[Guest]

This is one of my favorite puzzles..

Common sense and basic math should get you there =)

You have 10 bags of marbles, each bag has exactly 10 marbles in it. All 100 marbles look the same and have the same outer size. 9 bags have marbles that weigh exactly the same (90 marbles in total). The tenth bag has marbles that are a little bit lighter than the other 90 but the same as each other.

You only get one chance to weigh.. how do you find out the bag with the lighter marbles?

[Guest]

Hi,

I can solve this with 2 additional inputs... The weight for a standard marble & lighter marbles should be known. If yes then...

1. Take 1 marble from bag number 1, 2 from bag number 2, 3 from bag number 3 & so on for all 10 bags. You will have have 55 marbles in total (1+2+3+...10)

2. Since the weight of the standard marble and the light marble is known, just find difference between the measured weight of 55 marbles & standard marble weight X 55 will point to the bag no.

E.g.: If the standard marble weight is 10gm & light marble weight is 9gm, and the lighter marbles are in say 5th bag then the measured weight of 55 marbles will be less by 5gm to 55 X 10gm

In case there is any other way to solve this, without knowing the weights of standard & lighter marbles, I would be curious to know.

[Guest]

Assumption: We know the weight difference b/w the heavy and the light ball..

I'm assuming that we have a weighing machine with two bases( taraazu),, and the machine is advanced enough to tell the difference between the weights...

No. the bags as 1,2,3...10..

Bag 1- 1 ball.. Bag 3- 3 balls Bag 5- 5 balls Bag 7- 4 balls Bag 9- 2 balls

Bag2- 2 balls Bag 4- 4 balls Bag 6- 5 balls Bag 8- 3 balls Bag 10- 1 ball

Now, place the balls from bag 1-5 in 1 part and d balls from bag 6-10 in the other part,,,

We'll be using one reading as said in the question, and as we know the difference in the weights, we will be able to know which bag is the defaulter one...

[Guest]

y not put all bags in water and see which bag takes the longest time to sinks...thats the one!!!

[Guest]

I think the easiest thing would be...

I would be using scales to weigh the bags, so if you took two at a time and put them on opposite sides of the scales, then as soon as you have a light bag that side would rise a bit when you put both bags on. I have only two hands, so I think it makes sense to put the bags on the scales two at a time, while still satisfying the requirement of only one weighing, right? Otherwise I like Newbie's idea right above me!

By the way, did everyone forget to use spoilers now?

[Guest]

I just dont get the answer jpar1983...

[Quantum.Mechanic]

I just dont get the answer jpar1983...

putting bags on a scale is not an atomic, all-at-once step. Bags are put on one or two at a time.

Assume bags are put on in pairs, one on each side. As soon as the scale becomes unbalanced, you know which pair the light bag is in. The light bag will be on the light side of the scale.

[Guest]

you put all ten bags on the scale. take one bag out at a time untill one of the bags scale goes down a smaller number than the other bags.

example if 9 bags are 10kg each n 1 bag is 5kg. put all ten on the scale n itll read 95. take one out n itl read 85 then 75 then 65 (so its going down by tens) untill u one goes down by 5kg then that bag will be the light bag. that way we found out the light bag in one weighing.

the solution is simple. the Q says one chance to weigh so doing it in pairs wont work cuz it may take more than one weighing. here is the awnser.

Edited July 28, 2011 by ameer awad