NCERT Solutions Class 6 Maths Chapter 7 Fractions

NCERT solutions for class 6 - Maths chapter 7 Fractions includes topics like mixed fractions, improper fractions, proper fractions, and how to represent a fraction on a number line. A fraction is a numerical value that denotes a portion of a whole. Even in real life, you need to execute this topic, so you must know about every concept of fractions.  

The NCERT solution will be based on the latest CBSE syllabus. Different types of questions will be based on every concept of class 6 Science Chapter 7, so you can easily grasp all principles and fundamentals. It also ensures that students can practice and reinforce their knowledge. 

Therefore, in this article, we have come up with expertly crafted NCERT solutions for class 6 of chapter 7, Fractions. Each solution is detailed to help you understand it easily.   

1.0NCERT Solutions for Class 6 Maths Chapter 7 PDF

We are providing ALLEN NCERT Solutions a downloadable PDF for Chapter 7 of Class 6 Maths, which covers fractions. The solutions are detailed to make them clearer and easier to understand, helping students solve questions step by step. These solutions are especially important for revising and practising at home, making learning more convenient.

2.0NCERT Solutions for Class 6 Maths Chapter 7 Fractions: All Exercises

In this section, you will find solutions to all the exercises from Chapter 7 of the Class 6 Maths book. It includes detailed answers for each question, helping students understand how to add, compare, subtract, and simplify fractions. These solutions will help students grasp the concept of fractions and apply them to real-world problems.

3.0NCERT Questions with Solutions Class 6 Maths Chapter 7

EXERCISE 7.1

• Solve : (i) $2-\frac{3}{5}$ (ii) $4+\frac{7}{8}$ (iii) $\frac{3}{5}+\frac{2}{7}$ (iv) $\frac{9}{11}-\frac{4}{15}$ (v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$ (vi) $2\frac{2}{3}+3\frac{1}{2}$ (vii) $8\frac{1}{2}-3\frac{5}{8}$ Sol. (i) $2-\frac{3}{5}=\frac{2\times 5}{5}-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}$ (ii) $4+\frac{7}{8}=\frac{(4\times 8)+7}{8}=\frac{39}{8}=4\frac{7}{8}$ (iii) $\frac{3}{5}+\frac{2}{7}=\frac{3\times 7}{5\times 7}+\frac{2\times 5}{7\times 5}=\frac{21+10}{35}=\frac{31}{35}$ (iv) $\frac{9}{11}-\frac{4}{15}=\frac{9\times 15}{11\times 15}-\frac{4\times 11}{5\times 11}$ $=\frac{135-44}{165}=\frac{91}{165}$ (v) $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7}{10}+\frac{2\times 2}{5\times 2}+\frac{3\times 5}{2\times 5}$ $=\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$ (vi) $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{8\times 2}{3\times 2}+\frac{7\times 3}{2\times 3}$ $=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$ (vii) $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{17\times 4}{2\times 4}-\frac{29}{8}$ $=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$
• Arrange the following in descending order: (i) $\frac{2}{9},\frac{2}{3},\frac{8}{21}$ (ii) $\frac{1}{5},\frac{3}{7},\frac{7}{10}$ Sol. (i) $\frac{2}{9},\frac{2}{3},\frac{8}{21}$ Changing them to like fractions, we obtain $\frac{2}{9}=\frac{2\times 7}{9\times 7}=\frac{14}{63}\frac{2}{3}=\frac{2\times 21}{3\times 21}=\frac{42}{63}$ $\frac{8}{21}=\frac{8\times 3}{3\times 21}=\frac{24}{63}$ Since $42>24>14$, $\therefore \frac{2}{3}>\frac{8}{21}>\frac{2}{9}$ (ii) $\frac{1}{5},\frac{3}{7},\frac{7}{10}$ Changing them to like fractions, we obtain $\frac{1}{5}=\frac{1\times 14}{5\times 14}=\frac{14}{70}$ $\frac{3}{7}=\frac{3\times 10}{7\times 10}=\frac{30}{70}$ $\frac{7}{10}=\frac{7\times 7}{10\times 7}=\frac{49}{70}$ As $49>30>14$, $\therefore \frac{7}{10}>\frac{3}{7}>\frac{1}{5}$
• In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square ?

