50 Puzzles Commonly asked in HR Interviews
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- čas přidán 6. 06. 2024
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50 Puzzles Commonly asked in HR Interviews
00:00 - 8 Balls Weight Puzzle (Interview Puzzle) || 8 Identical Balls Problem || 8 Balls Problem
03:08 - Crossing Bridge Puzzle || Bridge and Torch Puzzle
06:33 - Gold Bar Puzzle || Gold Bar Distribution Puzzle
09:16 - 3 Ants and Triangle Problem || 3 Ants Problem
11:41 - Matchstick Puzzle (Hexagon to equilateral Triangle )
13:48 - Fox, Chicken, Corn Puzzle
16:21 - Defective Box Puzzle
19:23 - Supersonic Bee
25:46 - Find Ages of Daughters
28:46 - 81 Cows Distribution Puzzle
35:58 - Measure 1 KG Rice
43:52 - Red Hat vs Blue Hat
53:30 - Batteries and Torch Problem
57:27 - The Fox and Duck Puzzle
01:03:35 - Man fell in well puzzle
01:06:15 - Traveller and Coconut Puzzle
01:09:34 - Headshot Puzzle
01:12:33 - Burning Rope Puzzle
01:16:13 - Best Time to Escape
01:19:32 - 3 Glass and 10 Coins
01:21:52 - Three Ants Go Marching
01:23:05 - 4 Prisoners Puzzle
01:26:34 - 9 Dogs Fence Problem
01:27:46 - The Rabbit Problem
01:32:54 - 13 Caves and Thief Puzzle
01:36:43 - 100 People In a Circle
01:41:38 - Bat and Ball Puzzle
01:44:26 - Camel Race Puzzle
01:46:23 - Cat In a Square Room Puzzle
01:49:04 - Farmer Fencing Problem
01:51:09 - Grandma and Cakes Puzzle
01:54:55 - Lily Pond Puzzle
01:57:09 - Famous Egg Problem
01:59:23 - Hotel With 100 Rooms Riddle
02:01:19 - Blue Black and Red Socks Puzzle
02:02:32 - Quarantine Puzzle
02:04:21 - Teresa’s Daughter Puzzle
02:06:47 - An Island of Puzzles
02:09:56 - A Door of Fate and Logics
02:12:47 - 100 Doors Puzzle
02:17:27 - 240 Barrels of Wine Puzzle
02:24:05 - 2014 Bulbs Logical Puzzle
02:28:41 - 100 Floors and Egg Problem
02:38:07 - 5 Pirates and 1000 Coins Puzzle
02:53:37 - Heaven or Hell Puzzle
02:55:52 - How Much Money Initially Had ?
03:00:00 - Is Your Husband a Cheat ?
03:05:19 - Number Of Squares In ChessBoard
03:10:26 - Find The Fastest 3 Horses
03:15:39 - Flip The Triangle
03:18:33 - Measure 4L using 3 buckets
🔥🔥 50 Puzzles Commonly asked in HR Interviews 🔥🔥
In Island puzzle only 4 trips are needed.
1St trip (5,3,1)→and come back with 1
2nd trip (6,2,1)→ and come back with 1
3rd trip (8,7,1)→ and come back with 3
4th trip (9,4,3)→ completed
@@BipinKumar-ou7tj when u comeback it's also counted as a trip so total 7
@@dhirajlaha4927 oo..yes thanks
Awesome and really helpful content 👏 😇.....Thanks a lot 🙏 😊
At 28:46, the cow distribution problem's solution has been unnecessarily complicated. There is a simple and elegant solution. Arrange all numbers 1through 81 in a magic square - you need to have two such squares stacked one above the other.
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63
64 65 66 67 68 69 70 71 72
73 74 75 76 77 78 79 80 81
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27
28 29 30 31 32 33 34 35 36
37 38 39 40 41 42 43 44 45
46 47 48 49 50 51 52 53 54
55 56 57 58 59 60 61 62 63
64 65 66 67 68 69 70 71 72
73 74 75 76 77 78 79 80 81
Now, add the numbers diagonally from the 1st column in first row, 2nd column in second row and so on. It will add up to 369 and the columns of the diagonal will give you the cow's number. Unfortunately, YT doesn't let me copy the rich text format. However, you can see below the magic square based solution
1st son : 1, 11, 21, 31 ... 81
2nd son : 10, 20, 30 ... 9
...
9th son : 73,2,12,22 ... 72
Those are not magic squares.
