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?

Correct Option (4)

Let the cost of article Q be $x$x.

• Article R costs 20% more than Q. Therefore, the cost of R is $x + 0.20x = 1.2x = \frac{6x}{5}$x+0.20x=1.2x=56x​.
• Article P costs 25% more than R. Therefore, the cost of P is $R + 0.25R = 1.25R = \frac{5}{4}R$R+0.25R=1.25R=45​R.
• Substituting the value of R: Cost of P $= \frac{5}{4} \times \frac{6x}{5} = \frac{30x}{20} = \frac{3x}{2}$=45​×56x​=2030x​=23x​.

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

$\frac{3x}{2} + x + \frac{6x}{5} = 3330$23x​+x+56x​=3330

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

$\frac{15x}{10} + \frac{10x}{10} + \frac{12x}{10} = 3330$1015x​+1010x​+1012x​=3330

$\frac{15x + 10x + 12x}{10} = 3330$1015x+10x+12x​=3330

$\frac{37x}{10} = 3330$1037x​=3330

$37x = 3330 \times 10$37x=3330×10

$37x = 33300$37x=33300

$x = \frac{33300}{37} = 900$x=3733300​=900

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

The cost of article P is $\frac{3x}{2}$23x​:

Cost of P $= \frac{3 \times 900}{2} = \frac{2700}{2} = 1350$=23×900​=22700​=1350

Therefore, the cost of P is ₹ 1,350.

Incorrect Options:

• 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.