Solution:

Given, 3X - 4Y = 20

Let us find the relation between the mean and the quartile deviation,

Let Q1, Q2 and Q3 be the first, second and third quartiles of a set of observations X1, X2, …., Xn. Then the quartile deviation is given by

Q.D = (Q3 - Q1)/2

Let the observations be X1, X2, …., Xn and let their mean be X̄.

Then the first, second and third quartiles of the deviations (X1 - X̄), (X2 - X̄), …., (Xn - X̄) are

Q1 = -0.6745σ, Q2 = 0, Q3 = 0.6745σ

where σ is the standard deviation of the set of observations.

Therefore, the quartile deviation is given by

Q.D = (Q3 - Q1)/2 = 0.6745σ

Given that the quartile deviation of X is 12.

Therefore,

0.6745σx = 12

σx = 17.77

Let the quartile deviation of Y be Q.DY

Then, Q.DY = (Q3Y - Q1Y)/2

Now, let us find Q1Y and Q3Y

From the given relation,

3X - 4Y = 20

Y = (3X - 20)/4

Let us find the quartiles of Y in terms of the quartiles of X.

Q1Y = (3Q1X - 20)/4

Q3Y = (3Q3X - 20)/4

Substituting the value of Q1X and Q3X in the above equations, we get

Q1Y = (3(-0.6745σx) - 20)/4 = -9.045

Q3Y = (3(0.6745σx) - 20)/4 = 10.545

Therefore,

Q.DY = (Q3Y - Q1Y)/2 = 19.59/2 = 9.795

Hence, the quartile deviation of Y is approximately 9.8.

Therefore, the correct option is (d) 9.