1. ## The Cheating Grocer

Six grocers in a town each sell a different brand of tea in 4-ounce packets at 25 cents per packet. One of the grocers gives short weight, each packet of his brand weighing only 3.75 ounces.

If I can use a scale for only one weighing, (a) how can I find the cheating grocer and ([img]/forums/images/smilies/cool.gif[/img] what is the minimum amount of money I must spend to solve the issue?

2. ## Re: The Cheating Grocer

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>I will need to buy 21 bags \$0.25 = \$5.25
Buy 1 from grocer 1, 2 from grocer 2, 3 from grocer3, etc.

Cheating #/weight
1 / 83.75
2 / 83.5
3 / 83.25
4/ 83
5/ 82.75
6/ 82.5</font color=yellow></span hi>

Steve

3. ## Re: The Cheating Grocer

Steve ..... almost, but not quite.

4. ## Re: The Cheating Grocer

<span style="background-color: #FFFF00; color: #000000; font-weight: bold"><font color=yellow>Buy one less from each than Steve proposes: 0 from the first one, 1 from the second one, etc. So we spend 15 * \$ 0.25 = \$ 3.75. Rest of the solution is similar.</font color=yellow></span hi>

5. ## Re: The Cheating Grocer

I disagree on the logic of NOT buying from one on principles. I thought of this solution (yes it was later, which is why I did not mention it), but if the weight was correct (all 4) it does NOT prove that grocer 1 was shortchanging (yes your puzzle states that 1 is, but that is NOT enough!)

Especially since there will also be some "variation" in the measurement. In reality many companies always put in MORE than required so that the +/- 95% Confidence is > than the target!

Steve

6. ## Re: The Cheating Grocer

<hr>(yes your puzzle states that 1 is, but that is NOT enough!)<hr>

This time, Steve, it's enough! <img src=/S/smile.gif border=0 alt=smile width=15 height=15>

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