Maths Puzzles

By now you might have heard many, many, many puzzles of some people always telling truth, and others always telling lies. But in real life, we all neither knights, nor knaves. Anyway this puzzle is slightly different - there are couples and these couples both lie or both tell truth or one lies and other one tells truth. (We all like to believe ourselves to be the last one - wink wink) Dags, Eggs and Fens are being interrogated as one of them is a spy. One of the couples always tells truth. In another couple, both husband and wife always lie. And in the third couple, one of the spouses always lies and the other one always tells truth. Here are the statements by the three couples when interrogated. Mr. Dag:  I am not the spy Mrs. Dag:  Mr. Egg is the spy. Mr. Egg:  Mr. Dag is truthful. Mrs. Egg:  Mr. Fen is the spy. Mr. Fen:  I am not the spy. Mrs. Fen: Mr. Dag is the spy.     Which person is a spy? (This puzzle is taken from the book "The Godelian ...

 In a certain bank the positions of teller, manager and cashier are held by Jones, Smith and Brown, but not necessarily respectively.  The teller is the only child and he earns the least. Smith is married to Brown's sister and he earns more than the manager.  What position does each man hold? (This puzzle is from the book 101 puzzles in thought and logic by C.R.Wylie)    Show answer   Solution : As teller is the only child - Brown can't be teller. Teller earns the least, so teller can't be Smith as Smith earns more than manager.  So teller has to be Jones. Because Smith earns more than manager, Smith can't be manager. And he can't be teller as Jones is teller. So Smith must be cashier.  Brown who is the only one remaining should be manager.  

An artist has a strange idea that a canvas looks good only if its perimeter is equal to its area. Now the question is how many such rectangles are there with integer sides  with their perimeter equal to area? And why?  Show answer Solution :  Area = l * w Perimeter = 2l+2w area = perimter => l*w = 2l+2w Adding 4 => -2l-2w + lw+4 = 4 => (l-2)(w-2) = 4 The integer pairs with 4 as product are  (1,4),(2,2),(4-1) If l-2 = 1 and w-2=4, we get l = 3 and w = 6 If l-2 = 2 and w-2 = 2, we get l = 4 and w = 4 If l-2 = 4 and w-2 = 1, we again get l = 6 and w = 4 Solutions 1 and 3 are identical. So the only two rectangles with integers sides and area and perimeter being equal are rectangles with sides 3 and 6 and 4 and 4.  

 A magician standing next to a bridge offers to double the money to a man. But he also charges 40$ for crossing the bridge. The man happily agrees and keeps crossing the bridge.  But sadly after the third time he crossed the bridge, the man notices that he has only 40$ left in his pocket which he has to give to the magician. How much money did the man have originally?     Show Answer  Answer : 35 dollars. He had 35 dollars and the magician gave him 70  dollars after crossing the bridge. The man gave him 40$ as crossing fee. Now he has 30 dollars. After he crossed the bridge, the magician gave him 60 dollars out of which man gave the magician 40 dollars.  Now he has only 20 dollars left and when he crossed the bridge third time, magician gave him 40 dollars and man had to give all of it to magician.  

To the fraction 1/3, if you add 3 to both its numerator and denominator, its value doubles - it becomes 4/6 = 2/3. Find a fraction whose value triples when you add its numerator to both numerator and denominator. Find also the fraction whose value quadruples when you add the numerator to numerator and denominator.   Show Answer Answer : 1/5 and 1/7 To any fraction with odd denominator 2n-1 and 1 as numerator, adding numerator to numerator and denominator, increases its value by n. 

This puzzle is from "The Moscow Puzzles" by Boris Kodemsky is from another era - when there were things like clocks which need to be wound each day. If not they would stop. The puzzle goes like this. I have a wall clock which is the only source for knowing the time in my house. One day I forgot to wind the clock and it stopped. Luckily, I have a friend whose clock shows the correct time always.  I walk down to his house, see the time in his clock, stay for a while and come back. Now I know the time and adjust my clock. It is is showing the correct time again.  How did I do that?      Show Answer Answer: I wind up the clock before leaving the house and span style="font-family: Comic Neue;">and set it at 12 o'clock. It starts running. When I go to my friend's house I notice the time. And just before leaving the house, I also see the time. When I come back, I see the time on my clock. That will show the total time elapsed since I left home....

 There are 7 matches arranged in the form of a star. Your aim is to place 6 coins at the head of 6 of the matches. You should start from a match which does not have a coin at the end, count and place a coin at the third match. You should not skip any match while counting. Can you arrange the coins this way without placing two coins at the head of any match? (This puzzle is also taken from the book "The Moscow Puzzles" by Boris Kordemsky)   Show Answer   Answer: Count from one match say A and place a coin at the head of third match. Next select the match B such that the match A will be the third match and place a coin at A.  Now start from a match such that B will be third match and place a coin at B. Continue this process and you will place 6 coins. 

There are three piles of match sticks with 11 matches, 6 matches and 7 matches in them. Now you need to move the matches from one to another such that all of them have 8 matches each.  But there is a condition - the condition is that you can add as many matches to a pile as it contains. For example if a pile has 3 matches, you can add only 3 matches to it - no more, no less.  You are allowed three moves. How will you move the matches? (The puzzle is borrowed from the book "The Moscow Puzzles" by Boris Kordemsky)  Show answer Answer : First you move 7 matches from first pile and add it to the last pile.  Now you have 4 - 6 - 14 Next you move 6 matches from last pile to middle pile. Now you have 4 - 12 - 8 matches in the three piles. Lastly you move 4 matches from middle pile to first pile.  You have 8 - 8 - 8 matches in the piles.  

 On a table, three normal playing cards are placed face down. To the right of a king, there is a queen or queens. To the left of a queen there is a queen or queens. To the left of a heart, there is a spade or spades. To the right of a spade, there is a spade or spades.  Which three cards are on the table? (Borrowed from the book "The colossal book of short puzzles and problems" by Martin Gardner)  Show answer The first rule says that there is one king. Second rule says that there are two queens. (KQQ,QKQ) Third rule says that there is a heart and a spade. And the fourth rule says that there are two spades. (HSS,SHS,SSH) So the four possibilities are   KS, QS, QH KS, QH, QS QS, KS, QH QS, KH, QS The fourth is an impossible option as there can't be two queen of spades in a deck. So the three possible answers are  King of spades, Queen of Spades and Queen of hearts King of spades, Queen of hearts and queen of spades Queen of spades, king of spades...