# NCERT Solutions For Class 7 Maths Chapter 9 : Rational Numbers

## NCERT Solutions For Class 7 Maths Chapter 9 PDF Free Download

NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers is the best study materials for those students who are finding difficulties in solving problems on Rational Numbers. These solutions can help students to clear their doubts easily and help in understanding the topic effectively. These rational numbers solutions are available for free download in PDF format and provide solutions for all the questions provided in class 7 math NCERT textbook. Students can either practice online or download these solutions and have a quick review during their exams.

NCERT solutions for class 7 Chapter 9 Rational Numbers, is designed by experts of BYJU’S in accordance with CBSE syllabus (2019-2020) prescribed by the board itself. These solutions can also be used as worksheets for practice. Students are advised to solve class 7 sample papers and previous year question papers to know the types of questions asked from chapter Rational numbers. Also, students can use these materials for preparation of competitive exams such as maths olympiad, conducting in 2019 to practice and score well.

## Class 7 Maths NCERT Solutions Chapter 9- Rational Numbers

The important topics covered in chapter 9 Rational numbers by NCERT solutions are;

• Rational numbers
• Positive and Negative rational numbers
• Rational Numbers On A Number Line
• Rational Numbers In Standard Form
• Comparison Of Rational Numbers
• Operations On Rational Numbers
• Rational Numbers Between Two Rational Number.

With the help of NCERT Solutions for class 7 maths, one can easily find the solutions to the major problems that are present. It is also important to learn the different solutions for the problems that are available in the syllabus. One can easily understand the differences between the problems based on each topic.

### NCERT Solutions For Class 7 Maths Chapter 9 Exercises

For the full NCERT solutions of Rational Numbers of class 7 maths PDF, check it the below given links.

### NCERT Solutions For Class 7 Maths Chapter 9 Exercises 9.1

Q1:

List 9 natural numbers between:

(i) -1 and 0

(ii) -2 and -1

(iii) $-\frac{4}{5}\; and \;-\frac{2}{3}$

(iv) $-\frac{1}{2}\; and \; \frac{2}{3}$

Sol:

(i) -1 and 0

Let us write -1 and 0 as rational number with denominator 10.   (as we need 9 number between the  numbers given so 9+1=10).

$\Rightarrow -1=-\frac{10}{10}$ and $0=\frac{0}{10}$

$\ therefore \frac{-10}{10}<\frac{-9}{10}<\frac{-8}{10}<\frac{-7}{10}<\frac{-6}{10}<\frac{-5}{10}<\frac{-4}{10}<\frac{-3}{10}<\frac{-2}{10}<\frac{-1}{10}<0$

$\Rightarrow -1<\frac{-9}{10}<\frac{-4}{5}<\frac{-7}{10}<\frac{-3}{5}<\frac{-1}{2}<\frac{-2}{5}<\frac{-3}{10}<\frac{-1}{5}<\frac{-1}{10}<0$

Therefore 9 rational numbers between -1 and 0 would be

$\frac{-9}{10},\frac{-4}{5},\frac{-7}{10},\frac{-3}{5},\frac{-1}{2},\frac{-2}{5},\frac{-3}{10},\frac{-1}{5},\frac{-1}{10}$

(ii) -2 and -1

Let us write -2 and -1 as rational number with denominator 10.   (as we need 9 number between the  numbers given so 9+1=10).

$\Rightarrow -2=\frac{-20}{10}$ and $-1=\frac{-10}{10}$

$\ therefore \frac{-20}{10}<\frac{-19}{10}<\frac{-18}{10}<\frac{-17}{10}<\frac{-16}{10}<\frac{-15}{10}<\frac{-14}{10}<\frac{-13}{10}<\frac{-12}{10}<\frac{-11}{10}<\frac{-10}{10}$

$\Rightarrow -2<\frac{-19}{10}<\frac{-9}{5}<\frac{-17}{10}<\frac{-8}{5}<\frac{-3}{2}<\frac{-7}{5}<\frac{-13}{10}<\frac{-6}{5}<\frac{-11}{10}<-1$

Therefor 9 rational numbers between -2 and -1 would be,

$\frac{-19}{10},\frac{-9}{5},\frac{-17}{10},\frac{-8}{5},\frac{-3}{2},\frac{-7}{5},\frac{-13}{10},\frac{-6}{5},\frac{-11}{10}$ .