Sol. Along the first row, sum $=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}$ Along the second row, sum $=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{15}{11}$ Along the third row, sum $=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{15}{11}$ Along the first column, sum $=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{15}{11}$ Along the second column, sum $=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{15}{11}$ Along the third column sum $=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{15}{11}$ Along the first diagonal, sum $=\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{15}{11}$ Along the first diagonal, sum $=\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{15}{11}$ Since the sum of the numbers in each row, in each column, and along the diagonal is the same, it is a magic square.

• A rectangular sheet of paper is $12\frac{1}{2}\mathrm{cm}$ long and $10\frac{2}{3}\mathrm{cm}$ wide. Find its perimeter. Sol. Length $=12\frac{1}{2}\mathrm{cm}=\frac{25}{2}\mathrm{cm}$ Breadth $=10\frac{2}{3}\mathrm{cm}=\frac{32}{3}\mathrm{cm}$ Perimeter $=2\times$ (Length + Breadth $)$ $=2\times \left[\frac{25}{2}+\frac{32}{3}\right]=2\times \left[\frac{(25\times 3)+(32\times 2)}{6}\right]$ $=2\times \left[\frac{75+64}{6}\right]=2\times \frac{139}{6}=\frac{139}{3}$ $=46\frac{1}{3}\mathrm{cm}$
• Find the perimeters of (i) $△ABE$ (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

• Sol. (i) Perimeter of $△ABE=AB+BE+EA$ $=\left(\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\right)=\left(\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\right)$ $=\left(\frac{5\times 10}{2\times 10}+\frac{11\times 5}{4\times 5}+\frac{18\times 4}{5\times 4}\right)$ $=\frac{50+55+72}{20}=\frac{177}{20}=8\frac{17}{20}\mathrm{cm}$ (ii) Perimeter of rectangle $=2$ (Length + Breadth) Perimeter of rectangle $=2\left[\frac{11}{4}+\frac{7}{6}\right]$ $=2\left[\frac{11\times 3}{4\times 3}+\frac{7\times 2}{6\times 2}\right]=2\left[\frac{33+14}{12}\right]$ $=2\times \frac{47}{12}=\frac{47}{6}=7\frac{5}{6}\mathrm{cm}$ Perimeter of $△ABE=\frac{177}{20}\mathrm{cm}$ Changing them to like fractions, we obtain $\frac{177}{20}=\frac{177\times 3}{20\times 3}=\frac{531}{60}$ $\frac{47}{6}=\frac{47\times 10}{6\times 10}=\frac{470}{60}$, as $531>470$ $\therefore \frac{177}{20}>\frac{47}{6}$ Perimeter ( $△\mathrm{ABE}$ ) > Perimeter (BCDE)
• Salil wants to put a picture in a frame. The picture is $7\frac{3}{5}\mathrm{cm}$ wide. To fit in the frame the picture cannot be more than $7\frac{3}{10}\mathrm{cm}$ wide. How much should the picture be trimmed? Sol. Width of picture $=7\frac{3}{5}=\frac{38}{5}\mathrm{cm}$ Required width $=7\frac{3}{10}=\frac{73}{10}\mathrm{cm}$ The picture should be trimmed by $=\left(\frac{38}{5}-\frac{73}{10}\right)$ $=\left(\frac{38\times 2}{5\times 2}-\frac{73}{10}\right)=\frac{76-73}{10}=\frac{3}{10}\mathrm{cm}$
• Ritu ate $\frac{3}{5}$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much? Sol. Part of apple eaten by Ritu $=\frac{3}{5}$ Part of apple eaten by Somu =1 - Part of apple eaten by Ritu $=1-\frac{3}{5}=\frac{2}{5}$ Therefore, Somu ate $\frac{2}{5}$ part of the apple. Since $3>2$, Ritu had the larger share. Difference between the 2 shares Therefore, Ritu's share is larger than the share of Somu by $\frac{1}{5}$.
• Michael finished colouring a picture in $\frac{7}{12}$ hour. Vaibhav finished colouring the same picture in $\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer? Sol. Time taken by Michael $=\frac{7}{12}\mathrm{hr}$ Time taken by Vaibhav $=\frac{3}{4}\mathrm{hr}$ Converting these fractions into like fractions, we obtain $\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12}$ and, $\frac{7}{12}$ Since $9>7$, Vaibhav worked longer. Difference $=\frac{9}{12}-\frac{7}{12}=\frac{2}{12}=\frac{1}{6}$ hour

