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23 Jan 2023, 09:57
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:45) correct
32%
(02:43)
wrong
based on 195
sessions
History
Date
Time
Result
Not Attempted Yet
On a certain tour, half of the men are married, and the wives of 9/10 of them are also on the tour. If there are 3 more women on the tour than men, what is the least possible number of people who could be on the tour?
(A) 19
(B) 22
(C) 23
(D) 43
(E) 52
(A) 19
(B) 22
(C) 23
(D) 43
(E) 52
24 Jan 2023, 22:55
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Bunuel
We can represent the information in a tree diagram as shown below
Let the number of men in the tour = 2x
Let the number of women (excluding the wives) = y
Total women on the tour = \(\frac{9}{10}x + y \)
Total men on the tour = \(2x\)
Given
\(\frac{9}{10}x + y - 2x = 3\)
Solving further
\(10 y = 30 + 11x\)
\(y = 3 + \frac{11}{10} x \)
Note: x must be a multiple of 10 as y should be an integer
The minimum possible value of x = 10
At x = 10,
\(y = 3 + 11 = 14\)
Minimum number of people who could be on the tour = \(2x + \frac{9}{10} x + y\)
\(= 20 + 9 + 14 = 43\)
Option D
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23 Dec 2025, 16:24
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