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
56:05 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 - not 7 - tests to find a working pair of batteries.
Please give a close read to how the question is posed. The question asks how many pairs you need to TEST. you don't need to TEST the 7th pair {b7,b8}. Either you will have found a pair in 6 or fewer tests, or else you'll be GUARANTEED that {b7,b8} will get the torch on. tl;dr: {b7,b8} will not need to be TESTED.
But, technically, with 6 tries you just know that 7th and 8th batteries are good, but you need to do the 7th trial to switch the torch on! If you don't put the 7th and the 8th battery in the 7th trial, the torch will never turn on. Just that this time you are sure that the torch will definitely turn on.
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.
Q.50: Another approach Step1: Fill 5L bucket with 5L water. Step2- Pour the 5L water from 5L bucket to 8L bucket. Step3- Repeat step 1 & 2 again. Step4- you’re now left with 2L water in 5L bucket. Step5- pour 2L water from 5L bucket to 3L bucket. Step6- Fill 5L bucket with 5L water. Step7- since 3L bucket has already 2L water in it (step 5), therefore while filling the water from 5L bucket to 3L bucket, only 1L water will be poured to 3L bucket and 5L bucket will remained with 4L water.
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.
In the 50th Puzzle (scenario 1), I have a much more straightforward and quick solution. 👇 First, fill the 5-litre bucket entirely and pour 3 litres of water into a 3-litre bucket. Now, shift that remaining 2 litre of water into 8 Litre bucket. Do this step again, and you will get the 2+2 = 4 Litre water in the bucket of 8 Litre capacity 😌
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.
#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.
1:22:59 the ants are walking on a 3d shape like a branch, from their perspective they are following the straight path but in reality they are just walking on the lateral circumference of the branch.
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.
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
#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)
1:35:20 this is wrong, u can catc hhim in 7 days, also with the same explanation provided afterwards (in regards to the 3rd bulletpoint), because the cop in cave 12 can also move clockwise, because if he would switch with the thief, he would catch him. so there is no point for the other cop to stay stationary in cave 13
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.
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 #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 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
@@GeoffAlbertson Yup, but you don't need to do it twice as 2(0.5 x 0.5 x 0.5), because it is given that the remaining two must follow the direction of the first. So you can say the first is 1 and the other two are 0.5. It is a programming efficiency and computational efficiency resulting in an energy efficiency that I convey. The same answer yes. x+x+x+x+x+x+x+x+x+x or 10(x) ... which would you rather compute?
Dude i have got like 2 days to prepare for the interview which video from your channel should i refer so that i can atleast try wtver the interviewer throws at me
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.
Last puzzle solution is wrong. It can be done in 4 steps only if we use 5L bucket first. 1. Fill the 5L bucket and emty it in 8L bucket 🪣 2. Again fill the 5L bucket and trasfer 3 L water in 8L bucket as it already has 5L water already in it. 3. 2L Remaining water in 5L bucket. Emapty it in 3L bucket. 4. Again fill 5L bucket and trasfer 1 L to 3 L 🪣 and thats it. You have 4 litter water
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.
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.
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.
#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.
@@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?
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.
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
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
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.
You can choose any 2 balls and put them aside. Maybe one of those is the heavy one and maybe it isnt. If one of those is the heavy one, the first balance test will show that all six balls on the scale are the same by staying balanced- then those six can be eliminated and one more test of the two remaining balls will show which is heavier. If the one heavy ball is among the six that are put on the balance the first time, then the balance will show which set of three balls contains the one heavy ball and the other three can be eliminated leaving three balls including the one heavy ball. Put any two of those on the balance and if the scale stays balanced then the one that was not "weighed" is the heavy one and if the balance tips one way or the other then that will show which one is heavier.
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.
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.
Agree, tough concept I would have missed in an interview. After seeing the explanation, I get it. Reduce population until landing on even power of 2. Try with a smaller number like 10. Did you get the 5th guy?
