Question

The variable x varies directly as the cube of y, and y varies directly as the square root of z. If x equals 1 when z equals 4, what is the value of z when x equals 27?

Answer 1

Answer:

$z=36$

Step-by-step explanation:

According to the question,

$x$∝$y^3$          .......(1)

$y$∝$\sqrt{z}$    .......(2)

From equation 1,2 let constant of proportionality be $k1,k2$ respectively.

⇒$x=k1(y^3)$            .......(3)

⇒$y=k2(\sqrt{z} )$    .......(4)

From the above equations putting 4 into 3,

$x=k1((k2\sqrt{z})^3) =k1.k2^3.(\sqrt{z})^3$

Let the new constant to the above equation be $k3$,

$x=k3(\sqrt{z})^3$

Given,if x=1, when z=4

$1=k3(\sqrt{4} )^3=k3(8)$

⇒$k3=\frac{1}{8}$

Now if x=27, then z=?

$27=\frac{1}{8} (\sqrt{z} )^3$

⇒$(\sqrt{z} )^3=27(8)$

⇒$\sqrt{z}=3(2)=6$

$z=36$