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A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
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28 Apr 2026, 01:45
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GMAT Club Official Solution:
A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
Let T be the total number of food trucks that actually applied. If 2 more food trucks had applied, the total would have been T + 2. So the number of possible 3-truck groups would have been:
(T + 2)C3 = 120
(T + 2)!/(3!(T - 1)!) = 120
(T + 2)(T + 1)T/6 = 120
There is no need to waste time trying to solve this equation algebraically. Notice that, as T increases, this expression increases. So there can be only one value of T that satisfies this equation. Therefore, statement (1) determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
If 2 fewer food trucks had applied, the total would have been T - 2. So the number of possible 3-truck groups would have been:
(T - 2)C3 = 20
(T - 2)! / [3!(T - 5)!] = 20
(T - 2)(T - 3)(T - 4) / 6 = 20
Again, as T increases, this expression increases. So there can be only one value of T that satisfies this equation. Therefore, statement (2) also determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
Answer: D.
A festival organizer is putting together the final set of food trucks for a weekend street-food event. Several food trucks applied, and exactly 3 of them will be chosen. How many different groups of 3 food trucks could be chosen?
(1) If 2 more food trucks had applied, the number of possible 3-truck groups would have been 120.
Let T be the total number of food trucks that actually applied. If 2 more food trucks had applied, the total would have been T + 2. So the number of possible 3-truck groups would have been:
(T + 2)C3 = 120
(T + 2)!/(3!(T - 1)!) = 120
(T + 2)(T + 1)T/6 = 120
There is no need to waste time trying to solve this equation algebraically. Notice that, as T increases, this expression increases. So there can be only one value of T that satisfies this equation. Therefore, statement (1) determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
(2) If 2 fewer food trucks had applied, the number of possible 3-truck groups would have been 20.
If 2 fewer food trucks had applied, the total would have been T - 2. So the number of possible 3-truck groups would have been:
(T - 2)C3 = 20
(T - 2)! / [3!(T - 5)!] = 20
(T - 2)(T - 3)(T - 4) / 6 = 20
Again, as T increases, this expression increases. So there can be only one value of T that satisfies this equation. Therefore, statement (2) also determines exactly one actual number of applicants, and hence exactly one value for the number of 3-truck groups in the original question.
Sufficient.
Answer: D.
General Discussion
20 Apr 2026, 23:38
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Statement 1:
Take nC3 = 120.
Solving for n and subtracting 2 we get our actual number of items.
SUFFICIENT.
Statement 2:
Same process nc3 = 20.
n+2 is our actual number of items.
SUFFICIENT.
Answer: Option D.
Take nC3 = 120.
Solving for n and subtracting 2 we get our actual number of items.
SUFFICIENT.
Statement 2:
Same process nc3 = 20.
n+2 is our actual number of items.
SUFFICIENT.
Answer: Option D.
Bunuel
