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Raffle tickets and ratios

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by LulaBrazilia » Thu Dec 19, 2013 8:34 am
A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, the average number sold was 84, and among the male members, the average 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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Source: — Problem Solving |

by Patrick_GMATFix » Thu Dec 19, 2013 8:46 am
This solution is taken from the GMATFix App. My signature below has more info.

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by theCodeToGMAT » Thu Dec 19, 2013 8:52 am
Total Average = 92

Female,F = 84
Male, M= 96

96(M) + 84(F) = 92(M+F)
4M = 8F
M : F = 2 : 1

[spoiler]{D}[/spoiler]
R A H U L
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by GMATGuruNY » Fri Dec 20, 2013 5:50 am
LulaBrazilia wrote:A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, the average number sold was 84, and among the male members, the average 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
Average for the men: 96.
Average for the women: 84.
Average for the MIXTURE of men and women: 92.

The following approach is called ALLIGATION -- a very efficient way to handle MIXTURE PROBLEMS.

Step 1: Plot the 3 averages on a number line, with averages for the men and women on the ends and the average for the mixture in the middle.
M 96-----------92-----------84 W

Step 2: Calculate the distances between the averages.
M 96-----4-----92----8-----84 W

Step 3: Determine the ratio in the mixture.
The required ratio of men to women is equal to the RECIPROCAL of the distances in red.
M:W = 8:4 = 2:1.

The correct answer is D.

An alternate approach is the PLUG IN THE ANSWERS, which represent the ratio of men to women.
Since the average for the mixture (92) is closer to the average for the men (96) than to the average for the women (84), there must be MORE MEN than women.
Eliminate A, B and C.
Of D and E, the correct ratio must yield an average of 92 tickets per member.

Answer choice D: 2:1
Total sales for 2 men and 1 woman = 2*96 + 1*84 = 276.
Average sales per member = 276/3 = 92.
Success!

The correct answer is D.

For two similar problems, check here:

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by Mathsbuddy » Fri Dec 20, 2013 6:25 am
Total sold:
84F + 96M = 92(F + M)
So 4M = 8F
i.e. M = 2F

So M:F = 2:1
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hi

by Scott@TargetTestPrep » Mon Dec 18, 2017 7:54 am
LulaBrazilia wrote:A club sold an average (arithmetic mean) of 92 raffle tickets per member. Among the female members, the average number sold was 84, and among the male members, the average 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

2/1= m/f

Answer: D

Scott Woodbury-Stewart
Founder and CEO
[email protected]

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