Van ede & partners Interview Questions, Process, and Tips

B. Wheat

( a ) BQDCJCMF
“TERMINAL” split it into the half we get “TERM” “INAL”

Now the first half is decreased by one and the next half is
increased by one, so we get:

“SDQLJOBM” (pls note in the ques we have “SDQIJOBM” where
‘L’ shud have come instead of ‘I’)

so “CREDIBLE” is to “BQDCJCMF”

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.

friendly environment, supportive colleagues and place where i’ll have to work.

A

five multiplication & three addtion

let A goes up and B comes down.
assume A takes 1 step in 1sec.
hence,B takes 5 steps in 1sec.
50 steps ll be moved by A in 50/1=50secs
125 steps ll be moved by B in 125/5=25 secs
hence speed of escalator=(difference of no of
steps)/(difference of time)
speed=(125-50)/(50-25)= 3 steps/sec
hence during upward journey of 50secs by A,total steps=50 by
A and 150 by escalator..total=150+50=200
during downward journey of 25 secs by B,total steps=125 by B
and 75 by escalator..total=125+75=200steps.
ANS:200 steps

126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126

the answer is E

Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5

So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.

Hog

Answer: 10
Reason:
The best case happens ONLY when each rat dies just as they
taste the FIRST bottle given to them (you can imagine it a
miracle 😀 )
In this case, the very first 10 attempts reveal the
poisonous bottles. So the answer is 10.

27

– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.

– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m

To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec

In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m

at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m

If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)

Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.

For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.

This will take an additional (9.6 + 2) seconds.

Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds

3 days

15 min 100 line
30 min 200 lines + 15 min rest total 45 min +5 min think +5 min write + 5 min rest
250 line code will be done