Addis Interview Questions, Process, and Tips

562.

2 Leap years
+2 to tuedsay
Thursday is the correct answer

when know that 13 is factor of 39 so it will also not be able to divide it differently. so the remainder will be same 20 but as we know that 20 is greater then 13 so we can divide it once again and in the casE of 13 the remainder will be 20-13=7

⇒ First part = Rs. 828

66.66%

12

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​

if rs 12 and rs 15 per meter then 750*12+750*15

2000

Let the number be X,
Instead of dividing the number by 10, i.e.,X*10.
he got the result as 100,
so, (X*10)=100
X=100/10
X=10.
BUt he has to divide X by 10,
so, X/10=10/10=1.
obviously the answer is 1.

X = 4

(60/100)*60 + (40/100)*40
=52%