(iii) $-\frac{4}{5}\; and \;-\frac{2}{3}$

Let us write $-\frac{4}{5}\; and \;-\frac{2}{3}$ as rational number with the same denominator.

$\Rightarrow \frac{-4}{5} = \frac{-72}{90}\; and \;\frac{-2}{3}=\frac{-60}{90}$.

$\ therefore \frac{-72}{90}<\frac{-71}{90}<\frac{-70}{90}<\frac{-69}{90}<\frac{-68}{90}<\frac{-67}{90}<\frac{-66}{90}<\frac{-65}{90}<\frac{-64}{90}<\frac{-63}{90}<\frac{-60}{90}$

$\Rightarrow \frac{-4}{5}<\frac{-71}{90}<\frac{-7}{9}<\frac{-23}{30}<\frac{-34}{45}<\frac{-67}{90}<\frac{-11}{15}<\frac{-13}{18}<\frac{-32}{45}<\frac{-7}{10}<\frac{-2}{3}$

Therefor 9 rational numbers between  $-\frac{4}{5}\; and \;-\frac{2}{3}$ are

$\frac{-71}{90},\frac{-7}{9},\frac{-23}{30},\frac{-34}{45},\frac{-67}{90},\frac{-11}{15},\frac{-13}{18},\frac{-32}{45},\frac{-7}{10}$

(iv) $\frac{-1}{2}\; and \; \frac{2}{3}$

Let us write $\frac{-1}{2}\; and \; \frac{2}{3}$ as rational numbers having same denominator.

$\frac{-1}{2}=\frac{-6}{12}\; and \; \frac{2}{3}=\frac{8}{12}$

$\ therefore \frac{-6}{12}< \frac{-5}{12}< \frac{-4}{12}<\frac{-3}{12}<\frac{-2}{12}<\frac{-1}{12}<\frac{0}{12}<\frac{1}{12}<\frac{2}{12}<\frac{3}{12}<\frac{8}{12}$

$\Rightarrow \frac{-1}{2}< \frac{-5}{12}< \frac{-1}{3}<\frac{-1}{4}<\frac{-1}{6}<\frac{-1}{12}<0<\frac{1}{12}<\frac{1}{6}<\frac{1}{4}<\frac{2}{3}$

$\frac{-5}{12}, \frac{-1}{3},\frac{-1}{4},\frac{-1}{6},\frac{-1}{12},0,\frac{1}{12},\frac{1}{6},\frac{1}{4}$

Q2: With the pattern given below, write down four more rational number, which comes in continuation:

(i) $\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}, ……..$

(ii) $\frac{-1}{4},\frac{-2}{8},\frac{-3}{12}, ……..$

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

(iv) $\frac{-2}{3},\frac{-4}{6},\frac{-6}{9}, ……..$

Sol:

(i) $\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}, ……..$ $\Rightarrow \frac{-3}{5},\frac{-3\times 2}{5\times 2},\frac{-3\times 3}{5\times 3},\frac{-3\times 4}{5\times 4},………$

Therefore, the next four rational numbers following the pattern would be

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

(ii) $\frac{-1}{4},\frac{-2}{8},\frac{-3}{12}, ……..$ $\Rightarrow \frac{-1\times 1}{4\times 1},\frac{-1\times 2}{4\times 2},\frac{-1\times 3}{4\times 3}, ……..$

Therefore, the next four rational numbers following the pattern would be

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

(iii) $\frac{-1}{6},\frac{-2}{12},\frac{-3}{18},\frac{-4}{24}, ……..$ $\Rightarrow \frac{-1\times 1}{6\times 1},\frac{-1\times 2}{6\times 2},\frac{-1\times 3}{6\times 3},\frac{-1\times 4}{6\times 4}, ……..$

