Aptitude - Simplification - Discussion

Discussion Forum : Simplification - General Questions (Q.No. 2)
2.
There are two examinations rooms A and B. If 10 students are sent from A to B, then the number of students in each room is the same. If 20 candidates are sent from B to A, then the number of students in A is double the number of students in B. The number of students in room A is:
20
80
100
200
Answer: Option
Explanation:

Let the number of students in rooms A and B be x and y respectively.

Then, x - 10 = y + 10      x - y = 20 .... (i)

     and x + 20 = 2(y - 20)      x - 2y = -60 .... (ii)

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

The required answer A = 100.

Discussion:
75 comments Page 3 of 8.

Rama said:   1 decade ago
If you don't mine, somebody please explain this in clear cut. First of all I can't understand the question.

Ash said:   1 decade ago
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.

Swetha said:   1 decade ago
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.

Devicky said:   1 decade ago
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.

Anup said:   1 decade ago
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.

Prashant said:   1 decade ago
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.

Prashant said:   1 decade ago
Assume maximum capacity of both side 100 and subtract 20 from it a:b = 100:80.

Manish Kumar said:   1 decade ago
I couldn't get how we solve both equation please make it clear by multiply.

Idhaa said:   10 years ago
2 has come because there are two rooms in the question.

Chandrasekhar said:   9 years ago
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.
(1)


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