NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers - Download PDF

NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 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 $p=2andq=-3\ne 0$ .

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

Answer:

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

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

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

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

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

Hence, we have

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

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

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

$\Rightarrow \frac{-3}{7}=\frac{-3\times 2}{7\times 2}=\frac{-6}{14}$

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

$\Rightarrow \frac{-3}{7}=\frac{-3\times 2}{7\times 2}=\frac{-6}{14}$

Hence, we have

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

NCERT rational numbers class 7 solutions 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, $-8$ 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 a 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 :

$\frac{-18}{45}=\frac{-6\times 3}{15\times 3}=\frac{-2\times 3\times 3}{5\times 3\times 3}=\frac{-2}{5}$

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

We can make it in the standard form :

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

NCERT Solutions for Maths Chapter 9 Rational Numbers Class 7 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:

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

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

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

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

–2 and –1

Answer: To find five rational numbers between $-2and-1$ we will convert each rational numbers as a denominator $5+1=6$ , we have

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

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 $5\times 3=15$ , we have $(\because LCMof5and3=15)$

$\frac{-4}{5}=\frac{-4\times 3}{5\times 3}=\frac{-12}{15}and\frac{-2}{3}=\frac{-2\times 5}{3\times 5}=\frac{-10}{15}$

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

$\frac{-12}{15}=\frac{-12\times 3}{15\times 3}=\frac{-36}{45}and\frac{-10}{15}=\frac{-10\times 3}{15\times 3}=\frac{-30}{45}$

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:

$\frac{-3}{5}=\frac{-3\times 1}{5\times 1}$ $\frac{-6}{10}=\frac{-3\times 2}{5\times 2}$ $\frac{-9}{15}=\frac{-3\times 3}{5\times 3}$ $\frac{-12}{20}=\frac{-3\times 4}{5\times 4}$

Now, following the same pattern, we have

$\frac{-3\times 5}{5\times 5}=\frac{-15}{25}$ $\frac{-3\times 6}{5\times 6}=\frac{-18}{30}$ $\frac{-3\times 7}{5\times 7}=\frac{-21}{35}$ $\frac{-3\times 8}{5\times 8}=\frac{-24}{40}$

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:

$\frac{-1}{4}=\frac{-1\times 1}{4\times 1}$ $\frac{-2}{8}=\frac{-1\times 2}{4\times 2}$ $\frac{-3}{12}=\frac{-1\times 3}{4\times 3}$

Now, following the same pattern, we have

$\frac{-1\times 4}{4\times 4}=\frac{-4}{16}$ $\frac{-1\times 5}{4\times 5}=\frac{-5}{20}$ $\frac{-1\times 6}{4\times 6}=\frac{-6}{24}$ $\frac{-1\times 7}{4\times 7}=\frac{-7}{28}$

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:

$\frac{-1}{6}=\frac{-1\times 1}{6\times 1}$ $\frac{-2}{12}=\frac{-1\times 2}{6\times 2}$ $\frac{-3}{18}=\frac{-1\times 3}{6\times 3}$ $\frac{-4}{24}=\frac{-1\times 4}{6\times 4}$

Now, following the same pattern, we have

$\frac{-1\times 5}{6\times 5}=\frac{-5}{30}$ $\frac{-1\times 6}{6\times 6}=\frac{-6}{36}$ $\frac{-1\times 7}{6\times 7}=\frac{-7}{42}$ $\frac{-1\times 8}{6\times 8}=\frac{-8}{48}$

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:

$\frac{-2}{3}=\frac{-2\times 1}{3\times 1}$ $\frac{2}{-3}=\frac{2}{-3}=\frac{2\times 1}{-3\times 1}$ $\frac{4}{-6}=\frac{-2\times 2}{3\times 2}$ $\frac{-6}{9}=\frac{-2\times 3}{3\times 3}$

Now, following the same pattern, we have

$\frac{-2\times 4}{3\times 4}=\frac{-8}{12}or\frac{8}{-12}$ $\frac{-2\times 5}{3\times 5}=\frac{-10}{15}or\frac{10}{-15}$

$\frac{-2\times 6}{3\times 6}=\frac{-12}{18}or\frac{12}{-18}$ $\frac{-2\times 7}{3\times 7}=\frac{-14}{21}or\frac{14}{-21}$

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:

