A puzzle presents a challenge that tests the solver's rational thinking abilities. Crafted with clever language and logical elements, a welldesigned puzzle requires unraveling step by step. While often created for entertainment, puzzles can be a source of embarrassment if not successfully solved in public.
In the context of various exams, puzzlesolving serves a straightforward purpose: it aids in the development of cognitive and logical abilities, particularly for successfully navigating logical reasoning sets. Puzzles, being an excellent starting point, are valuable tools for students at all levels, fostering aptitude in logic.
The essence of this section is simple: we firmly believe that engaging in puzzlesolving enhances logical skills. In this section, we offer puzzles aimed at honing your logical abilities. New puzzles are regularly added, so make sure to check this section frequently for updates.
Q1: “Now, then, Mahesh, how old is Migo?” Anupama’s young man asked her brother.
“Well, five years ago,” was the youngster’s reply, “sister was four times older than the dog, but now she is only three times as old.” Can you tell Migo’s age?
Ans: Migo’s present age is ten years and Anupama’s thirty years. Five years ago their respective ages were five and twentyfive. Remember that we said “four times older than the dog,” which is the same as “five times as old.”
Q2: Maninder: “Do you know, dear, that in seven years’ time our combined ages will be sixtythree years?”
Megha: “Is that really so? And yet it is a fact that when you were my present age you were twice as old as I was then. I worked it out last night.”
Now, what are the ages of Megha and Maninder?
Ans: Maninder ‘s age must have been twentynine years and twofifths, and Megha’s nineteen years and threefifths. When Maninder was aged nineteen and threefifths, Megha was only nine and fourfifths; so Maninder was at that time twice her age.
Q3: The Singhs recently were visited by their favourite uncle, who had landed from America. The parents displayed all the five kids. First in line was Bunny and then can Geet. When the uncle asked how old they were, Mr. Singh said that the boy was exactly twice as old as the girl. Then in walked Harpreet, who told that the combined age of Geet and herself was twice that of Bunny. Then the next kid, Chatur came running in and all excited (possibly on account of the fact that he might receive some gifts), and the mother of the kids remarked that the combined ages of the two boys were exactly twice the combined ages of the two girls. The Uncle has barely absorbed all these facts when Jaspreet entered the room and announced, “Uncle you have actually arrived on my 21st birthday”. To this fact, Mr. Singh finally added: “The combined ages of my three girls is exactly equal to twice the combined ages of my two boys”. What were the ages of the kids, according to these facts provided to the uncle of the kids?
Ans: The ages were as follows: Bunny, 3½ years; Geet, 1¾ year; Harpreet, 5¼ years; Chatur, 10½; years; and Jaspreet, 21 years.
Sol: With the given information, we can build the some equations. By reading the puzzle carefully, we can establish that there are two boys and three girls: Boys : Bunny , Chatur Girls : Geet , Harpreet, Jaspreet
Now from the information given in the question, we form the equations to find the ages of these children:
B=2G…………………..I H+G=2B ………………II C+B =2(G+H)…………III J= 21…………………..IV (GIVEN ) J+G+H =2(C+B)………….V
From these equations, we find the respective ages of the kids.
First for the value of H: We have: B=2G Put this in the equation II We have: H+G=3B H+G=2(2G) H+G=4G H=4GG H=3G………………………………VI
We also know J+G+H=2(C+B) J+2B=2C+2B( from eq.II)
As we know the age of J is 21 so 21=2C C=21/2…………………………….VII
Now from eq.III and II C+B=2(2B) C+B=4B C=3B Now put value of c from VII B becomes 21/6=7/2 B=7/2………………………………..VIII Now let’s revisit the first equation now We have: B=2G 7/2=2G G=7/4……………………………….IX
From equation II we have: H+G=2B H=2(2G) – G H= 3G H=3X7/4=21/4………………………X
Thus, solving these equations, we arrive at the ages of all the kids: J = Jaspreet = 21 years C= Chatur = 21/2 years B= Bunny = 7/2 years G = Geet = 7/4 years H= Harpreet = 21/4 years.
