When x is added to each of 26, 40, 22 and 34, then the numbers so obtained, in this order, are in proportion. Then, if 4x : y :: y : (7x-6), and y > 0, what is the value of y?​

Correct option is B

Given:

After adding (x): (26+x):(40+x)::(22+x):(34+x)

Also 4x : y :: y :(7x-6) with  y > 0.

Concept Used:

If a:b:$$:c:d\implies \dfrac{a}{b}=\dfrac{c}{d}$$​ 

ad = bc.

​Solution:​

For 4x : y :: y : (7x-6):

product of extremes = product of means

$$(4x)(7x-6)=y^2.$$​​

From the first proportion:
(26+x)(34+x)=(40+x)(22+x)
​$$884+60x+x^2=880+62x+x^2$$ 

4 = 2x

x = 2 

Then,

$$(4x)(7x-6)=y^2 \implies (8)(14-6)=y^2 \implies 64=y^2.$$​​

Then,
Since (y>0), y = 8.