Aptitude - Simplification - Discussion

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.

In this a to b represents the equation x-10 = y+10.

In this b to a represents the equation x+20 = 2(y-20).

Solving 2 equation we get the answer.

First read the question 1 time.

A = x; B=y.

'A' sent 10 students to 'B' => x-10 = y+10.

We solve the equation as x-y = 20. It becomes equation 1.

Then 'B' sent 20 students to 'A' => x+20 = 2(y-20).

The number of students in 'A' is double the number of students in 'B'.

x+20 = 2(y-20).

We simply the equation as x+20 = 2y-40.

Bring 2y-40 to left side,

x-2y+20+40 = 0.

We solve the equation as x-2y = -60.......equation 2.

Now solve equation 1&2,

x-y = 20.

x-2y = -60.
_____________.

y = 80.

Now we substitute y = 80 in equation 1 or 2.

x-80 = 20.

x = 100.

First of all.

Let the no of students in room A = x; B=y.

A sent 10 students to b....i.e. x-10 = y+10.

Take y to left side and -10 to right side.

It becomes x-y = 20.....equation 1.

Then B sent 20 students to A....i.e. 2x+20 = 2(y-20).

In questions it is mentioned that after sending 20 students A have Double no of students.

That's why we are using 2 here now write equation as,

x+20 = 2(y-20).

x+20 = 2y-40.

Bring 2y-40 to left side,

x-2y+20+40 = 0.

We can write it as,

x-2y = -60.....equation 2.

Now solve equation 1&2,
x-y = 20.
x-2y = -60.
_____________
y = 80.

Now substitute y = 80 in equation 1 or 2.

x-80 = 20.

Therefore x = 100.

Hope you understand this @RAMA.

If you don't mine, somebody please explain this in clear cut. First of all I can't understand the question.