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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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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? (A) 1 : 1 (B) 1 : 2 (C) 1 : 3 (D) 2 : 1 (E) 3 : 1
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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, 04:26
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 Male : Average : Female 96 : 92 : 84 9284: 9692 8:4 2:1 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, 05:06
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
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 = \(\frac{(Total average  Female average)}{(Male average  Total Average)}\) = \(\frac{(9284)}{(9692)}\) =\(\frac{8}{4}\) = \(\frac{2}{1}\)
Answer 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, 09:10
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 [ 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.



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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, 20:54
Simple put nos In given option and you will be able to eliminate wrong ones and get the correct one as 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, 21:49
\(M:F = (9284) : (9692) = 8:4 = 2:1\). Ans  D.
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Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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30 Jun 2017, 23:19
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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Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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02 Nov 2018, 07:07
shouldn't the answer be 1:2, as it has asked about ratio of males to females?



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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 members to 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 can you explain



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Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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02 Nov 2018, 12:02
dave13, Female is X and male is Y in his solution... So male to female will be y:x which is what is used. 2:1 I think you're mixing the variables up. Try solving again from scratch with m & f Best G
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A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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Updated on: 02 Nov 2018, 12:49
Hi dave13This can be solved in a way similar to mixture questions using allegation method male = x Female = y Using allegation method, here the desired is the average Desired  y/x  desired So 92  84 = 8 96  92 = 4 x/y = 8/4 = 2/1 Refer to the link below https://gmatclub.com/forum/tipsandtri ... 51906.htmlPosted from my mobile device
Originally posted by Salsanousi on 02 Nov 2018, 12:17.
Last edited by Salsanousi on 02 Nov 2018, 12:49, edited 1 time in total.



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A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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02 Nov 2018, 12:39
priyamba wrote: shouldn't the answer be 1:2, as it has asked about ratio of males to females? priyamba , I got it thanks to great people on GC Female is X and male is Y in his solution... So male to female will be y:x which is what is used. 2:1 So when we get 4Y=8X to find ratio of male to female solve for Y > i.e. Y = \(\frac{8x}{4}\) > y = 2x hence ratio of males to females is 2x to X i.e. 2:1 thanks guys Salsanousi Gladiator59



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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: 92 = (84f + 96m)/(m + f) 92m + 92f = 84f + 96m 8f = 4m 2f = m 2 = m/f Answer: D
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Re: A club sold an average (arithmetic mean) of 92 raffle tickets per memb
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