Question

A shopkeeper gives a discount of 25% on the marked price of an article and still earns a profit of 20% on his outlay, if he had not given any discount, then the profit on his outlay would have been
(a) 45%
(b) 50%
(c) 55%
(d) 60%

Answer 1

Hint: To solve this question, first we will assume a variable for marked price and find the discounted price based on the given information. Then, from the question we know that the discounted price is 20% more than the selling price. With this, we find a relation between marked price and cost price and thus percentage increase of price from cost price to marked price.

Complete step-by-step answer:
Let us assume that the marked price is x.
It is given the shopkeeper gives a 25% discount on the marked price, that is x.
Therefore, 25% of x will be $\dfrac{25}{100}\text{x}$ = 0.25x.
Hence, the discounted price will be (marked price – 25% of marked price) = x – 0.25x = 0.75x.
If a shopkeeper sells the article at 0.75x, he still can earn 20% on the cost price.
Therefore, this 0.75x is (cost price + 20% of cost price).
Let the cost price be cp.
Therefore, 20% of the cost price will be $\dfrac{20}{100}\text{cp}$ = 0.20cp.
Thus, cp + 0.20cp = 0.75x
1.2cp = 0.75x
x = $\dfrac{1.20}{0.75}$ = 1.6cp…………………………………...……(1)
This means that the marked price is 1.6 times the cost price of the article.
To find the percentage change between the marked price and the cost price, we will use the equation \[\text{pi}=\dfrac{\text{mp}-\text{cp}}{\text{cp}}\times 100\], where pi is the percentage increase, mp is the marked price and cp is the cost price.
Therefore, \[\text{pi}=\dfrac{\text{x}-\text{cp}}{\text{cp}}\times 100\], but from (1), x = 1.6cp
Thus, \[\text{pi}=\dfrac{1.6\text{cp}-\text{cp}}{\text{cp}}\times 100\] = 60%
Hence, the marked price is 60% more than the cost price. This means, if he doesn’t give any discount, the shopkeeper will earn 60% profit.
So, option (d) is the correct option. \[\]

Note: Students are advised to be well versed with the concepts of percentage, so that profit and loss problems can be solved easily. Step by step approach for this question is very beneficial. Students can also use option verification by assuming some value for marked price, but it can be very tedious.