Question

A shopkeeper cheats to the extent of \[10\% \] while buying as well as selling, by using false weights. His total gain is:
A) \[10\% \]
B) \[11\% \]
C) \[20\% \]
D) \[21\% \]
E) \[22\dfrac{2}{9}\% \]

Answer 1

Hint:
We will solve this problem by assuming that the shopkeeper is buying \[100\] units of a product. Then, we will calculate the amount he buys and the amount he sells by cheating by using the given information. Finally, we will calculate the profit he has gained by subtracting the number of quantities he purchased by cheating and the number of quantities he sold by cheating. Using this we will find the required answer.

Complete step by step solution:
It is given that the shopkeeper cheats to the extent of \[10\% \] while buying as well as selling, by using false weights.
Let us first assume that the shopkeeper wants to buy 100 units of a product.
Since, he cheats to the extent of \[10\% \], the actual amount he buys will be \[10\% \] excessive than the 100 units. So,
Number of units the shopkeeper actually buys by cheating \[ = 100 + 10\% {\text{ of }}100\]
Converting percentage into fraction, we get
\[ \Rightarrow \] Number of units the shopkeeper actually buys by cheating \[ = 100 + \dfrac{{10}}{{100}} \times 100\]
Multiplying the terms, we get
\[ \Rightarrow \] Number of units the shopkeeper actually buys by cheating \[ = 100 + 10 = 110\]
This means that the shopkeeper actually buys 110 units of the product.
But, since he cheated, he has paid only for 100 units of the product.
Now, it is also given that he sells products by cheating up to \[10\% \].
Suppose he wants to sell the same product, then since he cheats up to \[10\% \], he will sell \[10\% \] more than the 110 units he originally bought.
Number of units the shopkeeper actually sells by cheating \[ = 110 + 10\% {\text{ of }}110\]
Converting percentage into a fraction, we get
\[ \Rightarrow \] Number of units the shopkeeper actually sells by cheating \[ = 110 + \dfrac{{10}}{{100}} \times 110\]
Multiplying the terms, we get
\[ \Rightarrow \] Number of units the shopkeeper actually sells by cheating \[ = 110 + 11 = 121\]
This means that the shopkeeper actually sells 121 units of the product.
So, the shopkeeper has bought 110 units and sold 121 units of the product.
But he has paid for 100 units while buying and received the price of 121 units while selling.
Thus, total gain \[ = 121 - 100 = 21\] units
Gain percentage \[ = \dfrac{{21}}{{100}} \times 100 = 21\% \]

Therefore, the correct option is D.

Note:
We have solved the question by using the given conditions on a small quantity like 100. Then, the results can be generalized based on the calculations. Instead of 100, a variable quantity like \[x\] can also be used. But this will complicate the calculations to a certain extent. Here we have found gain in terms of units and not in terms of rupees. We can make a mistake in finding a gain percentage if we take the quantity he purchased as 110. This is because, even though he purchased 110 units, he only paid for 100 units.