Question

Two vessel X and Y contain petrol and kerosene mixed in the ratio 3:5 and 7:1 respectively. In what ratio must these mixture to form a new mixture containing thrice as much petrol as kerosene.

Answer 1

Answer:

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Explanation:

pool : kerosene

3 : 2(initially)

2 : 3(after replacement)

Remaining / Initial =(1−Replaced /Total )

(for petrol) 23=(1−10k)

k = 30

Therefore the total quantity of the mixture in the container is 30 liters.

Let original quantity is LCM(3/2, 2/3, 10/1) = 30 L

If 10 L is removed, 10 x 2/5 = 4 litres of kerosene is removed and 8 litres of kerosene & 12 litres of Petrol remains.

If additional 10 L of kerosene is poured, total quantity becomes 30 L and kerosene becomes 18 L.

Ratio of new Mixture = 12:18 = 2:3

=> Original volume was 30 L.