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How much had really been ordered? [closed]
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The clerk misunderstood the order for rope. He reversed feet and inches, and the customer got only 30% of what he ordered.
How much rope had really been ordered?
mathematics
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
$endgroup$
closed as off-topic by TwoBitOperation, deep thought, A. P., athin, Omega Krypton Jan 31 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – TwoBitOperation, deep thought, A. P., athin, Omega Krypton
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
The clerk misunderstood the order for rope. He reversed feet and inches, and the customer got only 30% of what he ordered.
How much rope had really been ordered?
mathematics
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
$endgroup$
closed as off-topic by TwoBitOperation, deep thought, A. P., athin, Omega Krypton Jan 31 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – TwoBitOperation, deep thought, A. P., athin, Omega Krypton
If this question can be reworded to fit the rules in the help center, please edit the question.
8
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This would never happen in a metric rope shop!
$endgroup$
– Nuclear Wang
Jan 30 at 18:09
add a comment |
$begingroup$
The clerk misunderstood the order for rope. He reversed feet and inches, and the customer got only 30% of what he ordered.
How much rope had really been ordered?
mathematics
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
$endgroup$
The clerk misunderstood the order for rope. He reversed feet and inches, and the customer got only 30% of what he ordered.
How much rope had really been ordered?
mathematics
mathematics
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
edited Jan 30 at 13:09
JonMark Perry
20.6k64099
20.6k64099
asked Jan 30 at 12:42
JoanneJoanne
843
asked Jan 30 at 12:42
JoanneJoanne
843
asked Jan 30 at 12:42
JoanneJoanne
843
843
closed as off-topic by TwoBitOperation, deep thought, A. P., athin, Omega Krypton Jan 31 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – TwoBitOperation, deep thought, A. P., athin, Omega Krypton
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TwoBitOperation, deep thought, A. P., athin, Omega Krypton Jan 31 at 9:58
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – TwoBitOperation, deep thought, A. P., athin, Omega Krypton
If this question can be reworded to fit the rules in the help center, please edit the question.
8
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This would never happen in a metric rope shop!
$endgroup$
– Nuclear Wang
Jan 30 at 18:09
add a comment |
8
$begingroup$
This would never happen in a metric rope shop!
$endgroup$
– Nuclear Wang
Jan 30 at 18:09
8
8
$begingroup$
This would never happen in a metric rope shop!
$endgroup$
– Nuclear Wang
Jan 30 at 18:09
$begingroup$
This would never happen in a metric rope shop!
$endgroup$
– Nuclear Wang
Jan 30 at 18:09
add a comment |
5 Answers
5
active
oldest
votes
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Assume the order was for $x$ feet and $y$ inches, a total of $12x+y$ inches, and the customer got $y$ feet and $x$ inches, a total of $12y+x$ inches. We know that $12x+y=frac{10}{3}(12y+x)$, so that $36x+3y=120y+10x$, or $26x=117y$. As $gcd(26,117)=13$, we have $2x=9y$. So for any positive integer $k$, there is a solution $x=9k,y=2k$. As @Jaap points out, $x,ylt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
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1
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I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
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– Jaap Scherphuis
Jan 30 at 13:02
5
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Proof is more useful than brute force answers! +1
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– Krad Cigol
Jan 30 at 13:02
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@JaapScherphuis; good point - a semi-stupid tailor!
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– JonMark Perry
Jan 30 at 13:04
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Why x ( measurement of feet) should be less than 12?
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– Mea Culpa Nay
Jan 30 at 16:46
2
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It's pretty obvious what it is from the working out, but should the answer include the actual answer?
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– ZanyG
Jan 31 at 9:50
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show 5 more comments
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With some trial and error:
Seems he ordered:
9 feet and 2 inches and received only 2 feet and 9 inches
answered Jan 30 at 12:48
piratepirate
2,298733
$endgroup$
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You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
add a comment |
$begingroup$
First, we know that a foot is 12 inches.
Let the length of the rope be $x$ feet and $y$ inches.
In other words, the rope is $12x+y$ inches long.
