Everest Bank Interview Questions, Process, and Tips

A’s speed=6 mph ,B’s speed=8 mph
Let, after x hrs, they will meet.
so, the distance traveled by A in x hrs should be the same as the distance traveled by B in (x-1/2)hrs [as B started the journey after 30 min of A]
Thus, 6x=8(x-1/2)[as distance=speed*time]
=>8x-6x=4
=>2x=4
=>x=2
after 2 hrs they will meet so time=(9+2)=11.00 a.m

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​

b

let the speed limit be x.
speed by first rider S1 and 2nd rider be S2.
S1=x+10;
ans S2=x+2*10; and given S2=35.
solving we get x=15mph.

1. 43 venn diagram
2. 10 using venn diagram

done it roughly chances of having wrong are their..thank u.

Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.

If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.

Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.

Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).

Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.

6561 → 6560/2 [-1]
3280
1640
820
410
205 [+1] =206
103 → [-1] = 102
51 [+1] = 52
26
13 [-1] = 12
6
3 [+1] 4
2
1

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