How many minutes would it take to make 100 toys?
$begingroup$
7 machines take 7 min to make 7 identical toys. At the same rate how many minutes would it take for 100 machines to make 100 toys?
(A) 1 (B) 7
(C) 100 (D) 700
My trial: Given that
7 machines:7minutes:7 toys
1 machine:1minute:1 toy
100 machines:100 minutes:100 toys
therefore it would 100 minutes for 100 machines to make 100 toys.
But my book suggests that answer must be 7 minutes I don't know where I am wrong. I think there should be a correct method to solve such problems. Please give correct solution to this problem.
thanks
arithmetic
$endgroup$
add a comment |
$begingroup$
7 machines take 7 min to make 7 identical toys. At the same rate how many minutes would it take for 100 machines to make 100 toys?
(A) 1 (B) 7
(C) 100 (D) 700
My trial: Given that
7 machines:7minutes:7 toys
1 machine:1minute:1 toy
100 machines:100 minutes:100 toys
therefore it would 100 minutes for 100 machines to make 100 toys.
But my book suggests that answer must be 7 minutes I don't know where I am wrong. I think there should be a correct method to solve such problems. Please give correct solution to this problem.
thanks
arithmetic
$endgroup$
$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04
add a comment |
$begingroup$
7 machines take 7 min to make 7 identical toys. At the same rate how many minutes would it take for 100 machines to make 100 toys?
(A) 1 (B) 7
(C) 100 (D) 700
My trial: Given that
7 machines:7minutes:7 toys
1 machine:1minute:1 toy
100 machines:100 minutes:100 toys
therefore it would 100 minutes for 100 machines to make 100 toys.
But my book suggests that answer must be 7 minutes I don't know where I am wrong. I think there should be a correct method to solve such problems. Please give correct solution to this problem.
thanks
arithmetic
$endgroup$
7 machines take 7 min to make 7 identical toys. At the same rate how many minutes would it take for 100 machines to make 100 toys?
(A) 1 (B) 7
(C) 100 (D) 700
My trial: Given that
7 machines:7minutes:7 toys
1 machine:1minute:1 toy
100 machines:100 minutes:100 toys
therefore it would 100 minutes for 100 machines to make 100 toys.
But my book suggests that answer must be 7 minutes I don't know where I am wrong. I think there should be a correct method to solve such problems. Please give correct solution to this problem.
thanks
arithmetic
arithmetic
asked Feb 8 '18 at 18:28
jeanne clementjeanne clement
411315
411315
$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04
add a comment |
$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04
$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - $frac {7} {7}.100 $ minutes = 100 minutes
Or, 100 machines make 100 toys in - $ 100. frac {7}{100} = 7 $ minutes
The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.
$endgroup$
add a comment |
$begingroup$
$7$ machines in $7$ minutes make $7$ toys.
If we divide either the number of machines or the number of toy by seven we will get two correct statements:
Either $1$ machine takes $7$ minutes to make $1$ toy. OR
$7$ machines take $1$ minute to make $1$ toy.
Both are true.
Notice that we CAN'T divide BOTH machines AND time by $7$ because that is dividing by a factor of $7$ two times, or is actually dividing the entire system by $49$.
We divide the other term (time of machine) by seven and we get:
$1$ machine takes $1$ minute to make $frac 17$ toy.
Either multiply the machines by $100$ to get:
$100$ machines take $1$ minute to make $frac {100}7$ toys.
And then multiply the toys by $7$ to get:
$100$ machines take $7$ minutes to make $100$ toys.
Of multiply the toys by $700$ to get either:
$700$ machines take $1$ minute to make $100$ toys.
Or one machine takes $700$ minuts to make $100$ toys.
If you did that we can divide the machines and multiply the time (or vice versa) to get:
$100$ machines take $7$ minutes to make $100$ toys.
====
Now, if we were clever or if we had experience we might have noticed:
$7$ machines in $7$ minutes make $7$ toys means
$1$ machine in $7$ minutes make $1$ toy means
$x$ machines in $7$ minutes make $x$ toys.
So $100$ machine will take $7$ minutes to make $100$ toys.
That would have been faster and easier.
But... maybe not as immediately clear.
$endgroup$
add a comment |
$begingroup$
In some fixed amount of time, $7$ machines make $7$ toys. If all machines are identical, $1$ machine must in that time have made $1$ toy. So $100$ machines make the $100$ toys in that same amount of time.
$endgroup$
add a comment |
$begingroup$
This is a bogus "applied math" problem, whose "correct" solution is predicated on implicit, possibly erroneous, assumptions being made.
@Manish Kundu and others have arrived at the "correct" answer B (7 minutes) based on the implicit assumption that all machines operate independently to produce toys, and can be run in parallel at the same production rate. The folowing is not meant as a criticism to the answerers, but to the creator of the question.
Suppose that in fact 7 machines need to work together to produce ANY toys, and let's still suppose that all 7 or 100 machines are identical. Under this supposition, 14 sets of 7 machines can work in parallel to produce 98, not 100, toys in 7 minutes. Only at the completion of this initial 7 minutes can some set of 7 machines then be used to create another 7 (or maybe 2) toys, which as far as we know might take as long as another 7 minutes, even without any set up time being required after the first 7 minutes until the 2nd 7 minutes begins. So the answer could be 14 minutes, not 7 minutes. And given the assumptions still being made, even that may not be correct.
$endgroup$
add a comment |
$begingroup$
I had similar question on examination few days ago. The question is next:
7 robots will produce 7 toys in 7 minutes. How long it will take for
70 robots to produce 70 toys.
FIRST APPROACH:
It takes 7 minutes for 1 robot to produce 1 toy.
1 robot will produce 70 toys in = 70 toys * 7 minutes (number of minutes per toy) = 490 minutes
So, 1 robot will create 70 toys in 490 minutes.
If we divide number of minutes to produce 70 toys for one robot which is 490 minutes with 70 robots, which is question, it will take 7 minutes to produce 70 toys using 70 robots.
This also means that each robot will produce 1 toy.
In case that we have 100 robots to build 100 toys, and all parametars are same as above, 7 robots – 7 toys – 7 minutes.
You can easily apply this logic to solve this issue:
1 robot will produce 100 toys in 700 minutes
100 robots will produce 100 toys in 7 minutes
SECOND APPROACH:
Main question is, “How much toys each robot needs to produce?”
Answer on this question is next:
We have 7 robots to produce 7 toys – each robot will produce 1 toy. 1 toy * 7 minutes (number of minutes per toy) = 7 minutes for production.
In our case above we have 70 robots to produce 70 toys. 70 toys / 70 robots = 1 toy each robot. We will multiply number of minutes for 7 robots to build 7 toys which is 7 minutes with 1 and we will get 7 minutes.
Also if we have 35 robots to produce 70 toys. It means that 70 toys need to produce 35 robots, each robot will produce 1.5 toy.
If we multiply 1.5 * 7, we will get 10:30 for 35 robots to produce 70 toys.
And for 100 robots to produce 100 toys = 100 / 100 * 7 = 7 minutes.
Hope this is understandable and it will help someone to solve similar problems.
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add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes
Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in 7 mins
so 1 st machine takes 7 mins to produce a toy
2 nd machine takes 7 mins to produce a toy
soooo on
But all machines run at simultaneously so total time is 7 mins only
if you run 100 machines for 7 mins we can produce 100 toys
$endgroup$
add a comment |
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7 Answers
7
active
oldest
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7 Answers
7
active
oldest
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active
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$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - $frac {7} {7}.100 $ minutes = 100 minutes
Or, 100 machines make 100 toys in - $ 100. frac {7}{100} = 7 $ minutes
The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - $frac {7} {7}.100 $ minutes = 100 minutes
Or, 100 machines make 100 toys in - $ 100. frac {7}{100} = 7 $ minutes
The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - $frac {7} {7}.100 $ minutes = 100 minutes
Or, 100 machines make 100 toys in - $ 100. frac {7}{100} = 7 $ minutes
The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.
$endgroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - $frac {7} {7}.100 $ minutes = 100 minutes
Or, 100 machines make 100 toys in - $ 100. frac {7}{100} = 7 $ minutes
The general way to proceed is to keep all quantities constant, except one, and see how it varies with the quantity you require.
