# 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:

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* - 2*y* = -60 .... (ii)

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

The required answer A = 100.

Discussion:

74 comments Page 1 of 8.
Nonsocial said:
7 months ago

The students first sent from room A to B means 10 students sent from A to B so the number is equal between them just multiply;

i.e 10 × 10 = 100.

Then it's said from B to A they sent 20 students and A got double that B.

i.e 100 - 20 = 80.

i.e 10 × 10 = 100.

Then it's said from B to A they sent 20 students and A got double that B.

i.e 100 - 20 = 80.

(5)

JOEPAUL said:
2 years ago

Then the number of students in A is double the number of students in B. This is because 2 (y-20).

(1)

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.

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.

(7)

Jeet dutta said:
2 years ago

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

The required answer A = 100.

The required answer A = 100.

(2)

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.

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)

Jayashri said:
4 years ago

I disagree, if A is double the number of B.

How can 100 be the answer?

As B is 60 if we double it we get 160.

How can 100 be the answer?

As B is 60 if we double it we get 160.

(4)

Sylvester said:
4 years ago

If 2A=B, Also A=2B.

A=80, B=40, in other words: A=40*2=80.

A=80, B=40, in other words: A=40*2=80.

(2)

Alpesh Patel said:
4 years ago

x-y=20

x=20+y ---> (1)

x-2y=-60 ---> (2)

20+y-2y=-60 --->(3)

+y-2y=-60-20

-y=-80

y=80

Sub value of y in eqn 1, we get;

x=20+80 = 100.

x=20+y ---> (1)

x-2y=-60 ---> (2)

20+y-2y=-60 --->(3)

+y-2y=-60-20

-y=-80

y=80

Sub value of y in eqn 1, we get;

x=20+80 = 100.

(6)

Pratham said:
4 years ago

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

(2)

Suresh said:
4 years ago

In step 1 how you got 100?

(1)

Post your comments here:

Quick links

Quantitative Aptitude

Verbal (English)

Reasoning

Programming

Interview

Placement Papers