$\frac{-2}{7}=\frac{-2\times 2}{7\times 2}=\frac{-4}{14}$ $\frac{-2}{7}=\frac{-2\times 3}{7\times 3}=\frac{-6}{21}$

$\frac{-2}{7}=\frac{-2\times 4}{7\times 4}=\frac{-8}{28}$ $\frac{-2}{7}=\frac{-2\times 5}{7\times 5}=\frac{-10}{35}$

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:

$\frac{5}{-3}=\frac{5\times 2}{-3\times 2}=\frac{10}{-6}$ $\frac{5}{-3}=\frac{5\times 3}{-3\times 3}=\frac{15}{-9}$

$\frac{5}{-3}=\frac{5\times 4}{-3\times 4}=\frac{20}{-12}$ $\frac{5}{-3}=\frac{5\times 5}{-3\times 5}=\frac{25}{-15}$

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:

$\frac{4}{9}=\frac{4\times 2}{9\times 2}=\frac{8}{18}$ $\frac{4}{9}=\frac{4\times 3}{9\times 3}=\frac{12}{27}$

$\frac{4}{9}=\frac{4\times 4}{9\times 4}=\frac{16}{36}$ $\frac{4}{9}=\frac{4\times 5}{9\times 5}=\frac{20}{45}$

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:

$\frac{7}{8}$

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:

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

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:

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

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 $\frac{-7}{21}and\frac{3}{9}$

$\Rightarrow \frac{-7\times 9}{21\times 9}=\frac{-63}{189}$

$\Rightarrow \frac{3\times 21}{9\times 21}=\frac{63}{189}$

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

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 $\frac{-16}{20}and\frac{20}{-25}$

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

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

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

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 $\frac{-3}{5}and\frac{-12}{20}$

$\Rightarrow \frac{-3\times 20}{5\times 20}=\frac{-60}{100}$

$\Rightarrow \frac{-12\times 5}{20\times 5}=\frac{-60}{100}$

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

So, they represent the same rational number.

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

$\Rightarrow \frac{8\times 15}{-5\times 15}=\frac{120}{-75}$

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

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

So, they represent the same rational number.

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

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

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

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

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 $-5\ne 5$ .

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:

$\Rightarrow \frac{-8}{6}=\frac{-8/2}{6/2}=\frac{-4}{3}$ $\left[HCFof8and6is2\right]$

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

$\Rightarrow \frac{25}{45}=\frac{25/5}{45/5}=\frac{5}{9}$ $\left[HCFof25and45is5\right]$

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

$\Rightarrow \frac{-44}{72}=\frac{-44/4}{72/4}=\frac{-11}{18}$ $\left[HCFof44and72is4\right]$

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

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

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

(vii) $0◻\frac{-7}{6}$

Answer: (i) $\frac{-5}{7}◻\frac{2}{3}$

$\Rightarrow \frac{-5\times 3}{7\times 3}◻\frac{2\times 7}{3\times 7}$

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

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

(ii) $\frac{-4}{5}◻\frac{-5}{7}$

$\Rightarrow \frac{-4\times 7}{5\times 7}◻\frac{-5\times 5}{7\times 5}$

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

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

(iii) $\frac{-7}{8}◻\frac{14}{-16}$

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

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

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

(iv) $\frac{-8}{5}◻\frac{-7}{4}$

$\Rightarrow \frac{-8\times 4}{5\times 4}◻\frac{-7\times 5}{4\times 5}$

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

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

(v) $\frac{1}{-3}◻\frac{-1}{4}$

$\Rightarrow \frac{1\times 4}{-3\times 4}◻\frac{-1\times -3}{4\times -3}$

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

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

(vi) $\frac{5}{-11}◻\frac{-5}{11}$

$\Rightarrow \frac{5\times 11}{-11\times 11}◻\frac{-5\times -11}{11\times -11}$

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

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

(vii) $0◻\frac{-7}{6}$

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

$\Rightarrow \frac{2\times 2}{3\times 2},\frac{5\times 3}{2\times 3}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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, $-3<-2<-1$

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 $3,9and3=9$ .

Therefore, we have

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

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

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 $7,2and4=28$ .

Therefore, we have

$\frac{-3\times 4}{7\times 4},\frac{-3\times 14}{2\times 14},\frac{-3\times 7}{4\times 7}$

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

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 Chapter 9 Maths Class 7 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:

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

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

Question:(ii) Find: