ncert-solutions-class-7-maths-ch-9-rational-numbers

NCERT Solutions for Class 7 Maths Chapter 9Â Rational Numbers

NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.3

Question:1 Is the number $\frac{2}{-3}$ rational? Think about it.

Answer:

Yes , $\frac{2}{-3}$ is a rational number because it is written in the form: $\frac{p}{q}$ , where $p = 2\ and\ q = -3\neq 0$ .

Question:2 List ten rational numbers.

Answer:

Any ten rational numbers are:

$1, \frac{2}{3}, -\frac{1}{3}, -5, 0, 356, -\frac{39}{25}, -36, 1999, \frac{-1}{-2}$

Question: Fill in the boxes:

(i) $\frac{5}{4}=\frac{\square }{16}=\frac{25}{\square }=\frac{-15}{\square }$ (ii) $\frac{-3}{7}=\frac{\square }{14}=\frac{9}{\square }=\frac{-6}{\square }$

Answer:

(i) $\frac{5}{4}=\frac{\square }{16}=\frac{25}{\square }=\frac{-15}{\square }$

$\frac{5}{4}$ can be written as:

$\Rightarrow \frac{5}{4} = \frac{5\times4}{4\times4} = \frac{20}{16}$

$\Rightarrow \frac{5}{4} = \frac{5\times5}{4\times5} = \frac{25}{20}$

$\Rightarrow \frac{5}{4} = \frac{5\times-3}{4\times-3} = \frac{-15}{-12}$

Hence, we have

$\frac{5}{4}=\frac{20 }{16}=\frac{25}{20 }=\frac{-15}{-12 }$

(ii) $\frac{-3}{7}=\frac{\square }{14}=\frac{9}{\square }=\frac{-6}{\square }$

$\frac{-3}{7}$ can be written as:

$\Rightarrow \frac{-3}{7} = \frac{-3\times2}{7\times2} = \frac{-6}{14}$

$\Rightarrow \frac{-3}{7} = \frac{-3\times-3}{7\times-3} = \frac{9}{-21}$

$\Rightarrow \frac{-3}{7} = \frac{-3\times2}{7\times2} = \frac{-6}{14}$

Hence, we have

$\frac{-3}{7}=\frac{-6 }{14}=\frac{9}{-21}=\frac{-6}{14 }$

NCERT Solutions for class 7 maths chapter 9 topic 9.4

Question:1 Is 5 a positive rational number?

Answer:

Yes , 5 can be written as a positive rational number $\frac{5}{1}$ , where 5 and 1 are both positive integers and denominator not equal to zero.

Question:2 List five more positive rational numbers.

Answer:

Five more positive rational numbers are:

$4,\ \frac{1}{3},\ 56,\ \frac{99}{98}, 5$

Question:1 Is – 8 a negative rational number?

Answer:

Yes , is a negative rational number because it can be written as $\frac{-8}{1}$ , where the numerator is negative integer and denominator is a positive integer.

Question:2 List five more negative rational numbers.

Answer:

Five more negative rational numbers are:

$-1,\ -99,\ -\frac{2}{3},\ -\frac{5}{7}, \ -27$

Question: Which of these are negative rational numbers?

(i) $\frac{-2}{3}$ (ii) $\frac{5}{7}$ (iii) $\frac{3}{-5}$ (iv) 0 (v) $\frac{6}{11}$ (vi) $\frac{-2}{-9}$

Answer:

(i) $\frac{-2}{3}$ here, the numerator is -2 which is negative and the denominator is 3 which is positive.

Hence, the fraction is negative.

(ii) $\frac{5}{7}$ here, the numerator is 5 which is positive and the denominator is 7 which is also positive.

Hence, the fraction is positive.

(iii) $\frac{3}{-5}$ here, the numerator is 3 which is positive and the denominator is -5 which is negative.

Hence, the fraction is negative.

(iv) 0 zero is neither positive nor a negative number.

(v) $\frac{6}{11}$ here, the numerator is 6 which is positive and the denominator is 11 which is also positive.

Hence, the fraction is positive.

(vi) $\frac{-2}{-9}$ here, the numerator is -2 which is negative and the denominator is -9 which is also a negative integer.

Hence, the fraction is overall a positive fraction.

NCERT Solutions for class 7 maths chapter 9 topic 9.6

Question: Find the standard form of

(i) $\frac{-18}{45}$ (ii) $\frac{-12}{18}$

Answer:

(i) Given fraction $\frac{-18}{45}$ .

We can make it in the standard form :

$\frac{-18}{45} = \frac{-6\times3}{15\times3} = \frac{-2\times3\times3}{5\times3\times3} = \frac{-2}{5}$

(i) Given fraction $\frac{-12}{18}$ .

We can make it in the standard form :

$\frac{-12}{18} = \frac{-6\times2}{9\times2} = \frac{-2\times3\times2}{3\times3\times2} = \frac{-2}{3}$

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.8

Question: Find five rational numbers between

$\frac{-5}{7} and \frac{-3}{8}$

Answer:

LCM of 7 and 8 is 56.

Hence we can write given fractions as:

$\frac{-5}{7} = \frac{-5\times8}{7\times8} = \frac{-40}{56}$ and $\frac{-3}{8} = \frac{-3\times7}{8\times7} = \frac{-21}{56}$

Therefore, we can find five rational numbers between $\frac{-5}{7} and \frac{-3}{8}$ .

