Google interview questions – fun brain teasers!

08 Sep

1. How many golf balls can fit in a school bus?

2. You are shrunk to the height of a nickel and your mass is proportionally reduced so as to maintain your original density. You are then thrown into an empty glass blender. The blades will start moving in 60 seconds. What do you do?

3. How much should you charge to wash all the windows in Seattle?

4. How would you find out if a machine’s stack grows up or down in memory?

5. Explain a database in three sentences to your eight-year-old nephew.

6. How many times a day does a clock’s hands overlap?

7. You have to get from point A to point B. You don’t know if you can get there. What would you do?

8. Imagine you have a closet full of shirts. It’s very hard to find a shirt. So what can you do to organize your shirts for easy retrieval?

9. Every man in a village of 100 married couples has cheated on his wife. Every wife in the village instantly knows when a man other than her husband has cheated, but does not know when her own husband has. The village has a law that does not allow for adultery. Any wife who can prove that her husband is unfaithful must kill him that very day. The women of the village would never disobey this law. One day, the queen of the village visits and announces that at least one husband has been unfaithful. What happens?

10. In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?

11. If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?

12. If you look at a clock and the time is 3:15, what is the angle between the hour and the minute hands? (The answer to this is not zero!)

13. Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it’s only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?

14. You are at a party with a friend and 10 people are present including you and the friend. your friend makes you a wager that for every person you find that has the same birthday as you, you get \$1; for every person he finds that does not have the same birthday as you, he gets \$2. would you accept the wager?

15. How many piano tuners are there in the entire world?

16. You have eight balls all of the same size. 7 of them weigh the same, and one of them weighs slightly more. How can you find the ball that is heavier by using a balance and only two weighings?

17. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)

The original place I got these questions is apparently gone now, so here’s a new link:
Interview Cake list

Posted by on Saturday, September 8, 2007 in brain teasers, google interview questions

117 responses to “Google interview questions – fun brain teasers!”

1. Friday, November 16, 2007 at 4:38 am

Hi All
Answer 17: This is a classical question. and surprisingly the hint given is correct!
The answer is Pirate 5 98, Pirate 2 1 and Pirate1 1 gold coin.

Reasonning:
pirate 2 plan: pirate 5 dead, Pirate 4 dead, pirate 3 dead, pirate 2 0 , pirate 100 coin.
pirate 3 plan :pirate 5 dead, Pirate 4 dead, pirate 3 99 , pirate 2 1 coin, pirate 0 coin.
pirate 4: pirate 5 dead, Pirate 4 98 coin, pirate3 0 coin , pirate 2 1 coin, pirate 1 coin.

pirate 5: pirate 5 98 , Pirate 4 0 coin, pirate3 0 coin , pirate 2 1 coin, pirate 1 coin.

This will the plan of the fifith pirate.

• Monday, May 17, 2010 at 5:12 pm

If you give pirates more credit for their thinking skills and desire to not die, you get a different result:
pirate 1 plan: always vote no so he gets all 100.
pirate 2 plan: give pirate 1 100 coins so 2 doesn’t die.
pirate 3 plan: basically it’s not going to end well for 3. offer 1 all the coins and tell 2 he better agree or pirate 1 may have him killed anyway. hope for the best.
pirate 4 plan: hey 2 and 3, if i die, you both get nothing *and* you probably die. i’m taking all 100 coins, vote with me if you want to live.
pirate 5 plan: hey 2 and 3, look at plans 1-4: they leave you either dead or with nothing. i’ll give you each 1 coin and take 98.

2. Friday, January 11, 2008 at 2:03 pm

You’re a pirate.

3. Friday, January 18, 2008 at 3:30 pm

pirate 5’s plan ( top )
98 for himself(p5)
0 for p4
0 for p3
1 for p2
1 for p1

p1 & p2 have doubt that, they may not get anything if p4 or p3 puts the plan, hence they will be happy with even 1 coin.

4. Tuesday, February 5, 2008 at 6:44 pm

I didn’t like the way the question is asked first of all. It does not contain all the necessary information. Also the solutions posted here are not correct. You can find the solution in this link: http://en.wikipedia.org/wiki/Pirate_game

This problem is a game theory question and I think is not very appropriate for an interview where people struggle to answer many different questions under pressure.

• Monday, May 17, 2010 at 5:17 pm

I think the question here is different — in wikipedia’s pirate game, the proposer gets a vote. Here he doesn’t (at least, that’s the way I read it), but still needs half of the remaining pirates to agree with him.

5. Tuesday, March 18, 2008 at 4:05 am

6. Wednesday, May 14, 2008 at 2:15 pm

i cant find an answer to any of the brain teasers.

7. Friday, May 23, 2008 at 11:00 am

#1
About 500,000, assuming the bus is 50 balls high, 50 balls wide, and 200 balls long

#2
Use the measurement marks to climb out
Try to unscrew the glass
Risk riding out the air current

#3
Assuming 10,000 city blocks, 600 windows per block, five minutes per window, and a rate of \$20 per hour, about \$10 million

#4
Instantiate a local variable. Call another function with a local. Look at the address of that function and then compare. If the function’s local is higher, the stack grows away from address location 0; if the function’s local is lower, the stack grows towards address location 0. (If they’re the same, you did something wrong!)

#5
A database is a way of organizing information. It’s like a genie who knows where every toy in your room is. Instead of hunting for certain toys yourself and searching the whole room, you can ask the genie to find all your toy soldiers, or only X-Men action figures, or only race cars — anything you want.

#6
The clock hands don’t overlap at 1:05, 2:10, 3:15 etc. Because, by the time the minutes hand reaches where the hours hand was, the hours hand has moved on, as it doesn’t remain fixed through the entire hour. the answer can be worked out in this way:
In every hour, as the hour hand moves across 5 minute spaces, the minute hand moves across 60. That means the minute hand gains 55 min sp. over hour hand in 60 min. Thus, between the nth and (n+1)st hr, to overlap the minute hand must gain the 5n min sp. that exists between them when the clock strikes n hrs. Apply unitary method and you get the time in which it does so i.e. 60n/11 min. So, the clock hands overlap at n hrs 60n/11 min. Obtain the values by putting m = 0 to 11, and you will find that from 00:00:00 to 23:59:59 (please note that 23:59:60 is equivalent to 24:00:00 or 00:00:00), the hands overlap 22 times.

