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nislam
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At the morning session of an academic conference, the ratio of the num Updated on: 16 Nov 2021, 09:10
Originally posted by nislam on 16 Nov 2021, 00:39.
Last edited by nislam on 16 Nov 2021, 09:10, edited 1 time in total.
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Difficulty:

85% (hard)

Question Stats:

58% (02:56) correct 43% (02:44) wrong based on 80 sessions

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At the morning session of an academic conference, the ratio of the number of men in attendance to the number of women in attendance was 11 to 7. If, during the afternoon session, the number of men and the number of women in attendance were each 5 greater than in the morning, which of the following could NOT be the ratio of the number of men to the number of women in attendance during the afternoon session?

I) 3 to 2

II) 5 to 3

III) 31 to 20

A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
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Official Answer
D
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chetan2u
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At the morning session of an academic conference, the ratio of the num 16 Nov 2021, 02:25
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nislam
At the morning session of an academic conference, the ratio of the number of men in attendance to the number of women in attendance was 11 to 7. If, during the afternoon session, the number of men and the number of women in attendance were each 5 greater than in the morning, which of the following could NOT be the ratio of the number of men to the number of women in attendance during the afternoon session?

I) 3 to 2

II) 5 to 3

III) 31 to 20

A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III
Show more


Let the common ratio be x, then the number \(m:w = 11x:7x\)
The changed ratio = 11x+5:7x+5

I) 3 to 2
\(\frac{11x+5}{7x+5}=\frac{3}{2}\)
\(22x+10=21x+15........x=5\)
Possible

II) 5 to 3
\(\frac{11x+5}{7x+5}=\frac{5}{3}\)
\(33x+15=35x+25........2x=-10\)
x cannot be negative

III) 31 to 20
\(\frac{11x+5}{7x+5}=\frac{31}{20}\)
\(220x+100=217x+155........3x=55\)
x cannot be negative


Only II and III not possible


D
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Vishal120
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At the morning session of an academic conference, the ratio of the num 16 Nov 2021, 04:21
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chetan2u
nislam
At the morning session of an academic conference, the ratio of the number of men in attendance to the number of women in attendance was 11 to 7. If, during the afternoon session, the number of men and the number of women in attendance were each 5 greater than in the morning, which of the following could NOT be the ratio of the number of men to the number of women in attendance during the afternoon session?

I) 3 to 2

II) 5 to 3

III) 31 to 20

A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III


Let the common ratio be x, then the number \(m:w = 11x:7x\)
The changed ratio = 11x+5:7x+5

I) 3 to 2
\(\frac{11x+5}{7x+5}=\frac{3}{2}\)
\(22x+10=21x+15........x=5\)
Possible

II) 5 to 3
\(\frac{11x+5}{7x+5}=\frac{5}{3}\)
\(33x+15=35x+25........2x=-10\)
x cannot be negative

III) 31 to 20
\(\frac{11x+5}{7x+5}=\frac{31}{20}\)
\(220x+100=217x+155........3x=45.....x=9\)
Possible


Only II not possible


A
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last 3x= 55

Not possible

Posted from my mobile device
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Vishal120
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At the morning session of an academic conference, the ratio of the num 16 Nov 2021, 04:24
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Option D is possible

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chetan2u
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At the morning session of an academic conference, the ratio of the num 16 Nov 2021, 04:33
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Vishal120
chetan2u
nislam
At the morning session of an academic conference, the ratio of the number of men in attendance to the number of women in attendance was 11 to 7. If, during the afternoon session, the number of men and the number of women in attendance were each 5 greater than in the morning, which of the following could NOT be the ratio of the number of men to the number of women in attendance during the afternoon session?

I) 3 to 2

II) 5 to 3

III) 31 to 20

A) II only
B) III only
C) I and II only
D) II and III only
E) I, II and III


Let the common ratio be x, then the number \(m:w = 11x:7x\)
The changed ratio = 11x+5:7x+5

I) 3 to 2
\(\frac{11x+5}{7x+5}=\frac{3}{2}\)
\(22x+10=21x+15........x=5\)
Possible

II) 5 to 3
\(\frac{11x+5}{7x+5}=\frac{5}{3}\)
\(33x+15=35x+25........2x=-10\)
x cannot be negative

III) 31 to 20
\(\frac{11x+5}{7x+5}=\frac{31}{20}\)
\(220x+100=217x+155........3x=45.....x=9\)
Possible


Only II not possible


A
last 3x= 55

Not possible

Posted from my mobile device
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Thanks, silly error. I took it to be 45 rather than 55
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At the morning session of an academic conference, the ratio of the num 16 Mar 2025, 07:06
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