# 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
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:
74 comments Page 1 of 8.

Naga Invy said:   5 years ago
If 10 students sent from room A to B; then;

step 1: A -10 = B +10
Step 2 : If you change the B and -10 vice versa then the equation will be
A - B = 10 +10
therefore (A - B = 20) -----------------> Equation (1)

If 20 students sent from room B to A; then A becomes double as size of B
Step 3 : A + 20 = 2 (B - 20)
Please understand we are adding 2 in front of (B -20) because question states A becomes double as size B.

Now solve the step 3;
Step 4 : A + 20 = 2B -40
By changing the values vice versa, it becomes
Step 5 : A - 2B = (-20-40)
Step 6 A -2B = -60 -------------------> Equation (2)

Now solve both Equation 1 and 2
A- B = 20 ---------------> (1)
A- 2B = -60 ---------------> (2)

By changing the signs of equation 2, it becomes
A - B = 20 ---------------> (1)
- A + 2B = + 60 ---------------> (2)
----------------------
B = 80
---------------------
Solve the B value in equation ------> 1 to get an answer

A - B= 20.
then,
A - 80 = 20.
A = 20+80.
(2)

Ash said:   10 years 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.

Reagan said:   2 years ago
When 10 are sent from A to B. A loses and B gains.

A -10 = B+10...
= A - B = 20---> 1

When 20 are sent from B to A. A gains and B loses. But A gains twice as much as B has. Therefore A = 2B.
A+20 = 2(B-20)...
=> A-2B = -40-20 = A-2B=-60 ---> 2

Use elimination to solve equations a and b simultaneously
A-B = 20 (NOTE: When 2 -s meet, they become +)
-A-2B=-60
................
B = 80 (from 20--60 = 20+60)

Substitute B= 80 into eqtn 1
A - 80 = 20
A= 20 + 80 = 100.

Therefore, A =100, and B = 80.

If B has 80 students and 20 are taken to A which has 100, we get:
A+20 = B-20.
100+20 = 80-20.
120 = 60.
Therefore, room A has twice than B.
(11)

Manojkumar said:   8 years ago
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.

Koteswararao chimmili said:   7 years ago
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

Swetha said:   9 years 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.

x-y = 20 ------------(1).
x-2y = -60 ------------(2).

Solving (1) - (2),

x -y = 20----(1).
x -2y = -60----(2).
---------------
y = 80.
---------------
(If you want to solve two equations in any problem you must change sing of (2)nd equation without fail.)

Now you substitute the y value in (1)st or (2)nd equations as you like then you will get the value of x as 100.

x - y = 20-----(1).
y = 80.

so, x - 80 = 20.
x = 20+80.
x = 100.

Now you got it @Selvi.

Take these two cases individually let us look first through options 1st option is 20 so to apply both conditions is not possible so come to 80 so given if 10 people are sent to b then people in both class rooms are same so definetly b will be 60.

So again now apply second condition individually so by sending 20 people 4rm b A is twice that of B so this condition is not satisfied so now take 100 and check then both conditions satisfy and hence ans is 100

Ramesh pai said:   4 years ago
Let's c option C i.e. 100.
A exam hall ----------- B exam hall
100 ----------- 80
10 from A to B = 90 | 90.

1st condition satisfies:

Let's see the 2nd condition:
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 exam hall ----------- B exam hall
100 ----------- 80.

20from B to A = 120(100+20) | 60(80-20).

Therefore 100 is the right answer.
(3)

10 students moving from A to B

( decrease 10 from A & add 10 to B)
Then, A - 10 = B + 10 A - B = 20 .... (i)

20 Students moving B to A.(Note: After deducting 20 from B , A will be double) .

A + 20 = 2(B - 20)

A - 2B = -60 .... (ii)

====>

Solving (i) and (ii) we get: A = 100 , B = 80.

The required answer A = 100.