Hmg Interview Questions, Process, and Tips

C. 39

3*2+1 = 7
15*2+1 = *31*
31*2+1 = 63
63*2+1 = 127
127*2+1 = 255
255*2+1 = 511

A whole number

C

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!

X×x-x=272
X^2-x-272=0
(X-16) or (x-17)
X=16 or x=17

17^2-17=272
Answer Is 17

these are question tht are seem to be like it consumes
time. so dont see jus the question and run to the next .

ans. a) 12

if 9 balls are added then the ratio to the combination
becomes 2:4:3.

9 balls make the ratio 3 for grey balls in tht mixture.
so the factor is 9/3 = 3 (in tht mixture)

so the same factor has to be maintained through out the
ratio. so black balls is 4 * 3 = 12 and white balls is 2 *
3 = 6.

Answer: 4200

Explanation:

15:25 => 3:5

9600*10/100 = 960

9600 – 960 = 8640

8640*3/8 = 3240 + 960

= 4200

0,8x= y
y – 0,7y = 270

y= 270 / 0,3
y=900

0,8x=900
x=1125

Ratio is 3:2:1

Total=6

Reverse of ratio is 1/3:1/2:1/1

Portion of first part is=1/3*6=>2

Portion of second part is=1/2*6=>3

Portion of third part is 1*6=>6

∴So the new ratio is 2:3:6

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​

190.85