#7
The two most common way to tackle problem like this is using Depth-First Search (DFS)and Breadth-First Search (BFS).

In the example above, You drive and drive around until you hope to find the destination. Reach a deadend? Turn around and come back where you came from. It might not be the shortest way, but if you tried every possible road, you’ll know that maybe you can’t get to China from New York (And that’s a DFS!).

Understanding these leads to you more exotic things like A* Search, and it’s a fundamental exploring algorithm. Too many things can be reduced into a graph, and knowing the best way to search a graph will come in handy. DFS is basically a stack-based implementation, while BFS is a queue-based implementation, and each with its strength and weaknesses.

#8
Hash it. Take every shirt, and put it in a different drawer depending on the color of the shirt. Whenever you want a shirt, all you have to do is simply open the drawer with the corresponding color. That’s hashing in a nutshell.

#9
Since each wife never knows her husband has cheated on her, there is no way of killing an husband because she can’t prove it.

#10

Half the couples have boys first, and stop. The rest have a girl. Of those, half have a boy second, and so on.
So suppose there are N couples. There will be N boys. There will be an “infinite” sum of girls equal to N/2 + N/4 + N/8 + … As any college math student knows, this sum adds up to N. Therefore, the proportion of boys to girls will be pretty close to 1:1, with perhaps a few more boys than girls because the sum isn’t actually infinite.

#11
Since you know the probability of seeing a car in 30 minutes is 0.95, then look at the probability of seeing a car in the first 10 minutes, or the second 10 minutes or the third ten minutes.

So then it is like a disjunctive probability: P( A or B or C ), i.e. P( 1st 10m or 2nd 10m or 3rd 10m ).

The formula for a disjunctive probability is

P(A or B or C) = P(A) + P(B) + P(C)

Since the default probability is constant, then, P(A) = P(B) = P(C) = x, so

3x=0.95

x = 0.95/3

#12
The hour hand moves at 30 degrees per hour, so the angle is 7.5 degrees.

#13
Let us organize our thought… let’s say that we have 4 persons, each identified with a letter: A, B, C, or D. Let’s say that A is the fastest, while D is the slowest. Each person’s speed is thus:

A – 1 bridge/min

B – 1/2 bridge/min

C – 1/5 bridge/min

D – 1/10 bridge/min

The flashlight only lasts 17 minutes. If they take turns, allow a buddy to cross the bridge and only then start crossing the bridge, then they will take 1+2+5+10 = 18 minutes, which is 1 minute past the flashlight’s battery lifetime.

MY SOLUTION: based on intuition, I would propose that the fastest person (A) should cross first, hold the flashlight and illuminate the bridge for the others. Then the slowest one (D) and the remaining fastest one (B) would start crossing, and once B gets to the other side, then C starts crossing. This way, we would take only 11 minutes to cross the bridge, and we would never have more than 2 persons simultaneously on the bridge. To make it more visualizable:

[0,1] minutes -> A is crossing the bridge
[1,3] minutes -> B and D are crossing
[3,8] minutes -> C and D are crossing
[8, 11] minutes -> D is alone crossing

#14
There are 10 people: p1 p2 p3 p4 p5 p6 p7 p8 f y
F=Friend
Y=You
Thus taking the bet puts you at a -1\$ start assumeing that your friend doesn’t share your birthday. The chance of someone else having your birthday is 7.4% thus you would be at a huge disadvantige. You would end up in the hole roughly 19\$, Because you have your own birthday.

#15
First think about how many pianos there are and how often they are used. Pianos are typically owned by middle to upper class households, and are also found schools, churches, hotels, etc.

The US population is approx 300 million. Assuming that each household has 3 people in it, that leaves 100 mil lion households. Of those, 50 million are in the middle to upper class and may have pianos. Of those 50 million households, you can think that no more than 10 percent have pianos. (That is probably high…it may be more like 2-3%). Anycase, that leaves us with approximately 5 million households with pianos.

Now you need to think about how many piano tuners it takes to maintain those pianos. I would say a typical piano tuner could tune approx 20 pianos a week (2 hours each), and given a full year, that is approximately 1000 pianos.

Now, how often do pianos need to be tuned? Once a year maybe? If that is the case, then there would be 1 piano tuner for every 1000 pianos. That mean approx 5000 piano tuners in the US.

Now think about the rest of the world, it’s population, and it’s wealth. I would guess that the US has 1/3 of the world’s pianos. Therefore, I would guess the total number of piano tuners in the world is approximately 15000.

#16
Put 4 balls on either of pan, you will get to know which pan is heavier. Now take the four balls from heavier pan and split them in group of two and place in each pan of balance to get the heavier group. And you can physically feel and can distinguish the heavier of the 2 balls.

#17
P5 98 (High Ranking)
P4 1 (So as not to complain)
P3 1 (So as not to complain)
P2 0
P1 0

P1 and 2 can complain all they want, but as long as P3 or 4 dosent complain the high ranking pirate get to stay alive.

• Saturday, April 9, 2011 at 12:00 pm

#16
Weigh three balls each:
Scenario 1:
Balls are equal in weight (you know one of the other two is the heavy and weigh one each)

Scenario 2:
Balls are unequal. Take two of the three heavier balls and weigh. If they are equal the other is the heavy ball, else one of the balls will weigh heavier.

• Wednesday, January 15, 2014 at 5:52 am

@ justin; the question is , how many golf balls fit in school bus,? jst think logically and look at the spelling” schOOl bus, and your answer must be “2”..if am not wrong..

