When x is added to each of 20, 34, 11 and 16, then the numbers so obtained, in this order, are in proportion. Then, if 3x : y :: y : (2x-6), and y > 0, what is the value of y?

Correct option is C

Given:

Numbers 20 + x, 34 + x ,11 + x ,16 + x are in proportion.
Also, 3x : y :: y : (2x - 6) with y > 0.

Solution:
From $$\dfrac{20+x}{34+x}=\dfrac{11+x}{16+x}:$$​​
​$$(20+x)(16+x)=(34+x)(11+x)$$​​
​$$x^{2}+36x+320=x^{2}+45x+374$$​​
36x + 320 = 45x + 374

-54 = 9x

x = -6.

Now y$$^{2}=3x(2x-6)=3(-6)(2(-6)-6)=3(-6)(-18)=324.$$​​
Since y > 0 then, y = 18.