Given that z varies directly as x^2 and inversely as \sqrt{y}, and z = 12 when x = 2 and y… — JAMB UTME

Given that $z$ varies directly as $x^2$ and inversely as $\sqrt{y}$ , and $z = 12$ when $x = 2$ and $y = 9$ , what is the value of $z$ when $x = 3$ and $y = 16$ ?

$\frac{81}{4}$ Answer
B
$\frac{27}{2}$
C
$\frac{243}{16}$
D
$\frac{64}{9}$

Answer

\frac{81}{4}

The joint variation formula is

z = \frac{k x^2}{\sqrt{y}}

. Substituting the given values

x = 2, y = 9, z = 12

gives

12 = \frac{4k}{3}

, which yields

k = 9

. Evaluating

z

for

x = 3

and

y = 16

gives

z = \frac{9 \times 3^2}{\sqrt{16}} = \frac{81}{4}

Step-by-Step Solution

Set up the joint variation equation

z = \frac{k x^2}{\sqrt{y}}

Direct variation means

x^2

is in the numerator, and inverse variation means

\sqrt{y}

is in the denominator.

Substitute the initial values to solve for the constant of variation

k

12 = \frac{k (2)^2}{\sqrt{9}} \implies 12 = \frac{4k}{3} \implies 4k = 36 \implies k = 9

Using

x = 2

y = 9

, and

z = 12

allows us to find the constant

k

Calculate the new value of

z

using

x = 3

and

y = 16

z = \frac{9 (3)^2}{\sqrt{16}} = \frac{9 \times 9}{4} = \frac{81}{4}

Substitute

k = 9

x = 3

, and

y = 16

into the variation formula.

Key Concept

Joint Variation involving powers and roots

Estimated Time:1m 30s

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