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15 Jul 2026, 22:51
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
39% (02:12) correct
61%
(02:43)
wrong
based on 46
sessions
History
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Time
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Not Attempted Yet
A survey was performed on exchanging Christmas presents. Of the people surveyed, one eighth exchanged the most presents on December 24, five sixths exchanged the most on December 25, and 55 exchanged the most on January 6. In addition, exactly 22 percent of the people surveyed exchange presents on at least four days during Christmas. What is the smallest possible number of people surveyed?
A) 1,300
B) 1,320
C) 1,350
D) 1,400
E) 1,800
A) 1,300
B) 1,320
C) 1,350
D) 1,400
E) 1,800
17 Jul 2026, 10:34
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This question is mainly about the concept of least common multiple
18 Jul 2026, 01:18
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This is one of those "smallest possible number" questions where the real test is whether you notice there are two separate divisibility conditions hiding in the wording, not just one equation to solve.
1. Let N be the total surveyed. N/8 people had their most on Dec 24, 5N/6 had their most on Dec 25, and 55 had it on Jan 6.
2. For N/8 and 5N/6 to come out as whole people, N has to be divisible by both 8 and 6, so N must be a multiple of 24.
3. Separately, 22% of N also has to be a whole number of people. 22% is 11/50, and since 11 and 50 share no factors, N has to be a multiple of 50 too.
4. Put those together and N needs to be a multiple of the LCM of 24 and 50, which is 600.
5. Now check the constraint that the three named groups can't add up to more than the total: N/8 + 5N/6 + 55 ≤ N. Simplify and you get N/24 ≥ 55, so N ≥ 1,320.
6. The smallest multiple of 600 that clears 1,320 is 1,800, since 600 and 1,200 both fall short.
7. Check it: 1,800/8 = 225, 5(1,800)/6 = 1,500, plus 55 is 1,780, leaving 20 people for other days, and 22% of 1,800 is 396, a clean integer.
Answer is E.
I'll admit my first instinct was to just solve N/8 + 5N/6 + 55 = N assuming those three groups covered everybody, which spits out N = 1,320 and feels done. But 1,320 fails the 22% check completely (1,320 times 0.22 isn't close to a whole number), so B is a trap answer sitting right there for anyone who stops one step early.
One-liner: whenever a question mentions a percent AND a fraction of the same population, check both for divisibility before you touch the algebra.
1. Let N be the total surveyed. N/8 people had their most on Dec 24, 5N/6 had their most on Dec 25, and 55 had it on Jan 6.
2. For N/8 and 5N/6 to come out as whole people, N has to be divisible by both 8 and 6, so N must be a multiple of 24.
3. Separately, 22% of N also has to be a whole number of people. 22% is 11/50, and since 11 and 50 share no factors, N has to be a multiple of 50 too.
4. Put those together and N needs to be a multiple of the LCM of 24 and 50, which is 600.
5. Now check the constraint that the three named groups can't add up to more than the total: N/8 + 5N/6 + 55 ≤ N. Simplify and you get N/24 ≥ 55, so N ≥ 1,320.
6. The smallest multiple of 600 that clears 1,320 is 1,800, since 600 and 1,200 both fall short.
7. Check it: 1,800/8 = 225, 5(1,800)/6 = 1,500, plus 55 is 1,780, leaving 20 people for other days, and 22% of 1,800 is 396, a clean integer.
Answer is E.
I'll admit my first instinct was to just solve N/8 + 5N/6 + 55 = N assuming those three groups covered everybody, which spits out N = 1,320 and feels done. But 1,320 fails the 22% check completely (1,320 times 0.22 isn't close to a whole number), so B is a trap answer sitting right there for anyone who stops one step early.
One-liner: whenever a question mentions a percent AND a fraction of the same population, check both for divisibility before you touch the algebra.
