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 1 of 3.
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.
Nje said:
1 decade ago
Geeni you're a genius.
Rehana said:
1 decade ago
Its 40f/120f/m=1/3min
Hardik said:
1 decade ago
@Geeni
Please explain last step
11+(1/3)*57=51-(1/3)*63
Please explain last step
11+(1/3)*57=51-(1/3)*63
Deepti said:
1 decade ago
Can someone explain how you got 11+(1/3)*57b = b51-(1/3)*63?
Subhadip said:
1 decade ago
Just visualize the problem solve & understand last step.
Radhika K R said:
1 decade ago
We can also predict it though an train problem, two people who are travelling opposite direction:
Here total speed will be 63+57 (f/m) =120 f/m.
Total distance will be subtracted 51 -11.
So the time they will meet will be 40/120=1/3 min.
So the distance will be for their meet will be:
11+57*1/3= 30.
or
51-63*1/3=30.
Here total speed will be 63+57 (f/m) =120 f/m.
Total distance will be subtracted 51 -11.
So the time they will meet will be 40/120=1/3 min.
So the distance will be for their meet will be:
11+57*1/3= 30.
or
51-63*1/3=30.
(1)
Amit said:
1 decade ago
@Geeni.
Just implemented the concept of relative velocity.
1/3 is time at which david and albert meet. So he multiply that time to the velocity of david and calculate the distance travelled by david and finally he added that distance to david's initial position.
Just implemented the concept of relative velocity.
1/3 is time at which david and albert meet. So he multiply that time to the velocity of david and calculate the distance travelled by david and finally he added that distance to david's initial position.
Mani said:
1 decade ago
s = d/t.
In this problem time travel is same for both=> d1/s1 = d2/s2.
Lets assume they cross the paths at yth floor.
(y-11)/57=(51-y)/63.
Solve this,
y=30.
In this problem time travel is same for both=> d1/s1 = d2/s2.
Lets assume they cross the paths at yth floor.
(y-11)/57=(51-y)/63.
Solve this,
y=30.
Mukesh said:
10 years ago
@Mani.
You are cool, this is the best way and the easiest way.
You are cool, this is the best way and the easiest way.
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