(Paper) WIPRO Puzzles Questions

If each drink costs 50p, how much minimum money do you have to put into the machine to work out which button gives which selection?
Answer
You have to put just 50p.


Put 50p and push the button for Random. There are only 2 possibilities. It will give either Tea or Coffee.

• If it gives Tea, then the button named Random is for Tea. The button named Coffee is for Random selection. And the button named Tea is for Coffee.
  

• If it gives Coffee, then the button named Random is for Coffee. The button named Tea is for Random selection. And the button named Coffee is for Tea.
  

Thus, you can make out which button is for what by putting just 50p and pressing Random selection first.

You have 13 balls which all look identical. All the balls are the same weight except for one. Using only a balance scale, can find the odd one out with only 3 weighings?

Is it possible to always tell if the odd one out is heavier or lighter than the other balls?
Submitted by : Brett Hurrell

Answer

It is always possible to find odd ball in 3 weighings and in most of the cases it is possible to tell whether the odd ball is heavier or lighter. Only in one case, it is not possible to tell the odd ball is whether heavier or lighter.

• Take 8 balls and weigh 4 against 4.
  

  • If both are not equal, goto step 2
    

  • If both are equal, goto step 3
    

  
   

• One of these 8 balls is the odd one. Name the balls on heavier side of the scale as H1, H2, H3 and H4. Similarly, name the balls on the lighter side of the scale as L1, L2, L3 and L4. Either one of H's is heavier or one of L's is lighter. Weigh (H1, H2, L1) against (H3, H4, X) where X is one ball from the remaining 5 balls in intial weighing.
  

  • If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
    

    • If both are equal, L4 is the odd ball and is lighter.
      

    • If L2 is light, L2 is the odd ball and is lighter.
      

    • If L3 is light, L3 is the odd ball and is lighter.
      

    
     

  • If (H1, H2, L1) is heavier side on the scale, either H1 or H2 is heavier. Weight H1 against H2
    

    • If both are equal, there is some error.
      

    • If H1 is heavy, H1 is the odd ball and is heavier.
      

    • If H2 is heavy, H2 is the odd ball and is heavier.
      

    
     

  • If (H3, H4, X) is heavier side on the scale, either H3 or H4 is heavier or L1 is lighter. Weight H3 against H4
    

    • If both are equal, L1 is the odd ball and is lighter.
      

    • If H3 is heavy, H3 is the odd ball and is heavier.
      

    • If H4 is heavy, H4 is the odd ball and is heavier.
      

  
   

• One of the remaining 5 balls is the odd one. Name the balls as C1, C2, C3, C4, C5. Weight (C1, C2, C3) against (X1, X2, X3) where X1, X2, X3 are any three balls from the first weighing of 8 balls.
  

  • If both are equal, one of remaining 2 balls is the odd i.e. either C4 or C5. Weigh C4 with X1
    

    • If both are equal, C5 is the odd ball. But you can not tell whether it is heavier or lighter.
      

    • If C4 is heavy, C4 is the odd ball and is heavier.
      

    • If C4 is light, C4 is the odd ball and is lighter.
      

    
     

  • If (C1, C2, C3) is heavier side, one of C1, C2, C3 is the odd ball and is heavier. Weigh C1 and C2.
    

    • If both are equal, C3 is the odd ball and is heavier.
      

    • If C1 is heavy, C1 is the odd ball and is heavier.
      

    • If C2 is heavy, C2 is the odd ball and is heavier.
      

    
     

  • If (C1, C2, C3) is lighter side, one of C1, C2, C3 is the odd ball and is lighter. Weigh C1 and C2.
    

    • If both are equal, C3 is the odd ball and is heavier.
      

    • If C1 is light, C1 is the odd ball and is lighter.
      

    • If C2 is light, C2 is the odd ball and is lighter.
      

How many squares are there in a 5 inch by 5 inch square grid? Note that the grid is made up of one inch by one inch squares.
Submitted by : Kristin Monroe

Answer

There are 55 squares in a 5 by 5 grid.


There are 25 squares of one grid.
There are 16 squares of four grids i.e. 2 by 2
There are 9 squares of nine grids i.e. 3 by 3
There are 4 squares of sixteen grids i.e. 4 by 4
There is 1 square of twenty-five girds i.e. 5 by 5

Hence, there are total 25 + 16 + 9 + 4 + 1 = 55 squares.

You must have noticed one thing that total number squares possible of each size is always a perfact square i.e. 25, 16, 9, 4, 1