KMK Consulting Interview Questions, Process, and Tips

D

on the earth

1.75-1.68=0.07

28th February,1956

c

3;4

let the 4 digit number be ABCD.
First Digit is A :
Therefore; according to the question
A=B/3
B=3A
C=A+B=A+3A=4A
D=3B=3(3A)=9A
Since the last Digit is D and it can neither be double-digit nor 0 ;
Therefore ;
A=1;
B=3A=3(1)=3
C=A+B=1+3=4
D=9A=9(1)=9
Therefore, the 4 digit number is 1349.

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​

8/15 left.

1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15

1+0^3=1
1+1^3=2
2+2^3=10
10+3^3=37
37+4^3=101
101+5^3=226
1,2,10,37,101,226……..

a

Diagonal of square=1.414* a, where a is side of square.
ans=1.414*20
ans=28.28

He will work for 35 days in 60 days to earn Rs.170 as per the condition