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How many minutes does it take $n$ moles to dig $m$ holes.
Pigeon holes principleReference request basic logic/model theoryWhy is the numbering of computable functions significant?Universal languages are primitive recursive.Correctly using the construction method in proofsHow does one understand and resolve Zeno's paradox?How to prove that there must be a bijection between two different implementations of the same functionIntro to Logic - helpWhy we have 2 quantifiers in predicate logic?Why is residues modulo 2 a model for natural numbers
$begingroup$
Recently, I came across this problem:
5 moles dig 4 holes in 3 minutes. How many minutes will it take 9 moles to dig 6 holes?
I decided that I should just use proportions and calculate it.
$implies 5$ moles dig 1 hole in $frac34$ minutes.
$implies 9$ moles dig $frac95$ holes in $frac2720$ minutes.
I quickly realize that this won't work, because the proportions asked are not the same.
However, I can't think of any other way to calculate this.
I figured that this question can be in the form
If $n$ objects do $m$ things in $o$ time units, how many time units can $x$ objects do $y$ things?
How can I solve this type of problem?
Thanks for your help! Your help is appreciated!
Max0815
logic
asked Mar 17 at 22:47
Max0815Max0815
81418
$endgroup$
add a comment |
$begingroup$
Recently, I came across this problem:
5 moles dig 4 holes in 3 minutes. How many minutes will it take 9 moles to dig 6 holes?
I decided that I should just use proportions and calculate it.
$implies 5$ moles dig 1 hole in $frac34$ minutes.
$implies 9$ moles dig $frac95$ holes in $frac2720$ minutes.
I quickly realize that this won't work, because the proportions asked are not the same.
However, I can't think of any other way to calculate this.
I figured that this question can be in the form
If $n$ objects do $m$ things in $o$ time units, how many time units can $x$ objects do $y$ things?
How can I solve this type of problem?
Thanks for your help! Your help is appreciated!
Max0815
logic
asked Mar 17 at 22:47
Max0815Max0815
81418
$endgroup$
$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13
add a comment |
$begingroup$
Recently, I came across this problem:
5 moles dig 4 holes in 3 minutes. How many minutes will it take 9 moles to dig 6 holes?
I decided that I should just use proportions and calculate it.
$implies 5$ moles dig 1 hole in $frac34$ minutes.
$implies 9$ moles dig $frac95$ holes in $frac2720$ minutes.
I quickly realize that this won't work, because the proportions asked are not the same.
However, I can't think of any other way to calculate this.
I figured that this question can be in the form
If $n$ objects do $m$ things in $o$ time units, how many time units can $x$ objects do $y$ things?
How can I solve this type of problem?
Thanks for your help! Your help is appreciated!
Max0815
logic
asked Mar 17 at 22:47
Max0815Max0815
81418
$endgroup$
Recently, I came across this problem:
5 moles dig 4 holes in 3 minutes. How many minutes will it take 9 moles to dig 6 holes?
I decided that I should just use proportions and calculate it.
$implies 5$ moles dig 1 hole in $frac34$ minutes.
$implies 9$ moles dig $frac95$ holes in $frac2720$ minutes.
I quickly realize that this won't work, because the proportions asked are not the same.
However, I can't think of any other way to calculate this.
I figured that this question can be in the form
If $n$ objects do $m$ things in $o$ time units, how many time units can $x$ objects do $y$ things?
How can I solve this type of problem?
Thanks for your help! Your help is appreciated!
Max0815
logic
logic
asked Mar 17 at 22:47
Max0815Max0815
81418
asked Mar 17 at 22:47
Max0815Max0815
81418
asked Mar 17 at 22:47
Max0815Max0815
81418
asked Mar 17 at 22:47
Max0815Max0815
81418
asked Mar 17 at 22:47
Max0815Max0815
81418
81418
$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13
add a comment |
$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13
$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Just start with $x=3 (textminutes)cdot ldots$, where 3 is the number you start from. And $x$ is the time to dig 6 holes by 9 moles. Then build a fraction of the following pairs of numbers: $(5,9);(6,4)$
The more moles are digging the less time is needed. Thus the fraction has to be smaller than 1. That means the factor is $frac59$.
The more holes supposed to be dug the more time is needed. Thus the fraction has to be greater than 1. That means the factor is $frac64$.
Thus $x=3cdot frac59cdot frac64=2.5 (textminutes)$
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
$endgroup$
add a comment |
$begingroup$
You are on the right track. You are expected to compute how many mole-minutes it takes to dig one hole. Then given any number of holes, you know how many mole-minutes are required, so divide by the number of moles.
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
add a comment |
$begingroup$
The general approach to questions of this sort is to assume that a particular animal (or person) performs a certain task at a constant rate. In this problem we suppose all moles are the same, so doubling the number of moles will double the number of holes in a given length of time;
doubling the time without changing the number of moles also will double the number of holes; but doubling the time and doubling the number of moles will cause them to dig four times as many holes.
