Pinnacle investments, llc 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​

let original radius=100
original surface area lets say it x =4 * (22/7) * 100 *100
new radius=100 + 50% of 100=150
new surface area lets say it y = 4 * (22/7)* 150* 150
increase in surface area=((y-x)/x) * 100
i.e.
((4*(22/7)*150*150 – 4*(22/7)*100 *100) / (4*(22/7)*100*100))*100
=125
so the answer is 125%.
hope this helps you.

Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.

325

The number of ways of selecting a group of eight is
5 men and 3 women=5C5​×6C3​ =20
4 men and 4 women=5C4​×6C4​ =75
3 men and 5 women=5C3​×6C5​=60
2 men and 6 women=5C2​×6C6​=10
Thus the total possible cases is 20+75+60+10=165.

C-2896

None

i think it is a and b

d

( e ) Apple

38 / 2 = 19 – 5 = 14 years old

C is grand child of A

665