Question

&Complete the following tables given that $x$ varies directly as $y$

(i)
$\begin{array}{ccccc}x& 2.5& ....& ....& 15\\ y& 5& 8& 12& ....\end{array}$

(ii)
$\begin{array}{ccccccc}x& 5& ....& 10& 35& 25& ...\\ y& 8& 12& ...& ....& ....& 32\end{array}$

(iii)
$\begin{array}{cccccc}x& 6& 8& 10& ....& 20\\ y& 15& 20& ...& 40& ....\end{array}$

(iv)
$\begin{array}{ccccccc}x& 4& 9& ...& ....& 3& ....\\ y& 16& ...& 48& 36& ....& 4\end{array}$

(v)
$\begin{array}{ccccc}x& 3& 5& 7& 9\\ y& ...& 20& 28& ....\end{array}$

Answer 1

Here, $x$ and $y$ vary directly, So $x=ky$......(i)
(i) Given $x=2.5$ and $y=5$
Substitute above values in equation(i), we get $2.5=k\times 5$
$\Rightarrow k=\frac{2.5}{5}=0.5$
Thus, for $y=8$ and $k=0.5$,
we have: $x=ky$
$\Rightarrow x=8\times 0.5$
$\therefore x=4$
For $y=12$ and $k=0.5$,
we have: $x=ky$
$\Rightarrow x=12\times 0.5$
$\therefore x=6$
For $x=15$ and $k=0.5$,
we have: $x=ky$
$\Rightarrow 15=0.5\times y$
$\Rightarrow y=\frac{150}{5}=30$

(ii) Given $x=5$ 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: $x=ky$
$\Rightarrow x=12\times 0.625=7.5$
For $x=10$ and $k=0.625$
we have: $x=ky$
$\Rightarrow 10=0.625\times y$
$\Rightarrow y=\frac{10}{0.625}=16$
For $x=35$ and $k=0.625$,
we have: $x=ky$
$\Rightarrow 35=0.625\times y$
$\Rightarrow y=\frac{35}{0.625}$
$=56$
For $x=25$ and $k=0.625$, we have:$x=ky$
$\Rightarrow 25=0.625\times y$
$\Rightarrow y=\frac{25}{0.625}=40$
For $y=32$ and $k=0.625$,
we have: $x=ky$
$\Rightarrow x=0.625\times 32=20$
(iii) Given $x=6$ 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:$y=\frac{200}{4}=50$

(iv) Given $x=4$ 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:$9=\frac{1}{4}\times y$
$\Rightarrow y=4\times 9=36$
For $y=48$ and $k=\frac{1}{4}$,
we have: $x=ky$
$\Rightarrow x=\frac{1}{4}\times 48=12$
For $y=36$ and $k=\frac{1}{4},$
we have: $x=ky=\frac{1}{4}\times 36$
$=9$
For $x=3$ and $k=14$,
we have: $x=ky$
$\Rightarrow 3=\frac{1}{4}\times y$
$\Rightarrow y=12$
For $y=4$ and $k=\frac{1}{4}$,
we have: $x=ky=\frac{1}{4}\times 4=1$
(v) Given $x=5$ 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: $x=ky\Rightarrow 9=\frac{1}{4}\times y$
$\therefore y=9\times 4=36$