$\frac{-39}{56},\ \frac{-38}{56},\ \frac{-37}{56},\ \frac{-36}{56},\ and\ \frac{-35}{56}$

NCERT Solutions for class 7 maths chapter 9 rational numbers exercise 9.1

Question: 1(i) List five rational numbers between:

–1 and 0

Answer:

To find five rational numbers between $-1\ and\ 0$ we will convert each rational numbers as a denominator, we have

$-1 = \frac{-1\times6}{6} = \frac{-6}{6}\ and\ \frac{0\times6}{6} = \frac{0}{6}$

So, we have five rational numbers between $\frac{-6}{6}\ and\ \frac{0}{6}$

$\frac{-6}{6}<\frac{-5}{6}<\frac{-4}{6}<\frac{-3}{6}<\frac{-2}{6}<\frac{-1}{6}<\frac{0}{6}$

Hence, the five rational numbers between -1 and 0 are:

$\frac{-5}{6},\frac{-4}{6},\frac{-3}{6},\frac{-2}{6}\ and\ \frac{-1}{6}.$

Question: 1(ii) List five rational numbers between:

–2 and –1

Answer:

To find five rational numbers between $-2\ and\ -1$ we will convert each rational numbers as a denominator , we have

$-2 = \frac{-2\times6}{6} = \frac{-12}{6}\ and\ \frac{-1\times6}{6} = \frac{-6}{6}$

So, we have five rational numbers between $\frac{-12}{6}\ and\ \frac{-6}{6}$

$\frac{-12}{6}<\frac{-11}{6}<\frac{-10}{6}<\frac{-9}{6}<\frac{-8}{6}<\frac{-7}{6}<\frac{-6}{6}$

Hence, the required rational numbers are

$\frac{-11}{6},\frac{-5}{3},\frac{-3}{2},\frac{-4}{3}\ and\ \frac{-7}{6}.$

Question: 1(iii) List five rational numbers between:

$\frac{-4}{5}and \frac{-2}{3}$

Answer:

To find five rational numbers between $\frac{-4}{5}and \frac{-2}{3}$ we will convert each rational numbers with the denominator as $5\times3 =15$ , we have $\left ( \because LCM\ of\ 5\ and\ 3 = 15 \right )$

Since there is only one integer i.e., -11 between -12 and -10, we have to find equivalent rational numbers.

Now, we have five rational numbers possible:

$\therefore \frac{-36}{45}<\frac{-35}{45}<\frac{-34}{45}<\frac{-33}{45}<\frac{-32}{45}<\frac{-31}{45}<\frac{-30}{45}$

Hence, the required rational numbers are

$\frac{-7}{9},\frac{-34}{45},\frac{-11}{15},\frac{-32}{45}\ and\ \frac{-31}{45}.$

Question: 1(iv) List five rational numbers between:

$-\frac{1}{2} and \frac{2}{3}$

Answer:

To find five rational numbers between $-\frac{1}{2} and \frac{2}{3}$ we will convert each rational numbers in their equivalent numbers, we have

Making denominator as LCM(2,3)=6

that is

$\frac{-3}{6}\ and\ \frac{4}{6}$

Now, we have five rational numbers possible:

$\therefore \frac{-3}{6}<\frac{-2}{6}<\frac{-1}{6}<\frac{0}{6}<\frac{2}{6}<\frac{3}{6}<\frac{4}{6}$

Hence, the required rational numbers are

$\frac{-1}{3},\frac{-1}{6},0,\frac{1}{3}\ and\ \frac{1}{2}.$

Question: 2(i) Write four more rational numbers in each of the following patterns:

$\frac{-3}{5}, \frac{-6}{10}, \frac{-9}{15}, \frac{-12}{20}$

Answer:

We have the pattern:

$\frac{-3}{5} = \frac{-3\times1}{5\times1}$ $\ \frac{-6}{10} = \frac{-3\times2}{5\times2}$ $\frac{-9}{15} = \frac{-3\times3}{5\times3}$ $\frac{-12}{20} = \frac{-3\times4}{5\times4}$

Now, following the same pattern, we have

$\frac{-3\times5}{5\times5} = \frac{-15}{25}$ $\frac{-3\times6}{5\times6} = \frac{-18}{30}$ $\frac{-3\times7}{5\times7} = \frac{-21}{35}$ $\frac{-3\times8}{5\times8} = \frac{-24}{40}$

Hence, the required rational numbers are:

$\frac{-15}{25},\ \frac{-18}{30},\ \frac{-21}{35},and\ \frac{-24}{40}.$

Question: 2(ii) Write four more rational numbers in each of the following patterns:

$\frac{-1}{4},\frac{-2}{8}, \frac{-3}{12}....$

Answer:

We have the pattern:

$\frac{-1}{4} = \frac{-1\times1}{4\times1}$ $\frac{-2}{8} = \frac{-1\times2}{4\times2}$ $\frac{-3}{12} = \frac{-1\times3}{4\times3}$

Now, following the same pattern, we have

$\frac{-1\times4}{4\times4} = \frac{-4}{16}$ $\frac{-1\times5}{4\times5} = \frac{-5}{20}$ $\frac{-1\times6}{4\times6} = \frac{-6}{24}$ $\frac{-1\times7}{4\times7} = \frac{-7}{28}$

Hence, the required rational numbers are:

