Kingpin communications Interview Questions, Process, and Tips

required number =H.C.F of (73-25),(97-73) & (97-25)
=H.C.F of 48 , 24 and 72 = 4 (c)

assume the ratio 6:4 to be 60:40
boys taking lunch in canteen=60% of 60=36
Girls taking lunch in the canteen=40% of 40=16
Total boys and girls taking lunch in the canteen=52
Total students we assumed is 60+40=100
Therefore percentage of boys and girls taking lunch in the canteen=(52/100)*100=52%

4155

(x-2)^3==x^3-6x^2+12x-8
(x-2)^3==(2^(2/3)+2^(1/3))^3
therefore,by solving
Ans: x^3-6x^2+x== -8-5*2^(2/3)-5*2^(1/3)

5:6::7:10::6:5
=((5/6)/(7/10)/(6/5))
=((5/6)*(6/5)/(7/10))
=(1/(7/10))
=10/7

200/3

the sum of first 4 primes numbers =2+3+5+7 =>17

198.20

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​

In the given series 5,6,7,8,10,11,14…
There are two series
First is 5,7,10,14…
Second is 6,8,11….
First series
5+2=7
7+3=10
10+4=14
Second series
6+2=8
8+3=11
11+4=15
Hence complete series is 5,6,7,8,10,11,14,15…

B

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