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 March 28th, 2018, 07:08 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 99 Thanks: 1 Math Focus: Calculus Is there a quick method to find the right weight in this problem other than guessing? This is a problem I often see in exams, it involves calculating the least number of weighing operations needed to get the weight asked. It states this: Allison has a bag of rice which weighs more than 17 ounces and she has a two pan balance and 4 different weights being 3 oz, 4 oz, 7 oz and 11 oz respectively. How many times at least does she has to use the balance to get 17oz of rice? I tried to solve this problem by summing up all the weighs: $3\,\textrm{oz}+4\,\textrm{oz}+7\,\textrm{oz}+11\, \textrm{oz}=25\,\textrm{oz}$ Since what it is being asked is to get $17$ ounces what I tried to do is to find the possible combinations which put in each pan so that they sum up to $17$. But grouping either $11$ in one side and $3+4+7$ in the other the resulting weight is $3$. If I choose not to use one of the weights let's say $11$ and $4+7$ it cancels both sides, while $3+7$ produces $1$ ounce in the other side therefore it cannot be used, if it is $3+4$ it produces $4$ ounces. So it seems it is impossible to get the $17$ ounces at once. The other choice would be just using in the first attempt just $3$ and $7$, and in the second turn just the weight of $7$ ounce. Therefore the least number of times to use the balance would be $2$. But this answer is not correct. The method I tried to use I don't think it is right, it is prone to errors and more importantly is tedious which is something I can't use at an exam where time is limited. Can somebody help me to find a more orderly and logical method step by step to solve these kind of problems other than just guessing and starting to plug in numbers randomly as if it was trying to hit a target?. Please don't say just the answer alone or "put the weight and the object in one pan and the rest in the other pan", I know that and It's not what I'm looking for. What I need to know is if Is there anything that can be used with more success rate such as a diagram, a table or something that I can fill up to avoid double use the weights? or maybe a suggestion such as put the numbers from the highest to the lowest and group them together. I don't know, maybe building a system of two equations?.
March 28th, 2018, 07:50 PM   #2
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Quote:
 Originally Posted by Chemist116 This is a problem I often see in exams, it involves calculating the least number of weighing operations needed to get the weight asked. It states this: Allison has a bag of rice which weighs more than 17 ounces and she has a two pan balance and 4 different weights being 3 oz, 4 oz, 7 oz and 11 oz respectively. How many times at least does she has to use the balance to get 17oz of rice? I tried to solve this problem by summing up all the weighs: $3\,\textrm{oz}+4\,\textrm{oz}+7\,\textrm{oz}+11\, \textrm{oz}=25\,\textrm{oz}$ Since what it is being asked is to get $17$ ounces what I tried to do is to find the possible combinations which put in each pan so that they sum up to $17$. But grouping either $11$ in one side and $3+4+7$ in the other the resulting weight is $3$. If I choose not to use one of the weights let's say $11$ and $4+7$ it cancels both sides, while $3+7$ produces $1$ ounce in the other side therefore it cannot be used, if it is $3+4$ it produces $4$ ounces. So it seems it is impossible to get the $17$ ounces at once. The other choice would be just using in the first attempt just $3$ and $7$, and in the second turn just the weight of $7$ ounce. Therefore the least number of times to use the balance would be $2$. But this answer is not correct. The method I tried to use I don't think it is right, it is prone to errors and more importantly is tedious which is something I can't use at an exam where time is limited. Can somebody help me to find a more orderly and logical method step by step to solve these kind of problems other than just guessing and starting to plug in numbers randomly as if it was trying to hit a target?. Please don't say just the answer alone or "put the weight and the object in one pan and the rest in the other pan", I know that and It's not what I'm looking for. What I need to know is if Is there anything that can be used with more success rate such as a diagram, a table or something that I can fill up to avoid double use the weights? or maybe a suggestion such as put the numbers from the highest to the lowest and group them together. I don't know, maybe building a system of two equations?.
This is a very silly type of problem because grains of rice are not identical and there is no assurance that any whole number of grains is very close to 3 oz, 4 oz, 7 oz, 11 oz, or 17 oz. Moreover, it is not clear what a trial is. If I drop one grain into a pan is that a trial?

EDIT: You can get 14 oz with 7 + 3 + 4 in one pan and 14 oz of rice in the other. You can then get 3 oz of rice in one pan with a 3 oz weight in the other. The two pans of rice together equal 17 oz. That is one way to find 17 oz, but it is not the only way.

Last edited by JeffM1; March 28th, 2018 at 07:55 PM.

March 28th, 2018, 08:22 PM   #3
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Quote:
 Originally Posted by JeffM1 This is a very silly type of problem because grains of rice are not identical and there is no assurance that any whole number of grains is very close to 3 oz, 4 oz, 7 oz, 11 oz, or 17 oz. Moreover, it is not clear what a trial is. If I drop one grain into a pan is that a trial? EDIT: You can get 14 oz with 7 + 3 + 4 in one pan and 14 oz of rice in the other. You can then get 3 oz of rice in one pan with a 3 oz weight in the other. The two pans of rice together equal 17 oz. That is one way to find 17 oz, but it is not the only way.
Probably the person who made this problem assumed IDEAL conditions (which kind be some subjective), let's say the grains are in a bag with negligible weight i.e a paper bag so thin that it doesn't count, but since we're speaking of ounces. Does such material exists?. It also seems to implicitly consider that the trial or attempt is consider valid if the whole thing is put in the pan at once not just one small grain (hence inside into something or thrown in the pan so that all grains are the same shape, weight, texture, in other words identical) and extracted slowly so that it can balance to 17 oz.

