Exercise 9.1

"Rational Numbers" Chapter 9 - Introduction - NCERT Class 7th Maths Solutions

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Q 1, Ex 9.1 - Rational Numbers - Chapter 9 - Maths Class 7th - NCERT

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QUESTION 1
List five rational numbers between:
(i) -1 and 0
Sol :
Let us write — 1 and 0 as rational numbers with denominator 6
Here we take 6 because we want five rational number
⇒$-1 = \dfrac{-6}{\phantom{-}6}$ and $0=\dfrac{0}{6}$
⇒$\dfrac{-6}{\phantom{-}6}<\dfrac{-5}{\phantom{-}6}<\dfrac{-4}{\phantom{-}6}<\dfrac{-3}{\phantom{-}6}<\dfrac{-2}{\phantom{-}6}<\dfrac{-1}{\phantom{-}6}<\dfrac{0}{6}$
⇒$-1<\dfrac{-5}{\phantom{-}6}<\dfrac{-4}{\phantom{-}6}<\dfrac{-3}{\phantom{-}6}<\dfrac{-2}{\phantom{-}6}<\dfrac{-1}{\phantom{-}6}<0$

(ii) -2 and -1
Sol :
Let us write —2 and — 1 as rational numbers with denominator 6.
⇒$-2 = \dfrac{-12}{\phantom{-}6}$ and $-1=\dfrac{-6}{\phantom{-}6}$
⇒$\dfrac{-12}{\phantom{-}6}<\dfrac{-11}{\phantom{-}6}<\dfrac{-10}{\phantom{-}6}<\dfrac{-9}{\phantom{-}6}<\dfrac{-8}{\phantom{-}6}<\dfrac{-7}{\phantom{-}6}<\dfrac{-6}{\phantom{-}6}$
⇒$-2<\dfrac{-11}{\phantom{-}6}<\dfrac{-10}{\phantom{-}6}<\dfrac{-9}{\phantom{-}6}<\dfrac{-8}{\phantom{-}6}<\dfrac{-7}{\phantom{-}6}<-1$

(iii) $\dfrac{-4}{\phantom{-}5} \text{ and } \dfrac{-2}{\phantom{-}3}$
Sol :
Lets make the denominator same
On taking H.C.F of 3 and 5 is 15 . So , denominator have to be 15 and to make denominator 15 we multiply $\dfrac{3}{3}$ on L.H.S and by $\dfrac{5}{5}$ on R.H.S
⇒ $\dfrac{-4}{\phantom{-}5}\times \dfrac{3}{3}=\dfrac{-12}{\phantom{-}15}$ and $\dfrac{-2}{\phantom{-}3}\times \dfrac{5}{5}=\dfrac{-10}{\phantom{-}15}$
⇒ $\dfrac{-12}{\phantom{-}15}\times \dfrac{3}{3}$ and $\dfrac{-10}{\phantom{-}15}\times \dfrac{3}{3}$ [Multiplying both sides by $\dfrac{3}{3}$ to get five rational numbers ]
⇒ $\dfrac{-36}{\phantom{-}45}$ and $\dfrac{-31}{\phantom{-}45}$
Five rational numbers are
⇒ $\dfrac{-35}{\phantom{-}45},\dfrac{-34}{\phantom{-}45},\dfrac{-33}{\phantom{-}45},\dfrac{-32}{\phantom{-}45},\dfrac{-31}{\phantom{-}45}$

(iv) $\dfrac{1}{2} \text{ and } \dfrac{2}{3}$
Sol :
H.C.F of 2 and 3 is 6 . So , the denominator have to be 6
⇒ $\dfrac{1}{2}\times \dfrac{3}{3}=\dfrac{3}{6}$ and $\dfrac{2}{3}\times\dfrac{2}{2}=\dfrac{4}{6}$
⇒ $\dfrac{3}{6}\times \dfrac{6}{6}=\dfrac{18}{36}$ and $\dfrac{4}{6}\times \dfrac{6}{6}=\dfrac{24}{36}$
⇒ $\dfrac{19}{36},\dfrac{20}{36} , \dfrac{21}{36} , \dfrac{22}{36}, \dfrac{23}{36}$

Q 2, Ex 9.1 - Rational Numbers - Chapter 9 - Maths Class 7th - NCERT

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QUESTION 2
Write four more rational numbers in each of the following patterns:
(i) $\dfrac{-3}{\phantom{-}5},\dfrac{-6}{\phantom{-}10},\dfrac{-9}{\phantom{-}15},\dfrac{-12}{\phantom{-}20}\dots$
Sol : $\dfrac{-3}{\phantom{-}5}\times \dfrac{1}{1},\dfrac{-3}{\phantom{-}5}\times \dfrac{2}{2},\dfrac{-3}{\phantom{-}5}\times \dfrac{3}{3},\dfrac{-3}{\phantom{-}5}\times \dfrac{4}{4}\dots$
It can be observed that the numerator is a multiple of 3 while the denominator is a multiple of 5 and as we increase them further, these multiples are increasing. Therefore, the next four rational numbers in this pattern are
$\dfrac{-3}{\phantom{-}5}\times \dfrac{5}{5},\dfrac{-3}{\phantom{-}5}\times \dfrac{6}{6},\dfrac{-3}{\phantom{-}5}\times \dfrac{7}{7},\dfrac{-3}{\phantom{-}5}\times \dfrac{8}{8}\dots$
$\dfrac{-15}{\phantom{-}25},\dfrac{-18}{\phantom{-}30},\dfrac{-21}{\phantom{-}35},\dfrac{-24}{\phantom{-}40}\dots$