You made a mistake by multiplying all three numbers by $frac95.$ As shown in the examples above, the three numbers do not all go up or down in proportion.
The things that do go up or down proportionally are the number of holes and the product of the number of moles and the time:
$$ holes = constant times moles times minutes .
$$
answered Mar 17 at 23:21
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
Here is an easy way to see why your logic does not work:
Suppose 1 mole digs 1 hole in 1 minute
Then by your logic, it would follow that:
2 moles dig 2 holes in 2 minutes
But that's absurd: The 2 moles get twice as much done as 1 mole in the same time period, and so the 2 moles will dig 2 holes still in that same 1 minute.
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just start with $x=3 (textminutes)cdot ldots$, where 3 is the number you start from. And $x$ is the time to dig 6 holes by 9 moles. Then build a fraction of the following pairs of numbers: $(5,9);(6,4)$
The more moles are digging the less time is needed. Thus the fraction has to be smaller than 1. That means the factor is $frac59$.
The more holes supposed to be dug the more time is needed. Thus the fraction has to be greater than 1. That means the factor is $frac64$.
Thus $x=3cdot frac59cdot frac64=2.5 (textminutes)$
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
$endgroup$
add a comment |
$begingroup$
Just start with $x=3 (textminutes)cdot ldots$, where 3 is the number you start from. And $x$ is the time to dig 6 holes by 9 moles. Then build a fraction of the following pairs of numbers: $(5,9);(6,4)$
The more moles are digging the less time is needed. Thus the fraction has to be smaller than 1. That means the factor is $frac59$.
The more holes supposed to be dug the more time is needed. Thus the fraction has to be greater than 1. That means the factor is $frac64$.
Thus $x=3cdot frac59cdot frac64=2.5 (textminutes)$
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
$endgroup$
add a comment |
$begingroup$
Just start with $x=3 (textminutes)cdot ldots$, where 3 is the number you start from. And $x$ is the time to dig 6 holes by 9 moles. Then build a fraction of the following pairs of numbers: $(5,9);(6,4)$
The more moles are digging the less time is needed. Thus the fraction has to be smaller than 1. That means the factor is $frac59$.
The more holes supposed to be dug the more time is needed. Thus the fraction has to be greater than 1. That means the factor is $frac64$.
Thus $x=3cdot frac59cdot frac64=2.5 (textminutes)$
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
$endgroup$
Just start with $x=3 (textminutes)cdot ldots$, where 3 is the number you start from. And $x$ is the time to dig 6 holes by 9 moles. Then build a fraction of the following pairs of numbers: $(5,9);(6,4)$
The more moles are digging the less time is needed. Thus the fraction has to be smaller than 1. That means the factor is $frac59$.
The more holes supposed to be dug the more time is needed. Thus the fraction has to be greater than 1. That means the factor is $frac64$.
Thus $x=3cdot frac59cdot frac64=2.5 (textminutes)$
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
answered Mar 17 at 23:25
callculuscallculus
18.6k31428
18.6k31428
add a comment |
add a comment |
$begingroup$
You are on the right track. You are expected to compute how many mole-minutes it takes to dig one hole. Then given any number of holes, you know how many mole-minutes are required, so divide by the number of moles.
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
add a comment |
$begingroup$
You are on the right track. You are expected to compute how many mole-minutes it takes to dig one hole. Then given any number of holes, you know how many mole-minutes are required, so divide by the number of moles.
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
$endgroup$
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
add a comment |
$begingroup$
You are on the right track. You are expected to compute how many mole-minutes it takes to dig one hole. Then given any number of holes, you know how many mole-minutes are required, so divide by the number of moles.
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
$endgroup$
You are on the right track. You are expected to compute how many mole-minutes it takes to dig one hole. Then given any number of holes, you know how many mole-minutes are required, so divide by the number of moles.
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
answered Mar 17 at 22:50
Ross MillikanRoss Millikan
300k24200375
300k24200375
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
add a comment |
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
Would I do as @Mark Bennet hinted 3 times 15 then divide by $frac32$?
$endgroup$
– Max0815
Mar 17 at 23:00
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
$begingroup$
I don't understand the calculation you are suggesting. I don't see where $3$ or $frac 32$ come into it. He and I are suggesting the same thing.
$endgroup$
– Ross Millikan
Mar 17 at 23:03
2
2
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Max0815 Just to confirm that Ross and I are suggesting the same. The trip hazard here is that we can become used to thinking about the rates at which things happen in quantities per minute, and are tempted to divide when we shouldn't. If you get into your head the idea of the "mole-minute", which implies multiplication rather than division, you will see that the number of mole-minutes is the capacity for digging holes. Units like "man-days" arise in similar problems.
