Complite the following tables given that x varies directly as y.

Solution:

Here, $x$ and $y$ vary directly.

$\therefore x=k y$

(i) $x=2.5$ and $y=5$

i.e., $2.5=k \times 5$

$\Rightarrow k=\frac{2.5}{5}=0.5$

For $y=8$ and $k=0.5$, we have :

$x=k y$

$\Rightarrow x=8 \times 0.5=4$

For $y=12$ and $k=0.5$, we have :

$x=k y$

$\Rightarrow x=12 \times 0.5=6$

For $x=15$ and $k=0.5$, we have :

$x=k y$

$\Rightarrow 15=0.5 \times y$

$\Rightarrow y=\frac{15}{0.5}=30$

(ii) $x=5$ and $y=8$

i.e., $5=k \times 8$

$\Rightarrow k=\frac{5}{8}=0.625$

For $y=12$ and $k=0.625$, we have :

$x=k y$

$\Rightarrow x=12 \times 0.625=7.5$

For $x=10$ and $k=0.625$, we have :

$x=k y$

For $x=25$ and $k=0.625$, we have :

$x=k y$

$\Rightarrow 25=0.625 \times y$

$\Rightarrow y=\frac{25}{0.625}=40$

For $y=32$ and $k=0.625$, we have :

$x=k y$

$\Rightarrow x=0.625 \times 32=20$

(iii) $x=6$ and $y=15$

i.e., $6=k \times 15$

$\Rightarrow k=\frac{6}{15}=0.4$

For $x=10$ and $k=0.4$, we have :

$y=\frac{10}{0.4}=25$

For $y=40$ and $k=0.4$, we have :

$x=0.4 \times 40=16$

For $x=20$ and $k=0.4$, we have :

$y=\frac{20}{0.4}=50$

(iv) $x=4$ and $y=16$

i.e., $4=k \times 16$

$\Rightarrow k=\frac{4}{16}=\frac{1}{4}$

For $x=9$ and $k=\frac{1}{4}$, we have :

$9=k y$

$\Rightarrow y=4 \times 9=36$

For $y=48$ and $k=\frac{1}{4}$, we have :

$x=k y$

$=\frac{1}{4} \times 48=12$

For $y=36$ and $k=\frac{1}{4}$, we have :

$x=k y$

$=\frac{1}{4} \times 36=9$

For $x=3$ and $k=\frac{1}{4}$, we have :

$x=k y$

$\Rightarrow 3=\frac{1}{4} \times y$

$\Rightarrow y=12$

For $y=4$ and $k=\frac{1}{4}$, we have :

$x=k y$

$=\frac{1}{4} \times 4=1$

(v) $x=5$ and $y=20$

i.e., $5=k \times 20$

$\Rightarrow k=\frac{5}{20}=\frac{1}{4}$

For $x=3$ and $k=\frac{1}{4}$, we have :

$3=\frac{1}{4} \times y$

$\Rightarrow y=4 \times 3=12$

For $x=9, k=\frac{1}{4}$, we have :

$x=k y$

$\Rightarrow 9=\frac{1}{4} \times y$

$\Rightarrow y=9 \times 4=36$