 + 98 = 234).
• Therefore, there are only two possible values of EFG: 162 and 234. In both cases, the sum of E, F, and G is 9, so (1) is sufficient.
(2) INSUFFICIENT: If E = 2, then trial and error will yield a number of working sets of digits yielding different values for EFG. Examples: 58 + 68 + 78 = 204 (E + F + G = 6); 41 + 71 + 91 = 203 (E + F + G = 5); 49 + 69 + 89 = 207 (E + F + G = 9); and several others. Therefore, (2) is insufficient.
The correct answer is A; statement (1) alone is sufficient.

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