Therefore, the next four rational numbers following the pattern would be

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

(iv) $\frac{-2}{3},\frac{-4}{6},\frac{-6}{9}, ……..$ $\Rightarrow \frac{-2\times 1}{3 \times 1},\frac{-2\times 2}{3 \times 2},\frac{-2\times 3}{3 \times 3}, ……..$

Therefore, the next four rational numbers following the pattern would be

$\frac{-2\times 4}{3 \times 4},\frac{-2\times 5}{3 \times 5},\frac{-2\times 6}{3 \times 6}, \frac{-2\times 7}{3 \times 7}=\frac{-8}{12},\frac{-10}{15},\frac{-12}{18}, \frac{-14}{21}$

Q3: Write five rational number equivalent to the given fraction:

(i) $\frac{-2}{7}$

(ii) $\frac{-5}{3}$

(iii) $\frac{4}{9}$

Sol:

(i) $\frac{-2}{7}$ $\frac{-2\times 2}{7\times 2}=\frac{-4}{14},\frac{-2\times 3}{7\times 3}=\frac{-6}{21} ,\frac{-2\times 4}{7\times 4}=\frac{-8}{28} ,\frac{-2\times 5}{7\times 5}=\frac{-10}{35} ,\frac{-2\times 6}{7\times 6}=\frac{-12}{42}$

Thus five rational number equivalent to $\frac{-2}{7}$ are

$\frac{-4}{14},\frac{-6}{21}, \frac{-8}{28}, \frac{-10}{35} ,\frac{-12}{42}$

(ii) $\frac{-5}{3}$ $\frac{-5\times 2}{3\times 2}=\frac{-10}{6},\frac{-5\times 3}{3\times 3}=\frac{-15}{9},\frac{-5\times 4}{3\times 4}=\frac{-20}{12},\frac{-5\times 5}{3\times 5}=\frac{-25}{15},\frac{-5\times 6}{3\times 6}=\frac{-30}{18}$

Thus five rational number equivalent to $\frac{-5}{3}$ are

$\frac{-10}{6},\frac{-15}{9},\frac{-20}{12},\frac{-25}{15},\frac{-30}{18},$

(iii) $\frac{4}{9}$ $\frac{4\times 2}{9\times 2}=\frac{8}{18},\frac{4\times 3}{9\times 3}=\frac{12}{27},\frac{4\times 4}{9\times 4}=\frac{16}{36},\frac{4\times 5}{9\times 5}=\frac{20}{45},\frac{4\times 6}{9\times 6}=\frac{24}{54}$

Thus five rational number equivalent to $\frac{-5}{3}$ are

$\frac{8}{18},\frac{12}{27},\frac{16}{36},\frac{20}{45},\frac{24}{54}$

Q4: Draw the number line and represent the following rational numbers on it:

(i) $\frac{3}{4}$

(ii) $\frac{3}{4}$

(iii) $\frac{3}{4}$

(iv) $\frac{3}{4}$

Sol:

(i) $\frac{3}{4}$

(ii)  $\frac{3}{4}$

(iii) $\frac{3}{4}$

(iv) $\frac{3}{4}$

Q5: The point 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.

Sol:

Each part which is between the two numbers is divided into 3 parts.

Therefore   $A=\frac{6}{3},P=\frac{7}{3},Q=\frac{8}{3}\;and\; B=\frac{9}{3}$

Similarly  $T=\frac{-3}{3}, R=\frac{-4}{3},S=\frac{-5}{3}\; and\; U=\frac{-6}{3}$

Thus, the rational numbers represented $P,Q,R\; and\; S\; are\; \frac{7}{3},\frac{8}{3},\frac{-4}{3},\frac{-5}{3}$ respectively.