$\frac{-4}{16},\ \frac{-5}{20},\ \frac{-6}{24},and\ \frac{-7}{28}.$

Question: 2(iii) Write four more rational numbers in each of the following patterns:

$\frac{-1}{6}, \frac{-2}{12}, \frac{-3}{18}, \frac{-4}{24}....$

Answer:

We have the pattern:

$\frac{-1}{6} = \frac{-1\times1}{6\times1}$ $\frac{-2}{12} = \frac{-1\times2}{6\times2}$ $\frac{-3}{18} = \frac{-1\times3}{6\times3}$ $\frac{-4}{24} = \frac{-1\times4}{6\times4}$

Now, following the same pattern, we have

$\frac{-1\times5}{6\times5} = \frac{-5}{30}$ $\frac{-1\times6}{6\times6} = \frac{-6}{36}$ $\frac{-1\times7}{6\times7} = \frac{-7}{42}$ $\frac{-1\times8}{6\times8} = \frac{-8}{48}$

Hence, the required rational numbers are:

$\frac{-5}{30},\ \frac{-6}{36},\ \frac{-7}{42},and\ \frac{-8}{48}.$

Question: 2(iv) Write four more rational numbers in each of the following patterns:

$\frac{-2}{3}, \frac{2}{-3},\frac{4}{-6}, \frac{6}{-9}....$

Answer:

We have the pattern:

$\frac{-2}{3} = \frac{-2\times1}{3\times1}$ $\frac{2}{-3} =\frac{2}{-3} = \frac{2\times1}{-3\times1}$ $\frac{4}{-6} = \frac{-2\times2}{3\times2}$ $\frac{-6}{9} = \frac{-2\times3}{3\times3}$

Now, following the same pattern, we have

$\frac{-2\times4}{3\times4} = \frac{-8}{12}\ or\ \frac{8}{-12}$ $\frac{-2\times5}{3\times5} = \frac{-10}{15}\ or\ \frac{10}{-15}$

$\frac{-2\times6}{3\times6} = \frac{-12}{18}\ or\ \frac{12}{-18}$ $\frac{-2\times7}{3\times7} = \frac{-14}{21}\ or\ \frac{14}{-21}$

Hence, the required rational numbers are:

$\frac{8}{-12},\ \frac{10}{-15},\ \frac{12}{-18},and\ \frac{14}{-21}.$

Question: 3(i) Give four rational numbers equivalent to:

$\frac{-2}{7}$

Answer:

$\frac{-2}{7}$ can be written as:

$\frac{-2}{7} = \frac{-2\times2}{7\times2} = \frac{-4}{14}$ $\frac{-2}{7} = \frac{-2\times3}{7\times3} = \frac{-6}{21}$

$\frac{-2}{7} = \frac{-2\times4}{7\times4} = \frac{-8}{28}$ $\frac{-2}{7} = \frac{-2\times5}{7\times5} = \frac{-10}{35}$

Hence, the required equivalent rational numbers are

$\frac{-4}{14},\frac{-6}{21}, \frac{-8}{28},\ and\ \frac{-10}{35} .$

Question: 3(ii) Give four rational numbers equivalent to:

$\frac{5}{-3}$

Answer:

$\frac{5}{-3}$ can be written as:

$\frac{5}{-3} = \frac{5\times2}{-3\times2} = \frac{10}{-6}$ $\frac{5}{-3} = \frac{5\times3}{-3\times3} = \frac{15}{-9}$

$\frac{5}{-3} = \frac{5\times4}{-3\times4} = \frac{20}{-12}$ $\frac{5}{-3} = \frac{5\times5}{-3\times5} = \frac{25}{-15}$

Hence, the required equivalent rational numbers are

$\frac{10}{-6},\frac{15}{-9}, \frac{20}{-12},\ and\ \frac{25}{-20} .$

Question: 3(iii) Give four rational numbers equivalent to:

$\frac{4}{9}$

Answer:

$\frac{4}{9}$ can be written as:

$\frac{4}{9} = \frac{4\times2}{9\times2} = \frac{8}{18}$ $\frac{4}{9} = \frac{4\times3}{9\times3} = \frac{12}{27}$

$\frac{4}{9} = \frac{4\times4}{9\times4} = \frac{16}{36}$ $\frac{4}{9} = \frac{4\times5}{9\times5} = \frac{20}{45}$

Hence, the required equivalent rational numbers are

$\frac{8}{18},\frac{12}{27}, \frac{16}{36},\ and\ \frac{20}{45} .$

Question: 4(i) Draw the number line and represent the following rational numbers on it:

$\frac{3}{4}$

Answer:

Representation of $\frac{3}{4}$ on the number line,

Question: 4(ii) Draw the number line and represent the following rational numbers on it:

$\frac{-5}{8}$

Answer:

Representation of $\frac{-5}{8}$ on the number line,

Question: 4(iii) Draw the number line and represent the following rational numbers on it:

$\frac{-7}{4}$

Answer:

Representation of $\frac{-7}{4}$ on the number line,

Question: 4(iv) Draw the number line and represent the following rational numbers on it:

<img alt="\frac{7}{8}" height="38"

src="https://lh3.googleusercontent.com/hdFvFOUpqFeXVZ5q51jd9MXrsnpVZQmJerZr4xLLcnglNdRFCv5zpxGJBVRV7CUqDUifhs_mMsUDTWdBaO936vf2uwSLSRITVd8Qa39KQOwVFzIzRt5ceQXn3dwE1np9Tz18iYY" style="margin-left: 0px; margin-top: 0px;" width="9" />

Answer:

Representation of $\frac{7}{8}$ on the number line,

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.

