There are two checkpoints on a road namely Pt. A and Pt. B. There are 5 persons standing at the Pt. A namely P , Q , R , S, T. They have a car such that P , Q , R , S , T can travel to Pt. B in 1 , 3 , 6 , 8 , 12 seconds respectively. If only two persons can get into the car and person with more driving time drives also one person has to bring car back to Pt. A unless all have arrived Pt. B. How can you transport all the persons to Pt. B in 30 seconds or less.

  • 12
    $\begingroup$ This is a variation of a problem of four people who travel at rates of 1, 2, 5, and 10 minutes trying to cross a bridge in 17 minutes. $\endgroup$ Jul 3, 2016 at 0:02
  • 4
    $\begingroup$ Possible duplicate of Fastest way to cross a river $\endgroup$
    – mdc32
    Jul 4, 2016 at 4:20
  • 1
    $\begingroup$ If P does all the driving, it only takes 7 seconds $\endgroup$
    – cup
    Jul 4, 2016 at 12:18
  • 2
    $\begingroup$ "person with more driving time drives" makes no sense. At least in the bridge stuff there was a bit of consistency $\endgroup$
    – njzk2
    Jul 4, 2016 at 14:24
  • $\begingroup$ (plus, where is the time to get in and out of the car?) $\endgroup$
    – njzk2
    Jul 4, 2016 at 14:26

4 Answers 4


P and Q go, Q returns back alone. 6 seconds passed

S and T go. P returns back alone. 13 seconds passed.

P and R go, P returns alone. 7 seconds passed.

P and Q go. 3 seconds passed.

Total time: 29 seconds


No real-world car can accelerate faster than $5g$, so the maximum acceleration of the car is under $50m/s^2$. Even if the car instantly decelerates (against a wall or something), the fact that $P$ can do the trip in $1$ second means that $A$ and $B$ are at most $25m$ apart (more realistically $1m$), so running this distance takes less than five seconds at a very generous estimate.

Solution: $P$ and $Q$ use the car, and $R, S, T$ run. Total time: under $5$ seconds.

EDIT to merge the comments: Suppose that some of the people have much more trouble running, since they also seem to have trouble driving a car. In that case, we should also consider an average $0.4g$ car and not a $5g$ dragster. In this case, the maximum distance achievable in $1$ second is $2m$, so let's assume our car is placed sideways in a dark alley to ensure proper lithobraking.

Now let's look at our people. They may have more or less trouble walking, or getting into cars, or driving, but what we do know is that there is a lower bound on how fast they can get into and out of cars. Suppose they have to move $1m$ to enter the car, and $1m$ to leave again. Entering and leaving the car is part of driving the car from $A$ to $B$, so any person can therefore move at least $2m$ by foot in the time he needs to move $2m$ by car.

The optimal solution is now as follows: Everybody walks, and the car stays there. This takes at most $12$ seconds, potentially much less if the given people have much more trouble operating cars than walking.

  • 5
    $\begingroup$ Well, he's not wrong, haha. $\endgroup$
    – Josh
    Jul 2, 2016 at 16:15
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    $\begingroup$ Also, let's create a new puzzle changing 'seconds' to 'minutes'. $\endgroup$
    – Chaotic
    Jul 2, 2016 at 23:24
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    $\begingroup$ While I absolutely love answers that ring true even when they were not what the OP was thinking, I must point out a flaw. If you insist that the problem reflect reality, are you sure that T even can run? This person apparently weighs enough to slow a car to 1/12th of it's top speed. A 25m dash could induce a heart attack. :) $\endgroup$ Jul 3, 2016 at 6:00
  • 8
    $\begingroup$ How about a frictionless, perfectly spherical car? $\endgroup$ Jul 3, 2016 at 12:06
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    $\begingroup$ This is epic! So far, the best I have ever read on puzzling.se $\endgroup$
    – dryairship
    Jul 4, 2016 at 12:58

1 and 3 go. (3 seconds)
1 comes back. (4 seconds)
12 and 8 go (16 seconds)
3 comes back. (19 seconds)
1 and 6 go (25 seconds)
1 comes back (26 seconds)
1 and 3 go (29 seconds)

Total time:

29 seconds.


T and P gets in; T drives to pt.B and P drives back to pt. A(they now have 17secs left)

S gets in with P still inside; S drives to pt. B then P drives back (8secs left)

R gets in then drives to pt.B, P drives back (1sec remaining)

Q gets in then P drives them both to pt.B (0sec remaining)

Transporting them all took exactly 30secs

  • $\begingroup$ Actually it takes at least 31 s (P,Q takes 2 secs) $\endgroup$
    – user17008
    Jul 3, 2016 at 14:58
  • $\begingroup$ If P drives, it takes just a sec to arrive at pt. B... there would be no need to come back since all 5 of the would now be at pt. B $\endgroup$
    – Wells
    Jul 3, 2016 at 15:03
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    $\begingroup$ No. "person with more driving time drives." $\endgroup$
    – user17008
    Jul 3, 2016 at 15:23
  • $\begingroup$ Oh... I think I missed that :( $\endgroup$
    – Wells
    Jul 3, 2016 at 15:34
  • $\begingroup$ If the faster person were able to drive then you'd be able to improve your solution by just having P drive every leg. Why let T drive in the beginning when P is in the car and able to drive? This problem would obviously have a trivial solution which is part of the way to tell whether you might have missed something in any puzzle. $\endgroup$
    – Chris
    Jul 4, 2016 at 13:07

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