Excellent collection, best puzzles for interview preparation. Stop point for all good puzzles. Thanks bro 😎
Awesome, loved it
Thanks a lot.
#10 ... n² = (n²-1)+1 = (n+1)(n-1)+1, so if you do not have an even x even grid you can convert it to one with 1 remainder.
By subtracting one number from each row of (n-1) you have (n-1)(n+1) - (n-1) + 1 = (n-1) ((n+1) -1) + 1 = (n-1)(n-1+1) + 1 = (n-1)(n) + 1. So to add it back (n-1)(n) + 1 + (n-1). So it equals in this case (9)(9-1) + R + column of 8 numbers converted to a row = 9x9. This is just the math way of saying you can take one number from each row, and place it in a new row underneath (rotate 90 degrees) with the remainder and have an n x n matrix or grid. Drop down the remainder, n² (81), to the last row and slide the right hand section to fill the spaces. The first number of the nth row is the remainder.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
31 32 33 34 35 36 37 38 39 40 ← take this 10x8 + the Remainder 81 & convert to a 9x9
41 42 43 44 45 46 47 48 49 50
51 52 53 54 55 56 57 58 59 60
61 62 63 64 65 66 67 68 69 70
71 72 73 74 75 76 77 78 79 80
1 2 3 4 5 6 7 8 ↓ 10 ←
11 12 13 14 15 16 17 ↓ 19 20 ←
21 22 23 24 25 26 ↓ 28 29 30 ←
31 32 33 34 35 ↓ 37 38 39 40 ←
41 42 43 44 ↓ 46 47 48 49 50 ←
51 52 53 ↓ 55 56 57 58 59 60 ←
61 62 ↓ 64 65 66 67 68 69 70 ←
71 ↓ 73 74 75 76 77 78 79 80 ←
81 72 63 54 45 36 27 18 9 ← new row.
1 2 3 4 5 6 7 8 10
11 12 13 14 15 16 17 19 20
21 22 23 24 25 26 28 29 30
31 32 33 34 35 37 38 39 40
41 42 43 44 46 47 48 49 50
51 52 53 55 56 57 58 59 60
61 62 64 65 66 67 68 69 70
71 73 74 75 76 77 78 79 80
81 72 63 54 45 36 27 18 9
9 columns of 9 that add to 369.
I hope this posts correctly.
That hexagon had 8 sides!!!
Hectagon
Hexa -6
Figure shown is an octagon not the hexagon of the title
was gonna say that ... he even circled it lol.
amazing content
While solving #18 i was laughing in my mind that the solution i was trying to guess is completely wrong, but that was the correct solution 😅
56:38 given batteries b1, ..., b8 of which 4 work, you need only try at most 6 pairs of batteries - not 7 pairs! - in order to find a working pair:
1. if none of the pairs {b1,b2}, {b2,b3}, {b3,b1} work, then you know that at least 2 of {b1, b2, b3} are faulty.
2. likewise, if none of the pairs {b4,b5}, {b5,b6}, {b6,b4} work, then you know that at least 2 of {b4, b5, b6} are faulty.
3. therefore you either found a working pair among {b1, ..., b6} or else you know that all 4 of the faulty batteries are in {b1, ...,b6} - in which case the remaining b7, b8 must both work.
thus you only need at most 6 tests to find a working pair of batteries.
#4 Given a bug moves in one of two directions the other two have two choices 1x 1/2 x 1/2 = 1/4.
For 4 bugs 1 x 1/2 x 1/2 x 1/2 = 1/8.
nice work❤️
Puzzle 24. Thief and 13 caves: if instead of checking cave 13 each day, they check the next cave instead (ie Day1 the police check 12 & 13. Day 2 Police check 11 & 1... then on the 6th day they will check caves 7 & 5. Day 7 they have their thief. The answer is 7, not 12.
Tq u 🙏🙏
You can cut the gold bar into 4 pieces using a crossing cut (1/8, 1/8, 2/8. 4/8). 1/8 day 1 = 1/8, 1/8+1/8 day 2 = 2/8, trade a 2/8 for a 1/8 day 3 = 3/8, trade 4/8 for 1/8 & 2/8 day 4 = 4/8, 1/8 day 5 = 5/8, 1/8 day 6 = 6/8, trade 2/8 for 1/8 day 7 = 7/8. You keep 1/8 for yourself.
So I submit that the video's example is wrong, in that a King doesn't become a King his way. A King doesn't give away anymore than he has to, there is no reason to give away the entire bar.
To make 4 pieces, you need to cut thrice. You are allowed to cut only twice.
@@agytjax cut it in an x ... it makes 4.
With the ball and scale, works with 9 balls also
🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏 nice explainatio each q
About the thief in the cave one, is there a reason not to catch the guy in a sandwich by going C1 C13, then C2 C12, then C3 C11, etc? That would make it 7 days instead of 12.
thankyou for this helpful video !!!!!!!!! In a bee travelling between trains why we didn't took relative speed of bee to the trains ? won't it affect ?
For defective box puzzle, I see another solution too, according to the problem we've to use the weighing machine only once, but no rule to check the weight with gradual increase right? So instead of selecting in a Arithmatic progression, just select 1 ball from each box, We know ball weighs 10 gm but defective ball, so 10 boxes x 10 balls = should weight 100 gm. So if i gradually pick and add balls in the weighing machine I would be able to notice which box has the defective contains since the weight deviates from multiples of 10. Correct me if I'm wrong
13 caves and thief puzzle answer is not optimal. Optimal answer is 8 I think. Instead os always checking 13th cave, they can go clockwise and anticlockwise at the se time. i.e. c12 & c13 on 1st day, c11 & c1 on 2nd day, c10 and c2 on the 3d day, etc.
Hope this makes sense
Now a days, HR interviews never ask puzzles
The 10 gram ball question , I though 4 balls in each box as shown
Same, he left out a lot of important information in the question. He also did not say whether balls could be removed from boxes, or how much could be weighed at one time on the scale.
In the gold bar, king and worker question-- What if the worker sells the 1/7th bar which he gets after the day 1 of work and purchases something? The question shouldv'e mentioned that the worker does not uses the gold bar for till the end of the 7th day to make any purchase.
#1 ... why do you limit to 8 balls ? it works with 9 too (it's a divide by 3 problem, each use selects one of 3 groups ... with 3 uses you can find among 27 balls and in more general case with n uses among 3^n balls)
he limits it to 8 balls inorder to confuse people..if it's 9 the people could easily divide it into 3, 3, 3.
coz only 2 tries
For #24, I agree with Joe. Cops pick two adjacent caves and each day methodically check the next two caves moving around the circle in both directions. The MAXIMUM the thief can hold out is 7 days. There must be some confusion in the way the puzzle is presented. I don't understand what is meant by the third note.
Kevin, I agree with you and Joe, I had the same thought as you
(or a similar one at least).
Simply Logical has done us a service by assembling so mant puzzles and
demonstrating so many solutions. That is a lot of work.
It appears
that others have had the same thought about the caves.
Caroline and Tom made similar points.
The police could close in on the thief twice as quickly by moving BOTH
cave checks. The police could start out by looking in adjacent caves
and then move round the circle in opposte directions by one cave per
day until they eventually meet up again. The thief would be caught in
between them in an ever diminishing safe zone.
Perhaps this is what the third note was about. All the same, I do not
see what is different about one policeman and the other.
The puzzle to me about #24 is: What is the miscommunication about how the puzzle is presented? Why does his 12 day solution involve 1 cop staying in one cave? The solution I presented offers a maximum of 7 days, not a minimum. More likely less than 7 days. Looking up the problem shows many people concluding 12 days, more than those who say 7. Some say the thief can evade the cops forever. Something is missing in the presentation of this puzzle.
@@kevinevans3021 I see no reason why one police check should remain stationary while the other moves. Either the thief can pass through the police check--in contradiction to note three--or s/he cannot. Neither the direction of travel nor the movement of the police toward the thief should matter. Your solution of 7 days maximum stands if the thief cannot move into a cave checked by the police the day before.
In last puzzle (i.e the bucket one) we can also follow this approach first fill the 3l bucket put it into 8l bucket, then again fill 3l bucket and put it into 8l bucket, now 8l bucket has 6l of water, now put 5l of water from 8l bucket in 5l bucket this will give us only 1L of water left in 8l bucket now again fill 3l bucket and put it in 8L bucket
Or we can fill 3l bucket, put it into 5l bucket, fill 3l bucket again, put it into 5l bucket again, put the water from 5l bucket to 8l bucket, put 1l from 3l bucket to 5l bucket, fill 3l bucket again and put the 3l to 5l bucket and now we have 4l inside the 5l bucket.
The fox and duck.
