UN Mehta Institute of Cardiology & Research Ctre Interview Questions, Process, and Tips

125% of s = 60
=> 1.25s = 60
=> s = 60*4/5
=> s = 48

60 – s = 60 – 48 = 12

22

1/24S

I think it’s 32 not 37

D

40

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​

0.000256

Suppose the lengthier arm of weighing pan is of x cm and other arm is y cm .Also let weight if each melon be m kg.
so applying equilibrium of torque principles ,we get
case 1:-
1x-8my=0
case 2:-
2mx-1y=0
using case 1 equation , we substitute value of x into case 2 equation;
16mmy-1y=0
(16mm-1)y=0
y is length of weighing arm and cannot be 0,
therefore ,
16mm-1 =0
16mm=1
mm=1/16
m= square root (1/16)
m=+-1/4
m is weight of melon and cant be negative.
Hence m, weight of one melon is 1/4 kg