Half of the volume of petrol and kerosene mixture of ratio 7 : 5

Ques - Half of the volume of petrol and kerosene mixture of ratio 7 : 5 is converted into a mixture of ratio 3 : 1 by the substitution (or replacement) method. While the mixture of ratio 7 : 5 was formed from the mixture of 7 : 3 by adding the kerosene in it . If 240 litres petrol is required in the replacement method, what is the total amount of kerosene was added to prepare the mixture of 7 : 5 ?

(a) 100 litres

(b) 400 litres

(c) 50 litres

(d) 200 litres

Solution -

Initial quantity = P : K

                       7x : 3x

                              ↓

                             + kerosene (by addition)

Step 1 -           7y : 5y

                             ↓

                            (by replacement)

Step 2 -           3z : z

If,

240 litres petrol is required in the replacement method.

Now,

Using formula of replacement

(z/4z) = (5y / 12y) [1 - (240/12y)]

(1/4) = (5/12) [1 - (20/y)]

(1/4)(12/5) = [1 - (20/y)]

3/5 = 1 - 20/y

2/5 = 20/y

1/50 = 1/y

y = 50

Therefore,

Half of the initial amount = 7y + 3y

= 10y

= 10 × 50

= 500

Then,

Whole amount = 500 × 2

= 1000

Initial amount = 7x + 3x

= 10x

10x = 1000

x = 100 litres

Kerosene replayed = 2x

= 2 × 100

= 200 litres 

The total amount of kerosene was added to prepare the mixture of 7 : 5 is 200 litres.

Hence,

The correct answer is option (d) 200 litres.

Thank You