Outside the box Interview Questions, Process, and Tips

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.

WTO

Ans: 60 kph

Suppose Person meets the train everyday at 3 PM at Station A.

His speed is 12kph.

So normally he reaches 5 km before the meeting point (pt B) at (5/12 hr = 25 min before) 2:35PM.

But if he is late by 30 min, then he will reach that point (pt B) by 3:05 PM.

Train is traveling at its normal speed so it covers the distance of 5 Km in 5 min starting from Station A and reaches the meeting point (pt B) at 3:05 PM.

So speed of the train is 5KM/5min = 60 kph.

3 days

d=7s—–equation 1
d=5(s+12)
7s=5(s+12)
7s=5s+60
2s=60
s=30
from equation 1
d=7*30
d =210km

20 minutes

copy cat

At 680Km

((15 X m) + 23)/(m+1)= 16
solving , we get number of matches (m) as 7 (excluding last
match)
so he shud hit 39 runs in last innings to ake average to 18

8 games They played
A own 3 games = 18 Rs
B loss 3 Rs and own 1 games i,e 9(A own 3 Games)-6(B own
1 game)=3
c own 12 Rs and loss 4 games(A own 3+b own 1) and own 4
games i.e 24(own)-12(loss)
so totally A own 3 games
B own 1 game
c own 4 games
=8

Total 60 trees
4 rows each with 15 trees
5 rows each with 12 trees
6 rows each with 10 trees

2 minuts the speed ofthe train

Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.

If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.

Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.

Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).

Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.