Now the explanation is that same number of machines would take more time to make more toys, hence it is directly proportional. However, more machines would take less time to make same number of times,i.e, it is inversely proportional.
answered Feb 8 '18 at 18:33
Manish KunduManish Kundu
676215
676215
add a comment |
add a comment |
$begingroup$
$7$ machines in $7$ minutes make $7$ toys.
If we divide either the number of machines or the number of toy by seven we will get two correct statements:
Either $1$ machine takes $7$ minutes to make $1$ toy. OR
$7$ machines take $1$ minute to make $1$ toy.
Both are true.
Notice that we CAN'T divide BOTH machines AND time by $7$ because that is dividing by a factor of $7$ two times, or is actually dividing the entire system by $49$.
We divide the other term (time of machine) by seven and we get:
$1$ machine takes $1$ minute to make $frac 17$ toy.
Either multiply the machines by $100$ to get:
$100$ machines take $1$ minute to make $frac {100}7$ toys.
And then multiply the toys by $7$ to get:
$100$ machines take $7$ minutes to make $100$ toys.
Of multiply the toys by $700$ to get either:
$700$ machines take $1$ minute to make $100$ toys.
Or one machine takes $700$ minuts to make $100$ toys.
If you did that we can divide the machines and multiply the time (or vice versa) to get:
$100$ machines take $7$ minutes to make $100$ toys.
====
Now, if we were clever or if we had experience we might have noticed:
$7$ machines in $7$ minutes make $7$ toys means
$1$ machine in $7$ minutes make $1$ toy means
$x$ machines in $7$ minutes make $x$ toys.
So $100$ machine will take $7$ minutes to make $100$ toys.
That would have been faster and easier.
But... maybe not as immediately clear.
$endgroup$
add a comment |
$begingroup$
$7$ machines in $7$ minutes make $7$ toys.
If we divide either the number of machines or the number of toy by seven we will get two correct statements:
Either $1$ machine takes $7$ minutes to make $1$ toy. OR
$7$ machines take $1$ minute to make $1$ toy.
Both are true.
Notice that we CAN'T divide BOTH machines AND time by $7$ because that is dividing by a factor of $7$ two times, or is actually dividing the entire system by $49$.
We divide the other term (time of machine) by seven and we get:
$1$ machine takes $1$ minute to make $frac 17$ toy.
Either multiply the machines by $100$ to get:
$100$ machines take $1$ minute to make $frac {100}7$ toys.
And then multiply the toys by $7$ to get:
$100$ machines take $7$ minutes to make $100$ toys.
Of multiply the toys by $700$ to get either:
$700$ machines take $1$ minute to make $100$ toys.
Or one machine takes $700$ minuts to make $100$ toys.
If you did that we can divide the machines and multiply the time (or vice versa) to get:
$100$ machines take $7$ minutes to make $100$ toys.
====
Now, if we were clever or if we had experience we might have noticed:
$7$ machines in $7$ minutes make $7$ toys means
$1$ machine in $7$ minutes make $1$ toy means
$x$ machines in $7$ minutes make $x$ toys.
So $100$ machine will take $7$ minutes to make $100$ toys.
That would have been faster and easier.
But... maybe not as immediately clear.
$endgroup$
add a comment |
$begingroup$
$7$ machines in $7$ minutes make $7$ toys.
If we divide either the number of machines or the number of toy by seven we will get two correct statements:
Either $1$ machine takes $7$ minutes to make $1$ toy. OR
$7$ machines take $1$ minute to make $1$ toy.
Both are true.
Notice that we CAN'T divide BOTH machines AND time by $7$ because that is dividing by a factor of $7$ two times, or is actually dividing the entire system by $49$.
We divide the other term (time of machine) by seven and we get:
$1$ machine takes $1$ minute to make $frac 17$ toy.
Either multiply the machines by $100$ to get:
$100$ machines take $1$ minute to make $frac {100}7$ toys.
And then multiply the toys by $7$ to get:
$100$ machines take $7$ minutes to make $100$ toys.
Of multiply the toys by $700$ to get either:
$700$ machines take $1$ minute to make $100$ toys.
Or one machine takes $700$ minuts to make $100$ toys.
If you did that we can divide the machines and multiply the time (or vice versa) to get:
$100$ machines take $7$ minutes to make $100$ toys.
