# Top MCQs on Puzzles with Answers

Question 1 |

There are 25 horses among which you need to find out the fastest 3 horses. You can conduct race among at most 5 to find out their relative speed. At no point you can find out the actual speed of the horse in a race. Find out how many races are required to get the top 3 horses.

5 | |

6 | |

7 | |

8 |

**Explanation:**

The minimum no of races to be held is 7.

Make group of 5 horses and run 5 races. Suppose five groups are a,b,c,d,e and next alphabet is its individual rank in this group(of 5 horses).for eg. d3 means horse in group d and has rank 3rd in his group. [ 5 RACES DONE ]

a1 b1 c1 d1 e1

a2 b2 c2 d2 e2

a3 b3 c3 d3 e3

a4 b4 c4 d4 e4

a5 b5 c5 d5 e5

Now make a race of (a1,b1,c1,d1,e1).[RACE 6 DONE] suppose result is a1>b1>c1>d1>e1

which implies a1 must be FIRST.

b1 and c1 MAY BE(but not must be) 2nd and 3rd.

FOR II position, the horse will be either b1 or a2

(we have to fine top 3 horse therefore we choose horses b1,b2,a2,a3,c1 do racing among them [RACE 7 DONE].the only possibilities are :

c1 may be third

b1 may be second or third

b2 may be third

a2 may be second or third

a3 may be third

The final result will give ANSWER. suppose result is a2>a3>b1>c1>b2

then answer is a1,a2,a3,b1,c1.

HENCE ANSWER is 7 RACES

Source: http://www.geeksforgeeks.org/adobe-interview-set-4/

Question 2 |

A Lady (L) bought an item of Rs 100 from the Shopkeeper (C). She paid him with a 500 Rs Note. Realizing that he did not have a change, the shopkeeper C got change for that note from another shopkeeper (S) and paid Rs 400 to the Lady.

After a few days, S realized that the note is fake, And he railed at C and took 500 Rs back from him.

So in this whole process, how much money did C lose in the end?

100 | |

400 | |

600 | |

500 |

**Explanation:**

The total loss for shopkeeper = 500 ( given back to the person who had provided the change )

Consider a transaction box, the lady came with a counterfeit 500 Rs note which can be considered of 0 value.

Now the lady took the item (cost of the item 100 Rs ) and 400 Rs (the change given by shopkeeper(C) to the lady)

from the transaction box, total of 500 Rs. Now the equivalent amount should be lost by someone, thus shopkeeper(C) lost 500 Rs. Another shopkeeper(S) gave 500 Rs and took back the same amount hence no loss for him.

Question 3 |

A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) an unlimited number of times.

20000 | |

25000 | |

15000 | |

40000 |

**Explanation:**

Divide the lifetime of the spare tyre into 4 equal part i.e., 5000 and swap it at each completion of 5000 miles distance.

Let four tyres be named as A, B, C and D and spare tyre be S.

**5000 KMs:**Replace A with S. Remaining distances (A, B, C, D, S) : 15000, 15000, 15000, 15000, 20000.**10000 KMs:**Put A back to its original position and replace B with S. Remaining distances (A, B, C, D, S) : 15000, 10000, 10000, 10000, 15000.**15000 KMs:**Put B back to its original position and replace C with S. Remaining distances (A, B, C, D, S) : 10000, 10000, 5000, 5000, 10000.**20000 KMs:**Put C back to its original position and replace D with S. Remaining distances (A, B, C, D, S) : 5000, 5000, 5000, 0, 5000.**25000 KMs:**Every tyre is now worn out completely.

Question 4 |

Your friend said, “If yesterday was tomorrow, today would be Friday.”

On which day did your friend make this statement ?

Sunday | |

Wednesday | |

Friday | |

Thursday |

**Explanation:**

There are two ways to interpret this.

- If x + 1 == x – 1

Then

x == x – 2

x – 2 = Friday

x = Sunday - If x – 1 == x + 1

Then

x == x + 2

x + 2 = Friday

x = Wednesday

Option 1) makes sense and hence the answer.

Question 5 |

There are two trains(Train A and Train B) running on the same track towards each other at a speed of 100 km/hr. They enter a tunnel 200 km long at the same time. As soon as they enter, a supersonic bee flying at a rate of 1000 km/hr also enters the tunnel from one side (say Train A side). The bee flies towards the other Train B and as soon as it reaches the train B, it turns back and flies back to the Train A. This way it keeps flying to and fro between the Trains A and B. The trains collide after a certain point of time leading to a massive explosion. The task is to find the total distance travelled by the bee until the collision occurred.

100 | |

200 | |

1000 | |

500 |

**Explanation:**

This puzzle can be solved with the help of physics and observation.

- Speed of Train A and Train B =
**100 km/hr** - Length of the tunnel =
**200 km** - Since both the trains are travelling at the same speed, Hence they must have travelled equal distance before colliding, which means that they must have collided at the halfway of the tunnel.
- Therefore, distance travelled by each train before colliding =
**100 km** - Time taken by each train to travel 100 km and collide =
**Distance/Speed = 100/100 = 1 hr** - Therefore the bee must have travelled for 1 hr before colliding with the trains.
- Hence distance travelled by the bee in 1 hr before colliding =
**Speed * Time = 1000 * 1 = 1000 km**

Question 6 |

A farmer bought chickens for 4 unique clients on a selected day. Each customer buys half the amount of chicken left till his turn and half a chicken (i.e., if x chicken were left he buys x/2 + 1/2). The fourth customer buys a single chicken and after his turn, no chicken was left. Can you find the number of chickens the farmer bought on that day?

15 | |

17 | |

14 | |

16 |

**Explanation:**

Let, the customers are **C1, C2, C3, **and **C4** and the Number of chickens is **X**.

