2023 2024 MBA - Puzzles Asked In Elitmus

As you want here I am giving below sample Puzzles for preparation of Elitmus recruitment exam.

In a jar, there are some orange candies and some strawberry candies. You pick up two candies at a time randomly. If the two candies are of same flavor, you throw them away and put a strawberry candy inside. If they are of opposite flavors, you throw them away and put an orange candy inside.

In such manner, you will be reducing the candies in the jar one at a time and will eventually be left with only one candy in the jar.

If you are told about the respective number of orange and strawberry candies at the outset, will it be feasible for you to predict the flavor of the final remaining candy ?

Answer & Explanation

Solution:

At each draw, the number of strawberry candies are either decreasing by 2 or not decreasing at all. In the case of orange candies, at each draw, they are either increasing by 1 or decreasing by 1.

Thus on an assumed outset with at least one candy in the jar to begin with, if the number of strawberry candies are 0 or are even in numbers, they will finish off leaving an orange candy at the end. If otherwise, the remaining candy will be a strawberry one.

There are 100 doors. 100 strangers have been gathered in the adjacent room. The first one goes and opens every door. The second one goes and shuts down all the even numbered doors – second, fourth, sixth... and so on. The third one goes and reverses the current position of every third door (third, sixth, ninth… and so on.) i.e. if the door is open, he shuts it and if the door is shut, he switches opens it. All the 100 strangers progresses in the similar fashion.

After the last person has done what he wanted, which doors will be left open and which ones will be shut at the end?

Answer & Explanation

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.

You are given with two ropes with variable width. However if we start burning both the ropes, they will burn at exactly same time i.e. an hour. The ropes are non-homogeneous in nature. You are asked to measure 45 minutes by using these two ropes.

How can you do it?

Please note that you can’t break the rope in half as it is being clearly stated that the ropes are non-homogeneous in nature.

Answer & Explanation

Solution:

45 minutes

Explanation :
All you have to do is burn the first rope from both the ends and the second rope from one end only simultaneously. The first rope will burn in 30 minutes (half of an hour since we burned from both sides) while the other rope would have burnt half. At this moment, light the second rope from the other end as well. Where, the second rope would have taken half an hour more to burn completely, it will take just 15 minutes as we have lit it from the other end too.

Thus you have successfully calculated 30+15 = 45 minutes with the help of the two given ropes.

A thief was running from the police after the biggest theft the town saw. He took his guard in one of the thirteen caves arranged in a circle. Each day, the thief moves either to the adjacent cave or stay in the same cave. Two cops goes there daily and have enough time to enter any two of the caves out of them.

How will the cop make sure to catch the thief in minimum number of days and what are the minimum number of days?

Answer & Explanation

Solution:

One of the cop will move clockwise while the other will move anti-clockwise every day. In such a manner, they will catch the thief in minimum seven days.

Three college toppers are summoned by the inspecting faculty. To identify the best from them, the faculty takes them into a room and places one hat on each of their heads. Now all of them can see the hats on other’s heads but can’t see his own. There are two colored hats – green and red.

Now the faculty announces that he had made sure that the competition is extremely fair to all three of them. He also gives them a hint that at least one of them is wearing a red hat. Now the first one who is able to deduce his own hat color will be awarded the most intelligent student of all award. After a few minutes, one of them raises his hand and is able to deduce the color correctly.

How?

Answer & Explanation

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.