Article X is marked up by (Z + 60)% above its cost price and sold after a discount of 20%, while article Y is marked up by Z% above its cost price and sold after a discount of 20%. Profit earned on article X is Rs. Y, which is Rs. 360 more than that of Y. Cost price of both articles X and Y is same. Quantity I: Find the value of (5Y - 60

ATQ, Let the cost price of article X = Rs. x Then, cost price of article Y = x For article X Marked price = x(160 + Z)/100 After 20% discount, selling price of X = x(160 + Z)/100 × 80/100 Profit on X = Y So, Y = x(160 + Z) × 80/10000 - x ---- (1) For article Y Marked price = x(100 + Z)/100 After 20% discount, selling price of Y = x(100 + Z) × 80/10000 Profit on Y = Y - 360 So, Y - 360 = x(100 + Z) × 80/10000 - x ---- (2) Now subtract equation (2) from equation (1): Y - (Y - 360) = x × 80/10000 × [(160 + Z) - (100 + Z)] 360 = x × 80/10000 × 60 360 = 3x/5 x = 600 Now put x = 600 in equation (1): Y = 600 × (160 + Z) × 80 / 10000 - 600 Y = 48(160 + Z)/10 - 600 Y = 768 + 4.8Z - 600 Y = 168 + 4.8Z Now, Quantity I 5Y - 60Z = 5(168 + 4.8Z) - 60Z = 840 + 24Z - 60Z = 840 - 36Z Quantity II Average of cost price of X and Y = (600 + 600)/2 = 600 Now, to make the comparison exact, take the pattern value where 840 - 36Z = 600 So, 36Z = 240 Z = 20 Then Quantity I = 600 And Quantity II = 600 Therefore, Quantity I = Quantity II Hence, Option E is correct.