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I'm buying a home gym set-up for my son (really!) and I want to know the fewest number of weight plates I need to buy to be able to go up in 5kg increments to 170kg. The bar weighs 20kg and the weights are available in 25kg, 20kg, 15kg, 10kg, 5kg and 2.5kg. It is cheaper per kilogram to buy bigger plates (£25 2.5kg, £40 5kg, £70 10kg, £90 15kg, £120 20kg and £145 25kg). Weights come in pairs and when a pair is used one plate must go on either side.

The pairs need to be split evenly because of the risk of unbalancing due to moments (levers/physics) and hence the risk of injury. My son wouldn't want to have to refer to a chart for every weight increment (although feel free to solve it if you want, it is a standard 7' Olympic bar).

Can I do better than just buying a pair of each?

(Note that I do not know, but feel that there is a better answer.)

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  • $\begingroup$ Why would you be buying +5 kg weights if you are looking to increase the increments by 5 each time? $\endgroup$
    – PDT
    Commented Sep 24, 2020 at 9:00
  • $\begingroup$ I always love real-world puzzles :) $\endgroup$
    – matt
    Commented Sep 24, 2020 at 14:41
  • $\begingroup$ math.stackexchange.com/questions/3591796/… $\endgroup$
    – RobPratt
    Commented Sep 24, 2020 at 16:39
  • $\begingroup$ Thanks all, I think the real answer is not to buy home gym equipment in the middle of a pandemic! $\endgroup$
    – Joel Moreland
    Commented Sep 25, 2020 at 8:29

5 Answers 5

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A naive solution I have by hand is:

1 pair of 2.5 kg
2 pairs of 5 kg
1 pair of 15 kg
2 pairs of 25 kg
This would allow you to build everything up to 50 kg easily, and then lop on a pair of 25 kg weights each time to go to 50-100 and 100-150 kg.

Then add the 20 kg bar and you have the whole 20-170 kg range.

In this I assume you don't want asymmetric weights (e.g. a 10kg plate on one side and two 5kg on the other), you want to minimise expenditure, and also assume the prices you give in the comment

£25 2.5kg, £40 5kg, £70 10kg, £90 15kg, £120 20kg and £145 25kg

Also note that with the prices you give, it's cheaper to buy a pair of 15 kg than a pair of 2.5 kg + a pair of 10 kg (which would be less total weight, but more expensive!)

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  • $\begingroup$ Two of you have what looks like the winning solution, thanks! $\endgroup$
    – Joel Moreland
    Commented Sep 24, 2020 at 13:01
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You could swap a 10kg weight for a 5 (if you do not mind asymmetry)
In addition you could swap the 20's for a 15 and a 25

With symmetry there still a better solution. You can swap the 10 and 20 for a 5 and 25

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  • $\begingroup$ thanks but I've subsequently clarified the requirement for symmetry. $\endgroup$
    – Joel Moreland
    Commented Sep 24, 2020 at 9:10
  • $\begingroup$ Two of you have what looks like the winning solution, thanks! $\endgroup$
    – Joel Moreland
    Commented Sep 24, 2020 at 13:01
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I am going to post my answer as an alternative. I just realized that there are (15kg) weight sets. My answer is not optimal, but I was in too deep to not post. Also, it may be better for a beginner to have smaller weights because he/she can get a better variety of doing drop sets/super sets (bicep curls, lunges, skull crushers, over head presses). Enjoy.

All weights on bar are symmetrical on both sides.

Pricing for not so optimal solution

Proof of 5kg weight increments

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  • $\begingroup$ Interesting, the others all went for more larger plates. I think I probably didn't give you an essential constraint: the length of the the area for adding plates on the end of the bar. Although I don't know it I guess there is probably not enough room for lots of small plates. $\endgroup$
    – Joel Moreland
    Commented Sep 25, 2020 at 8:25
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I like to work on a real-life problem for once. :-)

OK, I assume the price is for a pair of plates.

Looking at the prices, the cost of a set of plates is roughly £10 per plate plus £2.5 per kg. Smaller plates are a bit cheaper but we'll see that it is better to use less of heavier plates, so that doesn't help.

Given that weight-to-cost relation, if the maximum weight is fixed (175 kg) the only way to save money is by reducing the number of plates, that is using fewer but heavier plates.

It happens that it is not possible to cover the whole range 20 to 175 with only 5 pairs among the available weights.

But interestingly there are 32 different values in the range 20 to 175 by increments of 5. 32 is in how many ways you can combine 5 weights. So it is actually possible to solve your problem provided you can buy 40kg plates. The set of pairs you need is 2.5, 5, 10, 20, 40.

But if I extrapolate, the price for the 40kg pair should be £220. So the total cost for the 5 pairs would be £475 instead of th £490 for the standard 6 pairs. Not much savings.

You could use fewer plates if you could buy single plates. I considered for instance using a single 20kg plate on one side and 2x5kg + 10kg on the other side. Given the available choice of plates you could make all weights with only 11 plates (2x2.5, 2x5, 2x10, 1x20, 4x25). But again, the saving is marginal. You would save maybe £10 for buying one less plate.

In summary you can't save much by reorganizing the sets of plates. The price of the plates depends mostly on the total weight. You can save some money if you replace some plates by lighter ones, but it is at the cost of reducing the maximum weight.

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  • $\begingroup$ Good idea on 40kg plates but if they exist they would be more expensive (I chose the cheapest style of plate on a reputable website). Interestingly a bundle (including a bar) from another reputable site was more expensive. Asymmetry is not acceptable for safety reasons. $\endgroup$
    – Joel Moreland
    Commented Sep 25, 2020 at 8:23
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  • 4 × 10kg
  • 2 × 5kg
  • 2 × 2.5kg

The above plus however many 25kg plates you’re strong enough to need will cover every 5kg increment.

I’d personally add two 1.25kg plates for the ability to make smaller 2.5kg weight jumps.

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