Math Is Fun Forum

Hannibal lecter

Member

From: Iraq

Registered: 2016-02-11

Posts: 408

weighing jelly babies puzzles

Hi, this puzzle:

https://www.mathsisfun.com/puzzles/weighing-jelly-babies-solution.html

is very complex puzzle I couldn't solve it or understand it

first I want to know did we put all the cartoons on the scale or only the 510 grams?
I mean maybe there is a lot of jelly babies around 1000g or more..

so are we measuring only the 51 ones that we take out or all of the boxes?

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Wisdom is a tree which grows in the heart and fruits on the tongue

Hannibal lecter

Member

From: Iraq

Registered: 2016-02-11

Posts: 408

Re: weighing jelly babies puzzles

The sentence in the puzzle says:

“Remove a different number of jelly babies from each box, weigh the boxes.”

That wording is misleading.

It really means:

“Remove jelly babies from the boxes and weigh what you removed.”

Not weigh the boxes themselves. right?

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Wisdom is a tree which grows in the heart and fruits on the tongue

Bob

Administrator

Registered: 2010-06-20

Posts: 10,826

Re: weighing jelly babies puzzles

If the boxes weighed nothing then it wouldn't matter.  So I agree with you that the removed jb should be weighed.

How does the solution work?

If you remove a different number from each box (carefully remembering how many from each box) then the weight of the jb removed gives a clue to which boxes contain the imitation ones.

But you have to choose the number removed with some advanced planning.

Let's take the simplest possible plan: take 1 from box 1; 2 from box 2 and so on.

That means you have taken 28 altogether (7 + 6 + 5 + 4 + 3 + 2 + 1). If none were imitation that amount would weigh 28 x 10 = 280.

But some a imitation so lets suppose the weight is 274. The underweight is 6 and that can only happen if the underweight ones came from box 1, box 2 and box 3, as 1+2+3 = 6.

But if the weight is 267 then there are (at least) two solutions 1 + 5 + 7 and 3 + 6 + 4.

To make sure that every weighing gives only one solution you need to choose a number to remove that will never have any repeat solutions. 

The answer suggests 0, 1, 2, 4, 7, 13, and 24.  Try it. Pick any three from this list and the total is unique. So you'll always be able to solve the problem with one weighing.

Other sets of 7 numbers are possible.  To really get to grips with this puzzle, try to find another.

Other issues with the solution.

(1) 24 is quite a large number. What happens if the total in a box is less than 24.

(2) Now you've mixed in the weighing tray a number of real and imitation jb  how do you put them back in the correct boxes?

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob