5.4 SPM Practice (Short Questions)

Posted on April 24, 2020 by user

Question 1:

It is given that y varies directly as the cube of x and y = 192 when x = 4. Calculate the value of x when y = – 24.

Solution:

y α x³
y = kx³

192 = k (4)³

192 = 64 k

k = 3

y = 3 x³

when y = – 24

– 24 = 3 x³

x³ = – 8

x = – 2

Question 2:
It is given that y varies directly as the square of x and y = 9 when x = 2. Calculate the value of x when y = 16.

Solution:

y α x²
y = kx²
9 = k (2)²

$\begin{array}{l} k = \frac{9}{4} \\ y = \frac{9}{4} x^{2} \\ When y = 16 \\ 16 = \frac{9}{4} x^{2} \\ x^{2} = \frac{64}{9} \\ x = \frac{8}{3} \end{array}$

Question 3:
Given that y varies inversely as w and x and y = 45 when w = 2 and x =

$\frac{1}{6}$ . Find the value of x when y = 15 and w =

$\frac{1}{3}$

Solution:

$\begin{array}{l} y α \frac{1}{w x} \\ y = \frac{k}{w x} \\ 45 = \frac{k}{(2) (\frac{1}{6})} \\ k = 45 \times \frac{1}{3} = 15 \\ y = \frac{15}{w x} \end{array}$

$\begin{array}{l} when y = 15, w = \frac{1}{3} \\ 15 = \frac{15}{(\frac{1}{3}) x} \\ \frac{x}{3} = 1 \\ x = 3 \end{array}$

Question 4:
Given that

$p α \frac{1}{q \sqrt{r}}$ and p = 3 when q = 2 and r = 16, find the value of p when q = 3 and r = 4.

Solution:

$\begin{array}{l} p α \frac{1}{q \sqrt{r}} \\ p = \frac{k}{q \sqrt{r}} \\ 3 = \frac{k}{(2) \sqrt{16}} \\ k = 24 \\ p = \frac{24}{q \sqrt{r}} \end{array}$

$\begin{array}{l} when q = 3, r = 4 \\ p = \frac{24}{3 \sqrt{4}} \\ p = \frac{24}{6} = 4 \end{array}$