Mathematical riddles and puzzles are frequently asked in various competitive exams. These are based on the basic concepts of probability, arithmetic, ratio & proportion, time & distance, etc. Here is a compilation of mathematical puzzles grouped into: easy, medium and difficult puzzles. Each article contains a set of 10 maths puzzles with answers. In this article, you will come across difficult maths puzzles devised to hone your mental aptitude.
Solve the given questions and check your mental abilities:
Q.1. An auditorium is 6 meters high, 20 meters long and 5 meters wide. A fly is present in the middle of one of the end walls (5 X 6 meters), one meter below the ceiling. In the middle of the other end wall one meter above the floor, a lizard wants to catch the fly. Find out the the shortest way for the lizard to catchhold its prey?
Generally most of the readers will answer this as (1 + 20 + 5) = 26 meters. Some may think of a path passing through the diagonal along the long wall and width and consequently get the answer as 25.3 metres. However, much shorter path is available and this is one along which, amazingly, the lizard will have to crawl over five (of the six) sides of the room.
The six sides of the room (including the floor and the ceiling) can be opened out into a plane surface. (In order to visualize this the reader is advised to imagine the process of opening out a cardboard box). When the room is considered as the development of this particular surface, straight line between the two points X and Y, which correspond to the positions of the lizard and the fly respectively, gives the shortest path.This path, being formed by the hypotenuse of a triangle whose sides are 22 and 11 meters, would be the square root of (222 + 112) or 24.6 meters long and traverses five sides of the auditorium, i.e. one end wall- the floor – a long wall- the ceiling- the other end wall.
Q.2. In 1991, a devastating flood ravaged many districts of Bihar. A group of villagers found itself confronted by a wide and deep flood water. However, they found two children who were rowing a boat. But the boat was so small that it can only carry either the 2 children, or 1 adult person and it can be rowed by single child or adult. Can the 246 villagers move across the flood water? If yes, what is the minimum number of movements that the boat has to make from shore to shore?
At least 984 movements are required. The two children move the boat to the opposite shore. One gets out and the other child brings the boat back. One villager rows across; villager gets out, and the child returns with boat. Thus, it takes 4 movements to get 1 person across and bring the boat back. Hence it takes 4*246 = 984 movements, to get the the entire group of villagers across the flood water.
Q.3. On a chessboard, one can choose 2 squares at random . According to you, what is the probability that they have a common side?
The total number of ways of choosing the two squares on a chessboard is 64 * 63 / 2 = 2016. If the first square happens to be any of the four corner ones, the second square can be chosen in 2 ways. If the first square happens to be any of the 24 squares on the side of the chess board, the second square can be chosen in 3 ways. If the first square happens to be any of the 36 remaining squares, the second square can be chosen in 4 ways. Hence the desired number of combinations = (4 * 2) + (24 * 3) + (36 * 4) = 224. But since we are counting each square two times (One - First X and then adjacent as Y & Second - First Y and then adjacent as X), thus favourable number of ways = 224/2 = 112. Therefore, the required probability = 112/2016 = 1/18.
Q.4. There are 12 identical balls having exactly equal weight except one ball which is heavier than the other 11 balls. How many minimum weighings are required to identify this different ball by a simple weighing balance?
Step I: Divide the balls in 3 groups of 4 balls each. Put two groups on the two sides of the weighing balance. If these two groups weigh equal, then the third group must be containing the heavier ball or else if one of these group is heavier than the other, then that group must be containing the heavier ball.
Step II: Now, in this heavier group, compare the weight of 2 balls each on the two sides of the weighing balance. One of the group which contains the hevier ball can be separated.
Step III: Now the 2 balls of this group can be further compared in the balance.
Hence the eavier ball can be determined in 3 steps.
Q.5. P Q R
S
T U V
W
X
Each of the digits 1,2,3,4,5,6,7,8, and 9 is:
i. Represented by a different letter in the figure above.
ii. Positioned in the figure above so that each of P+Q+R, R+S+T, T+U+V, and V+W+X is equal to 13.
Which digit does T represent?
One digit must be 9. Then, from [1] and [2], 9 must go with 1 and 3.
From the diagram no digit has been used in more than two sums. From this and the fact that 9 goes with 1 and 3:
Case I. If 8 goes with 1 and 4, then 7 goes with 2 and 4; then 6 goes with 2 and 5
Case II. If 8 goes with 2 and 3, then 6 goes with 2 and 5; then 7 goes with 1 and 5.
But Case II is impossible because the digit 4 does not occur. So Case I is correct and, from the diagram, T must be 4.
