Solving the Weighing Balls Problem in One Weighing | Precise Scale Method

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To solve the weighing balls problem, take a specific number of balls from each box, starting with one ball from the first box, two from the second, and so on, up to 20 balls from the last box, totaling 210 balls. Weigh these balls on a precise scale, which will indicate a total weight that deviates from 210 pounds based on whether the defective box contains heavier or lighter balls. If the scale reads less than 210 pounds, the difference in ounces indicates the box number with lighter balls, while a reading above 210 pounds indicates heavier balls. The method effectively identifies the defective box in one weighing by analyzing the weight discrepancy. This approach combines systematic selection and precise measurement to achieve the solution.
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Homework Statement


Consider a set of boxes, each containing 20 balls. Suppose every ball weighs one pound, except that the balls in one box are all one ounce too heavy or one ounce too light. A precise scale is available that can weigh to the nearest ounce (not a balance scale). By selecting some balls to place on the scale, explain how to determine in one weighing which is the defective box and whether its balls are too heavy or too light.

Homework Equations

The Attempt at a Solution


Take one ball from the first box, two from the second box, three from the third etc.
Hence, Σn, from n=1 to 20. I'm not sure how to determine which is the defective box and whether it's too heavy or too light. Please advise.
 
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How many balls are weighed ? What are possible outcomes ?
 
210 balls are weighed and they either weigh 3,360lbs and 1oz or 3359lbs and 15oz
 
Sorry, don't know what I was thinking there. The balls weigh greater that 210lbs if the defective balls are too heavy and less than 210lbs if they are too light.
 
You seem to know there are 20 boxes. Where did you get that information ? It's not in the problem statement.
210 * 20 = 4200 , why do you expect 3360 ##\pm## 1 ounce ?
 
BvU said:
You seem to know there are 20 boxes. Where did you get that information ? It's not in the problem statement.
210 * 20 = 4200 , why do you expect 3360 ##\pm## 1 ounce ?
Sorry, looks like I left that info out. So, if I sum from n=1 to 20 I get 210 balls which means their weight is greater than 210lbs if the defective balls are too heavy and less than 210lbs if they are too light.
 
Now suppose ALL the balls in box 8 are 1 ounce too heavy. What would the 'precise scale' say ?
 
8.5lbs
 
sorry, 210lbs 8oz
 
  • #10
My mistake for writing 210 * 20. Should have been 210 * 1. Never mind.

Now we turn it around: if the precise scale reads 209 pounds 3 ounces, which box is suspect ?
 
  • #11
20
 
  • #12
?
 
  • #13
actually, 1
 
  • #14
How do you deduce that ?
 
  • #15
ok, I think it's 13 because the scale is 13oz shy of 210lbs.
 
  • #16
Right. Do you get the picture now ?
 
  • #17
Am I correct in assuming that I don't need to figure out the actual defective box but rather a method for determining that box?
 
  • #18
Robb said:
I'm not sure how to determine which is the defective box and whether it's too heavy or too light
I tried to help you with that.
And I suspect you can now also describe the method.

PS was the
Robb said:
Take one ball from the first box, two from the second box, three from the third etc.
your attempt at solution or was it in the problem description ?
 
  • #19
BvU said:
I tried to help you with that.
And I suspect you can now also describe the method.

PS was the
your attempt at solution or was it in the problem description ?

Probably a little of both. I appreciate you taking the time! I'd buy you a draft if I could:)
 
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