The Percentage volume of alcohol in three solutions M, N, O form a geometric progression in that order. If we mix first , second , third solution in the volume of ratio of 2 : 3 : 4, we obatin a solution containing 32% alchohol . If we mix them in the ratio 3 : 2 : 1 by volume, we obatain a solution containing 22% of alcohol. What is the percentage of Alcohol in M?

Let percent of alcohal in mixture M, N and O mixture is 'a', 'ar' and 'a$$r^2$$' respectively.

Now, 

$$\frac{2a + 3ar + 4ar^2}{2 + 3 + 4} = 32$$

$$a(2 + 3r + 4r^2) = 288$$ .......... (1)

$$\frac{3a + 2ar + ar^2}{3 + 2 + 1} = 22$$

$$a(3 + 2r + r^2) = 132$$ .......... (2)

From (1) and (2):

$$\frac{a(2 + 3r + 4r^2)}{a(3 + 2r + r^2)} = \frac{288}{132} \frac{24}{11}$$

$$4r^2 - 3r - 10 = 0$$

r = 2

From equation (1):

$$a(2 + 6 + 16) = 288$$

a = 12

Ppercentage of Alcohol in M = a = 12%