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28 Jul 2021, 05:45
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
59% (02:25) correct
41%
(02:32)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
The number of boys in a school was 30 more than the number of girls. Subsequently, a few more girls joined the same school. Consequently, the ratio of boys and girls became 3:5. What is the minimum number of girls, who joined subsequently ?
A 31
B 51
C 52
D 55
E 56
A 31
B 51
C 52
D 55
E 56
28 Jul 2021, 06:11
Kudos
Bookmarks
Bunuel
Let the number of boys and girls be b and g respectively.
Given: b = g + 30
Let the subsequent joiners be a
So, \(\frac{b}{g+a} = \frac{3}{5}\)
\(\frac{g + 30}{g + a} = \frac{3}{5}\)
150 + 5g = 3g + 3a
150 + 2g = 3a
\(a = 50 + \frac{2}{3} g\)
The smallest possible value for g = 3. So, smallest possible value for a = 50 + 2 = 52. IMO, (C)!
28 Jul 2021, 06:45
Kudos
Bookmarks
Bunuel
I don't understand the question. The number of boys doesn't change, and if we start with 30 more boys than girls, then to get a 5 to 3 ratio of girls to boys, we need to first add 30 girls to make their number equal, then add 5/3 girls for each boy to get to a 5 to 3 ratio. So the fewer boys we have, the fewer girls we need to add. But the smallest number of boys we can have here is 30. Then 50 girls would need to join, so that's the right answer. But 50 isn't among the answer choices.
If we knew there was a nonzero number of girls to start with, then since the boys to girls ratio ends up being 3 to 5, the number of boys (which doesn't change) must be divisible by 3, and since we must have more than 30 boys if there is more than 0 girls, the minimum number of boys is then 33, in which case we start with 3 girls and end up with 55, for a difference of 52.
