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lexis
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arjtryarjtry
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LM
Ba/Ga = 3/4

Bb/Gb = 4/5

(Ba+Bb)/(Ga+Gb) = 17/22

(3Ga-2)/(Ga-1) = 34/11

Ga = 12

What was excessive and what is your approach?

We do not need the information shows 17/22 ratio.
I got 12 as well.
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lexis


The given information seems to be excessive. We do not need all constrains to solve problem.

Ba/Ga = 3/4
Bb/Gb = 4/5
(Ba + Bb)/(Ga + Gb) = 17/22

Bb = Ba - 1
Gb = Ga - 2

(2Ba - 1)/(2Ga - 2) = 17/22
Ba = (3/4)Ga

(2*(3/4)*Ga - 1)/(2Ga - 2) = 17/22
22*2*(3/4)*Ga - 22 = 17*2*Ga -34
33Ga - 22 = 34Ga - 34
Ga = 12
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IgnitedMind
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E... Agreed

I solved by back-solving, just plug in values to meet the conditions from the stem, 12 is the obvious choice.


If 12 girls in A then 9 boys in A

If 9 boys in A then 8 boys in B

If 12 girls in A then 10 girls in B

9+8=17
12+10=22

ratio =17/22

Time take - 45 seconds (varies based on which answer you select to back-solve, some might initially select a wrong answer, but is should be easy to tell at an early stage that its wrong)
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The information has one excessive fact anyway, but with those answer choices, the problem is absolutely straightforward:

We're told that class B has two fewer girls than class A. From the answer choices, class B must have 6, 7, 8, 9 or 10 girls. In class B, the ratio of boys to girls is 4 to 5. This guarantees that the number of girls in class B is divisible by 5. So the answer must be E, 12, because only 12-2 = 10 gives a multiple of 5.

So with those answer choices, the rest of the information in the question is completely unnecessary. E. What's the source?



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