SteelEye Interview Questions, Process, and Tips

13km

D=5, G=1

((15 X m) + 23)/(m+1)= 16
solving , we get number of matches (m) as 7 (excluding last
match)
so he shud hit 39 runs in last innings to ake average to 18

All books can be arranged in 10! ways. A single pair of books can be taken as a unit and arranged among the 8 others in 9! ways. The pair of books can also be interchanged and therefore rearranged in 2! ways. Thus the probability of the pair always being together is (9!*2!)/10!

900

224

n(n+1)/2
=10*11/2
55

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​

The answer is A)

y1 = 62 Rs/kg
y2 = 72 Rs/kg
y = 64.5 Rs/kg

y2 – y1 = 10 Rs/kg

The distance between the y and y1 is

y – y1 = 64.5 – 62 = 2.5

x1 = (y – y1)/(y2 – y1) = 2.5/10 = 0.25
x2 = 1 – x1 = 1 – 0.25 = 0.75

The target price is calculated by the lever method.

x1 * y2 + x2 * y1 = 0.25 * 72 + 0.75 * 62.5 = 64.5

The ratio is of y1 to y2 is

0.75 : 0.25

Divide by both by 0.25

3 : 1