Question

For those tired of answering questions and wanting to tickle their grey cells on a puzzle !!

A man buys 100 metal balls from a store. They r all the same shape and size and packed 10/bag.
BUT by mistake one bag contains balls which are all less by the same amount of weight as compared to the weight desired and available in other bags.

Shipment has to move out in 2 minutes when this discrepancy is noticed-- from the total weight of the shipment. However, an employee wins a reward from the boss by finding out the wrong bag by just one weighing (and it is not by fluke). Can u explain how ?????

Answer 1

I specifically said SINGLE weighing.

Even if you take them in groups and weigh, how r you going to locate that exact bag in one go !!!!

Answer 2

how many scale how we have? if we have ten scales :)

Answer 3

No, just one scale and one single weighing !!!!!

Answer 4

the bag is empty

Answer 5

Take one ball from the first bag, two balls from the second bag, and so on. In total we have 55 balls. Weight this collection of balls, call this weight W.

We know what the expected weight of the shipment is, call it E.

Then if the above collection of balls were all good, we would have a weight of 11E/20.

I'm hoping that we also know the weight of a good ball, call it G, and a bad ball, call it B.

The difference between 11E/20 and W divided by the difference between G and B, or (11E/20 - W)/(G-B), will give the number of the bad bag.

Answer 6

For example:
G = 2, B = 1.5, E= 200, 11E/20 = 110, bad bag = 5
W = 50*2 + 5*1.5 = 107.5
and (110-107.5)/0.5 = 5

Answer 7

sorry for reverting last but my broadband connection had "died" for an hour or so.....

cruffo, i did not get what 11E/20 is. pls clarify......

Answer 8

Well, if we expected the shipment to weigh E units, then each good ball should weigh E/100. The scheme of taking 1 ball from the first bag, 2 balls from the second, and so on would give us 1+2+3+...+10 = 55 ball.

If they were all good balls, then the total weight on the scale would be 55(E/100) = 11E/20

Answer 9

ok, I see... I think u hv the right answer

Simple explanation can be given as follows:

let the wt of good ball b = x
then as u said we will hv 55 balls by taking 1 ball from the first bag, two balls frm the second bag and so on..
so total IDEAL wt shud be 55x
let the wt by which one bad ball is less be d and let the ACTUAL WT of shipment be y

so we first find out the difference 55x - y = A

Then A divided by d gives us the number of balls having less wt and this in turn gives us the bag number containing the bad balls.

Well done Cruffo!!! Here's a medal for you :)

Answer 10

Wohoo!! I like you're answer better - less letters!

Answer 11

Yes this a more simple way of describing the process and the calculation required..

However, the main thing was how to do it in one weighing and that u got DEAD RIGHT !!!! Good show!!!