Answer:

Given TR = RS = SU and AP = PQ = QB then, we have

There are two rational numbers between A and B i.e., P and Q which are at equal distances hence,

The rational numbers represented by P and Q are:

$P =2+ \frac{1}{3} = \frac{7}{3}\ and\ Q = 2+\frac{2}{3} = \frac{8}{3}$

Also, there are two rational numbers between U and T i.e., S and R which are at equal distances hence,

The rational numbers represented by S and R are:

$S =-2+ \frac{1}{3} = \frac{-5}{3}\ and\ R = -2+\frac{2}{3} = \frac{-4}{3}$

Question: 6 Which of the following pairs represent the same rational number?

(i) $\frac{-7}{21}\ and\ \frac{3}{9}$ (ii) $\frac{-16}{20} \ and\ \frac{20}{-25}$

(iii) $\frac{-2}{-3} \ and\ \frac{2}{3}$ (iv) $\frac{-3}{5}\ and\ \frac{-12}{20}$

(v) $\frac{8}{-5}\ and\ \frac{-24}{15}$ (vi) $\frac{1}{3}\ and\ \frac{-1}{9}$

(vii) $\frac{-5}{-9}\ and\ \frac{5}{-9}$

Answer:

To compare we multiply both numbers with denominators:

(i) We have $\frac{-7}{21}\ and\ \frac{3}{9}$

$\Rightarrow \frac{-7\times9}{21\times9} = \frac{-63}{189}$

$\Rightarrow \frac{3\times21}{9\times21} = \frac{63}{189}$

$\Rightarrow \frac{-63}{189} \neq \frac{63}{189}$

Here, they are equal but are in opposite signs hence, $\frac{-7}{21}\ and\ \frac{3}{9}$ do not represent the same rational numbers.

(ii) We have $\frac{-16}{20} \ and\ \frac{20}{-25}$

$\Rightarrow \frac{-16\times-25}{20\times-25} = \frac{400}{-500}$

$\Rightarrow \frac{20\times20}{-25\times20} = \frac{400}{-500}$

$\Rightarrow \frac{400}{-500} = \frac{400}{-500}$

So, they represent the same rational number.

(iii) We have $\frac{-2}{-3} \ and\ \frac{2}{3}$

Here, Both represents the same number as these minus signs on both numerator and denominator of $\frac{-2}{-3} = \frac{2}{3}$ will cancel out and gives the positive value.

(iv) We have $\frac{-3}{5}\ and\ \frac{-12}{20}$

$\Rightarrow \frac{-3\times20}{5\times20} = \frac{-60}{100}$

$\Rightarrow \frac{-12\times5}{20\times5} = \frac{-60}{100}$

$\Rightarrow \frac{-60}{100} = \frac{-60}{100}$

So, they represent the same rational number.

(v) We have $\frac{8}{-5}\ and\ \frac{-24}{15}$

$\Rightarrow \frac{8\times15}{-5\times15} = \frac{120}{-75}$

$\Rightarrow \frac{-24\times-5}{15\times-5} = \frac{120}{-75}$

$\Rightarrow \frac{120}{-75} = \frac{120}{-75}$

So, they represent the same rational number.

(vi) We have $\frac{1}{3}\ and\ \frac{-1}{9}$

$\Rightarrow \frac{1\times9}{3\times9} = \frac{9}{27}$

$\Rightarrow \frac{-1\times3}{9\times3} = \frac{-3}{27}$

$\Rightarrow \frac{9}{27} \neq \frac{-3}{27}$

So, They do not represent the same rational number.

(vii) We have $\frac{-5}{-9}\ and\ \frac{5}{-9}$

Here, the denominators of both are the same but $-5 \neq 5$ .

So, $\frac{-5}{-9}\ and\ \frac{5}{-9}$ do not represent the same rational numbers.

Question: 7 Rewrite the following rational numbers in the simplest form:

(i) $\frac{-8}{6}$ (ii) $\frac{22}{25}$ (iii) $\frac{-44}{72}$ (iv) $\frac{-8}{10}$

Answer:

(i) $\frac{-8}{6}$ can be written as:

$\Rightarrow \frac{-8}{6} = \frac{-8/2}{6/2}= \frac{-4}{3}$ $\left [ HCF\ of\ 8\ and\ 6\ is\ 2 \right ]$

(ii) $\frac{25}{45}$ can be written in the simplest form:

$\Rightarrow \frac{25}{45} = \frac{25/5}{45/5}= \frac{5}{9}$ $\left [ HCF\ of\ 25\ and\ 45\ is\ 5 \right ]$

(iii) $\frac{-44}{72}$ can be written as in simplest form:

$\Rightarrow \frac{-44}{72} = \frac{-44/4}{72/4}= \frac{-11}{18}$ $\left [ HCF\ of\ 44\ and\ 72\ is\ 4 \right ]$

Question: 8 Fill in the boxes with the correct symbol out of >, <, and =.