====
Now, if we were clever or if we had experience we might have noticed:
$7$ machines in $7$ minutes make $7$ toys means
$1$ machine in $7$ minutes make $1$ toy means
$x$ machines in $7$ minutes make $x$ toys.
So $100$ machine will take $7$ minutes to make $100$ toys.
That would have been faster and easier.
But... maybe not as immediately clear.
$endgroup$
$7$ machines in $7$ minutes make $7$ toys.
If we divide either the number of machines or the number of toy by seven we will get two correct statements:
Either $1$ machine takes $7$ minutes to make $1$ toy. OR
$7$ machines take $1$ minute to make $1$ toy.
Both are true.
Notice that we CAN'T divide BOTH machines AND time by $7$ because that is dividing by a factor of $7$ two times, or is actually dividing the entire system by $49$.
We divide the other term (time of machine) by seven and we get:
$1$ machine takes $1$ minute to make $frac 17$ toy.
Either multiply the machines by $100$ to get:
$100$ machines take $1$ minute to make $frac {100}7$ toys.
And then multiply the toys by $7$ to get:
$100$ machines take $7$ minutes to make $100$ toys.
Of multiply the toys by $700$ to get either:
$700$ machines take $1$ minute to make $100$ toys.
Or one machine takes $700$ minuts to make $100$ toys.
If you did that we can divide the machines and multiply the time (or vice versa) to get:
$100$ machines take $7$ minutes to make $100$ toys.
====
Now, if we were clever or if we had experience we might have noticed:
$7$ machines in $7$ minutes make $7$ toys means
$1$ machine in $7$ minutes make $1$ toy means
$x$ machines in $7$ minutes make $x$ toys.
So $100$ machine will take $7$ minutes to make $100$ toys.
That would have been faster and easier.
But... maybe not as immediately clear.
answered Feb 8 '18 at 19:00
fleabloodfleablood
71.9k22686
71.9k22686
add a comment |
add a comment |
$begingroup$
In some fixed amount of time, $7$ machines make $7$ toys. If all machines are identical, $1$ machine must in that time have made $1$ toy. So $100$ machines make the $100$ toys in that same amount of time.
$endgroup$
add a comment |
$begingroup$
In some fixed amount of time, $7$ machines make $7$ toys. If all machines are identical, $1$ machine must in that time have made $1$ toy. So $100$ machines make the $100$ toys in that same amount of time.
$endgroup$
add a comment |
$begingroup$
In some fixed amount of time, $7$ machines make $7$ toys. If all machines are identical, $1$ machine must in that time have made $1$ toy. So $100$ machines make the $100$ toys in that same amount of time.
$endgroup$
In some fixed amount of time, $7$ machines make $7$ toys. If all machines are identical, $1$ machine must in that time have made $1$ toy. So $100$ machines make the $100$ toys in that same amount of time.
answered Feb 8 '18 at 21:43
M. VanM. Van
2,670311
2,670311
add a comment |
add a comment |
$begingroup$
This is a bogus "applied math" problem, whose "correct" solution is predicated on implicit, possibly erroneous, assumptions being made.
@Manish Kundu and others have arrived at the "correct" answer B (7 minutes) based on the implicit assumption that all machines operate independently to produce toys, and can be run in parallel at the same production rate. The folowing is not meant as a criticism to the answerers, but to the creator of the question.
Suppose that in fact 7 machines need to work together to produce ANY toys, and let's still suppose that all 7 or 100 machines are identical. Under this supposition, 14 sets of 7 machines can work in parallel to produce 98, not 100, toys in 7 minutes. Only at the completion of this initial 7 minutes can some set of 7 machines then be used to create another 7 (or maybe 2) toys, which as far as we know might take as long as another 7 minutes, even without any set up time being required after the first 7 minutes until the 2nd 7 minutes begins. So the answer could be 14 minutes, not 7 minutes. And given the assumptions still being made, even that may not be correct.
$endgroup$
add a comment |
$begingroup$
This is a bogus "applied math" problem, whose "correct" solution is predicated on implicit, possibly erroneous, assumptions being made.
