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 + 15) : 3x = 16 : 9

Or, 45x + 135 = 48x

So, x = 135/3 = 45

So, quantity of mixture removed = 480 - (5x + 3x) = 480 - 8 × 45 = 480 - 360 = 120

So, (120/480) × 100 = (m + n + 10)

So, 25 = m + n + 10

So, (m + n) = 15 ---- (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 = 15 - 10 = 5

Therefore, required value = (5m - 2n) = 5 × 5 - 2 × 10 = 25 - 20 = 5