(ii) $\dfrac{-1}{\phantom{-}4},\dfrac{-2}{\phantom{-}8},\dfrac{-3}{\phantom{-}12}\dots$
Sol :
$\dfrac{-1}{\phantom{-}4}\times \dfrac{1}{1}, \dfrac{-1}{\phantom{-}4} \times \dfrac{2}{2}, \dfrac{-1}{\phantom{-}4} \times \dfrac{3}{3} ,\dots$
The next four rational numbers in this patterns are
$ \dfrac{-1}{\phantom{-}4} \times \dfrac{4}{4} , \dfrac{-1}{\phantom{-}4} \times \dfrac{5}{5} , \dfrac{-1}{\phantom{-}4} \times \dfrac{6}{6} , \dfrac{-1}{\phantom{-}4} \times \dfrac{7}{7}$
$\dfrac{-4}{\phantom{-}16} , \dfrac{-5}{\phantom{-}20} , \dfrac{-6}{\phantom{-}24} , \dfrac{-7}{\phantom{-}28} \dots$

(iii) $\dfrac{-1}{\phantom{-}6},\dfrac{2}{-12},\dfrac{3}{-18},\dfrac{4}{-24}\dots$
Sol :
$\dfrac{-1}{\phantom{-}6}\times \dfrac{1}{1} , \dfrac{-1}{\phantom{-}6}\times \dfrac{2}{2} , \dfrac{-1}{\phantom{-}6}\times \dfrac{3}{3} , \dfrac{-1}{\phantom{-}6}\times \dfrac{4}{4}\dots$
The next four rational numbers in this pattern are
$\dfrac{-1}{\phantom{-}6}\times \dfrac{5}{5} , \dfrac{-1}{\phantom{-}6}\times \dfrac{6}{6} , \dfrac{-1}{\phantom{-}6}\times \dfrac{7}{7} , \dfrac{-1}{\phantom{-}6}\times \dfrac{8}{8}\dots $
$\dfrac{5}{-30} , \dfrac{6}{-36} , \dfrac{7}{-42} , \dfrac{8}{-48}\dots$

(iv) $\dfrac{-2}{\phantom{-}3},\dfrac{2}{-3},\dfrac{4}{-6},\dfrac{6}{-9}\dots$
Sol :
$\dfrac{-2}{\phantom{-}3} , \dfrac{2}{-3} , \dfrac{2}{-3}\times \dfrac{2}{2} , \dfrac{-2}{\phantom{-}3} \times \dfrac{3}{3}\dots$
The next four rational numbers in this pattern are
$\dfrac{2}{-3}\times \dfrac{4}{4} , \dfrac{2}{-3}\times \dfrac{5}{5} , \dfrac{2}{-3}\times \dfrac{6}{6} , \dfrac{2}{-3}\times \dfrac{7}{7} \dots$
$\dfrac{8}{-12} , \dfrac{10}{-15} , \dfrac{12}{-18} , \dfrac{14}{-21} \dots$

Q 3, Ex 9.1 - Rational Numbers - Chapter 9 - Maths Class 7th - NCERT

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QUESTION 3
Give four rational numbers equivalent to:
(i) $\dfrac{-2}{7}$
Sol :
$\dfrac{-2}{7}\times \dfrac{2}{2} , \dfrac{-2}{7} \times \dfrac{3}{3} , \dfrac{-2}{7} \times \dfrac{4}{4} , \dfrac{-2}{7} \times \dfrac{5}{5}$
$\dfrac{-4}{\phantom{-}14} , \dfrac{-6}{\phantom{-}21} , \dfrac{-8}{\phantom{-}28} , \dfrac{-10}{\phantom{-}35}$

(ii) $\dfrac{5}{-3}$
Sol :
$\dfrac{5}{-3}\times \dfrac{2}{2} , \dfrac{5}{-3} \times \dfrac{3}{3} , \dfrac{5}{-3} \times \dfrac{4}{4} , \dfrac{5}{-3} \times \dfrac{5}{5}$
$\dfrac{10}{-6} , \dfrac{15}{-9} , \dfrac{20}{-12} , \dfrac{25}{-15}$

(iii) $\dfrac{4}{9}$
Sol :
$\dfrac{4}{9}\times \dfrac{2}{2} , \dfrac{4}{9} \times \dfrac{3}{3} , \dfrac{4}{9} \times \dfrac{4}{4} , \dfrac{4}{9} \times \dfrac{5}{5}$

Q 4, Ex 9.1 - Rational Numbers - Chapter 9 - Maths Class 7th - NCERT

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QUESTION 4
Draw the number line and represent the following rational numbers on it:
(i) $\dfrac{3}{4}$
Sol :
This fraction represents 3 parts out of 4 equal parts. Therefore, each space between two integers on number line must be divided into 4 equal parts.

can be represented as

(ii) $\dfrac{-5}{\phantom{-}8}$
Sol :
This fraction represents 5 parts out of 8 equal parts. Negative sign represents that it is on the negative side of number line. Therefore, each space between two integers on number line must be divided into 8 equal parts.

can be represented as

(iii) $\dfrac{-7}{\phantom{-}4}$
Sol :
This fraction represents 1 full part and 3 parts out of 4 equal parts. Negative sign represents that it is on the negative side of number line. Therefore, each space between two integers on number line must be divided into 4 equal parts.

can be represented as

(iv) $\dfrac{7}{8}$
Sol :
This fraction represents 7 parts out of 8 equal parts. Therefore, each space between two integers on number line must be divided into 8 equal parts.

can be represented as

QUESTION 5
The points P, Q, R, S, T, U, A and B on the number line are such that,
TR = RS = SU and AP = PQ = QB. Name the rational numbers represented by P, Q, R and S.