@Manish Kundu and others have arrived at the "correct" answer B (7 minutes) based on the implicit assumption that all machines operate independently to produce toys, and can be run in parallel at the same production rate. The folowing is not meant as a criticism to the answerers, but to the creator of the question.
Suppose that in fact 7 machines need to work together to produce ANY toys, and let's still suppose that all 7 or 100 machines are identical. Under this supposition, 14 sets of 7 machines can work in parallel to produce 98, not 100, toys in 7 minutes. Only at the completion of this initial 7 minutes can some set of 7 machines then be used to create another 7 (or maybe 2) toys, which as far as we know might take as long as another 7 minutes, even without any set up time being required after the first 7 minutes until the 2nd 7 minutes begins. So the answer could be 14 minutes, not 7 minutes. And given the assumptions still being made, even that may not be correct.
$endgroup$
add a comment |
$begingroup$
This is a bogus "applied math" problem, whose "correct" solution is predicated on implicit, possibly erroneous, assumptions being made.
@Manish Kundu and others have arrived at the "correct" answer B (7 minutes) based on the implicit assumption that all machines operate independently to produce toys, and can be run in parallel at the same production rate. The folowing is not meant as a criticism to the answerers, but to the creator of the question.
Suppose that in fact 7 machines need to work together to produce ANY toys, and let's still suppose that all 7 or 100 machines are identical. Under this supposition, 14 sets of 7 machines can work in parallel to produce 98, not 100, toys in 7 minutes. Only at the completion of this initial 7 minutes can some set of 7 machines then be used to create another 7 (or maybe 2) toys, which as far as we know might take as long as another 7 minutes, even without any set up time being required after the first 7 minutes until the 2nd 7 minutes begins. So the answer could be 14 minutes, not 7 minutes. And given the assumptions still being made, even that may not be correct.
$endgroup$
This is a bogus "applied math" problem, whose "correct" solution is predicated on implicit, possibly erroneous, assumptions being made.
@Manish Kundu and others have arrived at the "correct" answer B (7 minutes) based on the implicit assumption that all machines operate independently to produce toys, and can be run in parallel at the same production rate. The folowing is not meant as a criticism to the answerers, but to the creator of the question.
Suppose that in fact 7 machines need to work together to produce ANY toys, and let's still suppose that all 7 or 100 machines are identical. Under this supposition, 14 sets of 7 machines can work in parallel to produce 98, not 100, toys in 7 minutes. Only at the completion of this initial 7 minutes can some set of 7 machines then be used to create another 7 (or maybe 2) toys, which as far as we know might take as long as another 7 minutes, even without any set up time being required after the first 7 minutes until the 2nd 7 minutes begins. So the answer could be 14 minutes, not 7 minutes. And given the assumptions still being made, even that may not be correct.
answered Jun 29 '18 at 15:57
Mark L. StoneMark L. Stone
1,93058
1,93058
add a comment |
add a comment |
$begingroup$
I had similar question on examination few days ago. The question is next:
7 robots will produce 7 toys in 7 minutes. How long it will take for
70 robots to produce 70 toys.
FIRST APPROACH:
It takes 7 minutes for 1 robot to produce 1 toy.
1 robot will produce 70 toys in = 70 toys * 7 minutes (number of minutes per toy) = 490 minutes
So, 1 robot will create 70 toys in 490 minutes.
If we divide number of minutes to produce 70 toys for one robot which is 490 minutes with 70 robots, which is question, it will take 7 minutes to produce 70 toys using 70 robots.
This also means that each robot will produce 1 toy.
In case that we have 100 robots to build 100 toys, and all parametars are same as above, 7 robots – 7 toys – 7 minutes.
You can easily apply this logic to solve this issue:
1 robot will produce 100 toys in 700 minutes
100 robots will produce 100 toys in 7 minutes
SECOND APPROACH:
Main question is, “How much toys each robot needs to produce?”
Answer on this question is next:
We have 7 robots to produce 7 toys – each robot will produce 1 toy. 1 toy * 7 minutes (number of minutes per toy) = 7 minutes for production.
In our case above we have 70 robots to produce 70 toys. 70 toys / 70 robots = 1 toy each robot. We will multiply number of minutes for 7 robots to build 7 toys which is 7 minutes with 1 and we will get 7 minutes.