• Monday, March 17, 2014 at 7:42 pm

# 13: Let 10 min and 2 min start to cross the bridge. Since they are walking at the pace of 10 min guy, it will take them 10 min to reach the other end. Once reached, ask the 2 min guy to turn around and simultaneously have the 5 min guy start walking.
It will take 5 mins to get the 5 min guy to the other end (by then the 2min guy would have already reached). Now the bridge is free and the clock has 2 mins still remaining.
With 2mins to go, the 2min guy and 1min guy can cross the bridge together. Total time: 17mins.

8. Saturday, June 7, 2008 at 8:41 pm

13. Try the following:
1. The 1min and 2min cross and 1 min returns. Total time=3min
2. The 5min and 10min cross and the 2 min returns. Total Time=12min.
3. The 1min and 2min cross. Total time =2min.

Total time to cross is 17min.

9. Monday, June 9, 2008 at 9:49 am

13. All the answers posted here are wrong. This is the only sequence.

a: 1 min and 2 min cross and 1 min returns = 2 min (not 3)
b: 5 min and 10 min cross and 5 min returns = 10 min
c: 1 min and 5 min. cross = 5 min
total 17 min

• Friday, July 17, 2009 at 9:09 pm

Question 13 is a classic brain teaser and is incomplete as listed here when compared to most writings of it. This site has left out a key part of the problem which is “each pair crosses the bridge at the SLOWEST pace for the pair”.

However, based on your example, you could infer that if 1 and 2 crossed at the same time and 1 returned but only took 2 minutes total to do so, that 2 would be alone on the bridge without a flashlight at the 1 minute mark and that violates the stated parameters.

Try again but this time include the slowest-pace constraint. Your example would take 23 minutes. It is possible in 17 minutes. Keep trying.

a: 1 min and 2 min cross and 1 min returns = 3 mins
b: 5 min and 10 min cross and 5 min returns = 15 mins
c: 1 min and 5 min. cross = 5 min
total 23 min

• Sunday, July 26, 2009 at 9:26 pm

You dare to right other people…
b: 5 min and 10 min cross and 5 min returns = 10 min

that’s 15 minutes in case you’re still wondering

• Sunday, September 7, 2014 at 8:38 pm

You’re forgetting about return time dummy.

10. Wednesday, June 18, 2008 at 10:49 pm

I had to take a moment and thank you for the post you wrote. I have been trying to understand about interview questions for a while, there is so much information out there, but your post helped me understand the concept.

Thanks!

11. Wednesday, June 25, 2008 at 12:51 am

From eight balls take each half and place it in balance one which weighs more will have the heaviest ball among them
then split half among them again continue .again pick balls which weighs higher than the other. now remaining two balls
one with heavier weight can be found in same way.

12. Monday, August 4, 2008 at 1:43 pm

concerning the balls question (#16):
split the balss to three groups, containing three,three and two balls.
compare the two groups of 3 balls:
a. if they are equal, the heavy ball is within the group of two, and you can find it by an additional comparison (two comparison in total).
b. otherwise, take the heavier group of 3 balls, and from it, take two arbitrary balls and compare them (your second comparison)
b1. if they are equal in weight, the third ball (the one you didn’t take) is the heavier
b2. if one is heavier than the other – it is the heavier ball.

13. Monday, August 4, 2008 at 2:11 pm

Most of these are bitten from “How Would You Move Mount Fuji,” a book about interviewing at Microsoft. 😉

14. Monday, August 4, 2008 at 2:32 pm

@9 – your numbers are wrong.

If the 1 & 2 minute campers cross it is going to take the slower of the times (2 minutes). The 1 minute camper then returns which equals 3 minutes total.

same with the 5 & 10 minute campers. If both cross the time it is going to take the 5 minute camper 10 minutes since the 10 minute camper can only cross in that amount of time. then if you return the 5 minute camper you have wasted 15 minutes.

you are right in the last one where if the 1 minute and 5 minute campers cross it will take 5 minutes.

15. Monday, August 4, 2008 at 2:41 pm

#16 – The solutions above take 3 weighings so I’d thought I’d take a stab at it.

Take 6 balls and place 3 on one side, 3 on the other.

If the scales says they weigh the same, take the last two balls and place one on each side. Which ever is heavier, is well, heavier 🙂

If the balance says one side is heavier, take the 3 balls from the heavier side. Select two of the balls from the 3 you now have and set one on each side. If the scale says they are the same, the third ball is heavier. If the scale says one of the two balls is heavier, then that one’s the heaviest.

16. Monday, August 4, 2008 at 2:43 pm

I don’t think any posts have gotten #16 right yet. My solution:
We have 8 balls:
oooooooo

First weighing:
ooo vs ooo, and keep oo aside.

If one of the ooo’s is heavier:
Second weighing:
o vs o, and keep o aside.
If one of the o’s is heavier, it’s that one, and if they are even, it’s the o you kept aside.

Else if ooo vs ooo yields even:
Second weighing:
o vs o, which ever one is heavier is it.

Fun questions!

17. Monday, August 4, 2008 at 3:44 pm

I don’t think these questions are prepared such that the answers involve trying to outsmart the facts presented. 13 isn’t like “oh, gee, just kind of shine it back so the others can see”, 16 isn’t like “oh, cheat and sneak a third measure, but, gosh, since you’re using your hands it’s okay!”

16. Group the eight balls into three groups, say 123, 456, and 78. Measure 123 vs. 456; there are three possible outcomes: 123 is heavier, 456 is heavier, or they are equal. If they are equal, measure 78 and find the heavier ball. If one side of the first measure is equal, take two balls from that trio and measure them; if they are equal, the odd ball out is heavier, other the, uh, heavier one is heavier.

18. Monday, August 4, 2008 at 4:03 pm

#16: Put 3 balls on each side of the balance, keep two in reserve. If the two sets of 3 are equal, weigh the two held in reserve, the heavier one is the outlier.

If the two sets of 3 are not equal, take the heavier set, and put 1 ball on each side of the balance, hold 1 in reserve. If they’re equal, then the 1 in reserve is the outlier, else one of the two weighed ball is heavier and the outlier.