(i) $\frac{-5}{7} \square \frac{2}{3}$ (ii) $\frac{-4}{5} \square \frac{-5}{7}$ (iii) $\frac{-7}{8} \square \frac{14}{-16}$

(iv) $\frac{-8}{5} \square \frac{-7}{4}$ (v) $\frac{1}{-3} \square \frac{-1}{4}$ (vi) $\frac{5}{-11} \square \frac{-5}{11}$

(vii) $0 \square \frac{-7}{6}$

Answer:

(i) $\frac{-5}{7} \square \frac{2}{3}$

$\Rightarrow \frac{-5\times3}{7\times 3}\square \frac{2\times7}{3\times7}$

$\Rightarrow \frac{-15}{21} < \frac{14}{21}$

Hence, $\frac{-5}{7} < \frac{2}{3}$

(ii) $\frac{-4}{5} \square \frac{-5}{7}$

$\Rightarrow \frac{-4\times7}{5\times 7}\square \frac{-5\times5}{7\times5}$

$\Rightarrow \frac{-28}{35} < \frac{-25}{35}$

Hence, $\frac{-4}{5} < \frac{-5}{7}$

(iii) $\frac{-7}{8} \square \frac{14}{-16}$

$\Rightarrow \frac{-7\times-16}{8\times -16}\square \frac{14\times8}{-16\times8}$

$\Rightarrow \frac{112}{-128} = \frac{112}{-128}$

Hence, $\frac{-7}{8} = \frac{14}{-16}$

(iv) $\frac{-8}{5} \square \frac{-7}{4}$

$\Rightarrow \frac{-8\times4}{5\times 4}\square \frac{-7\times5}{4\times5}$

$\Rightarrow \frac{-32}{20} > \frac{-35}{20}$

Hence, $\frac{-8}{5} > \frac{-7}{4}$

(v) $\frac{1}{-3} \square \frac{-1}{4}$

$\Rightarrow \frac{1\times4}{-3\times 4}\square \frac{-1\times-3}{4\times-3}$

$\Rightarrow \frac{4}{-12} < \frac{3}{-12}$

Hence, $\frac{1}{-3} < \frac{1}{-4}$

(vi) $\frac{5}{-11} \square \frac{-5}{11}$

$\Rightarrow \frac{5\times11}{-11\times 11}\square \frac{-5\times-11}{11\times-11}$

$\Rightarrow \frac{55}{-121} = \frac{55}{-121}$

Hence, $\frac{5}{-11} = \frac{-5}{11}$

(viI) $0 \square \frac{-7}{6}$

Zero is always greater than every negative number.

Therefore, $0 > \frac{-7}{6}$

Question: 9 Which is greater in each of the following:

(i) $\frac{2}{3}, \frac{5}{2}$ (ii) $\frac{-5}{6}, \frac{-4}{3}$

(iii) $\frac{-3}{4}, \frac{2}{-3}$ (iv) $\frac{-1}{4}, \frac{1}{4}$

(v) $-3\frac{2}{7}, -3\frac{4}{5}$

Answer:

(i) $\frac{2}{3}, \frac{5}{2}$

$\Rightarrow \frac{2\times2}{3\times2}, \frac{5\times3}{2\times3}$

$\Rightarrow \frac{4}{6}, \frac{15}{6}$

Since, $\frac{15}{6}> \frac{4}{6}$

So, $\frac{5}{2}> \frac{2}{3}.$

(ii) $\frac{-5}{6}, \frac{-4}{3}$

$\Rightarrow \frac{-5\times3}{6\times3}, \frac{-4\times6}{3\times6}$

$\Rightarrow \frac{-15}{18}, \frac{-24}{18}$

Since, $\frac{-15}{18}> \frac{-24}{18}$

So, $\frac{-5}{6}> \frac{-4}{3}.$

(iii) $\frac{-3}{4}, \frac{2}{-3}$

$\Rightarrow \frac{-3\times-3}{4\times-3}, \frac{2\times4}{-3\times4}$

$\Rightarrow \frac{9}{-12}, \frac{8}{-12}$

Since, $\frac{8}{-12}> \frac{9}{-12}$

So, $\frac{2}{-3}> \frac{-3}{4}$

(iv) $\frac{-1}{4}, \frac{1}{4}$

$\Rightarrow \frac{1}{4}> \frac{-1}{4}$

As each positive number is greater than its negative.

(v) $-3\frac{2}{7}, -3\frac{4}{5}$

$\Rightarrow \frac{-23}{7}, \frac{-19}{5} =\frac{-23\times5}{7\times5}, \frac{-19\times7}{5\times7}$

$\Rightarrow \frac{-115}{35} > \frac{-133}{35}$

So, $-3\frac{2}{7}> -3\frac{4}{5}$

Question: 10(i) Write the following rational numbers in ascending order:

$\frac{-3}{5}, \frac{-2}{5}, \frac{-1}{5}$

Answer:

(i) Here the denominator value is the same.

Therefore,

Hence, the required ascending order is

$\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}$

Question: 10(ii) Write the following rational numbers in ascending order:

$\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

Answer:

Given $\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

LCM of $3,9\ and\ 3 = 9$ .

Therefore, we have

$\frac{1\times3}{3\times3}, \frac{-2\times1}{9\times1}, \frac{-4\times3}{3\times3}$

$\Rightarrow \frac{3}{9}, \frac{-2}{9}, \frac{-12}{9}$

Since $\frac{-12}{9}<\frac{-2}{9}<\frac{3}{9}$

Hence, the required ascending order is

$\frac{-4}{3}<\frac{-2}{9}<\frac{1}{3}$

Question: 10(iii) Write the following rational numbers in ascending order:

$\frac{-3}{7},\frac{-3}{2}, \frac{-3}{4}$

Answer:

Given $\frac{-1}{3},\frac{2}{9}, \frac{-4}{3}$

LCM of $7,2\ and\ 4 =28$ .

