Aptitude - Simplification - Discussion

We can find out the answer by option if the 1st option is 80. Consider the student of class B and transfer 10 students then in both the class are equal so definitely, A class has a 100 student then tally it.

Exactly I have the same doubt as same as @Nandhakumar.

Can anyone help us?

If A is double then, Why 2 (y - 20)?

It is A which is double. So the equation should be 2 (x + 20) = y - 20. Right?

Friends you are doing upto x and why values. But we have to double the A also comparing to B.

2 has come because there are two rooms in the question.

If 10 students are sent from A to B,then the number of students in each room is the same
a-10 = b+10 ; ---- (1)

If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B.
a+20 = 2(b -20); ---- (2) ( suppose A=10,B=5 that means A=2B then 10=10 )

Equation 1

a-10 = b+10.
a-b = 20.

Equation 2

a+20 = 2(b-20)
a+20 = 2b -40
a-2b = -60.


Solving equation 1 and 2
a-b = 20
a-2b = -60 (-a+2b = +60)
_____________.
b = 80

b is substitute in equation 1
a - b = 20
a = 20 + b
a = 20 + 80
a = 100

The answer is a = 100

Assume maximum capacity of both side 100 and subtract 20 from it a:b = 100:80.

It is very simple assume maximum capacity of seats in both room is 100, now A:B = 100:80.

There is no need to think about 1st equation or second just do maximum-lesser no. of side.

Everyone is going directly from solve equation 1&2,

x-y = 20.
x-2y = -60.

And result is y = 80 and x = 100.

No one explain how to solve this. What is the best formula to solve this. If you have no idea how to solve those equation don't write the same word again and again in comments here.