Q6: Which of the following pairs represent the same rational numbers:

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

Sol:

(i)  $\frac{-7}{21}\; and \;\frac{3}{9}$ $\Rightarrow \frac{-7}{21}=\frac{-1}{3}\; and \; \frac{3}{9}=\frac{1}{3}$ $\ because \frac{-1}{3}\neq \frac{1}{3}$ $∴ \frac{-7}{21}\neq \frac{3}{9}$

(ii) $\frac{-16}{20}\; and \;\frac{20}{-25}$ $\Rightarrow \frac{-16}{20}=\frac{-4}{5}\; and \; \frac{20}{-15}=\frac{-4}{5}$ $\ because \frac{-4}{5}= \frac{-4}{5}$ $∴ \frac{-16}{20}= \frac{20}{-25}$

(iii) $\frac{-2}{-3}\; and \;\frac{2}{3}$ $\Rightarrow \frac{-2}{-3}=\frac{2}{3}\; and \;\frac{2}{3}$ $\ because \frac{2}{3}=\frac{2}{3}$ $∴ \frac{-2}{-3}=\frac{2}{3}$

(iv) $\frac{-3}{5}\; and \;\frac{-12}{20}$ $\Rightarrow \frac{-3}{5}\; and \;\frac{-12}{20}=\frac{-3}{5}$ $\ because \frac{-3}{5}= \frac{-3}{5}$ $∴ \frac{-3}{5}= \frac{-12}{20}$

(v) $\frac{-8}{5}\; and \;\frac{-24}{15}$ $\Rightarrow \frac{-8}{5}\; and \;\frac{-24}{15}=\frac{-8}{5}$ $\ because \frac{-8}{5}=\frac{-8}{5}$ $∴ \frac{-8}{5}=\frac{-24}{15}$

(vi) $\frac{1}{3}\; and \;\frac{-1}{9}$ $\Rightarrow \frac{1}{3}\; and \;\frac{-1}{9}$ $\ because \frac{1}{3} \neq \frac{-1}{9}$ $∴ \frac{1}{3} \neq \frac{-1}{9}$

(vii) $\frac{-5}{-9}\; and \;\frac{5}{-9}$ $\Rightarrow \frac{-5}{-9}= \frac{5}{9} \; and \;\frac{5}{-9}$ $\ because \frac{5}{9} \neq \frac{-5}{9}$ $∴ \frac{-5}{-9} \neq \frac{5}{-9}$

Q7: Rewrite the following rational numbers in the simplest form:

(i) $\frac{-8}{6}$

(ii) $\frac{25}{45}$

(iii) $\frac{-44}{72}$

(iv) $\frac{-8}{10}$

Sol:

(i) $\frac{-8}{6}$ = $\frac{-8}{6}=\frac{8\div 2}{6\div 2}=\frac{-4}{3}$  (as 8 and 6 has H.C.F. of 2)

(ii) $\frac{25}{45}$= $\frac{25}{45}=\frac{25\div 5}{45\div 5}=\frac{5}{9}$  (as 25 and 45 has H.C.F. of 5).

(iii) $\frac{-44}{72}$ = $\frac{-44}{72} =\frac{-44\div 4}{72\div 4}=\frac{-11}{18}$  (as 44 and 72 has H.C.F. of 4)

(iv) $\frac{-8}{10}$ = $\frac{-8}{10}=\frac{-8\div 2}{10\div 2}=\frac{-4}{5}$  (as 8 and 10 has H.C.F. of 2).

Q8: Fill in the blanks with the correct symbols 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) $0$ ___ $\frac{-7}{6}$

Sol:

(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) $0$ _>_ $\frac{-7}{6}$

Q9: Write the following rational number in ascending order:

(i) $\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}$

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

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

Sol:

(i) $\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}$ $\Rightarrow \frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}$

(ii)  $\frac{1}{3},\frac{-2}{9},\frac{-4}{3}$ $\frac{3}{9},\frac{-2}{9},\frac{-12}{9}$       (converting to same denominator)

$\frac{-12}{9}<\frac{-2}{9}<\frac{3}{9}$ $\Rightarrow \frac{-4}{3}<\frac{-2}{9}<\frac{1}{3}$

