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A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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29 Jun 2017, 04:21
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A
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D
E
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35% (medium)
Question Stats:
72% (01:29) correct 28% (01:47) wrong based on 292 sessions
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A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
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29 Jun 2017, 04:26
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Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
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29 Jun 2017, 05:06
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Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
Let x be the female members and y the male members.
Since we are given that -> the average number of raffle tickets sold among female members was 84 -> the average number of raffle tickets sold among female members was 96
From the question stem we can deduce the following : \(\frac{(84x + 96y)}{(x+y)} = 92\) \(84x + 96y = 92x + 92y\) \(4y = 8x\) \(\frac{y}{x} = \frac{2}{1}\)
Female members : Male members are in ratio 2:1(Option D) _________________
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A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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29 Jun 2017, 06:20
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The average of 92 is closer to 96 (male) than 84 (female) So there is more weightage on male number than female number which pulls the total average more towards Male average, so answer has to be D or E
using alligation: M:F = \(\frac{(Total average - Female average)}{(Male average - Total Average)}\) = \(\frac{(92-84)}{(96-92)}\) =\(\frac{8}{4}\) = \(\frac{2}{1}\)
Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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29 Jun 2017, 09:10
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Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
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Untitled.jpg [ 11.92 KiB | Viewed 5399 times ]
Thus, the ratio will be 8 : 4 = 2 : 1 Answer will be (D) _________________
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Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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29 Jun 2017, 09:20
Abhishek009 wrote:
Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
Attachment:
Untitled.jpg
Thus, the ratio will be 8 : 4 = 2 : 1 Answer will be (D)
awesome bro. need to develop this kind of thinking to attack problems. although I know the concept, I almost fail in application when required.
Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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30 Jun 2017, 23:19
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D. 84f + 96m =92(f+m) Divide the eqn on both sides by f Let (m/f) be x 84 + 96x = 92 + 92x X = 2:1 (D)
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A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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02 Nov 2018, 11:23
pushpitkc wrote:
Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
Let x be the female members and y the male members.
Since we are given that -> the average number of raffle tickets sold among female members was 84 -> the average number of raffle tickets sold among female members was 96
From the question stem we can deduce the following : \(\frac{(84x + 96y)}{(x+y)} = 92\) \(84x + 96y = 92x + 92y\) \(4y = 8x\) \(\frac{y}{x} = \frac{2}{1}\)
Female members : Male members are in ratio 2:1(Option D)
pushpitkc if question is to find the ratio of the number of male membersto the number of female members in the club?
And if \(x\) the number the female members and \(y\) is the# of male members.
how is it possible that answer is D ?
it should be vice versa 1 : 2 based on this \(4y = 8x\) ad not 2:1
Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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03 Nov 2018, 17:56
Bunuel wrote:
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?
(A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
We are given that a club sold an average (arithmetic mean) of 92 raffle tickets per member, that among the female members, the average number sold was 84, and that among the male members, the average number sold was 96. We can let f = the number of females and m = the number of males, and we can create the following weighted average equation:
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72% (01:29) correct 
