17 September
2015
A simple but elegant puzzle
Five friends at the gym came upon the new weighing machine recently installed there. They decided to take their weights in pairs (two at a time). The five of them can have ten pairings. They found the weights to be 259, 251, 264, 281, 268, 242, 257, 266, 253 and 275 pounds.
From this can you find out what were the individual weights of the friends?
If you are reading this on FB, message me if you have found the answers and your method.
First, Subramanian nailed it. Pradeep got it too using a very different approach (and Excel). Roy gave an approach but I am not sure of it. Although he is very sharp – so it might be me being the problem than his approach…
Solution:
First, note that none of the ten weights repeat. Therefore, we can safely conclude everybody has a unique weight.
Second, if we add all the ten combinations, we will have each of the friends included four times. So, if you add them up and divide by four you get the combined weight of the five friends as 654 pounds.
Third, let’s say the names are a b c d and e in increasing order of weight.
Note that the lowest combination – 242 has to be a+b and the second lowest combination 251 has to be a+c. You cannot say anything about the third lowest combination – it could be a+d or b+c. But we do not need that.
Conversely, the highest combination – 281 has to be d+e and second highest combination 275 has to be c+e. Like before you cannot go any further. And we do not need to.
Take a+b as 242 and d+e as 281 and take their sum from the combined 654. You get c as 131.
Take c from 251 which is a+c. You get a as 120. Then from a+b you get b as 122.
Similarly, d will be 137 and e will be 144.
Rajib, convinced about my approach? I inboxed it.
This was the easiest puzzle I have seen in your entire list. I did it in a minute and kept wondering if there was a catch.