Since the clerk misunderstood, the rope he got was $y$ feet and $x$ inches long, which can also be expressed as $12y+x$ inches.
With some trial and error, it is not hard to find that he ordered 9 feet and 2 inches of rope at first but received 2 feet and 9 inches of rope.
answered Jan 30 at 12:52
H_DH_D
15010
$endgroup$
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
add a comment |
$begingroup$
$0.3 times (12x + y)$ is what he received, with the ordered length being $xtext{'}ytext{"}$, $12y + x$ is what the clerk understood. $0 leq x < 12$ and $0 leq y < 12$
$0.3 times (12x + y) = 12y + x Rightarrow 3.6x + 0.3y = 12y + x Rightarrow 2.6x = 11.7y$
$y$ can further be limited because $frac{11.7}{2.6}y = 4.5y < 12$, leaving us with $0 leq y < 3$. $0$ and $1$ clearly don't work, which leaves us with $2$.
Set $y = 2 Rightarrow 2.6x = 23.4 Rightarrow x = 9$
$0.3 times (9text{'}2text{"}) = 33text{"} = 2text{'}9text{"}$
edited Jan 30 at 17:01
w l
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
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add a comment |
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If you do not place a restriction on inches, there are an infinite number of answers. If you place the restriction that it must be less than 12, but can be any real number > 0, I believe there are still an infinite number of solutions. If it must be an integer, then I would think there is only a finite number of solution in this case.
Either way, below outlines my process for deriving this solution:
12i+f = 0.3(12f+i) -> Left side is swapped inches and feet, right side is 30% of expected amount of rope12i+f = 0.3i+3.6f -> Reduction11.7i = 2.6f -> Reductionf=4.5i => i = 2/9f -> This is a relationship between the value for inches and feet that will satisfy the requirements of this problem.Arbitrarily pick a value for f => 27i = 2/9*27 = 6Verify solution: 12*6+27 = 0.3(12*27+6) -> True
This will be valid for any pick for f or i.
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
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2
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I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
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– Timbo
Jan 30 at 15:17
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@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
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– blakeoft
Jan 30 at 15:55
1
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@blakeoft wouldn't you just specify 11" in that case?
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– Baldrickk
Jan 30 at 16:18
1
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@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
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– blakeoft
Jan 30 at 20:03
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@Timbo, sure, but the question did not place such a restriction.
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– Brandon Dixon
Jan 31 at 16:33
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show 2 more comments
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assume the order was for $x$ feet and $y$ inches, a total of $12x+y$ inches, and the customer got $y$ feet and $x$ inches, a total of $12y+x$ inches. We know that $12x+y=frac{10}{3}(12y+x)$, so that $36x+3y=120y+10x$, or $26x=117y$. As $gcd(26,117)=13$, we have $2x=9y$. So for any positive integer $k$, there is a solution $x=9k,y=2k$. As @Jaap points out, $x,ylt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
1
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
5
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
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@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
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Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
2
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
|
show 5 more comments
$begingroup$
Assume the order was for $x$ feet and $y$ inches, a total of $12x+y$ inches, and the customer got $y$ feet and $x$ inches, a total of $12y+x$ inches. We know that $12x+y=frac{10}{3}(12y+x)$, so that $36x+3y=120y+10x$, or $26x=117y$. As $gcd(26,117)=13$, we have $2x=9y$. So for any positive integer $k$, there is a solution $x=9k,y=2k$. As @Jaap points out, $x,ylt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
1
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
5
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
$begingroup$
@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
$begingroup$
Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
2
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
|
show 5 more comments
$begingroup$
Assume the order was for $x$ feet and $y$ inches, a total of $12x+y$ inches, and the customer got $y$ feet and $x$ inches, a total of $12y+x$ inches. We know that $12x+y=frac{10}{3}(12y+x)$, so that $36x+3y=120y+10x$, or $26x=117y$. As $gcd(26,117)=13$, we have $2x=9y$. So for any positive integer $k$, there is a solution $x=9k,y=2k$. As @Jaap points out, $x,ylt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
$endgroup$
Assume the order was for $x$ feet and $y$ inches, a total of $12x+y$ inches, and the customer got $y$ feet and $x$ inches, a total of $12y+x$ inches. We know that $12x+y=frac{10}{3}(12y+x)$, so that $36x+3y=120y+10x$, or $26x=117y$. As $gcd(26,117)=13$, we have $2x=9y$. So for any positive integer $k$, there is a solution $x=9k,y=2k$. As @Jaap points out, $x,ylt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
edited Jan 31 at 9:54
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
answered Jan 30 at 12:59
JonMark PerryJonMark Perry
20.6k64099
20.6k64099
1
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
5
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
$begingroup$
@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
$begingroup$
Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
2
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
|
show 5 more comments
1
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
5
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
$begingroup$
@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
$begingroup$
Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
2
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
1
1
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
$begingroup$
I think it is safe to assume x<12 and y<12, because any more than 11 inches would be converted to feet. So there really is only one solution, with k=1.
