Showsec Interview Questions, Process, and Tips

189

24 times

Lets assume total LCM(5,8) = 40units.
As, 5 men or 8 women do equal amount of work in a day,
1 Man does 8units/day and 1 Woman does 5units/day.
3M and 5W in 10 days do (3*8 + 5*5)*10 = 490units
To do 490 units in 14 days, number of Women required = 490/(14*5) = 7

419

Because 2 daughter should have same age

C-2896

10m

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​

1261

NON

0

first 10 odd num are 1,3,5,7,9,11,13,15,17,19
sum=100
avg=100/10
avg of 10 odd num =10

Ans is 10%

Printer price = x;
Computer price = 3x
Total overall price = 20x + 60 x 3x = 200x

Printer percent = (20x/200x ) x 100 = 10%

Second

8 cubes