Aptitude - Simplification - Discussion
Discussion Forum : Simplification - General Questions (Q.No. 15)
15.
David gets on the elevator at the 11th floor of a building and rides up at the rate of 57 floors per minute. At the same time, Albert gets on an elevator at the 51st floor of the same building and rides down at the rate of 63 floors per minute. If they continue travelling at these rates, then at which floor will their paths cross ?
Answer: Option
Explanation:
Suppose their paths cross after x minutes.
Then, 11 + 57x = 51 - 63x 
120x = 40
| x = | 1 |
| 3 |
| Number of floors covered by David in (1/3) min. = |  |
1 | x 57 |  |
= 19. |
| 3 |
So, their paths cross at (11 +19) i.e., 30th floor.
Discussion:
28 comments Page 2 of 3.
DAg said:
2 years ago
Thanks for the answer @ Finara.
Anil Sah said:
5 years ago
Not understanding, Please anyone help me to get it.
Albert Einstein said:
7 years ago
Well said, thanks for your explanation @Radhika K R.
Raj said:
7 years ago
Thank you so much for explaining the solution of this.
Shivani said:
7 years ago
Thanks for your solution @Mani.
Geeni said:
1 decade ago
Consider david to be stationary then difference speed is
= 57+63 floor/minute
= 120f/m.
Total difference of floor is 51-11=40 floors.
Therefore minutes is (120f/m) /40f=1/3m.
So 11 + (1/3) * 57 = 51-(1/3)*63 = 30.
= 57+63 floor/minute
= 120f/m.
Total difference of floor is 51-11=40 floors.
Therefore minutes is (120f/m) /40f=1/3m.
So 11 + (1/3) * 57 = 51-(1/3)*63 = 30.
Saba khan said:
9 years ago
You are very talented @Geeni.
Prashant said:
9 years ago
@Kumar
Because at Yth position both of them meet so the distance from David is (y-11) and that of Albert is (51-y).
Because at Yth position both of them meet so the distance from David is (y-11) and that of Albert is (51-y).
Kumar said:
9 years ago
@Mani, How is Y value same for both but different in speeds?
Rabia said:
10 years ago
@Geeni.
Please explain last step.
Please explain last step.
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