Job Venture Interview Questions, Process, and Tips

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.

43

D

The 3-digit number can be written as the sequence [n, 2n, 3n]
n = 1 → [1, 2, 3] → valid
n = 2 → [2, 4, 6] → valid
n = 3 → [3, 6, 9] → valid

n = 4 → [4, 8, 12] is not valid because this would lead to a 4-digit number
any value of n > 4 would also produce invalid answer

Answer: Three numbers: {123, 246, 369}

1267

127.179(app)
given distance of the train along the wind is 695
and againt the wind is 498
and time = distance/speed
as we know that time is equal in both the cases hence equate
695/s1=498/s2———–(1);
where s1=speed of the plane + speed of the wind
and s2=speed of the plane -speed of the wind
given that speed of the wind is 21k/h
s1=sp+21
s2=sp-21
substitu in eq 1
we get the answer as 27.17(app)

b

37

39 hrs 15 mins..
note:
he will not be having .45 mins for his lunch on saturday..

let a = 9 * 9 * 9;
let b = 12 * 12 * 12;
let c = 15 * 15 * 15;

sum = a + b + c;

edge = Math.cbrt(sum);

console.log(edge);

CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….

45

The definition of left and right depends on the observer and
is reversed when facing the opposite direction. The
definition of up and down does not depend on the orientation
of the observer.

D