19. Monday, August 4, 2008 at 4:53 pm

you’re all wrong about the ball question.

weigh three balls against three
if same, weigh remaining two
if different, take heavier set of three and weigh two of them

20. Monday, August 4, 2008 at 6:14 pm

The insight with the scales is it is a trinary serach, not a binary one – so each measurement should have 3 possible outcomes, else you are wasting information, and you need all the information you can get, since 2 weighings, each giving a trit of information can only give you 9 (3*3) combinations, and you know there are 8 possible combinations, so you can’t waste more than 1!

21. Monday, August 4, 2008 at 7:01 pm

You only get 2 weighings of the balls.

Weigh 3 vs. 3. If they are equal, throw them out and weight the remaining 2 to find the heaviest.

If one side of the 3 vs. 3 weighing is heavier, then weigh 2 of the heavier side. If they are equal, the heaviest is the third, or if they are unequal, you’ve found the heaviest.

22. Monday, August 4, 2008 at 8:14 pm

For #16:

Take 6, (3 on one side, 3 on the other). If it’s balanced, put the remaining two on. If it’s not balanced, take the 3 balls from the heavier side. Put one on each side and keep one in the hand. If it’s balanced, the one in the hand is heaviest.

23. Monday, August 4, 2008 at 8:52 pm

Rohit you are no good! The bridge crossing problem……

When two of the people cross they can only walk at the slowest person’s pace! So, the person above you “Anonymous” is right…..silly person you is.

24. Monday, August 4, 2008 at 9:22 pm

#16
This is a deceptive question, but the correct solution is as follows:
Put balls 1, 2, and 3 on one side of the balance and balls 4, 5, and 6 on the other. If the balance is equal, then for your second weighing, put ball 7 in one side and 8 in the other.
If 1, 2, and 3 are heavier, then put 1 on one side and 2 on the other. If the balance is equal, then 3 is the heavier ball.
(similar procedure if 4, 5, and 6 are heavier)

The key is to realize that the balance can give three readings: side 1 is heavier, side 2 is heavier, or both sides are equal. All three possibilities should give useful information.

(also, Anonymous at 8 has the correct answer to #13)

25. Monday, August 4, 2008 at 9:42 pm

For the balls you weigh three balls and three balls. If they weigh the same as each other than the heavy ball is one of the two left out, weigh them and see which one is heavier. If either of the three together are heavier take those three, weigh two of them and leave one on the side. If either weighed ball is heavier that’s the answer, if they are equal than you know the left out ball of the group of three was the heavier.

26. Monday, August 4, 2008 at 10:00 pm

#16:

Grab six balls and weigh them 3 vs 3.
– easy case – if they balance, just weigh the remaining 2

otherwise, put the balls in the lighter pan aside and pick two of the balls from the heavier pan. Weigh them against one another. In the case that both are equal, the ball you excluded from the second weighing is heavier.

27. Monday, August 4, 2008 at 10:34 pm

you call these fun brain teasers ???
No way. Google has one of the worse and most ridiculous politic regarding hiring.
They ask you weird questions in order to decide whether you are fit to fulfill or not the jobs they have to offer.
As if working for Google was super cool and like no other company or job could be better …

28. Monday, August 4, 2008 at 11:29 pm

#16: we can weigh them only twice. so first take 3-3 balls in the balance and find out which is heavier. if they both are equal, the rest 1-1 balls can be weight for the 2nd time and the heavier of them is found. if the first 3-3 balance is not equal, then take the heavier ‘3’ and from those three, put 1-1 balls in the balance(this would be the 2nd attempt). if they turn up equal, the remaining 3rd is the heavier. else the answer is obvious. check again ‘justin’ 😉

#13: rohit above is getting confused. what ‘anonymous’ said just above him is the absolute answer.

#9: justin above could be wrong. all wives can make some declarations as follows, they can easily find the cheaters:
1) take each man and put him in the doc.
2) all women who ‘find’ him cheated, should raise hands.
so, if a man stands in the doc and his wife is the only woman, who is not raising her hand, then he is ‘cheating’ on her. this is sufficient proof 🙂

#17: needs more tuning. pl correct it according to the original question.

nice page. enjoyed it. thanks

29. Monday, August 4, 2008 at 11:42 pm

13. They wait until morning, and then don’t need the flashlight.

30. Monday, August 4, 2008 at 11:51 pm

From eight balls, take six and weigh them, three vs. three. If they weigh the same, weigh the two remaining balls to find the heavier one. If one of the groups of three is heavier, take two and weigh them against each other. If one of the two is heavier you have found the ball, if they weigh the same the heavy ball is the one not weighed.

31. Tuesday, August 5, 2008 at 12:00 am

Problem 16:
take 2 sets of 3 balls and balance them

case 1:
if they are equal => heavier is in remaining 2
weigh them and find the heavier

case 2:
one set of 3 is heavier than the other => take the heavier 3 balls
pick 2 balls from this set. balance them
case2.1: if equal => remaining ball is heavier
case 2.2 : heavier of the two is the odd ball.

32. Tuesday, August 5, 2008 at 12:07 am

#16: Weigh 3 on 3 balls
Case 1: They’re equal. Weigh the remaining two and you have the heaviest ball.
Case 2 : One bowl’s heavier. Take 2 out of the three balls in the heavier bowl, and weigh them.
Case 2.1 : They’re equal. The third ball is the heaviest.
Case 2.2 : One bowl’s heavier. This is the heaviest ball.

This way, you actually *only* use the two weighings as in the riddle, and not some hand-weigh-guessing.

33. Tuesday, August 5, 2008 at 12:13 am

@11: two weighings, genius.

34. Tuesday, August 5, 2008 at 1:38 am

regarding the 16th question : divide the 8balls into 3 groups :
A= 3 balls …. B=3 balls …C =2 balls
now weigh and compare set A and B…..{first weighing} if they r equal …then in the second weighing use the 2balls of set C and u will know the heavier one.
if in the first weighing A or SET B is heavier the take 2 balls from that heavier set and weigh …if they equal then the reamining ball is the odd one ..if they r unequal the u have ur answer …just in 2 weighings!!!!!

