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A group consists of both men and women. The average (arithmetic mean)

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Bunuel
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A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 00:31
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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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B
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Beminhae
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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 00:57
C. 2:1
(66x + 72y)/(x+y) = 70
66x + 72y =70x + 70y
66x-70x = 70y-72y
-4x=-2y
2:1

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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 00:58
C. 2:1
(66x + 72y)/(x+y) = 70
66x + 72y =70x + 70y
66x-70x = 70y-72y
-4x=-2y
2:1

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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 01:05
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Bunuel wrote:
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


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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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 06:20
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Bunuel wrote:
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...........
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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 06:41
Bunuel wrote:
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

\(66w + 72m = 70m + 70w\)

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) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 07:08
Bunuel wrote:
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


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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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink] A group consists of both men and women. The average (arithmetic mean)   30 Aug 2018, 07:11
selim wrote:
Bunuel wrote:
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...........



thanks for reminding me alligator method :)
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Re: A group consists of both men and women. The average (arithmetic mean) [#permalink]
30 Aug 2018, 07:11
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