NCERT Solutions For Class 7 Maths Chapter 9

NCERT Solutions Class 7 Maths Rational Numbers

Ncert Solutions For Class 7 Maths Chapter 9 PDF Download

NCERT Solutions for Class 7 Maths Chapter 9 is the best study materials for those students who are finding difficulties in solving problems on Rational Numbers. These solutions include several example problems and are mainly explained by the team of our subject experts to helps students in understanding the topic easily and effectively. NCERT Solutions for Class 7 Maths Rational Numbers are available for free in PDF format provides solutions for all the questions provided in class 7 math NCERT textbooks. Students can either practice online or download these NCERT Solutions for chapter 9 of Class 7 Maths pdf files from our website BYJU’S and have a quick review during their exams. 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 learn the differences between the problems. For the full NCERT solutions of Rational Numbers of class 7 maths, check it out below!

NCERT Solutions For Class 7 Maths Chapter 9 Exercises

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

\(\frac{-12}{28},\frac{-42}{28},\frac{-21}{28}\)

\(\frac{-42}{28}<\frac{-21}{28}<\frac{-12}{28}\)

\(\Rightarrow \frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}\)

 

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

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

The above solution is the NCERT Solutions for Class 7 Maths Chapter 9, it is essential to have the different solutions for the chapters. There are quite a few different chapters that are also present in the class 7 maths syllabus, some of these major topics are Integers, Fractions and Decimals, Simple Handling, Comparing quantities etc. These are most of the major topics that a student can be quizzed on during their main examinations. Some of the other major NCERT solutions can be learnt through with the help of NCERT solutions. Thus, these are some of the methods of finding the best solutions for class 7 maths solutions.

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