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19 Dec 2019, 02:21
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Difficulty:
75%
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Question Stats:
53% (02:12) correct
47%
(02:27)
wrong
based on 215
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Is the number of the boys in the class is more than the number of the girls?
(1) The Ratio of the number of boys to the total number of students is more than the ratio of the total number of the girls to the total number of boys.
(2) The ratio of the number of boys to the number of girls in the class is more than the ratio of the number of the girls to the number of the students in the class.
Are You Up For the Challenge: 700 Level Questions
(1) The Ratio of the number of boys to the total number of students is more than the ratio of the total number of the girls to the total number of boys.
(2) The ratio of the number of boys to the number of girls in the class is more than the ratio of the number of the girls to the number of the students in the class.
Are You Up For the Challenge: 700 Level Questions
19 Dec 2019, 03:31
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Let Boys - B, Girls - G and Total student - T. We need to find whether B > G?
(1) \(\frac{B}{T} > \frac{G}{B}\) - B will always be greater than G.
Take examples:
a. Let T-10 , B -7, G-3 .... \(\frac{7}{10} > \frac{3}{7}\) - Yes .. If we go below than this, we can get B > G but will not satisfy the statement 1. Check below
b. Let T-10 , B -6, G-4 .... \(\frac{6}{10} > \frac{4}{6}\) - No
A D /B C E
(2) \(\frac{B}{G} > \frac{G}{T}\) - INSUFFICIENT
Take examples:
a. Let T-10 , B-6, G-4 .... \(\frac{6}{4} > \frac{4}{10}\) - satisfy but Boys are more than Girls
b. Let T-10 , B-4, G-6 .... \(\frac{4}{6} > \frac{6}{10}\) - satisfy but Girls are more than Boys
'A' is the winner.
(1) \(\frac{B}{T} > \frac{G}{B}\) - B will always be greater than G.
Take examples:
a. Let T-10 , B -7, G-3 .... \(\frac{7}{10} > \frac{3}{7}\) - Yes .. If we go below than this, we can get B > G but will not satisfy the statement 1. Check below
b. Let T-10 , B -6, G-4 .... \(\frac{6}{10} > \frac{4}{6}\) - No
A D /
(2) \(\frac{B}{G} > \frac{G}{T}\) - INSUFFICIENT
Take examples:
a. Let T-10 , B-6, G-4 .... \(\frac{6}{4} > \frac{4}{10}\) - satisfy but Boys are more than Girls
b. Let T-10 , B-4, G-6 .... \(\frac{4}{6} > \frac{6}{10}\) - satisfy but Girls are more than Boys
'A' is the winner.
Bunuel
07 Feb 2020, 09:40
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Bunuel
total=b+g
b,g,total>0
(1) b/(b+g)>g/b sufic
\(b/(b+g)>g/b…b^2>gb+g^2…b^2-g^2>gb\)
\((gb>0)…b^2-g^2>0…b^>g^2…b>g\)
(2) b/g>g/(b+g) insufic
\(b/g>g/(b+g)…b^2+bg>g^2…b^2-g^2>-bg…b^2-g^2>negative\)
Ans (A)