(iii) $\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}$ $\frac{-12}{28},\frac{-42}{28},\frac{-21}{28}$ $\frac{-42}{28}<\frac{-21}{28}<\frac{-12}{28}$ $\Rightarrow \frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}$

### NCERT Solutions For Class 7 Maths Chapter 9 Exercises 9.2

Q.1.(i) $\frac{5}{4}+(\frac{-11}{4})$

(ii)$\frac{5}{3}+\frac{3}{5}$

(iii)$e \frac{-9}{10}+\frac{22}{15}$

(iv)$\frac{-3}{-11}+\frac{5}{9}$

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

(vi)$\frac{-2}{3}+0$

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

Solution:

(i)$\frac{5}{4}+(\frac{-11}{4})$ $\frac{5}{4}+(\frac{-11}{4})= \frac{5-11}{4}=\frac{-6}{4}=\frac{-3}{2}$

(ii)$\frac{5}{3}+\frac{3}{5}$ $\frac{5}{3}+\frac{3}{5}\\ \\ = \frac{5\times 5}{3\times 5}+\frac{3\times 3}{5\times 3}\\ \\ =\frac{25}{15}+\frac{9}{15}\\ \\ =\frac{25+9}{15}=\frac{34}{15}=2\frac{4}{15}$

(iii)$\frac{-9}{10}+\frac{22}{15}$ $\frac{-9}{10}+\frac{22}{15}\\ \\ = \frac{-9\times 3}{10\times 3}+\frac{22\times 2}{15\times 2}\\ \\ =\frac{-27}{30}+\frac{44}{30}\\ \\ =\frac{-27+44}{30}=\frac{17}{30}$

(iv)$\frac{-3}{-11}+\frac{5}{9}$ $\frac{-3}{-11}+\frac{5}{9}\\ \\ = \frac{-3\times 9}{-11\times 9}+\frac{5\times 11}{9\times 11}\\ \\ =\frac{27}{99}+\frac{55}{99}\\ \\ =\frac{27+55}{99}=\frac{82}{99}$

(v)$\frac{-8}{19}+\frac{(-2)}{57}$ $\frac{-8}{19}+\frac{(-2)}{57}\\ \\ = \frac{-8\times 3}{19\times 3}+\frac{(-2)\times 1}{57\times 1}\\ \\ =\frac{-24}{57}+\frac{(-2)}{57}\\ \\ =\frac{-24-2}{57}=\frac{-26}{57}$

(vi)$\frac{-2}{3}+0$ $\frac{-2}{3}+0=\frac{-2}{3}$

(vii)$-2\frac{1}{3}+4\frac{3}{5}$ $2\frac{1}{3}+4\frac{3}{5}\\ \\ =\frac{-7}{3}+\frac{23}{5}\\ \\ = \frac{-7\times 5}{3\times 5}+\frac{23\times 3}{5\times 3}\\ \\ =\frac{-35}{15}+\frac{69}{15}\\ \\ =\frac{-35+69}{15}=\frac{34}{15}=2\frac{4}{15}$

Q.2. Find:

(a)$\frac{7}{24}-\frac{17}{36}$

$\frac{7}{24}-\frac{17}{36}\\ \\ = \frac{7\times 3}{24\times 3}-\frac{17\times 2}{36\times 2}\\ \\ =\frac{21}{72}-\frac{34}{72}\\ \\ =\frac{21-34}{72}=\frac{-13}{72}$

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

$\frac{5}{63}-(\frac{-6}{21})\\ \\ = \frac{5\times 1}{63\times 1}-(\frac{-6\times 3}{21\times 3})\\ \\ =\frac{5}{63}-(\frac{-18}{63})\\ \\ =\frac{5+18}{63}=\frac{23}{63}$

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

$\frac{-6}{13}-(\frac{-7}{15})\\ \\ = \frac{-6\times 15}{13\times 15}-(\frac{-7\times 13}{15\times 13})\\ \\ =\frac{-90}{195}-(\frac{-91}{195})\\ \\ =\frac{-90-(-91)}{195}=\frac{-90+91}{195}=\frac{1}{195}$