The arrangement of the digits is shown below :
9 3 1
8
4 7 2
5
6
Hint : The Numbers corresponding to the alphabet which are Repeating should not be assumed large number i.e 9 8 7 : The alphabet such as R , T , V should be assumed lesser in value . So Try to assume the values from 1 2 3 4
Q.6. Khurana would like to take a new apartment on rent. The owner asks him: "Please tell me how many children you have." Khurana answers: "I have three of them." The owner: "What are the ages of your children?". He answers: "The product of the ages is equal to 72." The owner replies: "This is not enough information dear!". "Sorry that I was a little bit unclear, but the sum of the ages is equal to the house number in front of your apartment," says Khurana. The owner: "This still isn't enough information!". Khurana replies: "My oldest child loves chocolate." The owner: "Thanks for your cooperation, I now know the ages." Are you as smart as the owner? Then give the ages of the children.
The product of the ages is 72. Using this one can make the following combination of ages:
1,72, 1 sum = 74
1,36, 2 sum = 39
1,24, 3 sum = 28
1,18, 4 sum = 23
1,2, 18 sum = 21
12,6, 1 sum = 19
12,2, 3 sum = 17
9,8, 1 sum = 18
9, 4, 2 sum = 15
8, 3, 3 sum = 14
6, 6, 2 sum = 14
6, 4, 3 sum = 13
After the man had said that the product of the ages is equal to 72, the owner didn't have enough information. Then he was told that the sum is equal to house number in front of the apartment. He replied by saying that this still isn't enough information. So the sum of the ages should be 14, because otherwise he would have known the ages immediately. The last statement is that the oldest child loves chocolate. So there is an oldest child. Hence the owner concludes that the ages of the children are 8, 3 and 3 years.
Q.7. Ramlakhan, while leaving for school looks at the clock in the mirror. Since the clock has no number indication, he makes a mistake in interpreting the time. Assuming the clock must be out of order, he cycles to school, where he arrives after twenty minutes. At that moment the clock at school shows a time that is four and a half hours later than the time that Ramlakhan saw on the clock at home. At what time did he reach school?
The difference between the real time and the time of the mirror image is four hours and ten minutes (four and a half hours, minus the twenty minutes of cycling). Therefore, the original time on the clock at home that morning could only have been five minutes past eight 8:05 or 3:55 as a mirror image.
The difference between these clocks is exactly 4 hours and ten minutes
Conclusion: The boy reaches school at five minutes past eight plus twenty minutes of cycling, which is twenty-five minutes past eight.
Q.8. What is the minimum number of weighing operations required to measure 31 kg of wheat if only one weight of 1 kg is available?
When 1 weight of 1 kg is there with that we can weigh 1 kg. Next time we can weigh 2 kg(weight+1 kg wheat). Next time we can weight 4 kg. Next time 8 kg. can be weighed. After that we can weight 16 kg ..So after 5 weighing operations the whole wheat would be weighed.
Q.9. Mr. Ravinder Gulati has 3 timepieces in his house - a wall clock, an alarm clock and a wristwatch. The wrist watch is always accurate, whereas the wall clock gains 2 minutes every day and the alarm clock loses 2 minutes every day. At exactly midnight last night, all three watches were showing the same time. If today is 23 July 2007, then on which date all three clocks will show the same time again?
A clock finishes one round in 12x60 i.e. 720 minutes. If a clock gains 2 minutes everyday, then it would be 720 minutes ahead after 360 days. Thus, after 360 days, it will show the same time again. Similarly, if a clock loses 2 minutes everyday, then it would be 720 minutes behind after 360 days. Thus, after 360 days, it will show the same time again. Thus, after 360 days all three clocks will show the same time again i.e. midnight between 17 July 2008 and 23 July 2008.
Q.10. There are 3 persons- A, B and C. On some day, A lent tractors to B and C as many as they had. After a month B gave as many tractors to A and C as many as they have. After a month C did the same thing. At the end of this transaction each one of them had 16. Find the tractors each originally had?
One way to solve it is by making 3 equations and solve them simultaneously. But there is rather easier way to solve it using back tracking.
It's given that at the end, each had 16 tractors (16, 16, 16) i.e. after C gave tractors to A & B as many as they had. It means that after getting tractors from C their tractors got doubled. So before C gave them tractors, they had 8 tractors each and C had 32 tractors i.e. (8, 8, 32).
Similarly, before B gave tractors to A & C, they had 4& 16 tractors respectively and B had 42 tractors i.e. (4, 28, 16)
Again, before A gave tractors to B & C, they had 14& 8tractors respectively and A had 39 tractors i.e. (26,14, 8)
Hence, initially A had 26 tractors, B had 14 tractors and C had 8 tractors.
