A tank contains 480 litres mixture of petrol and diesel

Solution

After removing (m + n + 10) % of the mixture, Let the remaining quantity of petrol and diesel be 5x litres and 3x litres respectively. ATQ; (5x + 24) : 3x = 17 : 9 Or, 45x + 216 = 51x So, x = 216/6 = 36 So, quantity of mixture removed = 480 - (5x + 3x) = 480 - 8 × 36 = 480 - 288 = 192 So, (192/480) × 100 = (m + n + 10) So, 40 = m + n + 10 So, (m + n) = 30 ---- (I) If after removing n% of the mixture, let the quantity of petrol and diesel be 5y litres and 3y litres respectively. ATQ; (5y + 15) : 3y = 16 : 9 Or, 45y + 135 = 48y So, y = 135/3 = 45 So, quantity of mixture removed = 480 - (5y + 3y) = 480 - 8 × 45 = 480 - 360 = 120 So, (120/480) × 100 = n So, n = 25 ---- (II) From (I) and (II) m = 30 - 25 = 5 Therefore, required value = (2m + n) = 2 × 5 + 25 = 10 + 25 = 35