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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?

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    $\begingroup$ This would never happen in a metric rope shop! $\endgroup$ – Nuclear Wang Jan 30 at 18:09
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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, $x,y\lt12$ can be assumed, therefore $k=1$, and so $x=9, y=2$.

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    $\begingroup$ 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
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    $\begingroup$ 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! $\endgroup$ – JMP 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
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    $\begingroup$ 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
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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

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  • $\begingroup$ You should have listed out the trial and error process. $\endgroup$ – H_D Feb 4 at 2:54
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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.

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$0.3 \times (12x + y)$ is what he received, with the ordered length being $x\text{'}y\text{"}$, $12y + x$ 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 $y = 2 \Rightarrow 2.6x = 23.4 \Rightarrow x = 9$

$0.3 \times (9\text{'}2\text{"}) = 33\text{"} = 2\text{'}9\text{"}$

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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 rope

12i+f = 0.3i+3.6f -> Reduction

11.7i = 2.6f -> Reduction

f=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 => 27

i = 2/9*27 = 6

Verify solution: 12*6+27 = 0.3(12*27+6) -> True

This will be valid for any pick for f or i.

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    $\begingroup$ 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
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    $\begingroup$ @blakeoft wouldn't you just specify 11" in that case? $\endgroup$ – Baldrickk Jan 30 at 16:18
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    $\begingroup$ @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

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