Correct Option (4)
Let the cost of article Q be x.
- Article R costs 20% more than Q. Therefore, the cost of R is x+0.20x=1.2x=56x.
- Article P costs 25% more than R. Therefore, the cost of P is R+0.25R=1.25R=45R.
- Substituting the value of R: Cost of P =45×56x=2030x=23x.
The total cost of the three articles is ₹ 3,330. Thus, the sum of their costs is:
23x+x+56x=3330
To solve for x, find a common denominator (10):
1015x+1010x+1012x=3330
1015x+10x+12x=3330
1037x=3330
37x=3330×10
37x=33300
x=3733300=900
This value represents the cost of article Q (₹ 900).
The cost of article P is 23x:
Cost of P =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.