Brain Teaser

 Ques: You are given 2 eggs. You have access to a 100-storey building. Eggs can be very hard or very fragile means it may break if dropped from the first floor or may not even break if dropped from 100 th floor. Both eggs are identical.  You need to figure out the highest floor of a 100-storey building an egg can be dropped without breaking. Now the question is how many drops you need to make. You are allowed to break 2 eggs in the process

Ans: Let x be the answer we want, the number of drops required.

So if the first egg breaks maximum we can have x-1 drops and so we must always put the first egg from height x. So we have determined that for a given x we must drop the first ball from x height. And now if the first drop of the first egg doesn’t breaks we can have x-2 drops for the second egg if the first egg breaks in the second drop.

Taking an example, let’s say 16 is my answer. That I need 16 drops to find out the answer. Lets see whether we can find out the height in 16 drops. First we drop from height 16,and if it breaks we try all floors from 1 to 15.If the egg don’t break then we have left 15 drops, so we will drop it from 16+15+1 =32nd floor. The reason being if it breaks at 32nd floor we can try all the floors from 17 to 31 in 14 drops (total of 16 drops). Now if it did not break then we have left 13 drops. and we can figure out whether we can find out whether we can figure out the floor in 16 drops.

Lets take the case with 16 as the answer

1 + 15 16 if breaks at 16 checks from 1 to 15 in 15 drops
1 + 14 31 if breaks at 31 checks from 17 to 30 in 14 drops
1 + 13 45 .....
1 + 12 58
1 + 11 70
1 + 10 81
1 + 9 91
1 + 8 100 We can easily do in the end as we have enough drops to accomplish the task


Now finding out the optimal one we can see that we could have done it in either 15 or 14 drops only but how can we find the optimal one. From the above table we can see that the optimal one will need 0 linear trials in the last step.

So we could write it as

(1+p) + (1+(p-1))+ (1+(p-2)) + .........+ (1+0) >= 100.

Let 1+p=q which is the answer we are looking for

q (q+1)/2 >=100

Solving for 100 you get q=14.
So the answer is: 14
Drop first orb from floors 14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100... (i.e. move up 14 then 13, then 12 floors, etc) until it breaks (or doesn't at 100)

Ques:  Baseball bat and ball cost $50. If the bat cost $49 more than the ball. What is the cost of each?

Ans: Ball costs $0.5 and bat costs $49.5

Ques: Four people need to cross a dark river at night. They have only one torch and the river is too risky to cross without the torch. if all people cross simultaneously then torch light won’t be sufficient. Speed of each person of crossing the river is different. Cross time for each person is 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the river?

Ans: The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Ques: How many times do a clock's hands overlap in a day?

Ans: 22 times

am

12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55

pm
12:00
1:05
2:11
3:16
4:22
5:27
6:33
7:38
8:44
9:49
10:55

Ques: You can place weights on both side of weighing balance and you need to measure all weights between 1 and 1000. For example if you have weights 1 and 3,now you can measure 1,3 and 4 like earlier case, and also you can measure 2,by placing 3 on one side and 1 on the side which contain the substance to be weighed. So question again is how many minimum weights and of what denominations you need to measure all weights

Ans: For this answer is 3^0, 3^1, 3^2... That is 1,3,9,27,81,243 and 729

Ques: The puzzle is if the shopkeeper can only place the weights in one side of the common balance. For example if shopkeeper has weights 1 and 3 then he can measure 1, 3 and 4 only. Now the question is how many minimum weights and names the weights you will need to measure all weights from 1 to 1000.

Ans:
This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512

Ques: Four people need to cross a rickety bridge at night. Unfortunately, they have only one torch and the bridge is too dangerous to cross without one. The bridge is only strong enough to support two people at a time. Not all people take the same time to cross the bridge. Times for each person: 1 min, 2 mins, 7 mins and 10 mins. What is the shortest time needed for all four of them to cross the bridge?

Ans: 17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Ques: There are 2 sand hourglasses.
The small one can measure 5 hours and the large one can measure 7 hours.

How can we measure 16 hours with 2 sand hourglasses running together ?

Ans: Steps mentioned below
 Start both sand hourglasses at the same time.
 You flip over the small sand hourglass when it finishes the first 5 hours.
 Flip over the large one when it finishes its first 7 hours.
 When the small sand hourglass runs to 10 hours, flip over the small and large sand hourglass at the same time. It means the large sand hourglass will run 3 hours again. When the large sand hourglass run to 13 hours (7 + 3 + 3), flip both the small and large sand hourglass over again to run another 3 hours

 
Ques: You are a king of an empire. You have a servant working in your palace. He works all the seven days and you only pay him in the form of gold bar. You must pay the worker for his work every day at the end of the day.

If you are only able to make two breaks in the gold bar, how will you pay the servant if the servant works for the equal time every day and thus equal amount must be paid at the end of the day?

Ans: If you are thinking of how to cut in such a way that you are able to make seven equal pieces, you are thinking in the wrong direction. You should give more stress on how to make the transaction with the worker.