$endgroup$
– Jaap Scherphuis
Jan 30 at 13:02
5
5
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
$begingroup$
Proof is more useful than brute force answers! +1
$endgroup$
– Krad Cigol
Jan 30 at 13:02
$begingroup$
@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
$begingroup$
@JaapScherphuis; good point - a semi-stupid tailor!
$endgroup$
– JonMark Perry
Jan 30 at 13:04
$begingroup$
Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
$begingroup$
Why x ( measurement of feet) should be less than 12?
$endgroup$
– Mea Culpa Nay
Jan 30 at 16:46
2
2
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
$begingroup$
It's pretty obvious what it is from the working out, but should the answer include the actual answer?
$endgroup$
– ZanyG
Jan 31 at 9:50
|
show 5 more comments
$begingroup$
With some trial and error:
Seems he ordered:
9 feet and 2 inches and received only 2 feet and 9 inches
answered Jan 30 at 12:48
piratepirate
2,298733
$endgroup$
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
add a comment |
$begingroup$
With some trial and error:
Seems he ordered:
9 feet and 2 inches and received only 2 feet and 9 inches
answered Jan 30 at 12:48
piratepirate
2,298733
$endgroup$
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
add a comment |
$begingroup$
With some trial and error:
Seems he ordered:
9 feet and 2 inches and received only 2 feet and 9 inches
answered Jan 30 at 12:48
piratepirate
2,298733
$endgroup$
With some trial and error:
Seems he ordered:
9 feet and 2 inches and received only 2 feet and 9 inches
answered Jan 30 at 12:48
piratepirate
2,298733
answered Jan 30 at 12:48
piratepirate
2,298733
answered Jan 30 at 12:48
piratepirate
2,298733
answered Jan 30 at 12:48
piratepirate
2,298733
2,298733
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
add a comment |
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
$begingroup$
You should have listed out the trial and error process.
$endgroup$
– H_D
Feb 4 at 2:54
add a comment |
$begingroup$
First, we know that a foot is 12 inches.
Let the length of the rope be $x$ feet and $y$ inches.
In other words, the rope is $12x+y$ inches long.
Since the clerk misunderstood, the rope he got was $y$ feet and $x$ inches long, which can also be expressed as $12y+x$ inches.
With some trial and error, it is not hard to find that he ordered 9 feet and 2 inches of rope at first but received 2 feet and 9 inches of rope.
answered Jan 30 at 12:52
H_DH_D
15010
$endgroup$
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
add a comment |
$begingroup$
First, we know that a foot is 12 inches.
Let the length of the rope be $x$ feet and $y$ inches.
In other words, the rope is $12x+y$ inches long.
Since the clerk misunderstood, the rope he got was $y$ feet and $x$ inches long, which can also be expressed as $12y+x$ inches.
With some trial and error, it is not hard to find that he ordered 9 feet and 2 inches of rope at first but received 2 feet and 9 inches of rope.
answered Jan 30 at 12:52
H_DH_D
15010
$endgroup$
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
add a comment |
$begingroup$
First, we know that a foot is 12 inches.
Let the length of the rope be $x$ feet and $y$ inches.
In other words, the rope is $12x+y$ inches long.
