# How much had really been ordered? [closed]

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?

• This would never happen in a metric rope shop! – Nuclear Wang Jan 30 '19 at 18:09

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 $$x=9, y=2$$.

• 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. – Jaap Scherphuis Jan 30 '19 at 13:02
• Proof is more useful than brute force answers! +1 – Krad Cigol Jan 30 '19 at 13:02
• @JaapScherphuis; good point - a semi-stupid tailor! – JMP Jan 30 '19 at 13:04
• Why x ( measurement of feet) should be less than 12? – Mea Culpa Nay Jan 30 '19 at 16:46
• It's pretty obvious what it is from the working out, but should the answer include the actual answer? – ZanyG Jan 31 '19 at 9:50

With some trial and error:

Seems he ordered:

9 feet and 2 inches and received only 2 feet and 9 inches

• You should have listed out the trial and error process. – H_D Feb 4 '19 at 2:54

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.

$$0.3 \times (12x + y)$$ 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 $$y = 2 \Rightarrow 2.6x = 23.4 \Rightarrow x = 9$$

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

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.

• 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. – Timbo Jan 30 '19 at 15:17
• @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. – blakeoft Jan 30 '19 at 15:55
• @blakeoft wouldn't you just specify 11" in that case? – Baldrickk Jan 30 '19 at 16:18
• @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. – blakeoft Jan 30 '19 at 20:03
• @Timbo, sure, but the question did not place such a restriction. – Brandon Dixon Jan 31 '19 at 16:33