Therefore, we have

$\frac{-3\times4}{7\times4}, \frac{-3\times14}{2\times14}, \frac{-3\times7}{4\times7}$

$\Rightarrow \frac{-12}{28}, \frac{-42}{28}, \frac{-21}{28}$

Since $\frac{-42}{28}<\frac{-21}{28}<\frac{-12}{28}$

Hence, the required ascending order is

$\frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}$

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.1

Question: Find:

$\frac{-13}{7}+\frac{6}{7},\frac{19}{5}+\left ( \frac{-7}{5} \right )$

Answer:

For the given sum: $\frac{-13}{7}+\frac{6}{7}$

Here the denominator value is same that is 7 hence we can sum the numerator as:

$\frac{-13}{7}+\frac{6}{7}= \frac{-13+6}{7} = \frac{-7}{7} = -1$

For the given sum: $\frac{19}{5}+\left ( \frac{-7}{5} \right )$

Here also the denominator value is the same and is equal to 5 hence we can write it as:

$\frac{19}{5}+\left ( \frac{-7}{5} \right ) = \frac{19-7}{5} = \frac{12}{5}$

Question:(i) Find:

$\frac{-3}{7}+\frac{2}{3}$

Answer:

Given sum: $\frac{-3}{7}+\frac{2}{3}$

Taking LCM of 7 and 3 we get; 21

Hence we can write the sum as:

$\Rightarrow \frac{-3\times3}{7\times3}+\frac{2\times7}{3\times7}$

$\Rightarrow \frac{-9}{21}+\frac{14}{21} = \frac{5}{21}$

Question:(ii) Find:

$\frac{-5}{6}+\frac{-3}{11}$

Answer:

Given sum: $\frac{-5}{6}+\frac{-3}{11}$

Taking LCM of 6 and 11 we get; 66

Hence we can write the sum as:

$\Rightarrow \frac{-5\times11}{6\times11}+\frac{-3\times6}{11\times6}$

$\Rightarrow \frac{-55}{66}+\frac{-18}{66} = \frac{-73}{66}$

NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.9.2

Question:1 What will be the additive inverse of $\frac{-3}{9} ?, \frac{-9}{11} ?, \frac{5}{7} ?$$

Answer:

The additive inverse of $\frac{-3}{9} \ is \ \frac{3}{9}$

The additive inverse of $\frac{-9}{11} \ is \ \frac{9}{11}$

The additive inverse of $\frac{5}{7} \ is \ \frac{-5}{7}$

Question:2 Find $(i)\frac{7}{9}-\frac{2}{5} \ \ \ \ \ (ii) 2 \frac{1}{5}-\frac{(-1)}{3}$

Answer:

NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.9.3

Question: What will be

(i) $\frac{-3}{5}\times 7$ (ii) $\frac{-6}{5}\times (-2)$

Answer:

(i) $\frac{-3}{5}\times 7$

We can write the product as:

$\frac{-3}{5}\times 7 = \frac{-3}{5}\times\frac{7}{1} = \frac{-21}{5}$

(i) $\frac{-6}{5}\times (-2)$

We can write the product as:

$\frac{-6}{5}\times (-2) = \frac{-6}{5}\times\frac{-2}{1} = \frac{12}{5}$

Question:(i) Find:

$\frac{-3}{4}\times \frac{1}{7}$

Answer:

Given product:

$\frac{-3}{4}\times \frac{1}{7} = \frac{-3\times1}{4\times7} = \frac{-3}{28}$

Question:(ii) Find:

$\frac{2}{3}\times \frac{-5}{9}$

Answer:

Given product: $\frac{2}{3}\times \frac{-5}{9}$

$\Rightarrow \frac{2}{3}\times \frac{-5}{9} = \frac{2\times-5}{3\times9} = \frac{-10}{27}$

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.4

Question: What will be the reciprocal of

$\frac{-6}{11} and \frac{-8}{5}?$

Answer:

The reciprocal of $\frac{-6}{11}$ will be:

$1\div \frac{-6}{11} = \frac{1}{\frac{-6}{11}} =\frac{11}{-6}$

The reciprocal of $\frac{-8}{5}$ will be:

<img alt="1\div \frac{-8}{5} = \frac{1}{\frac{-8}{5}} =\frac{5}{-8}" height="45"

src="https://lh6.googleusercontent.com/R8VumcHnzXVNeB5tbYiuWsru99Ndev_znjKQ9dZmLUlRwjwHUr4QdXVE6hiLGBGzX3JKnwD3_l0batqGrMCP5Ppe1qUIlABNPNvfJEQhzR8rLFo3r7dTeBtLRrzHWAtxldIqsmE" style="margin-left: 0px; margin-top: 0px;" width="155" />

NCERT Solutions for class 7 maths chapter 9 rational numbers exercise: 9.2

Question: 1(i) Find the sum:

$\frac{5}{4}+\left (\frac{-11}{4} \right )$

Answer:

Given sum: $\frac{5}{4}+\left (\frac{-11}{4} \right )$

Here the denominator is the same which is 4.

Question: 1(ii) Find the sum:

$\frac{5}{3}+\frac{3}{5}$

Answer:

Given sum: $\frac{5}{3}+\frac{3}{5}$

Here the LCM of 3 and 5 is 15.

