Puzzle Challenge: Can You Solve This One Like I Did?

Hey guys,

Solve this puzzle. I have tried and got the answer too.

Hi, what is the answer to this? I am not able to cross all the guys across the bridge. I know 30 is the sum of their speeds, and there is a 30-second time limit. What is the trick?

Goodbye.

Hi Saritha,

Good puzzle! I can't solve it yet, but I have saved it and will try to solve it later.

Manoj

Not possible :)

Let's say the 5 members of the family are A (1), B (3), C (6), D (8), and E (12).

Now, since a lamp is necessary for the trip... Minimum time taken will be:

- AE go, A comes back = 12 + 1 = 13
- AD go, A comes back = 8 + 1 = 9
- AC go, A comes back = 6 + 1 = 7
- AB go finally = 3 seconds

Minimum time required = 13 + 9 + 7 + 3 = 32 Seconds :idea:

Any other combination should take more than this time... So well... what am I missing? (nothing :?:)

So, was this an HR question? :lol:

It's a solvable puzzle. Whoever is asking for the answers is just that you gave up so easily. The clue to the quiz is that once you finish solving it... 1 sec is left! Happy Solving! In case you are not able to get it, then I will post... Don't give up. Cheers, Mallet

I solved this in 3 attempts...

Here, I am giving the answer for those who tried really hard or those who assume it is mathematically impossible. We have 5 characters with speeds of 1, 3, 6, 8, 12, right!! And at a time, two can go.

Try this one (We name characters according to their moving speed):
1. 1 and 3 will go (3 seconds)
2. 1 will come back (1 second)
3. 8 and 12 will go (12 seconds)
4. 3 will come back (3 seconds)
5. 1 and 6 will go (6 seconds)
6. 1 will come back (1 second)
7. 1 and 3 will go (3 seconds)

Total: 29 seconds and 1 second remains.

Cheers!!!
Manoj

Friend :)

It's a nice exercise; if you don't want to try, I can help with the answer. But let me share with you a hint for one more try.

Hint: The two of the slowest can go together, i.e., 12, 8. (Because anyway, you will require 12 seconds to cross the chap regardless of who accompanies him, and then you require someone of 1/3 to be there to bring back the lamp). This should solve this puzzle. :D

Kudos!!!!

Hi folks,
I did this way:
Step1->send 1-6 first together
Step2à send back 1
Step3à send 3 and 1 together
Step4à send back 1
Step5à send 12 and 8(this is a critical step)
Step6à send back 3
Step 7àsend 1 and 3(game over)
In total 29 seconds
easy naa!!!

Hey guys,

Anybody can solve this puzzle. Actually, out of 5 rules, 1 is wrong: "A pair must walk together at the rate of the slower person." In this way, we get conditioned that the pair has to walk together, so we can't solve it.

While solving the puzzle, it is not so. I have solved it in 29 seconds.

Regards,
Nidhi

I solved this in 3 attempts... Here giving the answer for those who tried really hard or assumed it is mathematically impossible.

Here we have 5 characters with speeds of 1, 3, 6, 8, and 12. At a time, two can go. Try this one (we name characters according to their moving speeds):

1. 1 and 3 will go (3 seconds)
2. 1 will come back (1 second)
3. 8 and 12 will go (12 seconds)
4. 3 will come back (3 seconds)
5. 1 and 6 will go (6 seconds)
6. 1 will come back (1 second)
7. 1 and 3 will go (3 seconds)

Total time taken is 29 seconds, and 1 second remains.

Cheers!!! 😂😂😂

Hi,

First, AB goes, B comes back = 3 + 3 = 6.
DE goes, A comes back = 12 + 1 = 13.
AC goes, A comes back = 1 + 6 = 7.
AB goes 3 + 1 = 4.
Total 30 seconds.

Very interesting, enjoyed a lot. Thanks for sharing.

Hello,

Can you post the answer for me? I tried a lot but could not get the answer.

Regards,
Jyothi

Hi Jyothi,

If A and B move, then leave A and B should come back, 3 + 3 = 6 seconds.
Let D and E move, then A will come back, 12 + 1 = 13 seconds.
When A and C go, then A will come back, 6 + 1 = 7 seconds.
If A and B go, it takes 3 seconds.
Therefore, 1 second will remain.