Answer The sum of the digits od d is 1

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  Only 2 adjacent unlike coins can be moved at any one time.
  
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  The moved coins must be in contact with at least one other coin in line. i.e. no pair of coins is to be moved and placed away from the remaining ones.
  
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  No coin pairs can be reversed i.e. a S-C combination must remain in that order in its new positionwhen it is moved.
  

Show all moves.

The fly is flying at 90 km/h. If the fly flies 180 km before the trains meet, how far apart were the trains initially?

The fly is flying at the speed of 90 km/h and covers 180 km. Hence, the fly flies for 2 hours after trains started.

It's obvious that trains met 2 hours after they started travelling towards each other. Also, trains were travelling at the speed of 60 km/h. So, each train traveled 120 km before they met.

Hence, the trains were 240 km apart initially.

Example: To form 4884, you would need 2 4s & 2 8s. 4822 requires a 4, a 8, & 2 2s, but you would not count the numbers again that you had already counted from making 4884.

You will need 3 of numbers 0, 7, 8 & 9, & 4 of numbers 1-6.

A drinks machine offers three selections - Tea, Coffee or Random (Either tea or Coffee) but the machine has been wired up wrongly so that each button does not give what it claims.

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?

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

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  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.
  
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  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.
  

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.

1. 
   Take 8 balls and weigh 4 against 4.
   
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     If both are not equal, goto step 2
     
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     If both are equal, goto step 3
     

2. 
   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.
   
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     If both are equal, one of L2, L3, L4 is lighter. Weigh L2 against L3.
     
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       If both are equal, L4 is the odd ball and is lighter.
       
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       If L2 is light, L2 is the odd ball and is lighter.
       
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       If L3 is light, L3 is the odd ball and is lighter.
       

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  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.
    

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  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.
    

3. 
   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.
   
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     If both are equal, one of remaining 2 balls is the odd i.e. either C4 or C5. Weigh C4 with X1
     
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       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.
       

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  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.
    

There are 25 squares of one grid.

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

For a grid of N by N, the possible number of squares are

For 1 by 1 grid, total squares = 1² = 1

Five horses ran in the race.

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  There were no ties.
  
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  Sikandar did not come first.
  
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  Star was neither first nor last.
  
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  Mughal Glory came in one place after Sikandar.
  
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  Zozo was not second.
  
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  Rangila was two place below Zozo.
  

It's simple.

Let's find the possible places horses can finish. Possibilities are:
Sikandar - 2,3,4 (not 5th as Mughal Glory came one place after him)
Star - 2,3,4
Mughal Glory - 3,4,5
Zozo - 1,3 (not 4th & 5th as Rangila is two place after him)
Rangila - 3,5

So the result is:

The reason why is because 200-299 each begins with a 2!

How many e-mails would have been sent by the time it did 30 sets?

First person sent the mail to 2 persons. Those 2 sent the mail to 2 persons each, total 4 persons. Now, those 4 person sent mail to total 8 persons, then 8 to 16 persons, 16 to 32 persons and so on.... Hence, it a series of 2, 4, 8, 16, 32 upto 30 numbers

It is a Geometric series with common ratio 2 and first number is also 2. Summation of such series is given by A * (Rⁿ - 1) / (R - 1) where
A = First term
R = Common Ratio
n = total numbers

So total number of times mail sent by the time it did 30 sets

The answer is the number of letters in the word spoken by the door warden.

"Twelve" contains "Six" letters i.e. T, W, E, L, V, E
"Six" contains "Three" letters i.e. S, I, X
Similarly, "Four" contains "Four" letters i.e. F, O, U, R

There is a perfect sphere of diameter 40 cms. resting up against a perfectly straight wall and a perfectly straight floor i.e. the wall and the floor make a perfect right angle.

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