Hence, we can write the sum as:

$\Rightarrow \frac{5}{3}+\frac{3}{5} = \frac{5\times5}{3\times5}+\frac{3\times3}{5\times3}$

$\Rightarrow \frac{25}{15}+\frac{9}{15} = \frac{34}{15}$

Question: 1(iii) Find the sum:

$\frac{-9}{10}+\frac{22}{15}$

Answer:

Given sum: $\frac{-9}{10}+\frac{22}{15}$

Taking the LCM of 10 and 15, we have 30

$\Rightarrow \frac{-9\times3}{10\times3}+\frac{22\times2}{15\times2}$

$\Rightarrow \frac{-27}{30}+\frac{44}{30} = \frac{-27+44}{30} = \frac{17}{30}$

Question: 1(iv) Find the sum:

$\frac{-3}{-11}+\frac{5}{9}$

Answer:

Given sum: $\frac{-3}{-11}+\frac{5}{9}$

Taking LCM of 11 and 9 we have,

$\Rightarrow \frac{-3\times9}{-11\times9}+\frac{5\times11}{9\times11}$

$\Rightarrow \frac{-27}{-99}+\frac{55}{99}=\frac{27+55}{99} = \frac{82}{99}$

Question: 1(v) Find the sum :

$\frac{-8}{19}+\frac{(-2)}{57}$

Answer:

Given sum: $\frac{-8}{19}+\frac{(-2)}{57}$

Taking LCM of 19 and 57, we have 57

We can write the sum as:

$\frac{-8\times3}{57}+\frac{(-2)}{57} = \frac{-24-2}{57} = \frac{-26}{57}$

Question: 1(vi) Find the sum:

$\frac{-2}{3}+0$

Answer:

Given sum: $\frac{-2}{3}+0$

Adding any number to zero we get, the number itself

Hence, $\frac{-2}{3}+0 = \frac{-2}{3}$

Question: 1(vii) Find the sum:

$-2\frac{1}{3}+4\frac{3}{5}$

Answer:

Given the sum: $-2\frac{1}{3}+4\frac{3}{5}$

Taking the LCM of 3 and 5 we have: 15

$\Rightarrow \frac{-7\times5}{3\times5}+\frac{23\times3}{5\times3} = \frac{-35}{15}+\frac{69}{15} = \frac{34}{15}$

Question: 2(i) Find

$\frac{7}{24} -\frac{17}{36}$

Answer:

Given sum: $\frac{7}{24} -\frac{17}{36}$

We have LCM of 24 and 36 will be, 72

Hence,

$\Rightarrow \frac{7\times3}{24\times3} -\frac{17\times2}{36\times2} = \frac{21}{72} - \frac{34}{72} = \frac{-13}{72}$

Question: 2(ii) Find

$\frac{5}{63}-\left (\frac{-6}{21} \right )$

Answer:

Given $\frac{5}{63}-\left (\frac{-6}{21} \right )$ :

LCM of 63 and 21 is 63,

Then we have;

$\Rightarrow \frac{5}{63}-\left (\frac{-6\times3}{21\times3} \right ) = \frac{5+18}{63} =\frac{23}{63}$

Question: 2(iii) Find

$\frac{-6}{13}-\left (\frac{-7}{15} \right )$

Answer:

Given $\frac{-6}{13}-\left (\frac{-7}{15} \right )$ :

We have, LCM of 13 and 15 is 195.

Then,

Question: 2(iv) Find

$\frac{-3}{8}-\frac{7}{11}$

Answer:

Given $\frac{-3}{8}-\frac{7}{11}$ :

LCM of 8 and 11 is 88, then

$\Rightarrow \frac{-3}{8}-\frac{7}{11} = \frac{-3\times11}{8\times11} - \frac{7\times8}{11\times8}$

$\Rightarrow \frac{-33}{88} - \frac{56}{88} = \frac{-33-56}{88} = \frac{-89}{88}$

Question: 2(v) Find

$-2\frac{1}{9}-6$

Answer:

Given: $-2\frac{1}{9}-6$

$\Rightarrow -2\frac{1}{9}-6 = \frac{-19}{9} - 6 = \frac{-19}{9} - \frac{6}{1}$

LCM of 9 and 1 will be, 9

Hence,

$\Rightarrow \frac{-19}{9} - \frac{6}{1} = \frac{-19}{9} - \frac{6\times9}{1\times9}$

$\Rightarrow \frac{-19}{9} - \frac{54}{9} = \frac{-73}{9}.$

Question: 3(i) Find the product:

$\frac{9}{2}\times\left ( \frac{-7}{4} \right )$

Answer:

Given product: $\frac{9}{2}\times\left ( \frac{-7}{4} \right )$

$\Rightarrow \frac{9}{2}\times\left ( \frac{-7}{4} \right ) = \frac{9\times(-7)}{2\times4}$

$\Rightarrow \frac{-63}{8}$

Question: 3(ii) Find the product:

$\frac{3}{10}\times (-9)$

Answer:

Given $\frac{3}{10}\times (-9)$

$\Rightarrow \frac{3}{10}\times\frac{-9}{1} = \frac{3\times(-9)}{10\times1}$

So the value

$\Rightarrow \frac{-27}{10}$

Question: 3(iii) Find the product:

$\frac{-6}{5}\times \frac{9}{11}$

Answer:

