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A dishonest dealer sells his goods at $10\%$ loss on cost price but uses $20\%$ less weight. What is his profit or loss percentage?
(a) $12\%$
(b) $22.5\%$
(c) $13.9\%$
(d) $12.5\%$
(e) None

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: Let us assume that the cost price of the goods are “x” and the weight corresponding to that cost price is “w”. Now, the dealer sells the goods at a loss of $10\%$ so find the selling price of the goods by incorporating the loss of $10\%$. It is also given that he uses $20\%$less weight so the weight he uses is $w-\dfrac{20}{100}w$. Now, solve this expression in “w” then we get the weight of goods in “w” that the dealer sells. Then the selling price that we have calculated above after incurring loss is for this new weight of goods. Now, by unitary method find the actual price of “w” goods. The price corresponding to the “w” goods that we have just calculated is the selling price and cost price we have assumed as x then see if selling price is greater than cost price then gain has occurred and in the other way round situation loss has occurred.

Complete step-by-step answer:
Let us assume that the cost price of the goods is “x”. This cost price is corresponding to the weight of woods “w”.
Now, the dealer sells his goods at $10\%$ loss and the weight of the goods that he sold on this loss is $20\%$ less than the actual weight.
We are going to find the reduced weight which is the subtraction of $20\%$ of “w” from “w”.
$w-\dfrac{20}{100}w$
Taking “w” as common from the above expression we get,
$\begin{align}
  & w\left( 1-\dfrac{20}{100} \right) \\
 & =w\left( \dfrac{100-20}{100} \right) \\
 & =w\left( \dfrac{80}{100} \right) \\
 & =\dfrac{4}{5}w \\
\end{align}$
Now, $10\%$ loss incurred on this new reduced weight is given by:
$\begin{align}
  & x-\dfrac{10}{100}x \\
 & =x\left( 1-\dfrac{10}{100} \right) \\
 & =x\left( \dfrac{100-10}{100} \right) \\
 & =x\left( \dfrac{90}{100} \right) \\
 & =\dfrac{9}{10}x \\
\end{align}$
Now, the price for the reduced weight $\dfrac{4}{5}w$ is $\dfrac{9}{10}x$. Here, $\dfrac{4}{5}w$is the actual weight that the dealer is selling but as the dealer is a dishonest man so he is selling this reduced weight goods as “w” weight of goods.
From this reduced weight and price relation which says that the price for the reduced weight $\dfrac{4}{5}w$ is $\dfrac{9}{10}x$so by unitary method the price for “w” weight goods is equal to $\dfrac{9}{10}x\left( \dfrac{5}{4} \right)$. Solving this we get the price for “w” weight of goods is equal to $\dfrac{9}{8}x$.
The cost price is for “w” weight of weights so selling price is also for “w” weight of goods which is equal to $\dfrac{9}{8}x$
Now, you can see that the selling price is greater than cost price so profit has been incurred for the dealer.
We know that:
$\ Profit=\dfrac{S.P.-C.P.}{C.P.}\times 100$
Substituting S.P. as $\dfrac{9}{8}x$ and cost price as x in the above equation we get,
\[\begin{align}
  & \ Profit=\dfrac{\dfrac{9}{8}x-x}{x}\times 100 \\
 & \Rightarrow \ Profit=\dfrac{\dfrac{9x-8x}{8}}{x}\times 100 \\
 & \Rightarrow \ Profit=\dfrac{x}{8x}\times 100 \\
\end{align}\]
In the above equation, x will be cancelled out from the numerator and the denominator.
\[\begin{align}
  & \ Profit=\dfrac{1}{8}\times 100 \\
 & \Rightarrow \ Profit=12.5\% \\
\end{align}\]
Hence, the correct option is (d).

Note: You might find the language of the problem quite confusing. The point in the problem where you can go wrong is you might have considered the selling price as only the price that we incurred from $10\%$ loss on the cost price because here we have to take into account the reduced weight also.
You can also make the mistake by considering the selling price on behalf of the reduced weight. For instance we know the cost price of goods as x and the reduced weight that we have calculated in the above solution is $\dfrac{4}{5}w$ so you will use the unitary method to find the price for $\dfrac{4}{5}w$which we have shown below.
The cost price for “w” weight of goods is x and the price for $\dfrac{4}{5}w$ weight of goods is $\dfrac{4}{5}x$.
This way of solving the problem is absolutely wrong because we have given that $10\%$ loss is incurred on this reduced weight so we have to incorporate this loss also.