On Friday I posted this puzzle….

John weights much more than 20 stone.  He goes to the shops to buy some scales, but the scales there only go up to 20 stone.  However, he comes up with a way of accurately weighing himself every day using the scales.  What was his plan?

If you have not tried to solve it, have a go now.  For everyone else the answer is after the break.

John buys 2 scales and places one foot on each!  Did you solve it?  Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called PUZZLED and is available for the Kindle(UKhere and USA here) and on the iBookstore (UK here in the USA here). You can try 101 of the puzzles for free here.


  1. He could do it with one scales and a lever (plank). He would put the scales at one end of the lever, a fulcrum at the other and then stand half way along.

    1. The difficulty with this method is that to get any accuracy you would have to get your centre of gravity exactly over the midpoint of the plank. I think there might be difficulties if the plank was materially distorted by you standing on it. Also scales are often quite fussy about how the force is transmitted to their surface, so you’d probably have to design an end to the plank. Also you’d need to know the weight of the plank and subtract half of that.

  2. If he weights over 40 stone then he stands a quarter of the way down the lever and multiplies the displayed weight by 4.

    1. If he weighs over 40 stone I think we might be able to assume that he’s lost interest in how much he weighs… unless he wants to get on Record Breakers, of course. 🙂

    1. No, that way all scales would take the whole weight, plus the weight of any of the scales above them.

    2. Modern bathroom scales work on springs – if you put scales on a mattress you will ‘weigh’ considerably less as they measure the compression – three sets of these, or indeed three of the ‘hanging’ style spring scales one uses for luggage, would each measure 1/3 of the weight on the top one, plus the weight of 1 set of scales.

      Now I am not saying this would be precise, or that you could accurately tell from the top scale, but if these were sprung scales this would theoretically work.

    3. Respectfully, Manley, I admire your spirit, but no. Your reference to the answer not being accurate might have given you a bit of a clue. 🙂

    4. or he could squat on one scale while holding other in both hands above his head.

      in this start position the lower scale reads “too heavy” while the upper scale is zero.

      Now he stands upright very quickly while noting the reading on each scale.

      The lower one will fall as his weight moves upwards while the upper one increases as he pressed on it (weight being same as accelerate).

      then simply add the two readings.

    5. You keep saying no, but if the scales were identical why would it not work? If these were springs which are stretched then Hooke’s law applies – these are compressed fluids, so can be considered in the same way as compressed springs and . . . Hooke’s law still applies and I am an idiot.

    6. Manley,

      The fault in your thinking is simple physics. Regardless of the measurement method (spring, load cell, etc.) mass is mass. The fact that one scale is compressed does not lessen the force on the next scale below it. Each scale receives the full mass of everything stacked above it.

      If your theory was correct, you could carry a 1 ton steel beam in your hand if you simply placed enough springs between your hand and the beam. Thinking of it this way should make the error obvious.

  3. He should find something big, around a stone or half a stone and weigh it. He’ll need to note it’s weight. While the thing is on the scales he should set the needle to zero, or toggle it. Then when he weighs himself he simply adds the weight of the item he used to set the scales, to the scales readout. That works, right?

    1. You can’t exceed a scale’s maximum weight by zeroing out some portion of it. Your idea started out fine though… weigh some heavy object… then attach it to yourself with a rope and set up a pulley so that it lifts a portion of your weight. Then step on the scale and add the weight of the object to what’s displayed.

      This sort of solution should work regardless of much you exceed the scale’s maximum… it’d be helpful to have a standard weight object to counter your own weight… like sandbags.

  4. one pair of scales, one visit to the local swimming baths.

    weigh self in deep end, completely submerged.

    apply maths for coefficient of relative density for chlorinated water vs human body.

    job done.

    1. He’d have to know his body density. There is no standard human body density, mostly because fat and muscle have very different densities, and people vary in how much fat and muscle they have.

    2. Unlikely to work.

      A man with that much body fat (I assume), will likely be positively buoyant in water, fat having a lower density that water.

      I suppose you could put the scales above him…

    3. body density can be implied by measuring the rise in water level. if that is difficult to measure in a large pool it can be implied first in a jackussi.

      yes, fatty man may float, so attach divers weights of known heaviness and then subtract from final number.

      after that, all is easy. many solns here are making this puzzle harder and more expensive than it need be.

  5. My solution involved a pivot point, a sturdy board, and a known weight. The rest is simple (and cheaper than buying a second set of scales).