Q4: Maninder: Kaval, do you have any children?
Kaval: Yes, I am blessed with three demons.
Maninder: How old are these blessed demons?
Kaval: Well, Maninder, the product of their ages is 36,
Maninder: Now that doesn’t help much. Any more clues?
Kaval: Two of them are the same age as the even number of equal pizza slices we have each had. You should be able to figure it out now at least.
Maninder: Can I have one more clue?
Kaval: Well, the oldest always wears a blue dress. Can Maninder do it, and if he can, what is the age of the three kids?
Ans:
Q5: As a new year gift, three boys got a bag of chocolates. To divide the 770 chocolates in the bag, the boys worked out a formula. They would divide the chocolates in proportion to their ages. Now the boys, whose sum of ages amounted to 17½ years, carried out the following division:
How many chocolates did each boy get, and what were their ages?
Ans: We can see that when Hitesh takes 12 chocolates, Rajan gets 9, and Chetan gets 14. Taken together, this amounts to 35. As 770 is 35 x 22, we simply need to multiply 12, 9, 14 by 22 to discover their share: Hitesh: 264 Rajan: 198 Chetan: 308 As the sum of their ages is 17½ years and they take chocolates in proportion to their ages, we can simply see that we need to divide 12,9, 14 to obtain their ages. Their ages are: 6, 4½, and 7 years. Still confused how we arrived at the solution? This is question about factors. Start from the bottom, that is the ages, and work upwards to the total number of chocolates. You would see how all the numbers fit in.
Q6: Two friends, Kaval and Prashant, meet after a long time.
Kaval: Did you marry at last or still single?
Prashant: As a matter of fact, I got married and I have three kids now.
Kaval: That’s awesome. How old are they?
Prashant: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Kaval: Arghhh. I need more than that to solve this riddle. I am no mathematician.
Prashant: My eldest kid, Niharika, just started with her tennis classes.
Kaval: Ah, that does it.
How old are Prashant’s kids?
Ans: Ages of the kids: 3,3,8 Let’s break it down. The product of their ages is 72. So what are the possible choices?
Product: 2, 2, 18 Sum of (2, 2, 18) = 22 Product: 2, 4, 9 Sum of (2, 4, 9) = 15 Product: 2, 6, 6 Sum of (2, 6, 6) = 14 Product: 2, 3, 12 Sum of (2, 3, 12) = 17 Product: 3, 4, 6 Sum of (3, 4, 6) = 13 Product: 3, 3, 8 Sum of (3, 3, 8 ) = 14 Product: 1, 8, 9 Sum (1,8,9) = 18 Product: 1, 3, 24 Sum (1, 3, 24) = 28 Product: 1, 4, 18 Sum (1, 4, 18) = 23 Product: 1, 2, 36 Sum (1, 2, 36) = 39 Product: 1, 6, 12 Sum (1, 6, 12) = 19
The sum of their ages is the same as your birth date. That could be anything from 1 to 39 but the fact that Kaval was unable to find out the ages, it means there are two or more combinations with the same sum. From the choices above, only two of them are possible now. 2, 6, 6 sum(2, 6, 6) = 14 3, 3, 8 sum(3, 3, 8 ) = 14 Since the eldest kid is taking piano lessons, we can eliminate combination 1 since there are two eldest ones. The answer is 3, 3 and 8
Q7: Factfile:
The combined age of Megha and Anamika is 44 years.
Megha was twice as old as Anamika when Megha was half as old an Anamika will be when Anamika is 3times as old as Megha was when Megha was three times as old as Anamika.
How old is Megha?