#4, the ants on the triangle- I thought that there are only 2 ways there would be no collision. All go Anticlockwise or all go Clockwise =2 There are 6 ways there could be one collision or another; AAC, ACA- CAC, CCA- ACA, AAC =6 therefore 2/6 or 1/3 ??? But I see that one AAB is the same as another AAB etc so yeah. cool
🔥🔥 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
Question 50
Just pour water from 5L to 3L bucket we got 2L repeat the same🙄 why this much complicated
This is a great collection of logic puzzles suitable for interviews. Thank you.
56:05 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 - not 7 - tests to find a working pair of batteries.
But it works on the 7th try 🙂 so the answer is 7
Please give a close read to how the question is posed. The question asks how many pairs you need to TEST. you don't need to TEST the 7th pair {b7,b8}. Either you will have found a pair in 6 or fewer tests, or else you'll be GUARANTEED that {b7,b8} will get the torch on.
tl;dr: {b7,b8} will not need to be TESTED.
But, technically, with 6 tries you just know that 7th and 8th batteries are good, but you need to do the 7th trial to switch the torch on!
If you don't put the 7th and the 8th battery in the 7th trial, the torch will never turn on. Just that this time you are sure that the torch will definitely turn on.
@@souvikghosh572the question asks how many tests are required to find a pair that would get the torch on
@@souvikghosh572 "but, technically" ...give it a rest. {b7,b8} is not a trial. you're not testing anything at that point.
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.
😊
Amazing solution!
Excellent observation!
@@souvikghosh572 - Thanks for noticing
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.
So you confirm the answer given.
Q.50: Another approach
Step1: Fill 5L bucket with 5L water.
Step2- Pour the 5L water from 5L bucket to 8L bucket.
Step3- Repeat step 1 & 2 again.
Step4- you’re now left with 2L water in 5L bucket.
Step5- pour 2L water from 5L bucket to 3L bucket.
Step6- Fill 5L bucket with 5L water.
Step7- since 3L bucket has already 2L water in it (step 5), therefore while filling the water from 5L bucket to 3L bucket, only 1L water will be poured to 3L bucket and 5L bucket will remained with 4L water.
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.
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..
16 kaise aaya ??? How we get 16??
Thanks a lot.
Question 50. 3:24:31
Just pour water from 5L to 3L bucket we got 2L repeat the same🙄 why this much complicated
In the 50th Puzzle (scenario 1), I have a much more straightforward and quick solution. 👇
First, fill the 5-litre bucket entirely and pour 3 litres of water into a 3-litre bucket. Now, shift that remaining 2 litre of water into 8 Litre bucket. Do this step again, and you will get the 2+2 = 4 Litre water in the bucket of 8 Litre capacity 😌
Without dumping, fill 3L, move to 5L. Fill 3L, move 2L to fill 5L. remaining 1L moves to 8L. Fill 3L, and add to 8L.
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.
#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.
The puzzles are so badly written, you come up with rules that were not specified...
Yes bro
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.
1:22:59 the ants are walking on a 3d shape like a branch, from their perspective they are following the straight path but in reality they are just walking on the lateral circumference of the branch.
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.
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
Nailed it, no reason to stay fixed in a location.
Awesome, loved it
#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
amazing content
Battery puzzle wrong. Only 6 batteries need to be checked. You are looking for 2 that work
Agree 6 tests that fail assure 7th pair are good.
🙏🙏🙏🙏🙏🙏🙏🙏🙏🙏 nice explainatio each q
Awesome and really helpful content 👏 😇.....Thanks a lot 🙏 😊
1:35:20 this is wrong, u can catc hhim in 7 days, also with the same explanation provided afterwards (in regards to the 3rd bulletpoint), because the cop in cave 12 can also move clockwise, because if he would switch with the thief, he would catch him. so there is no point for the other cop to stay stationary in cave 13
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.
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 ?
Please i want to know that which software you have used to create these pictures and animations
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.