  6. You could stack scales in a triangle shape (kind of pyramid).
    2 on the base, 1 on top
    or 3 on the base, 2 on top, and 1 on top
    or 4 on the base, 3 on top, 2 on top, 1 on top
    So theoretically you could weigh times 20 stones, so if there are 4 base scales you could weigh up to 80 stones, of course you would then have to subtract the weight of the scales themselves.

    In this way you could theoretically weigh any amount of stones, if you make the pyramid high enough, though some point in time you will get a statical problem and the “tower” will fall over.

  7. The lever board was the solution that came immediately to mind. Also,

    Archimedes principle: Fill bathtub ALL the way up. Place bathtub on *giant* tray to collect overflow water. Get into tub and submerge until floating. Then weigh the spill-over water. Et voila!

    [You could even do it w/o the giant tray by re-filling the tub afterwards, and weighing the water as it goes in. And if you go by the approximation that there is 1kg water per liter, you don’t need the scale AT ALL, just figure out how man liters of water you have to put in, and you have the weight in kilograms. Why anybody would be daft enough to measure weight in Stones is beyond me, so I leave the mathematic excercise of converting kg to Stones as an excercise for the reader (or google)]

    1. I distinctly remember doing that when I was a little kid. I couldn’t figure out why the scale still showed the same weight and not half the weight.

  8. I guessed correctly the two scale answer given and the one scale method with the lever explained better elsewhere.
    As an alternative however, he could remove his prosthetic legs and arm, weighting them and then the weight separately, adding the two arrive at his weight.
    Let the criticisms commence!

    1. I thought of that one too! I imagined our big chap like Swift’s prostitute, who takes all her accoutrements off or out (teeth etc) and is a shell of her former pretty self. The big chap is one step on from that and can remove parts of himself like Robocop!

    2. Okay Anne, you thought of it too but I think I prefer your explanation hands down (along with any other detachable limbs etc.) particularly the Robocop reference!

  9. With apologies to how to find the height of a building using a barometer:
    1) He buys one scale.
    2) He finds a place that has a scale which will weigh him however much he weighs.
    3) He tells the owner that he’ll give him this nice new scale if the owner lets him weigh himself each day.
    4) He repeats step 3 until he finds an owner willing to make the exchange.

    1. Richard just forgot to add “And John has just to sum-up the both displayed numbers of the scales”

  10. Thanks Matt ! That’s what I thought right away but I dismissed it because I cannot understand how it can be that “accurate”. I will have to try it to be convinced… 😉

  11. So simple. John buys one scale, gets a reading of 20 stone, does the sign of the cross and declares that Jesus came to him in a dream and gave him a new, Holy definition of “accurate” as being whatever the scale read. You sciency people think waaaaay to hard.

  12. Nobody seems to have thought of this one-scale solution, good for many daily weighings:
    Get a largish balloon, fill with He or H.
    Attach it to as many filled gallon jugs as it takes to just pull them off the ground.
    Weigh the jugs. Then get rid of them.
    Stand on scale, holding onto balloon.
    Add “jug weight” to displayed weight.

  13. Of course the balloon method is the equivalent of a rope, pulley and measured weight, but you don’t have the frictional losses using the balloon.
    The “real” solution took about 1 second, but it’s SO predictable; the many alternative solutions here are much more creative.

  14. I thought of that solution, what if he weighs more than 40 stone? Then it won’t work. Here is what I thought of:

    John puts a large bucket on the scale, and fills it up to 20 stone. Then he ties a rope around his waste. The rope goes up around a pulley on the ceiling, and then down to the bucket handle. Then the bucket hangs from the pulley, such that John is effectively made 20 stone lighter by the weight of the bucket pulling up upwards toward the ceiling. Then John weighs himself, with the rope tied around his waste and connected through the pulley to the 20-stone bucket.

    If John still maxes out at 20 stones, then he knows he weighs more than 40 stone, and has to add another 20-stone bucket to the original, He continues adding 20-stone buckets over and over until he comes in at less than 20 stone on the scale, at which point he simply adds the number on the scale to 20 times the number of buckets.

    Took about 5 minutes to write, about 10 seconds to dream up.

  15. He buys 5 scales and firmly attaches 4 of them to the ceiling in a square pattern. He runs a rope from each scale to a corner of a suspended tarp centrally located between the 4 scales. He then climbs on to the tarp and records the reading on each scale. He cuts down the tarp/rope apparatus and weights it on the fifth scale. He subtracts the tarp/rope weight from the accumulative reading of the 4 scales on the ceiling.

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