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A group consists of both men and women. The average (arithmetic mean)
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29 Aug 2018, 23:31
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74% (01:17) correct 26% (01:29) wrong based on 114 sessions

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A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2
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Beminhae
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A group consists of both men and women. The average (arithmetic mean)
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29 Aug 2018, 23:57
C. 2:1
(66x + 72y)/(x+y) = 70
66x + 72y =70x + 70y
66x-70x = 70y-72y
-4x=-2y
2:1

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Beminhae
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A group consists of both men and women. The average (arithmetic mean)
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29 Aug 2018, 23:58
C. 2:1
(66x + 72y)/(x+y) = 70
66x + 72y =70x + 70y
66x-70x = 70y-72y
-4x=-2y
2:1

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A group consists of both men and women. The average (arithmetic mean)
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30 Aug 2018, 00:05
Bunuel
A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2
Show more

Sum of heights of women=66*w
Sum of heights of men=72*m
Sum of heights of the group(men+women)= 70*(m+w)

To find:- w:m=?

Now, Sum of height of men+sum of height of women=Sum of heights of the group
or, 66w+72m=70*(m+w)
or, 2m=4w
Or, w:m=1:2

Ans. (B)
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A group consists of both men and women. The average (arithmetic mean)
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30 Aug 2018, 05:20
Bunuel
A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2
Show more

Using Allegation Method , one can solve this question in 20 sec...........
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A group consists of both men and women. The average (arithmetic mean)
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30 Aug 2018, 05:41
Bunuel
A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2
Show more
\(66w + 72m = 70m + 70w\)

Or, \(4w = 2m\)

Or, \(w : m = 1 : 2\), Answer must be (B)

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A group consists of both men and women. The average (arithmetic mean)
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30 Aug 2018, 06:08
Bunuel
A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2
Show more

let #number of males be M
let #number of females be F
let number of all people be X

average (arithmetic mean) height of the women is 66 inches \(\frac{X}{F} = 66\) --- > X= 66F (1)

average (arithmetic mean) height of the men is 72 inches \(\frac{X}{M}= 72\) ---- > X=72M (2)

average (arithmetic mean) height of all the people in the group is 70 inches \(\frac{X}{M+F} = 70\) (plug in X1 and X2 here :) )


\(\frac{66F+72M}{M+F} = 70\)

\(66F+72M = 70(M+F)\)

\(66F+72M = 70M+70F\)

\(2M=4F\)

\(1M=2F\)
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A group consists of both men and women. The average (arithmetic mean)
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30 Aug 2018, 06:11
selim
Bunuel
A group consists of both men and women. The average (arithmetic mean) height of the women is 66 inches, and the average (arithmetic mean) height of the men is 72 inches. If the average (arithmetic mean) height of all the people in the group is 70 inches, what is the ratio of women to men in the group?

(A) 1:1
(B) 1:2
(C) 2:1
(D) 2:3
(E) 3:2

Using Allegation Method , one can solve this question in 20 sec...........
Show more


thanks for reminding me alligator method :)
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