Given product: $\frac{-6}{5}\times \frac{9}{11}$

$\Rightarrow \frac{-6}{5}\times \frac{9}{11} = \frac{-6\times9}{5\times11}$

The value of given product is

$\Rightarrow \frac{-54}{55}$

Question: 3(iv) Find the product:

$\frac{3}{7}\times \left (\frac{-2}{5} \right )$

Answer:

Given product $\frac{3}{7}\times \left (\frac{-2}{5} \right )$

$\Rightarrow \frac{3}{7}\times(\frac{-2}{5}) = \frac{3\times(-2)}{7\times5} = \frac{-6}{35}$

Question: 3(v) Find the product:

$\frac{3}{11}\times \frac{2}{5}$

Answer:

Given product: $\frac{3}{11}\times \frac{2}{5}$

$\Rightarrow \frac{3}{11}\times \frac{2}{5} = \frac{3\times2}{11\times5} = \frac{6}{55}$

Question: 3(vi) Find the product:

$\frac{3}{-5}\times \frac{-5}{3}$

Answer:

Given product: $\frac{3}{-5}\times \frac{-5}{3}$

$\Rightarrow \frac{3}{-5}\times \frac{-5}{3} = \frac{3\times-5}{-5\times3} = \frac{-15}{-15} =1$

Question: 4(i) Find the value of:

$(-4)\div \frac{2}{3}$

Answer:

Given: $(-4)\div \frac{2}{3}$

Dividing by $\frac{2}{3}$ , we get

$\Rightarrow \frac{-4}{\frac{2}{3}} =\frac{-4\times3}{2} = \frac{-12}{2} = \frac{-6}{1}$

Question: 4(ii) Find the value of:

$\frac{-3}{5}\div 2$

Answer:

Given $\frac{-3}{5}\div 2$

Dividing $\frac{-3}{5}$ with 2 we get,

$\Rightarrow \frac{\frac{-3}{5}}{2} = \frac{-3}{10}$

Question: 4(iii) Find the value of:

$\frac{-4}{5}\div (-3)$

Answer:

Given: $\frac{-4}{5}\div (-3)$

So, dividing $\frac{-4}{5}$ with -3, we get

$\Rightarrow \frac{\frac{-4}{5}}{-3} = \frac{-4}{5\times-3} = \frac{-4}{-15} = \frac{4}{15}$

Question: 4(iv) Find the value of:

$\frac{-1}{8}\div \frac{3}{4}$

Answer:

Given: $\frac{-1}{8}\div \frac{3}{4}$

Simplifying it:

$\Rightarrow \frac{-1}{8}\times\frac{4}{3} = \frac{-1\times4}{8\times3}$

$\Rightarrow \frac{-4}{24} = \frac{-1}{6}$

Question: 4(v) Find the value of:

$\frac{-2}{13}\div \frac{1}{7}$

Answer:

Given: $\frac{-2}{13}\div \frac{1}{7}$

Simplifying it: we get

$\Rightarrow \frac{-2}{13}\times\frac{7}{1} = \frac{-14}{13}$

Question: 4(vi) Find the value of:

$\frac{-7}{12}\div \left (\frac{-2}{3} \right )$

Answer:

Given: $\frac{-7}{12}\div \left (\frac{-2}{3} \right )$

Simplifying it: we get

Question: 4(vii) Find the value of:

$\frac{3}{13}\div \left (\frac{-4}{65} \right )$

Answer:

Given: $\frac{3}{13}\div \left (\frac{-4}{65} \right )$

Simplifying it: we get

Responsive Table

Chapter No.Chapter Name

Chapter 1	NCERT solutions for class 7 maths chapter 1 Integers
Chapter 2	NCERT solutions for class 7 maths chapter 2 Fractions and Decimals
Chapter 3	NCERT solutions for class 7 maths chapter 3 Data Handling
Chapter 4	NCERT solutions for class 7 maths chapter 4 Simple Equations
Chapter 5	NCERT solutions for class 7 maths chapter 5 Lines and Angles
Chapter 6	NCERT solutions for class 7 maths chapter 6 The Triangle and its Properties
Chapter 7	NCERT solutions for class 7 maths chapter 7 Congruence of Triangles
Chapter 8	NCERT solutions for class 7 maths chapter 8 comparing quantities
Chapter 9	NCERT solutions for class 7 maths chapter 9 Rational Numbers
Chapter 10	NCERT solutions for class 7 maths chapter 10 Practical Geometry
Chapter 11	NCERT solutions for class 7 maths chapter 11 Perimeter and Area
Chapter 12	NCERT solutions for class 7 maths chapter 12 Algebraic Expressions
Chapter 13	NCERT solutions for class 7 maths chapter 13 Exponents and Powers
Chapter 14	NCERT solutions for class 7 maths chapter 14 Symmetry
Chapter 15	NCERT solutions for class 7 maths chapter 15 visualising-solid-shapes

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NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.3

NCERT Solutions for class 7 maths chapter 9 topic 9.4

NCERT Solutions for class 7 maths chapter 9 topic 9.6

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.8

NCERT Solutions for class 7 maths chapter 9 rational numbers exercise 9.1

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.1

NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.9.2

NCERT Solutions for class 7 maths chapter 9 rational numbers topic 9.9.3

NCERT solutions for class 7 maths chapter 9 rational numbers topic 9.9.4

NCERT Solutions for class 7 maths chapter 9 rational numbers exercise: 9.2

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