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:
76 comments Page 2 of 8.
Jeet dutta said:
4 years ago
By Solving (i) and (ii) we get: x = 100, y = 80.
The required answer A = 100.
The required answer A = 100.
(3)
Musa said:
2 years ago
Anyone, please help me to get the answer.
(3)
Manojkumar said:
9 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.
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.
(1)
Chandrasekhar said:
10 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)
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.
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.
(1)
Sarvesh said:
7 years ago
Thank you @Naga.
(1)
Chandu said:
7 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)
Suresh said:
6 years ago
In step 1 how you got 100?
(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)
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