Anyway (like the Fibonacci numbers which were thought from a rabbit mating problem, which of course is unreal to happen in nature) this problem also considers in some way these things.

Now returning to the question. As you mentioned we can get to $17$ by doing these combinations. My question was if there was a way to get the right combination faster other than just trying randomly. Hence why I asked if there was a table or a graphical approach. Because when you put the answer so straightforward the mental process to get the answer remains hidden to the observer. In other words to be more explicit than whoever posed this question. Therefore this part is where I need help. Where to begin with, choose the highest number and subtract it from which or what?.

 March 28th, 2018, 10:15 PM #4 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 I suppose that you can think of it as an iterative algorithm. 1 trial You can get 3, 4, 7, or 11 with a single weight. You can get 7, 10, 11, 14, 15, or 18 with a pair of weights. You can get 14, 18, 21, or 22 with a triplet of weights. You get 25 with all four weights. What are the deficiencies corresponding to these combinations. 14, 13, 10, 6 10, 7, 6, 3, 2 3 Notice that we can get 14, 7, or 3 in a single trial. So if we pick a 14, 10, or 3 oz bag and the right number of weights on the first trial and pick a 3, 7, or 14 oz bag and the right number of weights on the second trial, we shall get two bags that sum to 17 oz in just two trials. It is ultimately a problem in combinatorics. I still think it is very contrived.
 March 28th, 2018, 11:35 PM #5 Senior Member   Joined: Apr 2014 From: UK Posts: 953 Thanks: 340 Clearly, some of you have never sandpapered rice.... I had a go at the question by following a systematic rule, use the big numbers first and hope the result falls out. So, in the left pan I put 11 oz. Then I put in the 7oz to give me 18oz. This leaves me with the 3 and 4oz and I'm over by 1 oz, well that's easy since 4-3 =1, so put the 3oz in the left pan to give me 21oz and put the 4oz in the right pan, now I need to add 17oz of rice to make it balance. The question doesn't state the type of rice, but if it's long grain then one grain will have to have about 1/3 cut off and sanded down Thanks from JeffM1
 March 29th, 2018, 05:42 AM #6 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 Dave Very nice. I was not thinking of enough combinations. 11 + 7 + 3 in one pan and 4 in the other pan is equivalent to 11 + 7 + 3 - 4 = 17 in the first pan. In that case, you can weigh 1 = 4 - 3 3 = 3 4 = 4 6 = 7 + 3 - 4 7 = 7 8 = 11 - 3 10 = 11 + 3 - 4 11 = 11 12 = 11 + 4 - 3 14 = 7 + 4 + 3 15 = 11 + 7 - 3 17 = 11 + 7 + 3 - 4 18 = 11 + 7 19 = 11 + 7 + 4 - 3 21 = 11 + 7 + 3 22 = 11 + 7 + 4 25 = 11 + 7 + 4 + 3 I do not see how to get 2, 5, 9, 13, 16, 20, 23, or 24.
March 29th, 2018, 06:41 PM   #7
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Quote:
 Originally Posted by weirddave Clearly, some of you have never sandpapered rice.... I had a go at the question by following a systematic rule, use the big numbers first and hope the result falls out. So, in the left pan I put 11 oz. Then I put in the 7oz to give me 18oz. This leaves me with the 3 and 4oz and I'm over by 1 oz, well that's easy since 4-3 =1, so put the 3oz in the left pan to give me 21oz and put the 4oz in the right pan, now I need to add 17oz of rice to make it balance. The question doesn't state the type of rice, but if it's long grain then one grain will have to have about 1/3 cut off and sanded down
Yes I have never done that. Actually I tried to do your approach first but I got tangled with the combinations later as I did not know how to subtract exactly 1 oz from the other pan without affecting the 18 oz too much. Most of the time I got errors in exams by incorrectly assuming additional trials because of that reason. Is there any suggestion on how to avoid this?.

March 30th, 2018, 12:37 AM   #8
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Quote:
 Originally Posted by JeffM1 I do not see how to get 2, 5, 9, 13, 16, 20, 23, or 24.
5 = 11 - 7 + 4 - 3

Quote:
 Originally Posted by Chemist116 Yes I have never done that. Actually I tried to do your approach first but I got tangled with the combinations later as I did not know how to subtract exactly 1 oz from the other pan without affecting the 18 oz too much. Most of the time I got errors in exams by incorrectly assuming additional trials because of that reason. Is there any suggestion on how to avoid this?.
Play with numbers, a lot.
Here in the UK we've had this for 36 years:

Note that you're not subtracting 1oz, so I don't quite understand what you mean by "without affecting the 18 oz too much". Both pans have 21oz in them.
There is a method you could follow. There are 3 options for each number, either add it, subtract it, or miss it out. This gives 3^4 combinations to try.

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