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 - 2y = -60 .... (ii)
Solving (i) and (ii) we get: x = 100 , y = 80.
The required answer A = 100.
Discussion:
75 comments Page 2 of 8.
Sylvester said:
5 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)
Sarvesh said:
6 years ago
Thank you @Naga.
(1)
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.
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)
Chandu said:
6 years ago
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.
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.
(1)
Swathi hj said:
6 years ago
Why we take x-10 and y+10, can anyone tell me?
(1)
Hello said:
6 years ago
Where is Ramesh? The one who explained the previous question very simply in this same chapter.
(1)
Suresh said:
5 years ago
In step 1 how you got 100?
(1)
Srinu said:
7 years ago
Nice explanation. Thanks.
Shiv Mohan Sharma said:
8 years ago
Explain how to solve equ.1 and equ.2?
Akshay said:
8 years ago
Nice explanation, Thanks @Manojkumar.
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