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 NCERT Solutions for Class 7 Maths Chapter 9 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 , -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 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\ 0we will convert each rational numbers as a denominator5+1 =6, 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 5+1 =6 , 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, -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 3,9\ 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 -4 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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