G.l. bajaj institute of technology and management Interview Questions, Process, and Tips

33.6

Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:

1. Weigh ABC versus DEF.

Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.

2. Weigh G versus H.

Scenario a.i: If these (2) balance, the oddball is I.

Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.

Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.

2. Weigh A versus B.

Scenario b.i: If these (2) balance, the oddball is C.

Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.

answer is maximum of 2.

360 times

a/g

(x+2)+(x)+(x-2)
X+2+x+x-2=18
3x=18
X=6
The 3digit number is 864

tan 30=h/x
tan 15=h/100+x
so h/tan 30=(h/tan 15)-100
=33.33app

6 inch

Soldier have to move 0.75 miles to West and then 0.375 miles to South to reach the camp.
How:-
Firstly soldier moving 1 mile to East from camp and then 1/2 miles= 0.50 miles to North .

Then it is moving 1/4 miles =0.25 miles to west
And then 1/8 miles =0.125 miles to South

Now we just have to count the difference and minus the value of camp to East with the value of North to west = 1mile – 0.25miles(1/4) = 0.75 miles

Same case with north and south = 0.50 miles(1/2) – 0.125 miles(1/8) = 0.375 miles

Hence proves to return camp the soldier have to move 0.75 miles to west and 0.375 miles to south

exactly