Manipal technologies ltd. (formerly manipal press ltd.) Interview Questions, Process, and Tips

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​

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

2
6*2=12
12*6=72
72*12=864

C

100

Assume Q can do the work alone in Q days:

work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.

Ratio of diameters=1:2
Ratio of radius=1:2
Ratio of volumes=(4/3*3.14*r1^3)/(4/3*3.14*r2^3)
=r1^3/r2^3=1^3/2^3=1/8

225

56.25 % is the right answer.
area of square = side ^ 2
at first, it was 100% , means 1
then, a side was enlarged to 25%, means now, 125% means new side is 1.25
taking square will give 1.5625
1.5625 – 1 = 0.5625
so, 56.25% increment in area.

390

17

6

Sunil Shetty

100