Vessels 'P' and 'Q' contain (4x - 10) litres and (x +

Solution

ATQ, Total cost of petrol in vessel 'P' = Rs. (4x - 10) × 24 Total cost of petrol in vessel 'Q' = Rs. (x + 10) × (2x - 8) ATQ; (4x - 10) × 24 = (x + 10) × (2x - 8) 96x - 240 = 2x² + 20x - 8x - 80 96x - 240 = 2x² + 12x - 80 2x² - 84x + 160 = 0 x² - 42x + 80 = 0 x² - 40x - 2x + 80 = 0 x(x - 40) - 2(x - 40) = 0 (x - 40) (x - 2) = 0 'x' = 40 or 2 For 'x' = 2, (2x - 8) = -4, but since the price per litre cannot be negative, So, 'x' = 40 Quantity of petrol in vessel 'P' = 4x - 10 = 150 litres Quantity of petrol in vessel 'Q' = x + 10 = 50 litres Per litre price of petrol in vessel 'Q' = 2x - 8 = Rs. 72 New quantity of petrol in vessel 'Q' after addition = 50 + 0.2 × 150 = 80 litres New total cost of petrol in vessel 'Q' = 50 × 72 + 30 × 24 = Rs. 4320 Therefore, required cost price = (4320/80) = Rs.54