A person buys three articles P, Q and R for ₹ 3,330. If P costs 25% more than R and R costs 20% more than Q, then what is the cost of P?

EXPLANATION

Let the cost of article Q be $x$ .

Article R costs 20% more than Q. Therefore, the cost of R is $x + 0.20 x = 1.2 x = \frac{6 x}{5}$ .
Article P costs 25% more than R. Therefore, the cost of P is $R + 0.25 R = 1.25 R = \frac{5}{4} R$ .
Substituting the value of R: Cost of P $= \frac{5}{4} \times \frac{6 x}{5} = \frac{30 x}{20} = \frac{3 x}{2}$ .

The total cost of the three articles is ₹ 3,330. Thus, the sum of their costs is:

$\frac{3 x}{2} + x + \frac{6 x}{5} = 3330$

To solve for $x$ , find a common denominator (10):

$\frac{15 x}{10} + \frac{10 x}{10} + \frac{12 x}{10} = 3330$

$\frac{15 x + 10 x + 12 x}{10} = 3330$

$\frac{37 x}{10} = 3330$

$37 x = 3330 \times 10$

$37 x = 33300$

$x = \frac{33300}{37} = 900$

This value represents the cost of article Q (₹ 900).

The cost of article P is $\frac{3 x}{2}$ :

Cost of P $= \frac{3 \times 900}{2} = \frac{2700}{2} = 1350$

Therefore, the cost of P is ₹ 1,350.

Options 1 (₹ 1,000), 2 (₹ 1,200), and 3 (₹ 1,250) are incorrect. If any of these values were the cost of P, the corresponding costs for Q and R, derived from the given percentage relationships, would not sum up to the total purchase price of ₹ 3,330. These values are inconsistent with the established proportional relationships between the articles' costs and their aggregate sum.