Puzzles

Gold for 7 Days of Work

Question: You’ve got someone working for you for seven days and a gold bar to pay him. The gold bar has a marks for each unit. The wage of worker is one unit of the gold bar for day. You must pay the worker for their work at the end of every day. If you are only allowed to make two breaks in the gold bar, how do you pay your worker? (Assuming equal amount of work is done during each day thus requiring equal amount of pay for each day)

Answer: The trick is not to try and how to cut in such a way to make 7 equal pieces but rather to make transactions with the worker. Make two cuts on the gold bar such that you have the following sizes of bars.

1/7, 2/7 and 4/7. For convenience sake, I would just refer to the bars as 1, 2 and 4.

At the end of Day 1: Give Bar 1 (You- 2 and 4, Worker- 1)

At the end of Day 2: Give Bar 2, Take back Bar 1 (You- 1 and 4, Worker- 2)

At the end of Day 3: Give Bar 1 (You- 4, Worker- 1 and 2)

At the end of Day 4: Give Bar 4, Take back Bar 1 and Bar 2 (You- 1 and 2, Worker- 4)

At the end of Day 5: Give Bar 1 (You- 2, Worker- 1 and 4)

At the end of Day 6: Give Bar 2, Take back Bar 1 (You- 1, Worker- 2 and 4)

At the end of Day 7: Give Bar 1 (You- Empty, Worker- 1, 2 and 4)

That should take care of everything.

Red and Blue Marbles

Question: You have 50 red marbles, 50 blue marbles and 2 jars. One of the jars is chosen at random and then one marble will be chosen from that jar at random. How would you maximize the chance of drawing a red marble? What is the probability of doing so? All 100 marbles should be placed in the jars.

Answer: Seems tricky at first right? Given that the number of red and blue marbles are the same, you would tend to think that the odds are 50-50. You would try different combinations, such as 25 of each colored marble in a jar or putting all red marbles in one jar and all the blue in the other. You would still end up with a chance of 50%.

So lets think of a better way to distribute the marbles. What if you put a single red marble in one jar and the rest of the marbles in the other jar? This way, you are guaranteed at least a 50% chance of getting a red marble (since one marble picked at random, doesn’t leave any room for choice). Now that you have 49 red marbles left in the other jar, you have a nearly even chance of picking a red marble (49 out of 99).

So let’s calculate the total probability.

P( red marble ) = P( Jar 1 ) * P( red marble in Jar 1 ) + P( Jar 2 ) * P( red marble in Jar 2 )
P( red marble ) = 0.5 * 1 + 0.5 * 49/99
P( red marble ) = 0.7474

Thus, we end up with ~75% chance of picking a red marble.

Camel and Bananas

Question: 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 kilometer. 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 kilometer it travels.

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

Answer: 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.

Find children age

Question: Two old friends, Jack and Bill, meet after a long time.
Three kids
Jack: Hey, how are you man?
Bill: Not bad, got married and I have three kids now.
Jack: That’s awesome. How old are they?
Bill: The product of their ages is 72 and the sum of their ages is the same as your birth date.
Jack: Cool… But I still don’t know.
Bill: My eldest kid just started taking piano lessons.
Jack: Oh now I get it.

How old are Bill’s kids?

Answer: Let’s break it down. The product of their ages is 72. So what are the possible choices?

2, 2, 18 – sum(2, 2, 18) = 22
2, 4, 9 – sum(2, 4, 9) = 15
2, 6, 6 – sum(2, 6, 6) = 14
2, 3, 12 – sum(2, 3, 12) = 17
3, 4, 6 – sum(3, 4, 6) = 13
3, 3, 8 – sum(3, 3, 8 ) = 14
1, 8, 9 – sum(1,8,9) = 18
1, 3, 24 – sum(1, 3, 24) = 28
1, 4, 18 – sum(1, 4, 18) = 23
1, 2, 36 – sum(1, 2, 36) = 39
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 31 but the fact that Jack 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.

The Ant Problem

Question: Three ants are sitting at the three corners of an equilateral triangle. Each ant starts randomly picks a direction and starts to move along the edge of the triangle. What is the probability that none of the ants collide?

Answer: So let’s think this through. The ants can only avoid a collision if they all decide to move in the same direction (either clockwise or anti-clockwise). If the ants do not pick the same direction, there will definitely be a collision. Each ant has the option to either move clockwise or anti-clockwise.

P(No collision) = 2/8 = 0.25

Trains and Bird

Question: A train leaves City X for City Y at 15 mph. At the very same time, a train leaves City Y for City X at 20 mph on the same track. At the same moment, a bird leaves the City X train station and flies towards the City Y train station at 25 mph. When the bird reaches the train from City Y, it immediately reverses direction. It then continues to fly at the same speed towards the train from City X, when it reverses its direction again, and so forth. The bird continues to do this until the trains collide. How far would the bird have traveled in the meantime?

Answer: Yes, you read it right. The bird is actually the fastest moving object in the problem!

