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Question inspired by :- A dishonest shopkeeper claims he sells at cost but his 1 kg weight actually measures 800 grams when selling. Find his profit/loss % , posted as a new question as the the conditions here are getting varied and I am not able to apply that same learning from the prior question over here

As a sales gimmick, a cloth seller began measuring the cotton cloth with a yard stick that was x inches longer, because of which his profit reduced from 50% of his cost to 20% of his cost, find x. (Given: 1 Yard = 36 inches)

Need help in understanding the below equation:-

Solution given by the author :- $CP*(6/5)*(36+x)$ = $CP*(3/2)*36$ , where CP is the cost price of the cloth per inch

My attempt at understanding this is :-

When a customer comes to buy 36 inches of cloth, the shop keeper mistakenly gives him $(36+x)$ inches. On the price tag of the 36 inch cloth the price is written to be $CP*(3/2)*36$ and this is the actual amount that the shopkeeper collects from the customer.

Now what does $CP*(6/5)*(36+x)$ represent and why are these 2 equations being equated , how do they value the same ?

Fin27
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  • $6/5 = 1.2$ which represents $100$% + $20$% = $120%$ of $CP$. $CP$ is not the price/yard the customer pays, but the price/yard the cloth seller pays to acquire said cloth which they resell. – joseville Mar 19 '23 at 22:14

2 Answers2

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The most important similarity between the first question and this question is that the total Selling Price can be equated. There, we equated what you gave to the shopkeeper and what he received from you. Here, we will equate the seller's $SP$ when he used a normal yard stick, and the $SP$ when he used the modified yard stick, since he never changed his selling price. You have understood correctly that the amount is written on the pricetag. I will use the same terminology. Here, $CP$ is the cost price of one inch for the seller.

When he uses a normal yard stick:
Total Cost of 36 inches (1 yard)? Well, that's just $36CP$.
He gets $50\%$ profit, so profit = $\frac 12 \times 36CP$.
Add this to the total cost price to get the total selling price of $36$ inches. $$36CP + \frac 12 36CP = 36CP\left(1+\frac 12\right) = \color{blue}{\left(\frac 32\right)36CP}$$


Now, the shopkeeper changes the yard stick, but keeps the pricetag same. So, he sells $(36+x)$ inches for the same price obtained above. Now, his profit is $20\%$ of $(36+x)CP$ (see the last part of this answer if you're confused why we took $(36+x$). That is $(20/100) \times (36+x)CP$. Add this to the total cost price to get the total selling price:
$$(36+x)CP+(20/100) \times (36+x)CP = (36+x)CP\left(1+\frac 15\right) = \color{blue}{(6/5) \times (36+x)CP}$$

The two things (in blue) are the amounts written on the pricetag. So they must be equal. Notice that when you equate them, the $CP$ just cancels out, leaving you with a simple linear equation in $x$.

Just in case you want to know why we took $36+x$ in the second part:
He measures cloth a $36+x$ inches long stick and calls it $36$ inches. However, he has brought it from a dealer at $(36+x)CP$ (because the dealer is using a proper yardstick). But now he tells you it's $36$ inches, and sells at the same amount mentioned on the price tag for $36$ inches. Due to this stupidity, his profits reduce.

D S
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  • as a side note, I'm really curious as to what economical advantage the seller was aiming to achieve by increasing the length of his yardstick. – D S Mar 20 '23 at 11:13
  • Thanks a ton @D S :) yeah I too agree this is a strange problem considering in real world scenario xD ; I had a question on how could we assume that the selling price remains constant ? Unlike in the last question I had posted it was specifically mentioned that the shopkeeper sells at cost, but here nothing of that sort is mentioned that the "total" selling price remains constant, can we assume it if its not mentioned ? – Fin27 Mar 20 '23 at 12:24
  • "yardstick that was x inches longer, because of which his profit reduced from 50% of his cost to 20% of his cost". There is no mention of the selling price. If the selling price would've changed, the question would've mentioned that. – D S Mar 20 '23 at 12:28
  • Thanks @D S , your explanation worked like wonders for me , I was struggling a lot in understanding these kind of questions, I have one more question but with new variation which I'll post in some time along with my understanding and learnings that I acquired from this question and the previous one, could you please review it whenever you get time, I'll share the link to the problem in comment to this question. Thanks again :) – Fin27 Mar 20 '23 at 12:36
  • @D S , please check out the links for similar kind of questions :- one of them was asked about a years back , I believe it finally get resolved with your help, Question 1:- https://math.stackexchange.com/questions/4662754/profit-loss-faulty-weights-even-at-the-time-of-buying, Question 2:- https://math.stackexchange.com/questions/4312650/profit-loss-faulty-measurements$ – Fin27 Mar 20 '23 at 13:08
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Suppose one yard of material normally costs $P$ dollarydoos. One yard is $36$ inches.

The shopkeeper uses a "yord stick" which is one yard plus $x$ inches. The stick is $36 + x$ inches.

One yard is exactly $\frac{36}{36+x}$ yords. Note this is less than $1$ since a yord is longer than a yard.

So using the yord stick the new shopkeeper sells fabric at $P' = \frac{36}{36+x} P$ dollarydoos per yard.

The old shopkeeper has a $50\%$ profit rate. Meaning he sells things for $1.5 = 3/2$ times what he pays. Let $Q$ be the price the shopkeeper pays for a yard of fabric. We have $P = (3/2) Q$.

The new shopkeeper has a $20\%$ profit rate. Meaning he sells things for $1.2 = 6/5$ times what he pays. We have $P' = (6/5) Q$. So we have the simultaneous equations

$$P = \frac{3}{2} Q $$

$$\frac{36}{36+x} P = \frac{6}{5} Q $$

You should be able to recover the given solution by eliminating $Q$ from the simultaneous equations.

Daron
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