Since the clerk misunderstood, the rope he got was $y$ feet and $x$ inches long, which can also be expressed as $12y+x$ inches.
With some trial and error, it is not hard to find that he ordered 9 feet and 2 inches of rope at first but received 2 feet and 9 inches of rope.
answered Jan 30 at 12:52
H_DH_D
15010
$endgroup$
First, we know that a foot is 12 inches.
Let the length of the rope be $x$ feet and $y$ inches.
In other words, the rope is $12x+y$ inches long.
Since the clerk misunderstood, the rope he got was $y$ feet and $x$ inches long, which can also be expressed as $12y+x$ inches.
With some trial and error, it is not hard to find that he ordered 9 feet and 2 inches of rope at first but received 2 feet and 9 inches of rope.
answered Jan 30 at 12:52
H_DH_D
15010
answered Jan 30 at 12:52
H_DH_D
15010
answered Jan 30 at 12:52
H_DH_D
15010
answered Jan 30 at 12:52
H_DH_D
15010
15010
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
add a comment |
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
$begingroup$
Thanks for upvoting!
$endgroup$
– H_D
Jan 30 at 12:54
add a comment |
$begingroup$
$0.3 times (12x + y)$ is what he received, with the ordered length being $xtext{'}ytext{"}$, $12y + x$ is what the clerk understood. $0 leq x < 12$ and $0 leq y < 12$
$0.3 times (12x + y) = 12y + x Rightarrow 3.6x + 0.3y = 12y + x Rightarrow 2.6x = 11.7y$
$y$ can further be limited because $frac{11.7}{2.6}y = 4.5y < 12$, leaving us with $0 leq y < 3$. $0$ and $1$ clearly don't work, which leaves us with $2$.
Set $y = 2 Rightarrow 2.6x = 23.4 Rightarrow x = 9$
$0.3 times (9text{'}2text{"}) = 33text{"} = 2text{'}9text{"}$
edited Jan 30 at 17:01
w l
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
$endgroup$
add a comment |
$begingroup$
$0.3 times (12x + y)$ is what he received, with the ordered length being $xtext{'}ytext{"}$, $12y + x$ is what the clerk understood. $0 leq x < 12$ and $0 leq y < 12$
$0.3 times (12x + y) = 12y + x Rightarrow 3.6x + 0.3y = 12y + x Rightarrow 2.6x = 11.7y$
$y$ can further be limited because $frac{11.7}{2.6}y = 4.5y < 12$, leaving us with $0 leq y < 3$. $0$ and $1$ clearly don't work, which leaves us with $2$.
Set $y = 2 Rightarrow 2.6x = 23.4 Rightarrow x = 9$
$0.3 times (9text{'}2text{"}) = 33text{"} = 2text{'}9text{"}$
edited Jan 30 at 17:01
w l
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
$endgroup$
add a comment |
$begingroup$
$0.3 times (12x + y)$ is what he received, with the ordered length being $xtext{'}ytext{"}$, $12y + x$ is what the clerk understood. $0 leq x < 12$ and $0 leq y < 12$
$0.3 times (12x + y) = 12y + x Rightarrow 3.6x + 0.3y = 12y + x Rightarrow 2.6x = 11.7y$
$y$ can further be limited because $frac{11.7}{2.6}y = 4.5y < 12$, leaving us with $0 leq y < 3$. $0$ and $1$ clearly don't work, which leaves us with $2$.
Set $y = 2 Rightarrow 2.6x = 23.4 Rightarrow x = 9$
$0.3 times (9text{'}2text{"}) = 33text{"} = 2text{'}9text{"}$
edited Jan 30 at 17:01
w l
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
$endgroup$
$0.3 times (12x + y)$ is what he received, with the ordered length being $xtext{'}ytext{"}$, $12y + x$ is what the clerk understood. $0 leq x < 12$ and $0 leq y < 12$
$0.3 times (12x + y) = 12y + x Rightarrow 3.6x + 0.3y = 12y + x Rightarrow 2.6x = 11.7y$
$y$ can further be limited because $frac{11.7}{2.6}y = 4.5y < 12$, leaving us with $0 leq y < 3$. $0$ and $1$ clearly don't work, which leaves us with $2$.
