When x is added to each of 30, 15, 21 and 11, then the numbers so obtained, in this order, are in proportion. Then, if 6x : y :: y : (3x - 9), and y > 0, what is the value of y?

Correct option is A

Given:

When x is added to each of 30, 15, 21, and 11, the numbers so obtained are in proportion.

6x : y :: y : (3x - 9), with y > 0.

We need to find the value of y

Solution:

From the proportion:

​$$\frac{30 + x}{15 + x} = \frac{21 + x}{11 + x}$$​​

(30 + x)(11 + x) = (15 + x)(21 + x)

​$$330 + 30x + 11x + x^2 = 315 + 15x + 21x + x^2$$​​

330 + 41x = 315 + 36x

330 - 315 = 36x - 41x

x = -3

Now, substituting x = -3 into the second proportion:

​$$\frac{6(-3)}{y} = \frac{y}{3(-3) - 9}$$​​

​$$\frac{-18}{y} = \frac{y}{-9 - 9}$$​​

​$$\frac{-18}{y} = \frac{y}{-18}$$​​

​$$-18 \times -18 = y \times y$$​​

​$$324 = y^2$$​​

y = $$\sqrt{324}$$​ = 18