(d)$\frac{-3}{8}-\frac{7}{11}$

$\frac{-3}{8}-\frac{7}{11}\\ \\ = \frac{-3\times 11}{8\times 11}-\frac{7\times 8}{11\times 8}\\ \\ =\frac{-33}{88}-\frac{56}{88}\\ \\ =\frac{-33-56}{88}=\frac{-89}{88}=-1\frac{1}{88}$

(e)$-2\frac{1}{9}-6$

$-2\frac{1}{9}-6\\ \\ =\frac{-19}{9}-\frac{6}{1}\\ \\ = \frac{-19\times 1}{9\times 1}-\frac{6\times 9}{1\times 9}\\ \\ =\frac{-19}{9}-\frac{54}{9}\\ \\ =\frac{-19-54}{9}=\frac{-73}{9}=-8\frac{1}{9}$

Q.3.(i)$\frac{9}{2}\times(\frac{-7}{4})$

$\frac{9}{2}\times(\frac{-7}{4})=\frac{9\times (-7)}{2\times 4}=\frac{-63}{8}=-7\frac{7}{8}$

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

$\frac{3}{10}\times(-9)=\frac{3\times (-9)}{10}=\frac{-27}{10}=-2\frac{7}{10}$

(iii)$\frac{-6}{5}\times\frac{9}{11}$

$\frac{-6}{5}\times\frac{9}{11}=\frac{(-6)\times 9}{5\times 11}=\frac{-54}{55}$

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

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

(v)$\frac{3}{11}\times\frac{2}{5}$

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

(vi)$\frac{3}{-5}\times(\frac{-5}{3})$ $\frac{3}{-5}\times(\frac{-5}{3})=\frac{3\times (-5)}{-5\times 3}=1$

Q.4. Find the value of:

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

$(-4)\div \frac{2}{3}=(-4)\times \frac{3}{2}=(-2)\times 3=-6$

(ii)$\frac{-3}{5}\div2$

$\frac{-3}{5}\div2=\frac{-3}{5}\times \frac{1}{2}=\frac{(-3)\times 1}{5\times 2}=\frac{-3}{10}$

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

$\frac{-4}{5}\div(-3)=\frac{(-4)}{5}\times \frac{1}{(-3)}=\frac{(-4)\times 1}{5\times (-3)}=\frac{4}{15}$

(iv)$\frac{-1}{8}\div\frac{3}{4}$

$\frac{-1}{8}\div\frac{3}{4} =\frac{-1}{8}\times \frac{4}{3}=\frac{(-1)\times 1}{2\times 3}=\frac{-1}{6}$

(v) $\frac{-2}{13}\div\frac{1}{7}$

$\frac{-2}{13}\div\frac{1}{7} =\frac{-2}{13}\times \frac{7}{1}=\frac{(-2)\times 7}{13\times 1}=\frac{-14}{13}=1\frac{-1}{13}$

(vi) $\frac{-7}{12}\div(\frac{-2}{13})$

$\frac{-7}{12}\div(\frac{-2}{13}) =\frac{-7}{12}\times \frac{13}{(-2)}=\frac{(-7)\times 13}{12\times (-2)}=\frac{-91}{24}=3\frac{19}{24}$

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

$\frac{3}{13}\div(\frac{-4}{65}) =\frac{3}{13}\times \frac{65}{(-4)}=\frac{3\times (-5)}{1\times 4}=\frac{-15}{4}=-3\frac{3}{4}$

It is essential for students to have a deep understanding of different solutions for the chapters. There are solutions for different chapters apart from rational numbers that are present in class 7 maths syllabus. Some of these major chapters are Integers, Fractions and Decimals, data handling, Comparing quantities etc. These are the most important topics based on which questions can be asked on during their main examinations. These topics can be learned with the help of NCERT solutions in a better way and students can easily understand the concepts and logic behind each topic.

Learn from BYJU’S and also get tips and tricks to prepare for exams in your class 7 final exams to score good marks and download BYJU’S app to experience a new method of learning and also learn about rational numbers with the help of Visuals.