EXERCISE 2.2

• Which of the drawings (a) to (d) show : (i) $2\times \frac{1}{5}$ (ii) $2\times \frac{1}{2}$ (iii) $3\times \frac{2}{3}$ (iv) $3\times \frac{1}{4}$ (a)

• (b)

• (c)

• (d)

• Sol. (i) $2\times \frac{1}{5}$ represents addition of 2 figures, each representing 1 shaded part out of 5 equal parts. Hence, $2\times \frac{1}{5}$ is represented by (d). (ii) $2\times \frac{1}{2}$ represents addition of 2 figures, each representing 1 shaded part out of 2 equal parts. Hence, $2\times \frac{1}{2}$ is represented by (b) (iii) $3\times \frac{2}{3}$ represents addition of 3 figures, each representing 2 shaded parts out of 3 equal parts. Hence, $3\times \frac{2}{3}$ is represented by (a). (iv) $3\times \frac{1}{4}$ represents addition of 3 figures, each representing 1 shaded part out of 4 equal parts. Hence, $3\times \frac{1}{4}$ is represented by (c).
• Some pictures (a) to (c) are given below. Tell which of them show : (i) $3\times \frac{1}{5}=\frac{3}{5}$ (ii) $2\times \frac{1}{3}=\frac{2}{3}$ (iii) $3\times \frac{3}{4}=2\frac{1}{4}$ (a)

• (b)

• (c)

• Sol. (i) $3\times \frac{1}{5}$ represents the addition of 3 figures, each representing 1 shaded part out of 5 equal parts $\frac{3}{5}$ represents 3 shaded parts out of 5 equal parts. Hence, $3\times \frac{1}{5}=\frac{3}{5}$ is represented (c). (ii) $2\times \frac{1}{3}$ represents the addition of 2 figures, each representing 1 shaded part out of 3 equal parts and $\frac{2}{3}$ represents 2 shaded parts out 3 equal parts. Hence, $2\times \frac{1}{3}=\frac{2}{3}$ is represented by (a) (iii) $3\times \frac{3}{4}$ represents the addition of 3 figures, each representing 3 shaded parts out of 4 equal parts and $2\frac{1}{4}$ represents 2 fully shaded figures and one figure having 1 part as shaded out of 4 equal parts. Hence, $3\times \frac{3}{4}=2\frac{1}{4}$ is represented by (b)
• Multiply and reduce to lowest form and convert into a mixed fraction : (i) $7\times \frac{3}{5}$ (ii) $4\times \frac{1}{3}$ (iii) $2\times \frac{6}{7}$ (iv) $5\times \frac{2}{9}$ (v) $\frac{2}{3}\times 4$ (vi) $\frac{5}{2}\times 6$ (vii) $11\times \frac{4}{7}$ (viii) $20\times \frac{4}{5}$ (ix) $13\times \frac{1}{3}$ (x) $15\times \frac{3}{5}$ Sol. (i) $7\times \frac{3}{5}=\frac{21}{5}=4\frac{1}{5}$ (ii) $4\times \frac{1}{3}=\frac{4}{3}=1\frac{1}{3}$ (iii) $2\times \frac{6}{7}=\frac{12}{7}=1\frac{5}{7}$ (iv) $5\times \frac{2}{9}=\frac{10}{9}=1\frac{1}{9}$ (v) $\frac{2}{3}\times 4=\frac{8}{3}=2\frac{2}{3}$ (vi) $\frac{5}{2}\times 6=15$ (vii) $11\times \frac{4}{7}=\frac{44}{7}=6\frac{2}{7}$ (viii) $20\times \frac{4}{5}=16$ (ix) $13\times \frac{1}{3}=\frac{13}{3}=4\frac{1}{3}$ (x) $15\times \frac{3}{5}=9$
• Shade: (i) $\frac{1}{2}$ of the circles in box (a) (ii) $\frac{2}{3}$ of the triangles in box (b) (iii) $\frac{3}{5}$ of the squares in box (c)