Also if we have 35 robots to produce 70 toys. It means that 70 toys need to produce 35 robots, each robot will produce 1.5 toy.
If we multiply 1.5 * 7, we will get 10:30 for 35 robots to produce 70 toys.
And for 100 robots to produce 100 toys = 100 / 100 * 7 = 7 minutes.
Hope this is understandable and it will help someone to solve similar problems.
$endgroup$
add a comment |
$begingroup$
I had similar question on examination few days ago. The question is next:
7 robots will produce 7 toys in 7 minutes. How long it will take for
70 robots to produce 70 toys.
FIRST APPROACH:
It takes 7 minutes for 1 robot to produce 1 toy.
1 robot will produce 70 toys in = 70 toys * 7 minutes (number of minutes per toy) = 490 minutes
So, 1 robot will create 70 toys in 490 minutes.
If we divide number of minutes to produce 70 toys for one robot which is 490 minutes with 70 robots, which is question, it will take 7 minutes to produce 70 toys using 70 robots.
This also means that each robot will produce 1 toy.
In case that we have 100 robots to build 100 toys, and all parametars are same as above, 7 robots – 7 toys – 7 minutes.
You can easily apply this logic to solve this issue:
1 robot will produce 100 toys in 700 minutes
100 robots will produce 100 toys in 7 minutes
SECOND APPROACH:
Main question is, “How much toys each robot needs to produce?”
Answer on this question is next:
We have 7 robots to produce 7 toys – each robot will produce 1 toy. 1 toy * 7 minutes (number of minutes per toy) = 7 minutes for production.
In our case above we have 70 robots to produce 70 toys. 70 toys / 70 robots = 1 toy each robot. We will multiply number of minutes for 7 robots to build 7 toys which is 7 minutes with 1 and we will get 7 minutes.
Also if we have 35 robots to produce 70 toys. It means that 70 toys need to produce 35 robots, each robot will produce 1.5 toy.
If we multiply 1.5 * 7, we will get 10:30 for 35 robots to produce 70 toys.
And for 100 robots to produce 100 toys = 100 / 100 * 7 = 7 minutes.
Hope this is understandable and it will help someone to solve similar problems.
$endgroup$
add a comment |
$begingroup$
I had similar question on examination few days ago. The question is next:
7 robots will produce 7 toys in 7 minutes. How long it will take for
70 robots to produce 70 toys.
FIRST APPROACH:
It takes 7 minutes for 1 robot to produce 1 toy.
1 robot will produce 70 toys in = 70 toys * 7 minutes (number of minutes per toy) = 490 minutes
So, 1 robot will create 70 toys in 490 minutes.
If we divide number of minutes to produce 70 toys for one robot which is 490 minutes with 70 robots, which is question, it will take 7 minutes to produce 70 toys using 70 robots.
This also means that each robot will produce 1 toy.
In case that we have 100 robots to build 100 toys, and all parametars are same as above, 7 robots – 7 toys – 7 minutes.
You can easily apply this logic to solve this issue:
1 robot will produce 100 toys in 700 minutes
100 robots will produce 100 toys in 7 minutes
SECOND APPROACH:
Main question is, “How much toys each robot needs to produce?”
Answer on this question is next:
We have 7 robots to produce 7 toys – each robot will produce 1 toy. 1 toy * 7 minutes (number of minutes per toy) = 7 minutes for production.
In our case above we have 70 robots to produce 70 toys. 70 toys / 70 robots = 1 toy each robot. We will multiply number of minutes for 7 robots to build 7 toys which is 7 minutes with 1 and we will get 7 minutes.
Also if we have 35 robots to produce 70 toys. It means that 70 toys need to produce 35 robots, each robot will produce 1.5 toy.
If we multiply 1.5 * 7, we will get 10:30 for 35 robots to produce 70 toys.
And for 100 robots to produce 100 toys = 100 / 100 * 7 = 7 minutes.
Hope this is understandable and it will help someone to solve similar problems.
$endgroup$
I had similar question on examination few days ago. The question is next:
7 robots will produce 7 toys in 7 minutes. How long it will take for
70 robots to produce 70 toys.
FIRST APPROACH:
It takes 7 minutes for 1 robot to produce 1 toy.