$endgroup$
– Mark Bennet
Mar 18 at 5:39
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
$begingroup$
@Mark Bennet I understand
$endgroup$
– Max0815
Mar 18 at 17:11
add a comment |
$begingroup$
The general approach to questions of this sort is to assume that a particular animal (or person) performs a certain task at a constant rate. In this problem we suppose all moles are the same, so doubling the number of moles will double the number of holes in a given length of time;
doubling the time without changing the number of moles also will double the number of holes; but doubling the time and doubling the number of moles will cause them to dig four times as many holes.
You made a mistake by multiplying all three numbers by $frac95.$ As shown in the examples above, the three numbers do not all go up or down in proportion.
The things that do go up or down proportionally are the number of holes and the product of the number of moles and the time:
$$ holes = constant times moles times minutes .
$$
answered Mar 17 at 23:21
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
The general approach to questions of this sort is to assume that a particular animal (or person) performs a certain task at a constant rate. In this problem we suppose all moles are the same, so doubling the number of moles will double the number of holes in a given length of time;
doubling the time without changing the number of moles also will double the number of holes; but doubling the time and doubling the number of moles will cause them to dig four times as many holes.
You made a mistake by multiplying all three numbers by $frac95.$ As shown in the examples above, the three numbers do not all go up or down in proportion.
The things that do go up or down proportionally are the number of holes and the product of the number of moles and the time:
$$ holes = constant times moles times minutes .
$$
answered Mar 17 at 23:21
David KDavid K
55.4k344120
$endgroup$
add a comment |
$begingroup$
The general approach to questions of this sort is to assume that a particular animal (or person) performs a certain task at a constant rate. In this problem we suppose all moles are the same, so doubling the number of moles will double the number of holes in a given length of time;
doubling the time without changing the number of moles also will double the number of holes; but doubling the time and doubling the number of moles will cause them to dig four times as many holes.
You made a mistake by multiplying all three numbers by $frac95.$ As shown in the examples above, the three numbers do not all go up or down in proportion.
The things that do go up or down proportionally are the number of holes and the product of the number of moles and the time:
$$ holes = constant times moles times minutes .
$$
answered Mar 17 at 23:21
David KDavid K
55.4k344120
$endgroup$
The general approach to questions of this sort is to assume that a particular animal (or person) performs a certain task at a constant rate. In this problem we suppose all moles are the same, so doubling the number of moles will double the number of holes in a given length of time;
doubling the time without changing the number of moles also will double the number of holes; but doubling the time and doubling the number of moles will cause them to dig four times as many holes.
You made a mistake by multiplying all three numbers by $frac95.$ As shown in the examples above, the three numbers do not all go up or down in proportion.
The things that do go up or down proportionally are the number of holes and the product of the number of moles and the time:
$$ holes = constant times moles times minutes .
$$
answered Mar 17 at 23:21
David KDavid K
55.4k344120
answered Mar 17 at 23:21
David KDavid K
55.4k344120
answered Mar 17 at 23:21
David KDavid K
55.4k344120
answered Mar 17 at 23:21
David KDavid K
55.4k344120
55.4k344120
add a comment |
add a comment |
$begingroup$
Here is an easy way to see why your logic does not work:
Suppose 1 mole digs 1 hole in 1 minute
Then by your logic, it would follow that:
2 moles dig 2 holes in 2 minutes
But that's absurd: The 2 moles get twice as much done as 1 mole in the same time period, and so the 2 moles will dig 2 holes still in that same 1 minute.
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
Here is an easy way to see why your logic does not work:
Suppose 1 mole digs 1 hole in 1 minute
Then by your logic, it would follow that:
2 moles dig 2 holes in 2 minutes
But that's absurd: The 2 moles get twice as much done as 1 mole in the same time period, and so the 2 moles will dig 2 holes still in that same 1 minute.
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
$endgroup$
add a comment |
$begingroup$
Here is an easy way to see why your logic does not work:
Suppose 1 mole digs 1 hole in 1 minute
Then by your logic, it would follow that:
2 moles dig 2 holes in 2 minutes
But that's absurd: The 2 moles get twice as much done as 1 mole in the same time period, and so the 2 moles will dig 2 holes still in that same 1 minute.
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
$endgroup$
Here is an easy way to see why your logic does not work:
Suppose 1 mole digs 1 hole in 1 minute
Then by your logic, it would follow that:
2 moles dig 2 holes in 2 minutes
But that's absurd: The 2 moles get twice as much done as 1 mole in the same time period, and so the 2 moles will dig 2 holes still in that same 1 minute.
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
answered Mar 18 at 1:29
Bram28Bram28
63.9k44793
63.9k44793
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$begingroup$
Hint: 4 holes take 15 mole minutes.
$endgroup$
– Mark Bennet
Mar 17 at 22:50
$begingroup$
You technically need to know that this is some sort of linear relationship too I guess.
$endgroup$
– Boots
Mar 17 at 23:13