Ussa Interview Questions, Process, and Tips

441=21^2

C

Answer: Pile II

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“One of my hobbies is playing club sports. … Playing sports is my favorite pastime and my hobby because it’s not just the games I love, but also the socializing part that comes with it. I love to meet new people and go to new places for tournaments.”

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

8*1024+3*64+3
8192+192+3
8387
1000/0011/1000/0111
So ie 7. 1s here

3

In 3 hours thinker candle becomes half
In the same time thinner candle becomes one fourth. Now the
required condition is satisfied. So 3 Hours

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​

ANS = 24.
16 = 2^4
24 = 2^3 x 3
H.C.F of 16 and 24 = 8
L.C.M will be 48

She is My grandmother

A = 20k x 2y, B = 15k x 2y, C = 20k x 1.5y => 100Ky
B contribution = (15k x 2y) /100ky
B will get 0.3 from 25k which is 7.5k = 7500

Simple interest Formula:
A=P(1+rt)
Therefore,
815=P(1+3r)
P+3r=815

854=P(1+4r)
P+4r=854
Solve the sums using Elimination method.
P+4r=854
P+3r=815
r=39.
Toget Principal
P+4r=854
P+4*39=854
P=854-117
P=698

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