Puzzles for you - Answers

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Puzzle Editions 7  
Simple Puzzle1 #
How many minutes after 3:00 PM or AM, will the 2 hands (hour hand and minute hand) be perpendicular again?

Answer:
The relative speed between hour hand and minute hand is 5.5 degrees per minute. To be perpendicular again after 3:00'O Clock, they need to move total of 180 degrees between them. Hence it will take 180/5.5 minutes for them to be perpendicular again which will be 3:32 and some seconds 43 seconds. 

Simple puzzle 2# 
In a day, how many times will the hour hand and minutes hand cross over each other?

Answer:
22 times. Using the same principle as above, it takes 360/5.5 minutes for them to meet each other once. So they will meet t(24 * 60) / (360/5.5) times in a day which is equal to 22.

Puzzle 6:  12 Identical Coins..
Problem : There are 12 identical coins (in look and weight). Only one of them is either slightly heavy or light. We have a scale that has two pans and we are allowed 3 trials to find the odd one and also tell if it is heavy or light. How would you do it ?  Click here to send your answers.. or click here to check if your answer is correct..
Puzzle Sent by Sridhar MV.

Solution :
Make 3 groups of 4 coins each. Place 1 group each into a pan. 

CASE 1 : Pans weigh equal
If they weigh equal, we know the odd one is in third group. Retain on one pan 3 good coins and place the 3
coins from the bad group in the other pan. If the pan with odd coin goes down/up, we know the odd coin is heavy/light. Now we have found out the nature of the odd coin's weight. In this case, take out the good coins, from the bad group, place one coin each on the pans and keep the third one aside. If they weigh equal, the one that was kept aside is either heavy/light (in this case, just place a good coin on one pan and the odd one on the other pan and we will know whether it is heavy or light). If they do not weigh equal then one of them is heavy or light as the case may be. 

CASE 2: Pans do not weigh equal 
In this case we know among these 8 coins one is heavy/light. Mark all the coins in the pan that went down as H and likewise mark all coins in the oter pan as L. All good coins are marked as G. (let us say)

Place 3 G's, 1 L in one pan. Place 2 H's and 2 L's in other pan.

case 2.1
Now if pan with 3 G's and one L goes down, we immediately know the 2 L's in the other pan have the odd one. We have one more trail and if we place these 2 L's one in each pan, whichever goes up is the odd one and it is light (3rd trial)

case 2.2
If the pan with 3 G's and one L goes up, we have that this L or the 2H's in the other Pan have the odd one.  For 3rd trial, if we place one H each in a pan, if one of the pan goes down we know the odd one is heavy. Otherwise (they weigh equal) we know odd one is light and which one it is.

case 2.3
In the second trial, if pans weigh equal, we have 2 H and one L outside of pans that may have the odd one in
them. Place from these one H each in a pan and if they weigh equal, the L is odd one and it is light. Otherwise whichever goes down is odd one and it is heavy. 

Contributors:
Shridhar MV: OurKarnataka.Com team knows Shridhar for a long time. Shridhar is into puzzles since his childhood and is very active in that area. He is a Design engineer in one of the Telecom Startups in Milpitas, CA

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