Question

x is partly constant and partly varies as the square root of y.when x=3,y=13.when x=7,y=27.find the value of y when x=5.

Answer 1

x is partly constant and partly varies as the square root of y
----> x = c + k√y
when x=3,y=13
3 = c + k√13
when x=7,y=27
7 = c + k√27

subtract those two equations
4 = k√27 - k√13
k = 4/(√27-√13) = appr 2.5148

then in 3 = c + k√13
3 = c + √13(4/(√27-√13))
c = 3 - √13(4/(√27-√13)) = appr -6.0671

so when x = 5
5 = 3 - √13(4/(√27-√13)) + 4/(√27-√13)√y
solve for √y, then square it.

or
x = c + k√y
x-c = k√y
(x-c)^2 = k^2 y^2

when x=3,y=13
(3-c)^2 = 13k^2 ----> k^2 = (3-c)^2 /13 **
when x=7,y=27
(7-c)^2 = 27 k^2 ----> k^2 = (7-c)^2 / 27 ***

set ** = ***
(3-c)^2 /13 = (7-c)^2 / 27
take √ of both sides
(3-c)/√13 = (7-c)/√27
I solved this for c and got c = appr -6.0671

into **
k^2 = (3-c)^2 /13
find k, which I found to be appr 2.5148

so......
x = appr (-6.0671 + 2.5148√y)

plug in x = 5 to find y

What a messy question!! I have a feeling you have a typo, expected "nicer" numbers.