Set $y = 2 Rightarrow 2.6x = 23.4 Rightarrow x = 9$
$0.3 times (9text{'}2text{"}) = 33text{"} = 2text{'}9text{"}$
edited Jan 30 at 17:01
w l
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
edited Jan 30 at 17:01
w l
3,8861229
edited Jan 30 at 17:01
w l
3,8861229
edited Jan 30 at 17:01
w l
3,8861229
3,8861229
answered Jan 30 at 16:04
maltegmalteg
211
answered Jan 30 at 16:04
maltegmalteg
211
answered Jan 30 at 16:04
maltegmalteg
211
211
add a comment |
add a comment |
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If you do not place a restriction on inches, there are an infinite number of answers. If you place the restriction that it must be less than 12, but can be any real number > 0, I believe there are still an infinite number of solutions. If it must be an integer, then I would think there is only a finite number of solution in this case.
Either way, below outlines my process for deriving this solution:
12i+f = 0.3(12f+i) -> Left side is swapped inches and feet, right side is 30% of expected amount of rope12i+f = 0.3i+3.6f -> Reduction11.7i = 2.6f -> Reductionf=4.5i => i = 2/9f -> This is a relationship between the value for inches and feet that will satisfy the requirements of this problem.Arbitrarily pick a value for f => 27i = 2/9*27 = 6Verify solution: 12*6+27 = 0.3(12*27+6) -> True
This will be valid for any pick for f or i.
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
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2
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I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
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– Timbo
Jan 30 at 15:17
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@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
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– blakeoft
Jan 30 at 15:55
1
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@blakeoft wouldn't you just specify 11" in that case?
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– Baldrickk
Jan 30 at 16:18
1
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@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
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– blakeoft
Jan 30 at 20:03
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@Timbo, sure, but the question did not place such a restriction.
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– Brandon Dixon
Jan 31 at 16:33
|
show 2 more comments
$begingroup$
If you do not place a restriction on inches, there are an infinite number of answers. If you place the restriction that it must be less than 12, but can be any real number > 0, I believe there are still an infinite number of solutions. If it must be an integer, then I would think there is only a finite number of solution in this case.
Either way, below outlines my process for deriving this solution:
12i+f = 0.3(12f+i) -> Left side is swapped inches and feet, right side is 30% of expected amount of rope12i+f = 0.3i+3.6f -> Reduction11.7i = 2.6f -> Reductionf=4.5i => i = 2/9f -> This is a relationship between the value for inches and feet that will satisfy the requirements of this problem.Arbitrarily pick a value for f => 27i = 2/9*27 = 6Verify solution: 12*6+27 = 0.3(12*27+6) -> True
This will be valid for any pick for f or i.
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
$endgroup$
2
$begingroup$
I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
$endgroup$
– Timbo
Jan 30 at 15:17
$begingroup$
@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
$endgroup$
– blakeoft
Jan 30 at 15:55
1
$begingroup$
@blakeoft wouldn't you just specify 11" in that case?
$endgroup$
– Baldrickk
Jan 30 at 16:18
1
$begingroup$
@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
$endgroup$
– blakeoft
Jan 30 at 20:03
$begingroup$
@Timbo, sure, but the question did not place such a restriction.
$endgroup$
– Brandon Dixon
Jan 31 at 16:33
|
show 2 more comments
$begingroup$
If you do not place a restriction on inches, there are an infinite number of answers. If you place the restriction that it must be less than 12, but can be any real number > 0, I believe there are still an infinite number of solutions. If it must be an integer, then I would think there is only a finite number of solution in this case.
Either way, below outlines my process for deriving this solution:
12i+f = 0.3(12f+i) -> Left side is swapped inches and feet, right side is 30% of expected amount of rope12i+f = 0.3i+3.6f -> Reduction11.7i = 2.6f -> Reductionf=4.5i => i = 2/9f -> This is a relationship between the value for inches and feet that will satisfy the requirements of this problem.Arbitrarily pick a value for f => 27i = 2/9*27 = 6Verify solution: 12*6+27 = 0.3(12*27+6) -> True
This will be valid for any pick for f or i.
