Adobe Interview puzzles with answers:

The given puzzles are asked in many written exams conducted by varies IT companies like Google, Adobe, Amazon, yahoo, Microsoft etc…

Adobe Interview Puzzles :  Weight Puzzle

Puzzle 1 :  (check this is already posted)
             There are 8 Balls available in a box, From that only one odd ball being heavier or Lighter. question is to find out the minimum number of weighings required to spot the odd heavier ball among 8 identically looking balls using a common balance.

Solution : 

              This is a little trickier to solve.Let us name the 8 balls as A1,A2,A3,B1,B2,B3,C1 and C2.Now weigh with A1,A2,A3 on one side and B1,B2,B3 on the other side. If both weigh equal then the odd ball is among C1 and C2.Now we know that A's and B's are all normal ones we can weigh C1 with A1 and check whether C1 weighs the same as the normal ball. In this way, we can figure out in one weigh which of C1 and C2 is odd. Now if A's weighed more than B's then we know for sure C's are normal ones. Now let’s assume A's heavier than B's and we still don't know whether the odd is among A's or B's.
We know A1 A2 A3 < B1 B2 B3
Now we compare A1 B2 B3 to B1 C1 C2 . Now if the A1 B2 B3 is heavier than B1 C1 C2 Then it means the odd ball is among A1 and B1.If A1 B2 B3 is lighter than B1 C1 C2 then the odd ball is among B2 and B3.If A1 B2 B3 equal to B1 C1 C2 then odd ball is among A2 and A3.So we have zeroed down to 2 balls.Now its very easy as we can compare any of the normal ball with one of the 2 balls.So answer is 3,ie in 3 weighings we can find out the odd ball.

Puzzle 2: 
               This is same like the above weight puzzle But instead of 8 balls here You are given 13 balls. The odd ball may be either heavier or lighter.  Find out the odd ball in 3 weightings.

Puzzle 3: 
              You are given a cake; one of its corner is broken. How will u cut the rest into Two equal parts?

Solution :
Slice the cake(Horigentally).

Puzzle 4 :
           How it is possible to place four points that are equidistance from each other?

Solution :
          place points in the shape of a pyramid.

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