Question

Write each of the following rational numbers in standard form:

(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

Answer 1

(i) $\frac{35}{49}$
H.C.F. of 35 and 49 is 7.

Dividing the numerator and the denominator by 7:

$\frac{35÷7}{49÷7}=\frac{5}{7}$
So, $\frac{35}{49}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{5}{7}$ in the standard form.

(ii)$\frac{8}{-36}$
Denominator is -36, which is negative.
Multiplying both the numerator and the denominator by -1:

$\frac{8\times (-1)}{-36\times (-1)}=\frac{-8}{36}$


H.C.F. of 8 and 36 is 4.
Dividing its numerator and denominator by 4:

$\frac{-8÷4}{36÷4}=\frac{-2}{9}$

So, $\frac{8}{-36}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-2}{9}$ in the standard form.

(iii) $\frac{-27}{45}$


H.C.F. of 27 and 45 is 9.
Dividing its numerator and denominator by 9:
$\frac{-27÷9}{45÷9}=\frac{-3}{5}$
Hence, $\frac{-27}{45}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-3}{5}$ in the standard form.

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{-14}{-49} \\ \mathrm{The}\mathrm{denominator}\mathrm{is}\mathrm{negative}. \\ \mathrm{Multiplying}\mathrm{its}\mathrm{numerator}\mathrm{and}\mathrm{denominator}\mathrm{by}-1: \\ \frac{-14\times (-1)}{-49\times (-1)}=\frac{14}{49} \end{gathered}$$


H.C.F. of 14 and 49 is 7.
Dividing both the numerator and the denominator by 7.
$$\begin{gathered} \frac{14÷7}{49÷7}=\frac{2}{7} \\ \mathrm{Hence},\frac{-14}{-49}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{2}{7}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{form}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\frac{91}{-78} \\ \mathrm{The}\mathrm{denominator}\mathrm{is}\mathrm{negative}. \\ M\mathrm{ultiplying}\mathrm{its}\mathrm{denominator}\mathrm{and}\mathrm{denominator}\mathrm{by}-1: \\ \frac{91\times (-1)}{-78\times (-1)}=\frac{-91}{78} \end{gathered}$$


H.C.F. of 91 and 78 is 13.
Dividing both the numerator and the denominator by 13:
$$\begin{gathered} \frac{-91÷13}{78÷13}=\frac{-7}{6} \\ \mathrm{Hence},\frac{91}{-78}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-7}{6}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{form}. \end{gathered}$$

$\left(\mathrm{vi}\right)\frac{-68}{119}$


H.C.F. of 68 and 119 is 17.
Dividing both the numerator and the denominator by 17:
$$\begin{gathered} \frac{-68÷17}{119÷17}=\frac{-4}{7} \\ \mathrm{Hence},\frac{-68}{119}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-4}{7}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{form}. \end{gathered}$$

$\left(vii\right)\frac{-87}{116}$


H.C.F. of 87 and 116 is 29.
Dividing both the numerator and the denominator by 29:
$$\begin{gathered} \frac{-87÷29}{116÷29}=\frac{-3}{4} \\ \mathrm{Hence},\frac{-87}{116}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-3}{4}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{form}. \end{gathered}$$

$\left(\mathrm{viii}\right)\frac{299}{-161}$
The denominator is negative.
Multiplying both the numerator and denominator by -1:

$\frac{299\times (-1)}{-161\times (-1)}=\frac{-299}{161}$


H.C.F. of 299 and 161 is 23.
Dividing both the numerator and the denominator by 23:
$$\begin{gathered} \frac{-299÷23}{161÷23}=\frac{-13}{7} \\ \mathrm{Hence},\frac{299}{-161}\mathrm{is}\mathrm{equal}\mathrm{to}\frac{-13}{7}\mathrm{in}\mathrm{the}\mathrm{standard}\mathrm{form}. \end{gathered}$$