Knowing that the bird is the faster than both the trains, you would only imagine that theoretically, the bird could fly an infinite number of times between the trains before they collide. This is because you know that no matter how close the trains get, the bird will always complete its trip before the crash happens. At the time of the crash, the bird would probably get squashed between the trains!

I bet sometime in school, you learnt how to sum up an infinite series. But do we have to do that?

The concept of relative speed (rings a bell?) can work handy here. Let’s assume that the distance between City X and City Y is d miles. The trains are approaching each other at a relative speed of (20 + 15) = 35 mph. The sum of the distances covered by the trains when they collide is d (i.e. the distance between the cities). Since distance/speed gives us time, we know that the trains collide d/35 hours after they start.

Since the speed of the bird is constant at 25 mph, we know that the bird would have covered

25 * (d/35) miles = 5d/7 miles

before the trains collide.

A Box of Defective Balls

Question: You have 10 boxes of balls (each ball weighing exactly10 gm) with one box with defective balls (each one of the defective balls weigh 9 gm). You are given an electronic weighing machine and only one chance at it. How will find out which box has the defective balls?

Answer: For convenience sake, let’s name the boxes from 1 to 10. In order to solve this problem, you have to leverage the fact that you know exactly what each good ball is supposed to weigh and what each defective ball is supposed to weigh. Many of us instinctively will take one ball out of each box and try to find a way to make it work but the trick to take different number of balls from each box.

The number of balls you pick from each bag is equal to the box number. For example, pick 1 ball from box 1, 2 balls from box 2 and so on. In total you will have 55 balls. If all of the boxes have good balls, then the total weight of these balls would be 550gm.

If box 1 has defective balls, then the total weight should be 1gm less than expected (only one ball weighing 9 gm). If box 2 has defective balls, then the total weight should be 2gm less than expected (two balls weighing 9 gm). So once you weigh the set of chosen balls, find out the difference between the total expected weight and the total weight. That number represents the box number which contains the defective balls.

Four People on a Rickety Bridge

Question: 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?

Answer: 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

Boys and Girls

Question: In a country where everyone wants a boy, each family continues having babies till they have a boy. After some time, what is the proportion of boys to girls in the country? (Assuming probability of having a boy or a girl is the same)

Answer: Assume there are C number of couples so there would be C boys. The number of girls can be calculated by the following method.

Number of girls = 0*(Probability of 0 girls) + 1*(Probability of 1 girl) + 2*(Probability of 2 girls) + …
Number of girls = 0*(C*1/2) + 1*(C*1/2*1/2) + 2*(C*1/2*1/2*1/2) + …
Number of girls = 0 + C/4 + 2*C/8 + 3*C/16 + …
Number of girls = C
(using mathematical formulas; it becomes apparent if you just sum up the first 4-5 terms)

The proportion of boys to girls is 1 : 1.

Coins on the Table

Question: You are blind folded and there is a table on which a number of coins are placed. You also know that there are as many coins with Head up as many coins with Tail up. Now you have to divide the coins (number of coins is even) into two equal piles such that number of coins with Heads up and Tails up in either piles be the same. You cannot determine the sides (for sure) if you are blinded.

Answer:

Make 2 piles with equal number of coins. Now, flip all the coins in one of the pile.

How this will work? Let's take an example with 10 coins

So initially there are 5 heads, so suppose you divide it in 2 piles.

Case 1:

P1 : H H T T T
P2 : H H H T T

Now when P1 will be flipped
P1 : T T H H H

P1(Heads) = P2(Heads)

Another case:

P1 : H T T T T
P2 : H H H H T

Now when P1 will be flipped
P1 : H H H H T

P1(Heads) = P2(Heads)

Summary: Divide the coins in half by quantity (easy to count coins while blindfolded).

Then, flip all the coins in one pile over.

Is Your Husband a Cheat?

Question: A certain town comprises of 100 married couples. Everyone in the town lives by the following rule: If a husband cheats on his wife, the husband is executed as soon as his wife finds out about him. All the women in the town only gossip about the husbands of other women. No woman ever tells another woman if her husband is cheating on her. So every woman in the town knows about all the cheating husbands in the town except her own. It can also be assumed that a husband remains silent about his infidelity. One day, the mayor of the town announces to the whole town that there is at least 1 cheating husband in the town. What do you think happens?

Answer: Stumped? Let’s solve this methodically. Say there was only 1 cheating husband in the town. There will be 99 women who know exactly who the cheater is. The 1 remaining woman, who is being cheated on, would have assumed there are no cheaters. But now that the mayor has confirmed that there is at least one cheater, she realizes that her own husband must be cheating on her. So her husband gets executed on the day of the announcement.

Now let’s assume there are 2 cheaters in the town. There will be 98 women in the town who know who the 2 cheaters are. The 2 wives, who are being cheated on, would think that there is only 1 cheater in the town. Since neither of these 2 women know that their husbands are cheaters, they both do not report their husbands in on the day of the announcement. The next day, when the 2 women see that no husband was executed, they realize that there could only be one explanation – both their husbands are cheaters. Thus, on the second day, 2 husbands are executed.

