Examples (Revised) - Chapter 9 - Rational Numbers - Ncert Solutions class 7 - Maths

Share this to Friend on WhatsApp

Chapter 8 - Rational Numbers | NCERT Solutions for Class 7 Maths

Example 1

Reduce $\frac{-45}{30}$ to the standard form.
Solution

We have, $\frac{-45}{30}=\frac{-45 \div 3}{30 \div 3}=\frac{-15}{10}=\frac{-15 \div 5}{10 \div 5}=\frac{-3}{2}$
We had to divide twice. First time by 3 and then by 5 . This could also be done as
$
\frac{-45}{30}=\frac{-45 \div 15}{30 \div 15}=\frac{-3}{2}
$

In this example, note that 15 is the HCF of 45 and 30 .
Thus, to reduce the rational number to its standard form, we divide its numerator and denominator by their HCF ignoring the negative sign, if any. (The reason for ignoring the negative sign will be studied in Higher Classes)
If there is negative sign in the denominator, divide by ' $-\mathrm{HCF}$ '.

Example 2

Reduce to standard form:
(i) $\frac{36}{-24}$
(ii) $\frac{-3}{-15}$

Solution
(i) The HCF of 36 and 24 is 12 .
Thus, its standard form would be obtained by dividing by -12 .
$
\frac{36}{-24}=\frac{36 \div(-12)}{-24 \div(-12)}=\frac{-3}{2}
$
(ii) The HCF of 3 and 15 is 3 .
Thus, $\frac{-3}{-15}=\frac{-3 \div(-2)}{-15 \div(-3)}=\frac{1}{5}$

Example 3

Do $\frac{4}{-9}$ and $\frac{-16}{36}$ represent the same rational number?
Solution

Yes, because $\frac{4}{-9}=\frac{4 \times(-4)}{9 \times(-4)}=\frac{-16}{36}$ or $\frac{-16}{36}=\frac{-16+-4}{35 \div-4}=\frac{-4}{-9}$.

Example 4

List three rational numbers between-2 and -1 .
Solution

Let us write -1 and -2 as rational numbers with denominator 5. (Why?)
We have, $-1=\frac{-5}{5}$ and $-2=\frac{-10}{5}$
So, $\quad \frac{-10}{5}<\frac{-9}{5}<\frac{-8}{5}<\frac{-7}{5}<\frac{-6}{5}<\frac{-5}{5}$ or $-2<\frac{-9}{5}<\frac{-8}{5}<\frac{-7}{5}<\frac{-6}{5}<-1$
The three rational numbers between -2 and -1 would be, $\frac{-9}{5}, \frac{-8}{5}, \frac{-7}{5}$
(You can take any three of $\frac{-9}{5}, \frac{-8}{5}, \frac{-7}{5}, \frac{-6}{5}$ )

Example 5

Write four more numbers in the following pattern:
$
\frac{-1}{3}, \frac{-2}{6}, \frac{-3}{9}, \frac{-4}{12}, \ldots
$

Solution

Wehave,
or
$
\begin{aligned}
& \frac{-2}{6}=\frac{-1 \times 2}{3 \times 2}, \frac{-3}{9}=\frac{-1 \times 3}{3 \times 3}, \frac{-4}{12}=\frac{-1 \times 4}{3 \times 4} \\
& \frac{-1 \times 1}{3 \times 1}=\frac{-1}{3}, \frac{-1 \times 2}{3 \times 2}=\frac{-2}{6}, \frac{-1 \times 3}{3 \times 3}=\frac{-3}{9}, \frac{-1 \times 4}{3 \times 4}=\frac{-4}{12}
\end{aligned}
$

Thus, we observe a pattern in these numbers.
The other numbers would be $\frac{-1 \times 5}{3 \times 5}=\frac{-5}{15}, \frac{-1 \times 6}{3 \times 6}=\frac{-6}{18}, \frac{-1 \times 7}{3 \times 7}=\frac{-7}{21}$.

Example 6

Satpal walks $\frac{2}{3} \mathrm{~km}$ from a place P, towards east and then from there $1 \frac{5}{7} \mathrm{~km}$ towards west. Where will he be now from $\mathrm{P}$ ?

Solution

Let us denote the distance travelled towards east by positive sign. So, the distances towards west would be denoted by negative sign.

Thus, distance of Satpal from the point $P$ would be

$
\begin{aligned}
\frac{2}{3}+\left(-1 \frac{5}{7}\right) & =\frac{2}{3}+\frac{(-12)}{7}=\frac{2 \times 7}{3 \times 7}+\frac{(-12) \times 3}{7 \times 3} \\
= & \frac{14-36}{21}=\frac{-22}{21}=-1 \frac{1}{21}
\end{aligned}
$

Since it is negative, it means Satpal is at a distance $1 \frac{1}{21} \mathrm{~km}$ towards west of $\mathrm{P}$.