A tank contains 640 litres mixture of petrol and diesel in the ratio 5:3. If (m + n + 10)% of the mixture is removed and then 28 litres of petrol is added, the final ratio of petrol and diesel becomes 11:6. Instead of that, if n% of the mixture is removed and then 54 litres of petrol is added, the final ratio of petrol and diesel becomes 23:12. Find the value of (5m - 2n).

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 + 28) : 3x = 11:6 Or, 30x + 168 = 33x So, x = 168 ÷ 3 = 56 So, quantity of mixture removed = 640 - (5x + 3x) = 640 - 8 X 56 = 640 - 448 = 192 So, (192/640) X 100 = (m + n + 10) So, 30 = m + n + 10 So, (m + n) = 30 - 10 = 20 - (I) If after removing n% of the mixture, let the quantity of petrol and diesel be 5y litres and 3y litres respectively. ATQ; (5y + 54) : 3y = 23:12 Or, 60y + 648 = 69y So, y = 648 ÷ 9 = 72 So, quantity of mixture removed = 640 - (5y + 3y) = 640 - 8 X 72 = 640 - 576 = 64 So, (64/640) X 100 = n = 10 - (II) From (I) and (II), m = 20 - 10 = 10 Therefore, required value = (5m - 2n) = 5 X 10 - 2 X 10 = 50 - 20 = 30