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sorry this question is irrelevant.
Prob(choosing two people who are a couple from 6 couples) = 1/6
We have 5 couple left (10 people)
Prob(choosing a person A from 10 people) = 1/10
Prob(choosing 1 person who is not a couple of A) = 1/8
Prob(choosing 4 people that has exactly one couple) = 1/6*1/8*1/10 = 0.0021
#1: N = 1, f(N) = 1
#2: N = 199981, f(N) = 199981
#3: N = 199982, f(N) = 199982
#4: N = 199983, f(N) = 199983
#5: N = 199984, f(N) = 199984
#6: N = 199985, f(N) = 199985
#7: N = 199986, f(N) = 199986
#8: N = 199987, f(N) = 199987
#9: N = 199988, f(N) = 199988
#10: N = 199989, f(N) = 199989
#11: N = 199990, f(N) = 199990
#12: N = 200000, f(N) = 200000
#13: N = 200001, f(N) = 200001
#14: N = 1599981, f(N) = 1599981
#15: N = 1599982, f(N) = 1599982
If the bell rung when they r started is also counted then it
will be 5 otherwise 4
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
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