Maths Puzzles

   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 - rectangles with their perimeter equal to their area? And why?  Show answer Solution :  Area = l * w Perimeter = 2l+2w area = perimter => l*w = 2l+2w Moving 2l+2w to left hand side, we get 2l+2w+l*w = 0 Adding 4 to both sides we get=> -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 each time he crosses the bridge. But he also charges 40$ for crossing the bridge. The man happily agrees and keeps crossing the bridge again and again.  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 denominator to both numerator and denominator.  Find also the fraction whose value quadruples when you add the denominator to its numerator and denominator. Show Answer Answer : 1/5 and 1/7 If we have any fraction with odd denominator 2n-1 and 1 as numerator, when we add the denominator to both numerator and denominator, its value is multiplied by n.  So for n=3, the fraction is 1/2X3-1 = 1/5 and when we add 5 to both numerator and denominator, we get 6/10 = 3/5 Which is 3 times 1/5

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 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. Now since I know the time in my friend's house as soo...

 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: Start by choosing any matchstick—let us say A—and place a coin on the third matchstick counting from A. Next, select another matchstick—let us say B—so that A becomes the third matchstick when counting from B, and place a coin on A. Then choose a new starting matchstick so that B is the third matchstick from it, and place a coin on B. Continue this process, and you will place all the six coins.