** Customer 1:** The first customer will buy (X/2 + 1/2) chicken and the left amount will be X – (X/2 + 1/2) =

**(X – 1)/2**.

** Customer 2:** The amount of chicken the second customer will buy is (X-1)/4 + 1/2 = (X+1)/4. So the amount of chicken left will be

(X – 1)/2 – (X + 1)/4 =

**(X-3)/4**.

** Customer 3:** The amount of chicken bought by the third customer will be ((X-3)/4)/2 + 1/2 = (X-3)/8 + 1/2 = (X+1)/8. So the remaining amount of chicken is (X – 3)/4 – (X + 1)/8 =

**(X – 7)/8**.

** Customer 4:** The amount of chicken bought by the fourth customer is ((X-7)/8)/2 + 1/2 = (X – 7)/16 + 1/2 = (X + 1)/16. The amount of chicken left is (X – 7)/8 – (X + 1)/16 =

**(X – 15)/16**.

Given that the fourth customer purchased only single chicken and no chicken is left after his purchase:

(X – 15)/16 = 0

So X – 15 = 0

X = 15

Therefore, **the farmer sold 15 chickens on that particular day** to each customer as follows,

Customer 1: C1 = (15 + 1)/2 = 8 Chickens

Customer 2: C2 = (15 + 1)4 = 4Chickens

Customer 3: C3 = (15 + 1)/8 = 2 Chickens

Customer 4: C4 = (15 + 1)/16 = 1 Chicken

Question 7 |

Suppose there are four cards labeled with the letters A, B, C, and D and the numerals 3, 4, 5, and 6. It is known that every card has a letter on one side and a number on the other. The rule of the game is that a card with a vowel on it always has an even number on the other side. How many and which cards should be turned over to prove this rule to be true?

1 | |

2 | |

4 | |

3 |

**Explanation:**

**Step 1: **There are four cards labeled as A, B, C, and D.

**Step 2: **One side of each card has a letter and another side has a number. It is known that a card with a vowel has always an even number on the other side. There is only one vowel card in the deck i.e. A. This means that the number on the other side of A can be 4 or 6. These are the only two even numbers. Let’s assume we have number 4 on the other side of card A.

**Step 3: **It is very intuitive that the rule ” A card with the vowel has an even number on the other side” can be proven to be true by turning *two cards. *

**Step 4: **The first one will be the A(vowel) card to verify it does actually have an even number on the back and also the 3(odd number) card to verify that it does not have a vowel on the back.

**Step 5: **Turning only these two cards **prove this rule to be true.** We need not turn any more cards because the puzzle does not exclude the possibility of saying a consonant card having either an odd or an even number on the back.

Question 8 |

There are 3 jars, namely, A, B, C. **All of them are mislabeled**. Following are the labels of each of the jars:

- A: Candies
- B: Sweets
- C: Candies and Sweets (mixed in a random proportion)

You can put your hand in a jar and pick only one eatable at a time. Tell the minimum number of eatable(s) that has/have to be picked in order to label the jars correctly. Assume that the shape of the candies and sweets are identical and there is no way to differentiate them by touching alone.

1 | |

3 | |

4 | |

2 |

**Explanation:**

You have to pick only one eatable from jar C. Suppose the eatable is a candy, then the jar C contains candies only(because all the jars were mislabeled).

Now, since the jar C has candies only, Jar B can contain sweets or mixture. But, jar B can contain only the mixture because its label reads “sweets” which is wrong.

Therefore, Jar A contains sweets. Thus the correct labels are:

- A: Sweets.
- B: Candies and Sweets.
- C: Candies.

Question 9 |

Given a 8×8 chessboard, figure out the maximum number of kings that can be placed on the chessboard so that no two kings attack each other, i.e., none of the kings is under check. A king can move only one step at a time any direction on the chessboard (horizontally, vertically and diagonally).

32 | |

18 | |

8 | |

16 |

**Explanation:**

Whenever we have two kings within a 2×2 square, they are always under check.

By this observation, we can easily conclude that we can have at max one king within a 2×2 square.

We can imagine a 2×2 square as a hole (cage) for our pigeon i.e. kings. So, a 2×2 square occupies 4 sq units of area. The total area for the square chess board is 64 square units (assuming the size of chessboard to be 8 units × 8 units). So, we have 64/4 = 16 cages or holes in this scenario.

We can easily place one king each in these big squares of 2×2. After 16 kings are completed , we would always have a scenario where two kings are placed in a 2×2 square (by Pigeon Hole Principle). This violates our initial condition and hence we can have at max 16 kings which satisfy the above condition.

Even for a n×n square the cage size (2×2 square) remains the same and we can easily say that whenever you exceed the number of cages possible you will definitely have two kings within a 2×2 square. Hence the maximum number of cages is given by: floor((N*N)/4)

Question 10 |

Ram used to arrive at the railway station every day at 6 pm from work. As soon as Ram arrived at the station, his wife Sita too arrives at the station by car to pick him up. Both drive back home. One day Ram arrived at the station one hour early and thought of taking a walk back to home. He started walking when after covering a specific distance, he meets with Sita. Both drive back home and this time they reached 30 minutes earlier than the usual time. How long has Ram been walking?

It is known that Sita drove every day at a uniform speed.

15 | |

30 | |

45 | |

60 |

**Explanation:**

- Since they reached home 30 minutes earlier, Sita must have saved 15 minutes in each trip i.e., up and down trip, as the speed of the car was uniform.
- This means that had Sita drove for 15 more minutes, she would have met Ram at the station.
- As we know Ram arrived at the station at 5 that day, and they must meet at 6, it means that Sita meets Ram at 5:45.
- Therefore, Ram must have walked 45 minutes before meeting Sita.

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