When x is added to each of 22, 26, 19 and 21, then the numbers so obtained, in this order, are in proportion. Then, if 2x : y :: y : (4x-8), and y > 0, what is the value of y?​

Correct option is A

​Given:

(22 + x) : (26 + x) :: (19 + x) : (21 + x)

2x : y :: y : (4x - 8) and y > 0

Formula Used:

Proportion: $$\dfrac{a}{b}=\dfrac{c}{d}\implies ad=bc$$​​

For $$\dfrac{p+x}{q+x}=\dfrac{r+x}{s+x}\implies x=\dfrac{qr-ps}{p+s-q-r}.$$​​

Solution:
From $$\dfrac{22+x}{26+x}=\dfrac{19+x}{21+x}:$$​​

x=$$\frac{(26)(19)-(22)(21)}{(22+21)-(26+19)}$$​

​$$=\frac{494-462}{43-45}$$​

=$$\frac{32}{-2}=-16.$$​​
Now $$\dfrac{2x}{y}=\dfrac{y}{4x-8}\implies y^{2}=2x(4x-8)$$​​
For x = -16: 4x- 8 = -72, so
​$$y^{2}=2(-16)(-72)=2304\ \implies\ y=\sqrt{2304}=48\quad(y>0).$$​​