35. Tuesday, August 5, 2008 at 2:26 am

#16
put 3 balls in each of the pans. if they weigh the same then you can split the remaining two with one weighing. if one pan of 3 is heavier, you can then put 1 of these balls in each pan leaving one aside. if they weigh the same you have the heavier ball in your hand, otherwise the scales will tell you the rest. now give me a job, google.

36. Tuesday, August 5, 2008 at 2:29 am

All current solutions to the ball problem require 3 weighings, which is 1 more than necessary. Correct solution:

– Put 3 balls in left pan, 3 balls in pan. Leave 2 balls on the side.
– If left side is heavier, discard balls on side and balls in left pan.
– If right side is heavier, discard balls on side and balls in left pan.
– If scales are balanced, discard all balls in scales.

This single weighing leaves us with either 2 balls (scales balanced) or 3 balls (scales unbalanced). If we have 2 balls, just weigh them to determine the heavier ball. If we have 3 balls, repeat the above routine (1 in each pan, 1 left on side).

In *two* weighings, not three, we’ve solved the problem.

37. Tuesday, August 5, 2008 at 2:30 am

Nuts, made a complete mess of that. Try again:

– Put 3 balls in left pan, 3 balls in right pan. Leave 2 balls on the side.
– If left side is heavier, discard balls on side and balls in right pan.
– If right side is heavier, discard balls on side and balls in left pan.
– If scales are balanced, discard all balls in scales.

This single weighing leaves us with either 2 balls (scales balanced) or 3 balls (scales unbalanced). If we have 2 balls, just weigh them to determine the heavier ball. If we have 3 balls, repeat the above routine (1 in each pan, 1 left on side).

In *two* weighings, not three, we’ve solved the problem.

38. Tuesday, August 5, 2008 at 2:32 am

Oh, and Rohit:

“b: 5 min and 10 min cross and 5 min returns = 10 min”

10 minutes to cross the bridge plus 5 minutes to get back does not equal 10 minutes total journey time, unfortunately.

39. Tuesday, August 5, 2008 at 2:35 am

“Since each wife never knows her husband has cheated on her, there is no way of killing an husband because she can’t prove it.”

Since *every* woman knows that a man has cheated when he cheats except his husband, and the Queen has now appeared demanding to know which man it was, we can assume that all of the women present bar one will be aware that someone has cheated. All the Queen needs to do is ask the women which one of them is unaware that a man has cheated recently, and the woman who responds is the wife of the cheating husband.

40. Tuesday, August 5, 2008 at 2:47 am

#16 : The previous solutions don’t seem to respect the “only two weighings” part of the question. (“physically feel and can distinguish the heavier of the 2 balls”, “one with heavier weight can be found in same way”). That’s cheating !

Here is my try :
– Put 3 balls in each scale. If none weights more, the heavier is among the two remaining ones. Your other weighing will suffise to know which.
– If not, put in each scale one of the balls from the heavier scale, and keep the third. Either one scale is heavier, or the last ball is the one.

41. Tuesday, August 5, 2008 at 3:40 am

Justin.. your answer to #6 is wrong. They overlap 24 times a day. The minute hand makes a full circle in one hour, the hour hand but a 1/12 of that circle in that same hour – it doesn’t move fast enough to escape the minute hand.

42. Tuesday, August 5, 2008 at 3:47 am

Thats a good attempt by Justin on all the answers, but some of them are wrong.

#9
All the husbands are killed on day 100.

Assume there was only 1couple, then the wife would kill the husband on day 1, since the queen declared that there is atleast one unfaithful husband.

Suppose there were 2 couples (for clarity lets say couple A and couple B), then wife A knows husband B has cheated and so does wife B know about husband A. But neither is sure about their own husband. Hence no one is killed on day 1, but know wife A knows for sure that husband A has cheated. Because if husband A did not cheat, then wife B would have been sure that husband B cheated (since atleast one man has cheated) and would have killed him on day 1 itself. So, both husbands are killed on day 2

A similar argument holds for 3,4… and so on. Hence all 100 husbands will die on day 100

#13
Both Rohit and Justin are wrong, and Anonymous is correct

#16
Divide the balls into groups of 3,3,2.
Weigh 1: Put the two groups of 3 in each balance

Outcome 1a: One side is heavier. Weigh 2 : Take any two balls of the group and weigh them, one on each side. If one side goes down, that is the heavier ball, else the third unweighed ball is the heavier ball

Outcome 1b: Both sides are equal, the heavier ball is in the unweighed group of 2 balls. Weigh 2: Put one ball each of the unweighed group and find out which is heavier

43. Tuesday, August 5, 2008 at 3:53 am

All the 16 answers are wrong. As for, Rohit is wrong, while anonymous is correct.

16:
-put 3 balls on either scale.
–if they are equal then one of the remaining 2 is heavier, which can found out with the 2nd weighing.

–if not, you’ll know which group of 3 is heavier. take 2 from those 3 and put 1 on each end. if they are equal, the one you left out is the heaviest, else, the heavier one on the scale.

44. Tuesday, August 5, 2008 at 4:16 am

Shiiiiiiiiiiat!!

45. Tuesday, August 5, 2008 at 6:00 am

#4 Justin, answer to 16 is incorrect.
the proper way is: you put balls 1-3 and balls 4-6 on each pan. If :
– their weight is the same, then you use the balance to compare ball 7 and ball 8.
– pan 1 is heavier, then you now compare ball 1 and ball 2. If one is heaviest, you’ve found it, if they match, the heaviest ball is ball 3.
– pan 2 is heavier, likewise you compare ball 4 and ball 5.

that’s how you do it in 2 weighings.

for 11, the question is ambiguous. I understand that 0.95 is the chance of seeing at least a car in 30 minutes, vs the probability of seeing one and only one car.
In the former scenario, if p is the probability we are looking for, what we can tell is that if we haven’t seen a car in 30 minutes, then there hasn’t been one in any of the 3 10-minutes periods of this half hour. in other terms, 1-0.95 = (1-p)^3. that gives us p around 63%.

for 9, the text is a bit ambiguous. so here’s another possibility. Each wife knows of 99 extra-marital affairs. By telling that the queen requires one execution, so wives now have to prove that their husbands are innocent. If a wife claims that her husband is innocent, she’s denied by all the others. So that very day, each wife learns that their husband cheat on them.

then, it’s 100 women who try to kill 100 men. I think the logic outcome is that they abolish the law and avoid the bloodbath.