@14:59 chicken, fox, take back chicken, leave chicken and take corn, come back and take chicken.
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
#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.
That is the answer given, 1/4
@@GeoffAlbertson Yup, but you don't need to do it twice as 2(0.5 x 0.5 x 0.5), because it is given that the remaining two must follow the direction of the first. So you can say the first is 1 and the other two are 0.5. It is a programming efficiency and computational efficiency resulting in an energy efficiency that I convey. The same answer yes.
x+x+x+x+x+x+x+x+x+x or 10(x) ... which would you rather compute?
Dude i have got like 2 days to prepare for the interview which video from your channel should i refer so that i can atleast try wtver the interviewer throws at me
y didn't u go with 9 balls on the first puzzle? it would work the same
Excellent collection, best puzzles for interview preparation. Stop point for all good puzzles. Thanks bro 😎
And what if the worker orders a pizza and a netflix subscription on day 1, and can't give back his 1/7th of a bar?
nice work❤️
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.
Last puzzle solution is wrong. It can be done in 4 steps only if we use 5L bucket first.
1. Fill the 5L bucket and emty it in 8L bucket 🪣
2. Again fill the 5L bucket and trasfer 3 L water in 8L bucket as it already has 5L water already in it.
3. 2L Remaining water in 5L bucket. Emapty it in 3L bucket.
4. Again fill 5L bucket and trasfer 1 L to 3 L 🪣 and thats it. You have 4 litter water
13:30
I formed 4 triangles
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.
what job are these questions for?
Tq u 🙏🙏
can you share the pdf of the questions?
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 😅
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.
If b1 and b4 are bad, your example fails.
the hexagon problem. a hexagon has 6 sides. you show 8
I m such a dumb person 😭
Now a days, HR interviews never ask puzzles
oh really? how many interviews you have been into?
@@divyanksharma236 i am an interviewer and know this is just bull-shit method
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.
#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?
With the ball and scale, works with 9 balls also
anyone explain defective ball ques qn.8 at 18.10 min
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
#24:
If theif can't go to visited one, why check C13 daily. Do below:
C13:C12
C11:C1
C10:C2
C9:C3
C8:C4
C7:C5
C6
Caught in 7 days.
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.
some of the puzzle's solutions are either confusing or not well described.☹
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
U are wrong
@@abhishekvsagarnal2176 Okay care to enlighten me pls?! 😇
In gold bar puzzle, how will the worker return the bar? He needs money for daily expenses.
In the first puzzle … how much more heavy is the one compared to the others?
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
The fox and duck.
1 answer we can find heaiver
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.
Puzzle no1 . How we reach up to conclusion that 1 and 2 would be heavier...we didn't compare 3 with anyone.
You can choose any 2 balls and put them aside. Maybe one of those is the heavy one and maybe it isnt. If one of those is the heavy one, the first balance test will show that all six balls on the scale are the same by staying balanced- then those six can be eliminated and one more test of the two remaining balls will show which is heavier. If the one heavy ball is among the six that are put on the balance the first time, then the balance will show which set of three balls contains the one heavy ball and the other three can be eliminated leaving three balls including the one heavy ball. Put any two of those on the balance and if the scale stays balanced then the one that was not "weighed" is the heavy one and if the balance tips one way or the other then that will show which one is heavier.
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.
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
I cant solve the single question
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
Not able to understand 25th
Agree, tough concept I would have missed in an interview. After seeing the explanation, I get it. Reduce population until landing on even power of 2. Try with a smaller number like 10. Did you get the 5th guy?
Is that a HR questions are. Pretty low level
#4, the ants on the triangle- I thought that there are only 2 ways there would be no collision. All go Anticlockwise or all go Clockwise =2
There are 6 ways there could be one collision or another; AAC, ACA- CAC, CCA- ACA, AAC =6 therefore 2/6 or 1/3 ??? But I see that one AAB is the same as another AAB etc so yeah. cool
6+2=8