You have to break the gold bar two times so you have the following sizes of the bar:

1/7 ---- Bar A
2/7 ---- Bar B
4/7 ---- Bar C

When the day 1 ends:
Give Bar A (You remain with Bar B and Bar C; Worker possess - Bar A)

When the day 2 ends:
Give Bar B and take the Bar A back (You remain with Bar A and Bar C; Worker possess - Bar B)

When the day 3 ends:
Give Bar A (You remain with Bar C; Worker possess - Bar A and Bar B)

When the day 4 ends:
Give Bar C and take back Bar A and Bar B (You remain with Bar A and Bar B; Worker possess - Bar C)

When the day 5 ends:
Give Bar A (You remain with Bar B; Worker possess - Bar A and Bar C)

When the day 6 ends:
Give Bar B and take back Bar A (You remain with Bar A; Worker possess - Bar B and Bar C)

When the day 7 ends:
Give Bar A (You remain with no bar; Worker possess - Bar A, Bar B and Bar C)

Thus by the end of seven days, you have taken care of everything.

Ques: There are 100 bulbs lined up in a room. All are turned on in the first pass. Then all the even numbered light bulbs are switched off. After that, every third bulb is switched on. Then all those bulbs that were switched off are turned back on and all those that were lit are turned off. Then the same process is being carried with the fourth bulb and the fifth bulb.

How many bulbs are glowing after 100 passes?

Ans: 10 Bulbs

For this tricky puzzle, you must check how many light bulbs in the row are having an odd number of factors. The first one surely has odd number of factors, the second has even, four has odd. Thus the bulb four and one will remain lit. The bulbs that are going to remain lit are perfect squares as they have an odd number of factors - 1, 4, 9, 16.

Since there are 100 passes, you can go up to 10 times 10 i.e. the square of 10. There are 10 perfect squares available to you - one, two, three, four, five, six, seven, eight, nine and ten. They corresponds to the bulb number 1, 4, 9, 16, 25, 36, 49, 64, 81 and 100. All of them will remain lit and thus total ten bulbs will remain lit after 100 passes.

Ques: The scenario comprises of a wobbly bridge and four people. It is the night time and the people have only one torch. Without torch one may risk his life in crossing the bridge. Also we have a condition; the bridge is not quite strong and can hold only two persons together at one. The four people take different time to cross the bridge - 1 min, 2 min, 7 min and 10 min.

Since the torch is a necessity and the bridge can't hold more than two persons at a time, two persons must travel at a time out of which one must return with the torch so they don't risk their life crossing in the dark.

What is the shortest time that will be required for all of them to cross the bridge?


Ans: Were you thinking of using the fastest person who takes one minute to travel to and fro till everyone has reached the other end? Yes that can be done but it will take 10 + 1 + 7 + 1 + 2 = 21 minutes in total.

If we pay attention towards finding a way to merge the one taking 7 minutes and the one taking 10 minutes, we will have to acknowledge the fact that one of them will also have to return back which will take much more time than 21 minutes.

What if we use the person taking 2 minutes to escort the one taking 1 minutes across? Let's come to the conclusion:

1 and 2 cross the bridge
2 returns with the torch
7 and 10 cross the bridge
1 comes back
1 and 2 cross the bridge

In such a manner, the total time taken will be
2 + 2 + 10 + 1 + 2 = 17 minutes.

Ques: You are invited to a logical competition where some mini games are to be played among the participants that comprises of logic. What they do is make you stand in front of the water tank and give you two jars - one of 3 litres capacity and one of 5 litres capacity. Then they tell you to collect exactly 4 litres of water using the two jugs.

How will you do it?


Ans: You can do it by two methods. However, we are explaining only one and leaving one for you to figure out yourself.

What you will do is fill the 3 litres jar and pour it into the 5 litres jar. Fill the 3 litres jar again and pour it into the 5 litres jar till it's full. Now we know that the 5 litres jar was already having 3 litres of water, thus you would be able to fill 2 litres only. In that case, the 3 litres jar will be left with 1 litres water. Empty the 5 litres jar and then pour the remaining 1 litres from 3 litres jar to the 5 litres jar. Refill the 3 litres jar and pour it completely into the 5 litres jar which already had 1 litres water. Thus now the 5 litres jar have 4 litres of water.

Ques: Briana possess three boxes that vary in sizes. One is large, one is medium and one is small. She places 11 large boxes on her bed. Then, she places 8 medium sized boxes in some and leave some empty. Now she puts 8 small boxes in some of the medium ones and leave the others empty. The small boxes are all empty and now she 102 out of all the boxes on the bed are empty.

How many boxes did Briana used?


Ans: By keeping 8 boxes in a box, the number of boxes that are left empty increases by (8 - 1) i.e. 7.

Now let us assume that x is the number of times the 8 boxes have been put in a box, we will come to know that
11 + 7x = 102
Or x = 13.

Thus in total
11 + 13 * 8 = 115 boxes have been used.

Ques: The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometre. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometre it travels.

What is the largest number of bananas that can be delivered to the market?

Ans: At KM#0, we have 3000 bananas. The maximum bananas the camel can carry is 1000 so the camel must at least make 3 trips from the start point. (Leave #0, Return to #0, Leave #0, Return to #0, Leave #0).
If we move just 1km, we need 1 banana for each step mentioned above thus making a total of 5 bananas for each km.

We continue making 3 trips until we reach a banana count of 2000.
3000 – 5*d = 2000 => d = 200
At #200km, we will have 2000 bananas

At this point, we only need to make 2 trips (Leave #200, Return to #200, Leave #200). This will cost 1 banana for each step thus making a total of 3 bananas for each km.

We continue making 2 trips until we reach a banana count of 1000.
2000 – 3*d = 1000 => d = 333km
At#(200+333) = #534km, we will have 998 bananas

At this point, we need to make one trip so the camel just carries everything and marches toward the market.
Remaining km = 1000 – 534 = 466km. Bananas needed = 466.

Therefore, the bananas remaining once the camel reaches the market is 998 – 466 = 532 bananas