1 robot will produce 70 toys in = 70 toys * 7 minutes (number of minutes per toy) = 490 minutes
So, 1 robot will create 70 toys in 490 minutes.
If we divide number of minutes to produce 70 toys for one robot which is 490 minutes with 70 robots, which is question, it will take 7 minutes to produce 70 toys using 70 robots.
This also means that each robot will produce 1 toy.
In case that we have 100 robots to build 100 toys, and all parametars are same as above, 7 robots – 7 toys – 7 minutes.
You can easily apply this logic to solve this issue:
1 robot will produce 100 toys in 700 minutes
100 robots will produce 100 toys in 7 minutes
SECOND APPROACH:
Main question is, “How much toys each robot needs to produce?”
Answer on this question is next:
We have 7 robots to produce 7 toys – each robot will produce 1 toy. 1 toy * 7 minutes (number of minutes per toy) = 7 minutes for production.
In our case above we have 70 robots to produce 70 toys. 70 toys / 70 robots = 1 toy each robot. We will multiply number of minutes for 7 robots to build 7 toys which is 7 minutes with 1 and we will get 7 minutes.
Also if we have 35 robots to produce 70 toys. It means that 70 toys need to produce 35 robots, each robot will produce 1.5 toy.
If we multiply 1.5 * 7, we will get 10:30 for 35 robots to produce 70 toys.
And for 100 robots to produce 100 toys = 100 / 100 * 7 = 7 minutes.
Hope this is understandable and it will help someone to solve similar problems.
answered Oct 14 '18 at 7:07
Aleksandar RakićAleksandar Rakić
1011
1011
add a comment |
add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes
Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes
Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes
Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!
$endgroup$
7 machines make 7 toys in - 7 minutes.
So, 7 machines make 100 toys in - 77.100 minutes = 100 minutes
Or, 100 machines make 100 toys in - 100.7100=7 minutes
Simple!!!!!!
answered Jun 29 '18 at 14:48
Ritik ChoudharyRitik Choudhary
1
1
add a comment |
add a comment |
$begingroup$
7 machines make 7 toys in 7 mins
so 1 st machine takes 7 mins to produce a toy
2 nd machine takes 7 mins to produce a toy
soooo on
But all machines run at simultaneously so total time is 7 mins only
if you run 100 machines for 7 mins we can produce 100 toys
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in 7 mins
so 1 st machine takes 7 mins to produce a toy
2 nd machine takes 7 mins to produce a toy
soooo on
But all machines run at simultaneously so total time is 7 mins only
if you run 100 machines for 7 mins we can produce 100 toys
$endgroup$
add a comment |
$begingroup$
7 machines make 7 toys in 7 mins
so 1 st machine takes 7 mins to produce a toy
2 nd machine takes 7 mins to produce a toy
soooo on
But all machines run at simultaneously so total time is 7 mins only
if you run 100 machines for 7 mins we can produce 100 toys
$endgroup$
7 machines make 7 toys in 7 mins
so 1 st machine takes 7 mins to produce a toy
2 nd machine takes 7 mins to produce a toy
soooo on
But all machines run at simultaneously so total time is 7 mins only
if you run 100 machines for 7 mins we can produce 100 toys
answered Jan 5 at 8:23
jasperjasper
1
1
add a comment |
add a comment |
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$begingroup$
Because the time is "shared", every machine take 7 minute to make a toy
$endgroup$
– james watt
Feb 8 '18 at 18:32
$begingroup$
"1 machine:1minute:1 toy" WRONG! That would mean if you ran that machine for 7 minutes it would make 7 toys. And if you ran 7 machines for 7 minutes you'd have 49 toys. Just because words are pretty and make patterns doesn't mean the math will work. After all if I gave you 200 million dollars total, you'd be rich. But if a give everyone in the US 200 million dollars total, everyone would have 79 cents. Not the same thing at all.
$endgroup$
– fleablood
Feb 8 '18 at 18:36
$begingroup$
The thing is by dividing both machines and time both by $7$ you are actually dividing the "entire system" by $49$. You may divide one determining factor by a value to get a one-dimensional linearity, but if you divide two determining factors you divide by a geometric or "square" value.
$endgroup$
– fleablood
Feb 8 '18 at 19:04