Aptitude - Simplification - Discussion

By Solving (i) and (ii) we get: x = 100, y = 80.

The required answer A = 100.

Hello, can you please explain the last step i.e x=100 y=80?

I know everyone understood the first part, but in the second part, some might find difficult to understand why we are doubling in why part rather than the x part. Here is my view.

Read the question very very slowly.

Here they told, "If 20 candidates are sent from B to A, then the number of students in A is double the number in B". So x+20 = 2 (y - 20).

So for those who argue it should be 2 (x + 20) = y - 20, then the question would be below.

"If 20 candidates are sent from B to A, then the number of students in A is doubled (the remain details about B would have been ignored).

I hope it clear your doubts.

Where is Ramesh? The one who explained the previous question very simply in this same chapter.

We have to only subtract 10 students from A ie;A-10=B (OR) A=B+10.

But how do you add 10 to A sub 10 from B at a time that means 20 students are sending to B from A.

A-10=B,
B-20=2B=>B=20,
A-10=20,
A=30.

Thank you @Naga.

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.

Students sent from A to B = 10;

Then A-10 = B+10; ---- (i)

Students sent from b to a = 20; then,

B-20 = A+20; --- (ii).

A is double no.of students in B so,

2(B-20) = A+20.

Solve 1 & 2, we get,

A = 100.

Why we take x-10 and y+10, can anyone tell me?

In step 1 how you got 100?