Csep scpe Interview Questions, Process, and Tips

4/5

(0! + 0! + 0! + 0! + 0!)! = 120

Here ‘!’ symbol is used for factorial.
And 0! = 1
= (0! + 0! + 0! + 0! + 0!)!
= (1 + 1 + 1 + 1 + 1)!
= 5!
= 120

1:2

The phrase is usually ‘like looking for a needle in a haystack’. It describes a task that is virtually impossible because you would have to search a huge area

I would expect progression in my career through acquiring new skills

Let the number be x.

So x-2 =15/x, or

x^2–2x-15=0, or

(x-5)(x+3)=0

Hence x = 5 or -3.

Check: When x=5, 5–2=3 which is 15/5=3.

When x=-3, -3–2=-5 which is 15/-3=-5.

The number is 5 or -3.

It is assumed to be constant. Now, to cross past the pole, the train should cover a distance of x meters. Now, the time taken by the train to cross a platform of length 100 m is 25 seconds. Hence, the length of the train is 150 m.

3.14

consider the tank capacity as 90 litres.
to fill the tank in 3 hrs first pipe must flow at a rate of 30 litres/hr
2nd pipe has to flow at a rate of 45 litres/hr
if two pipes are opened at a time the flow rate will become 75 litres/hr
90/75= 6/5 = 1.2= (i.e) 1hr and 12 minutes 1 1/5 hr

Petroleum

x

Let A, B and C be the three 6-faced dice.
Then, according to the question,
Since two dices has to be equal, that value can be any of the 6 faces, i.e., 6C1​ cases.
Now for each case, 2 equal dices can be selected from 3 dices in 3C2​ i.e., 3 ways.
And for each of the above, the third dice can have any of the 5 remaining faces
The possible outcomes are P(A)=61​,P(B)=61​,P(C)=65​,P(A)=61​,P(B)=65​,P(C)=61​ and P(A)=65​,P(B)=61​,P(C)=61​
Hence the required probability = 61​×61​×65​×6×3=21690​=125​

D. 760