46. Tuesday, August 5, 2008 at 6:11 am

You can only do 2 weighings with the 8 balls. The difference in weight may be so slight that you cannot tell, but the balance can.

1) Pick 6 balls at random, placing 3 on each side of the balance.

2) If they balance, then the heavier ball is in the remaining 2. Place one on each side of the balance to determine which one.

3) If they don’t balance, take the 3 balls from the heavier set and pick 2 at random. Place one on each side of the balance. If they balance, then the ball you didn’t place on the balance is the heaviest. Otherwise the balance will indicate which ball is heavier.

47. Tuesday, August 5, 2008 at 6:14 am

#9:

Basically, if N husbands have cheated, then on the Nth day all of those husbands will die.

Assume only one husband (H1) cheated. Then all the other wives know this, but wife (W1) doesn’t. She assumes no one has cheated. So therefore she can prove her husband cheated, so she kills him.

But what if two husbansd cheated (H1 and H2)?

W1 thinks H2 is the only cheater, and W2 thinks H1 is the only cheater. Surely they’ll kill their husband that very day according to the scenario described above.

When they wake up the next day and the other husband is still alive, they now know that their husband must have cheated, because the other wife couldn’t “prove” it. So they kill their husbands on the 2nd day.

With 3 cheating husbands, each wife knows about the 2 other cheaters, and figures on the 2nd day things will sort themselves out. They wake up on the 3rd day and then …

And so on and so forth.

48. Tuesday, August 5, 2008 at 6:15 am

#16
Weigh 6 balls first.
If they’re equal, weigh the remaining two balls to find the heaviest.
If the 6 balls are not equal, balance 2 of the 3 balls from the heaviest side.
If those 2 balls balance, the ball that was left out of this weighing is the heaviest.
If those 2 balls don’t balance, the heaviest ball is the… heaviest.

49. Tuesday, August 5, 2008 at 6:16 am

#16. – All of these answers are wrong. the correct answer is:

Weigh 3 balls on either side of the scale, while holding 2 balls

If they equal out, then weigh the two balls in your hand and you know which is heavier

If one side is heavier, then take those 3 balls and put one on each side of the scale while retaining one ball in your hand.

If one side is heavier, you know which is the heavier ball. If they equal out, then the ball in your hand is the heavier ball.

50. Tuesday, August 5, 2008 at 6:17 am

#16:

Place 3 of the balls on each side of the balance.

IF THEY ARE EQUAL:
Place each of the remaining balls on either side of the balance. Find the heavier one.

IF THEY ARE UNEQUAL:
Take the 3 balls from the heavier side (A, B, and C). Place A and B on either side of the balance. If A and B are equal, C is the heavier ball. Otherwise, the heavier of A and B is the heavier ball.

51. Tuesday, August 5, 2008 at 6:26 am

#9 is a version of the Blue Eyes puzzle – http://xkcd.com/blue_eyes.html. The answer is that on the 100th day, all the wives will kill their husbands.

#16:
1. Weigh three balls on each pan. The heavier pan will contain the heavier ball (narrow down to three balls). Weigh any two of the three. If they have the same weight, the heavier ball is the third one.

2. If the first weighing is balanced, the heavier ball is one of the two that were not weighed. Weigh them against each other to find the heavier one.

For #13, comment 8 has the right solution.

52. Tuesday, August 5, 2008 at 6:38 am

Justin & renusri: you both are using too many weighings, you only get 2 (and weighing them in your hands counts as a weighing as well!)

Take 6 of the balls put 3 on each side of the scale. Leave 2 on the table. If sides balance then your heavier ball is in the 2 on the table, weigh them against each other.

If one of the sets of 3 is heavier discard the other set of 3 and the 2 on the table. Take the set containing the heavier ball and put 1 on each side of the scales and leave 1 on the table.

If 1 side of the scales is heavier that is the heavier ball, if both balance then the heavier ball is still sitting on the table.

53. Tuesday, August 5, 2008 at 6:46 am

Take 3 balls vs 3 balls, leaving 2. If equal, then balance other two…if unequal, select 2balls to balance…if equal, 3rd ball is heavy…you only have two balances (re-read the problem) and you can’t just say “feel it”

54. Tuesday, August 5, 2008 at 7:14 am

weigh 6 balls, 3 on one side, 3 on the other

if one side is heavier, then weigh two balls from that side,
if they are the same, then the third ball is the heaviest,

if the first 6 balls are all equal, then weigh the last 2

this allows you to do it in exactly 2 weighings

55. Tuesday, August 5, 2008 at 8:16 am

#16
Take any 6 balls and put 3 each in the pans for comparison.
1. If they weigh the same, the odd one is in the remaining two. You can weigh those and find the odd one.
2. If one of the pans is heavier, the discard the other pile and the remaining two. Out of the remaining 3 that contains the heavier ball, pick two balls and compare the weights. If the weight is same, then the remaining ball is the odd one. Otherwise, the heavier pan contains the heavier ball

56. Tuesday, August 5, 2008 at 10:01 am

So far none of the 8 balls solutions are correct:

Weigh 3 balls against 3 other balls. If the two sides are equal, the heavier ball is in the remaining two and weigh those to find the heavier ball.

If one set of 3 balls was heavier then then other, then take two balls from the 3 and weight them. If equal, the heaviest ball is the one that wasn’t weighed in the second weighing. If unequal, then you have the heaviest ball.

57. Tuesday, August 5, 2008 at 10:46 am

The posted answers for 16 are not quite correct…
@Justin: Since one is only slightly heavier, you cannot assume you can physically distinguish the two.
@renusri: Remember you can only weigh twice.