I suspect quite a bunch of these puzzles have a deficiency. E.g. puzzle #7. The solution is to take 1 ball from box 1, 2 balls from box 2 etc. But what if there aren't enough balls in the respective box? E.g. box 10 has less than 10 balls, box 9 has less than 9 balls and so on.
can you share the pdf of the questions?
In the first puzzle … how much more heavy is the one compared to the others?
in the defective box one puzzle, i feel we cant assume there exist enough balls in each box to carry out that solution? There could be 2 balls in each box or 3. it should be specified for completeness that there are 10 balls in each box
3Ants and Triangle-
Collision doesn’t happen only in two cases - All ants going in (a)clockwise (b)anti-clockwise
Every ant has two choices and there are total 2^3 possibilities = 8.
Out of 2^3 possibilities, only 2 don’t cause collision.
probability of ants doesn't collide is 2/8 =1/4 =0.25.
@14:59 chicken, fox, take back chicken, leave chicken and take corn, come back and take chicken.
Puzzle no1 . How we reach up to conclusion that 1 and 2 would be heavier...we didn't compare 3 with anyone.
what job are these questions for?
The puzzles are so badly written, you come up with rules that were not specified...
y didn't u go with 9 balls on the first puzzle? it would work the same
In gold bar puzzle, how will the worker return the bar? He needs money for daily expenses.
the hexagon problem. a hexagon has 6 sides. you show 8
In second puzzle you tells that pull is not so strong but when return only 1 single man come ..
This is I can't understand
basically I could understand only those puzzles which had visible text on the screen. Because of accent I couldn't understand almost half of other ones correctly
13:30
I formed 4 triangles
2:07:54 7 trips, like first-349 return 3(2 trips) -second time-153 return 3 (2 trips)- third time -736 and return 6 (2 trips) -finally 286 --totally 7 trips..
The slave and poison barrel one is wrong on so many levels ... THE worst being you don't have to test one of the barrels ... if no-one dies, then it's the poison barrel.
#1 - I did not know that we were supposed to assume that all of the lighter balls were of equal weight.
#2 - We were not informed that the faster people could be allowed to slow down.
Bro what do you think, we mean by identical balls ?
@@elitemaths4994Yeah. I don't know man. I was just as sideways on number 2. I was drinkin' a lot back then or something, so...
My apologies, I think?
1 answer we can find heaiver
The thief and 13 caves: the answer is 7
Day 1 check c1 and c13
Day 2 check c2 and c12
Day 3 check c3 and c11
Day 4 check c4 and c10
Day 5 check c5 and c9
Day 6 check c6 and c8
Day 7 catch thief as he exits cave 7.
some of the puzzle's solutions are either confusing or not well described.☹
The 13 thieves and cave puzzle makes no sense.
If the cops start from two caves adjacent, c1 and c13, they can check c2 and c12 the next day, then c3 and c11 and so on.
The thief wil not be able to pass the cops, and therefore will be caught in 7 days maximum in the middle cave c7.
And that doesn't take into account that the cops can literally catch him on day 1 if they're lucky, which makes the minimum... 1.
Battery puzzle wrong. Only 6 batteries need to be checked. You are looking for 2 that work
Don't rely on this much of puzzles only..
Try to prepare other puzzles as well.
e.g. Temple and magical pond puzzle..
Do you think these are the difficult ones or can you advise me any books for better puzzles questions
That kingdom sounds like Queensland; what a stupid kingdom! The woman could have turned back at the 4 minute mark so that she appeared to be 5 eighths of the way across coming into the kingdom having been seemingly been walking in the same direction for 5 minutes.
26:40
In torch and batteries puzzle.....
If we divide in 4 pairs of 2 batteries in each pair.....b1b2---b3b4---b5b6---b7b8 and test one by one. Worst case you need 4 attempts.That means each pair has a defective battery.
Then take any two pair suppose b1b2---b3b4.......and check by trying combinations b1b3 and b2b4.....i.e. 2 trials.
So total trials =4+2=6 (less than 7)
Correct me if I'm missing any case.
Now you have 2 pairs with two good batteries. You've already tried b1-b2 & b3-b4
You need to check b1-b3 & b1-b4 & b2-b3 & b2-b4 so that's 4+2+2=8.
Regarding #4 - Ants on a triangle, there seems a simpler solution.
Ant 1 moves in either direction. Chance Ant 2 won’t collide with Ant 1 = .5. Chance Ant 3 won’t collide with Ant 2 = .5.
.5 x .5 = .25.
Not able to understand 25th