Through induction, it can be proved that when this logic is applied to n cheating husbands, they all die on the n th day after the mayor’s announcement.

Matching Nuts And Bolts

Question:

You are given a collection of nuts of different size and corresponding bolts. You can choose any nut & any bolt together, from which you can determine whether the nut is larger than bolt, smaller than bolt or matches the bolt exactly. However there is no way to compare two nuts together or two bolts together. Suggest an algorithm to match each bolt to its matching nut.

Compute time complexity of your algorithm.

Answer:

As we can’t compare a nut with other nuts and so for bolts, we can compare a nut with bolts and vice-versa. Also keep in mind that if a Nut(N) is smaller than a Bolt(B) than N is fit only for bolts smaller than B. Similarly if a Bolt(B’) is smaller than a Nut(N’) than N is fit only for bolts smaller than B’. So we can go like following in solving this problem.

● Take a nut from the nuts pile

● Divide bolts around it in 2 parts, which are smaller and larger than this.

● Find a matching bolt to this nut.

● Divide nuts in 2 parts, which are smaller and larger than matching bolt.

Now we have 2 subsets of Nuts and Bolts. If we pick a nut from small size nuts pile, we will get a corresponding bolt in smaller size bolts pile and similarly for large size piles. At every step, we will be able to divide these piles in 2 halves and reduce complexity by a factor of 2 in average case. In this case time complexity will be O(nlogn).

In worst case we might have choose the smallest/largest nut/bolt as reference and our one pile will have zero nut/bolt and another pile will have all the remaining. In this case time complexity will be O(n^2).

Algorithm:

Take a nut; partition the bolts with respect to this nut. We will find the matching bolt for this nut. Now take the bolt; partition the nuts.

In this manner, we have divided the Original problems into 2 sub-problems.

T(n) = T( i ) + T(n-i ) + O(n)

Where we have chosen the ith largest nut from a particular pile.

Average time complexity = O(nlogn).

Worst time complexity = O(n^2).

100 Prisoners with Red/Black Hats

Question: 100 prisoners in jail are standing in a queue facing in one direction. Each prisoner is wearing a hat of color either black or red. A prisoner can see hats of all prisoners in front of him in the queue, but cannot see his hat and hats of prisoners standing behind him.
Question: The jailer is going to ask color of each prisoner’s hat starting from the last prisoner in queue standing on steps. If a prisoner tells the correct color, then is saved, otherwise executed. How many prisoners can be saved at most if they are allowed to discuss a strategy before the jailer starts asking colors of their hats.

Answer:

At-most 99 prisoners can be saved and the 100th prisoner has 50-50 chances of being executed.
The idea is that every prisoner counts number of red hats in front of him.

100th prisoner says red if the number of red hats is even. He may or may not be saved, but he coneys enough information to save 99th prisoner.

The 99’th prisoner decides his answer on the basis of answer of 100’th prisoner’s answer. There are following possibilities and 99’th prisoner can figure out color of his hat in every case.

If 100’th prisoner said ‘Red’ (There must have been even number of red hats in front of him)
a) If 99’th prisoner sees even number of red hats in front of him, then his color is black.
b) If 99’th prisoner sees odd number of red hats in front of him, then his color is red.

If 100’th prisoner said ‘Black’ (There must have been odd number of red hats in front of him)
a) If 99’th prisoner sees even number of red hats in front of him, then his color is Red.
b) If 99’th prisoner sees odd number of red hats in front of him, then his color is Black.

The 98’th prisoner decides his answer on the basis of answer of 99’th prisoner’s answer and uses same logic.

Similarly other prisoners from 97 to 1 are saved.

Find the fastest 3 horses

Question: 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 (5 race tracks only). 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.

Answer:

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 tis 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, horse will be either b1 or a2
(we have to find 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

Please note that the 7 races work for the case also when all top 3 horses are same group or any top two horses are in same group. The group which has top 3 horses would always have winner in 6th race. In 7th race, we consider 2nd and 3rd horses of the group whose horse is overall winner. We also consider 2nd horse of the group whose horse came 2nd in 6th race.

Find the Jar with contaminated pills

Question: You have 5 jars of pills. Each pill weighs 10 grams, except for contaminated pills contained in one jar, where each pill weighs 9 grams. Given a scale, how could you tell which jar had the contaminated pills in just one measurement?

Answer:

Take out 1 pill from jar 1, 2 pills from jar 2, 3 pills from jar 3, 4 pills from jar 4 and 5 pills from jar 5. Put all these 15 pills on scale. The correct wight is 150 (15*10). But one of the jars has contaminated pills. So the wight will definitely less than 150. If the wight is 149 then jar 1 has contaminated pills because the there is only one contaminated pill. If the wight is 148 then jar 2, if the wight is 147 then jar 3, if 146 then jar 4, if 145 then jar 5.