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
$endgroup$
If you do not place a restriction on inches, there are an infinite number of answers. If you place the restriction that it must be less than 12, but can be any real number > 0, I believe there are still an infinite number of solutions. If it must be an integer, then I would think there is only a finite number of solution in this case.
Either way, below outlines my process for deriving this solution:
12i+f = 0.3(12f+i) -> Left side is swapped inches and feet, right side is 30% of expected amount of rope12i+f = 0.3i+3.6f -> Reduction11.7i = 2.6f -> Reductionf=4.5i => i = 2/9f -> This is a relationship between the value for inches and feet that will satisfy the requirements of this problem.Arbitrarily pick a value for f => 27i = 2/9*27 = 6Verify solution: 12*6+27 = 0.3(12*27+6) -> True
This will be valid for any pick for f or i.
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
answered Jan 30 at 13:54
Brandon DixonBrandon Dixon
1212
1212
2
$begingroup$
I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
$endgroup$
– Timbo
Jan 30 at 15:17
$begingroup$
@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
$endgroup$
– blakeoft
Jan 30 at 15:55
1
$begingroup$
@blakeoft wouldn't you just specify 11" in that case?
$endgroup$
– Baldrickk
Jan 30 at 16:18
1
$begingroup$
@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
$endgroup$
– blakeoft
Jan 30 at 20:03
$begingroup$
@Timbo, sure, but the question did not place such a restriction.
$endgroup$
– Brandon Dixon
Jan 31 at 16:33
|
show 2 more comments
2
$begingroup$
I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
$endgroup$
– Timbo
Jan 30 at 15:17
$begingroup$
@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
$endgroup$
– blakeoft
Jan 30 at 15:55
1
$begingroup$
@blakeoft wouldn't you just specify 11" in that case?
$endgroup$
– Baldrickk
Jan 30 at 16:18
1
$begingroup$
@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
$endgroup$
– blakeoft
Jan 30 at 20:03
$begingroup$
@Timbo, sure, but the question did not place such a restriction.
$endgroup$
– Brandon Dixon
Jan 31 at 16:33
2
2
$begingroup$
I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
$endgroup$
– Timbo
Jan 30 at 15:17
$begingroup$
I don't think 27 feet and 6 inches can be confused with 6 feet and 27 inches. Nobody would use 6 feet and 27 inches. Same with 3 feet and 0.66 inches: Nobody would use 0.66 feet and 3 inches.
$endgroup$
– Timbo
Jan 30 at 15:17
$begingroup$
@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
$endgroup$
– blakeoft
Jan 30 at 15:55
$begingroup$
@Timbo 0.66 feet + 3 inches isn't that strange. Suppose you know that you have a fixed length of 0.66 feet, and for this particular project, you need to go 3 inches farther.
$endgroup$
– blakeoft
Jan 30 at 15:55
1
1
$begingroup$
@blakeoft wouldn't you just specify 11" in that case?
$endgroup$
– Baldrickk
Jan 30 at 16:18
$begingroup$
@blakeoft wouldn't you just specify 11" in that case?
$endgroup$
– Baldrickk
Jan 30 at 16:18
1
1
$begingroup$
@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
$endgroup$
– blakeoft
Jan 30 at 20:03
$begingroup$
@Baldrickk You certainly could, and most probably would. Still, they are two perfectly fine ways to represent the same measurement. Although, I'd almost definitely say 11 inches if I were asking someone to make a cut for me rather than doing it myself.
$endgroup$
– blakeoft
Jan 30 at 20:03
$begingroup$
@Timbo, sure, but the question did not place such a restriction.
$endgroup$
– Brandon Dixon
Jan 31 at 16:33
$begingroup$
@Timbo, sure, but the question did not place such a restriction.
$endgroup$
– Brandon Dixon
Jan 31 at 16:33
|
show 2 more comments
8
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This would never happen in a metric rope shop!
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– Nuclear Wang
Jan 30 at 18:09