Ans: The age of Megha to that of Anamika must be as 5 to 3. And as the sum of their ages was 44, Megha was 27½ and Anamika 16½. One is exactly 11 years older than the other. I will now insert in brackets in the original statement the various ages specified: “Megha is (27½) twice as old as Anamika was (13¾) when Megha was half as old (24¾) as Anamika will be (49½) when Anamika is three times as old (49½) as Megha was (16½) when Megha was (16½) three times as old as Anamika (5½).” Now, check this backwards. When Megha was three times as old as Anamika, Megha was 16½ and Anamika 5½ (11 years younger). Then we get 49½ for the age Anamika will be when she is three times as old as Megha was then. When Megha was half this she was 24¾. And at that time Anamika must have been 13¾ (11 years younger). Therefore Megha is now twice as old—27½, and Anamika 11 years younger—16½
Q8: A mountain hiker is carrying a backpack, as he travels in the hills to climb new peaks. In his bag, he has the complete mountaineering kit and hit stuff. His kit is three kilograms and half a kit heavy. Along with that, his essential utilities are four kilograms minus onefourth the weight of his kit. His food supplies account for twofifth the weight of his kit and essential utilities. To add to that, his repair kit (consisting of various things to repair his mountaineering kit) is half a kg and onefourth his repair kit heavy. His bag is barely onefourth the weight of his repair kit. Calculate the total weight on his shoulders.
Ans: Let the weight of the Kit be= K Now it is given that: K = 3 + K/2 Therefore, K= 6 Kg
Let the weight of essential utilities= U Now it is given that: U = 4 – K/4 U = 4 – 1.5 U = 2.5 Kg
Let the weight of his food supplies= F Now it is given that: F = 2/5 (K+U) F = 2/5 (8.5) F = 3.4 Kg
Let the weight of his repair kit= R Now it is given that: R = 0.5 + R/4 R = .67 Kg
Let the weight of his bag = B Now it is given that: B = ¼ R B = .167 Kg
Therefore the total weight on his shoulders is= K+U+F+R+B = 12.73 Kg
Q9: In Mr. Sharma’s family, there are one grandfather, one grandmother, two fathers, two mothers, one fatherinlaw, one motherinlaw, four children, three grandchildren, one brother, two sisters, two sons, two daughters and one daughterinlaw. How many members are there in Mr. Sharma’s family? Give minimal possible number.
Ans: There are 7 members in Mr. Sharma’s family. Mother & Father of Mr. Sharma Mr. & Mrs. Sharma, his son and two daughters.
Mother & Father of Mr. Sharma   Mr. & Mrs. Sharma   One Son & Two Daughters Check with the conditions given. The above relationships satisfy them all.
Q10: In a zoo:
I. There were 9 male chimps and their baby chimps.
II. There were 2 more female chimps than baby chimps.
III. The number of different malefemale chimp couples possible was 24. Note that if there were 7 male and 5 female chimps, then the total number of couples possible is 35.
Also, of the three groups – males, females and babies at the zoo:
IV. 4 were of one kind.
V. 6 were of another kind.
VI. 8 were of the third kind.
Out of 6 statements given above (labeled I to VI), one statement is false. Identify the false statement.
Ans: Statement (4) is false. There are 3 male chimps, 8 female chimps and 6 baby chimps. Assume that Statements (4), (5) and (6) are all true. Then, Statement (1) is false. But then Statement (2) and (3) both cannot be true. Thus, contradictory to the fact that exactly one statement is false. So Statement (4) or Statement (5) or Statement (6) is false. Also, Statements (1), (2) and (3) all are true. From (1) and (2), there are 11 male and female chimps. Then from (3), there are 2 possible cases – either there are 8 male and 3 female or there are 3 female and 8 male. If there are 8 male and 3 female, then there is 1 baby chimp. Then Statements (4) and (5) both are false, which is not possible. Hence, there are 3 male, 8 female and 6 baby chimps. Statement (4) is false.
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1. What is the purpose of a Puzzle Test? 
2. How can one prepare for a Puzzle Test? 
3. What are some common types of puzzles that may appear in a Puzzle Test? 
4. How can candidates improve their speed in solving puzzles during a Puzzle Test? 
5. Are there any online resources available for practicing Puzzle Tests? 
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