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Bunuel
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Seems like an interesting question but I need an alternate method to solve it. I was using fractions but obviously cannot solve it algebraically. Any help appreciated here.
I will post my solution here,
You can try and understand this,

In village A
M:F = 4:5
M:C = 3:2

Let Males = 12y
So F = 15y
and C = 8y

Similarly in village B
Let females = 5x
So males =20x and Children = 2x

Now Chidlren in both villages are equal so we get 2x = 8y, x= 4y

So males in village B are 80y

Now check the options, option A is 10 means y = 1/8, which is not possible because females on village B will become fraction. Check for all options in similar fashion. Answer is D.­
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Given

MA- Males in A
MB- Males in B

FA- Females in A
FB- Females in B

CA- Children A
CB- Children B

Using the above to make life simpler!!

Given things:
MA/FA= 4/5---(1)
MA/CA=3/2---(2)
MB=4FB---(3)
CB/FB=2/5---(4)

We need a value for MB

Using eq (1) and (2)
We get CA/FA=8/15--->CA=8FA/15
From 4 we have CB=2FB/5
Since CA=CB
8FA/15=2FB/5--->FA=3FB/4
Substituting MB/4 instead of FB, we have
FA=(3*MB)/16

NOW!! In these type of questions, we now have to plug in answer choice which will give us an integer
Only when MB=80, We will have an Integer value for FA.
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