The solution is weighing three random balls versus another three.
If they are equal: The heavier one must be one of the remaining two. Measure between the two to find the heavier one.
If they are unequal: Measure one randomly selected out of the three against another. By the same reasoning, if they are equal, the remaining ball must be the heavier one; if they are unequal, you have your answer.

58. Tuesday, August 5, 2008 at 11:28 am

After reading several of the responses here, specifically to the Weighing Eight Balls question and Crossing the Bridge question, I feel the need to toss my own two cents into the mix.

Justin – you make a huge assumption (and one contrary to the intent of the question) in claiming that the person on the far side of the bridge can illuminate the bridge for the others crossing. Anonymous has the correct solution in post #8.
Rohit – you ignore travel time of people returning back across the bridge. Your solution would actually take 23 minutes (1+2 for the first trip, 10+5 for the second, and 5 for the the third).

Now, on to the Weighing of the Eight balls… obviously you can’t discern just by holding a ball in each hand which one is the heavier of the two, or the whole premise of the question falls apart. To accomplish this feat in TWO weigh-ins on the balance, place 3 balls on each side of the balance, and set the remaining 2 balls aside. There are then two scenarios that could follow: 1) one side is heavier – in which case, discard the rest of the balls, because you know that one of these 3 balls is the culprit. Set 1 of these three down and weigh the other 2… if they’re the same, the culprit is the third ball which you just set down. If they’re not equal, then whichever side of the balance dips down is the heavier one. Or 2) both sides are the same weight, which means the heavier ball is one of the two balls you originally set aside; use your remaining weigh-in to determine which one it is.

59. Tuesday, August 5, 2008 at 11:34 am

Come on guys!
Use some more thought –
16 for example:
Only two weighings are permitted. Only one ball is different. You have to figure it out by the weighings.

Weighing one: weigh 6 balls – three to a side.
(option 1) one side was heavy
result: you now have three balls, one of which is heavy
(option 2) weight was even
result: you now have the two balls that you left out, one of which is the heavy

Weighing two:
option 1: weigh two of the three balls
result: you see which is the heavy or if balanced, the ball you left out is the heavy
option 2: weigh the two balls, you see which is the heavy

60. Tuesday, August 5, 2008 at 12:09 pm

I did not see a good answer for number 16, the “eight ball” question (for want of a better identifier!) So I am posting it here.

Of the eight balls, take six and put three in each arm of the balance. If they are the same, then the “odd ball” is in the remaining two and can be easily found out in the next weighing. Lets say that in the first weighing one of the arms tips down. Then take two of the three balls and place them on opposite sides of the balance. If either side tips, that is the heavier ball. If both are constant, then the ball that was left out is the heavier one.

61. Tuesday, August 5, 2008 at 12:48 pm

With eight balls, and two weighings, you can easily bring it down to 2 balls of dissimilar weights. (4 each, then 2 each)

To get it down to the one ball that’s heavier is a bit tougher. I haven’t got it to work all the time, but I can see it done 75% of the time as follows:

Divide the balls into two groups of 4. From the first group, split it into 2 groups of 2, and weigh them.

If one group is heavier, weigh that group (1 each side) and you’re done.

If they’re the same weight, discard that whole group of 4. Select 2 from the remaining group, and see if one of those two is heavier. If it is, you’re done.

If not, then you hit the 25% case and you’re no worse off then if you had done the 8-4-2 split.

(after a bit of thought)

A-ha! I’ve got it! – And it would even work up to 9 balls in 2 weighings.

From the 8, choose any 6, and weigh 3 on each side. If they’re the same weight, weigh the remaining 2, and you’re done.

If one of the groups is heavier, choose 2 from that group and weigh them. If one’s heavier, it’s the one you want, if neither is heavier, the remaining ball is the one.

Sweet.

gj Anonymous (#8) with the bridge one – that’s a great solution.

62. Wednesday, August 6, 2008 at 9:19 am

@ justin. Your answer is not the best. The weight difference may be too small to distinguish. More importantly if you can rely on your senses instead of the scale, why use the scale at all.

Instead put three balls on each side of the scale. If the scale balances then you know the heavy ball is in the other set of two. Put balance them on the scale and the heavy ball will be obvious.

If the scale doesn’t balance you know which set of three contains the heavy ball. Of these three put one on each side of the scale and hold one back. If the scales balance the ball you held back is the heavy one. If the scales don’t balance then the heavy ball will be obvious.

63. Wednesday, August 6, 2008 at 11:03 am

#6: 2times, Once at noon, once at midnight, all other times, the hour hand has moved a little.

64. Wednesday, September 24, 2008 at 12:13 am

ARE U A NOODLE

65. Wednesday, September 24, 2008 at 7:41 am

I’M A NOOOOOOODLE!!!!

66. Thursday, October 16, 2008 at 10:32 pm

oooohhhh Booooyyy That was tough!!!!!

http://www.mockquestions.com

67. Sunday, November 30, 2008 at 5:42 am

Excellent article.

68. Monday, December 1, 2008 at 8:32 pm

Out of 70 odd answers (sure not all of them are for the same question but I dare say more than 70% of them are!), most of them are only trying to correct ONE person’s wrong answer without finishing reading all the other answers! Hilarious!

Even more ridiculous to see people posting “all the answers are wrong” and then giving exactly the same answer that someone else has already posted some months ago (the right answer of course). Obviously these people didn’t read!

To whoever reading this before posting (and who has a habit of reading all posts before replying), I congratulate and salute you for your patience and logic. Together let’s laugh at those idiots.

