A tank contains 480 litres mixture of petrol and diesel

Solution

After removing (m + n + 15) % of the mixture, Let the remaining quantity of petrol and diesel be 5x litres and 3x litres respectively. ATQ; (5x + 18) : 3x = 17 : 9 Or, 45x + 162 = 51x So, x = 162/6 = 27 So, quantity of mixture removed = 480 - (5x + 3x) = 480 - 8 × 27 = 480 - 216 = 264 So, (264/480) × 100 = (m + n + 15) So, 55 = m + n + 15 So, (m + n) = 40 ---- (I) If after removing n% of the mixture, let the quantity of petrol and diesel be 5y litres and 3y litres respectively. ATQ; (5y + 27) : 3y = 11 : 6 Or, 30y + 162 = 33y So, y = 162/3 = 54 So, quantity of mixture removed = 480 - (5y + 3y) = 480 - 8 × 54 = 480 - 432 = 48 So, (48/480) × 100 = n So, n = 10 ---- (II) From (I) and (II) m = 40 - 10 = 30 Therefore, required value = (m + 2n) = 30 + 2 × 10 = 30 + 20 = 50