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In a group containing only men and women
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25 Mar 2025, 02:53
25 Mar 2025, 02:53
Expert Reply
00:00
Question Stats:
0% (00:00) correct
100% (01:19) wrong based on 1 sessions
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In a group containing only men and women, $\(\frac{1}{3}\)$ of the total people are men. If 2 women leave the group, the number of men would become $\(\frac{2}{5}\)$ of the remaining people. How many people were present originally in the group?
(A) 8
(B) 10
(C) 12
(D) 15
(E) 25
(A) 8
(B) 10
(C) 12
(D) 15
(E) 25
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Re: In a group containing only men and women
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29 Mar 2025, 15:21
29 Mar 2025, 15:21
1
Let number of men be x and number of women y
Total number of people = x + y
Men x + y/3
Two women leave, total becomes x + y - 2
Number of men
2/5 ( x + y - 2 ) = x + y/3
6x + 6y - 12 = 5x + 5y
x + y = 12.
Total number of people = x + y
Men x + y/3
Two women leave, total becomes x + y - 2
Number of men
2/5 ( x + y - 2 ) = x + y/3
6x + 6y - 12 = 5x + 5y
x + y = 12.
Re: In a group containing only men and women
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31 Mar 2025, 04:00
31 Mar 2025, 04:00
Expert Reply
M=2/5 T
Picking one answer choice multiple of 5 is 15
So m would 6
6=1/3T (of the original number
T=18
So T is at least 18
E is the answer
Picking one answer choice multiple of 5 is 15
So m would 6
6=1/3T (of the original number
T=18
So T is at least 18
E is the answer