69. Wednesday, December 3, 2008 at 4:17 pm

Regarding the first problem, as I think not many people will know an answer. The answer can be found in atomic packings, it can be shown that the most compact cubic packing has a packing factor of 0.74 (74%) (this is for an FCC lattice, but also an HCP lattice). This means that on an overall volume of a cube, only 74% can be filled with balls. So actually, if one would have a bus with dimensions 200x50x50 (in terms of the diameter of the balls), you could indeed squish in 500 000 balls if each ball would have a thin cube around itself. But without that virtual cube around it, the theoretical maximum amount of balls is 706647 balls (this can be calculated by taking 74% of the total volume of the bus and dividing it by the volume of one ball). The real amount is somewhere between that 500 000 and 706 647, but rather close to that last one. By the way, both the HCP and FCC stackings are quite familiar: try stacking oranges or cannon balls in a ‘natural’ way (by putting the balls/oranges in the free spaces of the level below it, instead of trying to equilibrate the oranges exactly on top of each other; one gets and HCP lattice or FCC lattice (depending on whether one buts the bottom layer in a hexagonal (HCP) pattern or square pattern (FCC)

But as you can see, there is a rather big difference between 500k and 700k. 500k is a great lower bound (you can without any doubt fit in 500k balls), 706k is a rather optimistic upper bound (for any large bus, this value will get better and better; but for a small bus one will end up with a packing factor lower than 74% (as this packing factor is calculated for a lattice with infinite dimensions).

Regarding question 6: the hands will overlap 24 times a day, if my count is right. The first time the hands will overlap is at 00:00, then the minute hand will leap ahead of the hour hand, until a little after 01:05 (that little bit is for the amount the hour hand moves during that time). Repeat for every hour of the day, and we get 24, not 22.

For question 12, the answer is indeed 7.5 degrees. During an hour, the minute hand rotates 360°, while the hour hand moves 360°/12. At 3:15, the minute hand is at a quarter (360°/4). One would expect the hour hand to be also at that position, but due to the continuous rotation of the hour hand, it has moves (360°/12)/4 further, which is 7.5°.

70. Thursday, January 29, 2009 at 9:49 am

Golf balls on a bus: we cannot calculate this. What is this question asking more specifically? How many golf balls would fill the cuboid dimensions of the bus (including or not including the space between where the tires touch the ground and the bottom of the bus)? How many golf balls would fill just the seating compartment? How many golf balls would fill every nook and cranny of the bus (that’s with all the components like the engine in place)? …this question needs to be more specific, hence it cannot be answered as is.

The probability question (#11, .95 seeing a car): another answer could be… using the normal distribution (assuming that sighting these cars is normally distributed), the .95 level represents 2 standard deviations. So, +20 minutes (50 minutes would be .99 prob or 3 standard deviations), and -20 (10 minutes) would be at the .66 prob (or 1 standard deviation). Hence, my answer would be .66 probability.

In general, though, I also believe that asking questions like these in an interview process, and then strictly relying on ONE solidly correct answer, is wrong. The purpose should be to get insight on a candidate’s way of thinking. For example, I like the 100 crows on a fence question used to determine whether or not a person is a realist or lives in the theoretical world: “100 crows are sitting on a fence. A farmer with a double-barrelled shot gun comes by, and blasts 2 of them dead into feathers. How many crows are left on the fence?” The theoretical answer is: 100 minus 2 equals 98. The real world answer is: somewhere between 0 and 98 (some will get scared from the blast, others may have bigger balls). Both answers are correct.

71. Sunday, February 1, 2009 at 2:46 am

vc

72. Friday, March 13, 2009 at 9:31 am

I am with the cat with the long somewhat grammatically questionable name: it is insane how many of the responses were the exact same (correct) weighing algorithm. Yes, the first guy was wrong, yes we need trinary search; however, YOU FAIL. On the level that uses this page as any vague measure of aptitude, your decision to post the correct answer claiming “all the other posts are wrong” having read only the first 6 posts makes you look like an idiot. So I sure as hell wouldn’t hire you. Though you know the answer to the question, you don’t know what you are talking about

73. Friday, April 10, 2009 at 12:09 pm

I am so glad 74 and 79 finally mentioned the fact that every person posting a solution to question 16 is an idiot. I don’t know how many people stated that every other solution to 16 was wrong, but then would offer the exact same solution as the multiple posts prior to theirs. End it now…16 is solved…but the real geniuses here are 74 and 79 for realizing that this “puzzling” question has been correctly answered numerous times and does not need any further explanation.

74. Sunday, May 3, 2009 at 4:22 am

what are you pooing?

75. Monday, June 1, 2009 at 12:34 am

If you need more interview questions.
Php Interview Qeustions

C Interview Qeustions

Java Interview Qeustions

76. Tuesday, September 29, 2009 at 11:01 am

I love

77. Saturday, October 24, 2009 at 7:13 am

Excellent list! Thanks for sharing!

– Seattle Interview Coach

78. Thursday, November 26, 2009 at 11:22 am

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• Monday, March 26, 2012 at 12:09 am

lolz

79. Thursday, December 3, 2009 at 12:09 pm

Wow, I haven’t seen the easiest solution to #6 yet.

#6:
-In one day the minute mark makes 24 revolutions (24 hours)
-In the same time the hour mark makes 2 revolutions.

24-2=22.

Has nobody read around the world in 80 days?

80. Monday, January 4, 2010 at 10:12 am

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81. Wednesday, January 27, 2010 at 3:40 pm

Pirates:
p5=98, p4=0, p3=1, p2=0, p1=1

82. Monday, March 22, 2010 at 12:22 pm

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83. Friday, April 9, 2010 at 4:09 pm

2. Start masturbating.

84. Saturday, May 22, 2010 at 4:44 pm

Ai Bang Mai Ne: I bumped into the coffee table

85. Tuesday, August 3, 2010 at 12:16 pm

question14: no unless you wanna loss £19 oh i mean \$19 ok

86. Sunday, October 10, 2010 at 12:13 pm

Ah just have fun as my friend Hoyt Boender play only free games in your free time.

87. Tuesday, October 12, 2010 at 10:31 am

Nice questions…

88. Saturday, October 1, 2011 at 8:21 pm

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89. Monday, February 27, 2012 at 7:49 am

Since the eight balls question has so many arguments, let me ask a slightly harder problem:
12 balls, one weights slightly heavy or light, but not known which,
only give three weightings.
If you tackle this: try the upgrade version: your method has to be non adaptive, that is it cant based on the result of the previous weightings.
It is solvable.

90. Friday, June 28, 2013 at 8:55 pm

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