• (a)

• (b)

• (c) Sol. (i) It can be observed that there are 12 circles in the given box. We have to shade $\frac{1}{2}$ of the circles in it. As $12\times \frac{1}{2}=6$, therefore we will shade any 6 circles of it.

• (ii) It can be observed that there are 9 triangles in the given box. We have to shade $\frac{2}{3}$ of the triangles in it. As $9\times \frac{2}{3}=6$, therefore, we will shade any 6 triangles of it.

(iii) It can be observed that there are 15 squares in the given box. We have to shade $\frac{3}{5}$ of the squares in it. As $\frac{3}{5}\times 15=9$, therefore, we will shade any 9 squares of it.

• Find : (a) $\frac{1}{2}$ of (i) 24 (ii) 4 (b) $\frac{2}{3}$ of (i) 18 (ii) 27 (c) $\frac{3}{4}$ of (i) 16 (ii) 36 (d) $\frac{4}{5}$ of (i) 20 (ii) 35 Sol. (a) (i) $\frac{1}{2}\times 24=12$ (ii) $\frac{1}{2}\times 46=23$ (b) (i) $\frac{2}{3}\times 18=12$ (ii) $\frac{2}{3}\times 27=18$ (c) (i) $\frac{3}{4}\times 16=12$ (ii) $\frac{3}{4}\times 36=27$ (d) (i) $\frac{4}{5}\times 20=16$ (ii) $\frac{4}{5}\times 35=28$

• Multiply and express as a mixed fraction : (a) $3\times 5\frac{1}{5}$ (b) $5\times 6\frac{3}{4}$ (c) $7\times 2\frac{1}{4}$ (d) $4\times 6\frac{1}{3}$ (e) $3\frac{1}{4}\times 6$ (f) $3\frac{2}{5}\times 8$ Sol. (a) $3\times 5\frac{1}{5}=3\times \frac{26}{5}=\frac{78}{5}=15\frac{35}{5}$ (b) $5\times 6\frac{3}{4}=5\times \frac{27}{4}=\frac{135}{4}=33\frac{3}{4}$ (c) $7\times 2\frac{1}{4}=7\times \frac{9}{4}=\frac{63}{4}=15\frac{3}{4}$ (d) $4\times 6\frac{1}{3}=4\times \frac{19}{3}=\frac{76}{3}=25\frac{1}{3}$ (e) $3\frac{1}{4}\times 6=\frac{13}{4}\times 6=\frac{78}{4}=\frac{39}{2}=19\frac{1}{2}$ (f) $3\frac{2}{5}\times 8=\frac{17}{5}\times 8=\frac{136}{5}=27\frac{1}{5}$
• Find (a) $\frac{1}{2}$ of (i) $2\frac{3}{4}$ (ii) $4\frac{2}{9}$ (b) $\frac{5}{8}$ of (i) $3\frac{5}{6}$ (ii) $9\frac{2}{3}$ Sol. (a) (i) $\frac{1}{2}\times 2\frac{3}{4}=\frac{1}{2}\times \frac{11}{4}=\frac{11}{8}=1\frac{3}{8}$ (ii) $\frac{1}{2}\times 4\frac{2}{9}=\frac{1}{2}\times \frac{38}{9}=\frac{19}{9}=2\frac{1}{9}$ (b) (i) $\frac{5}{8}\times 3\frac{5}{6}=\frac{5}{8}\times \frac{23}{6}=\frac{115}{48}=2\frac{19}{48}$ (ii) $\frac{5}{8}\times 9\frac{2}{3}=\frac{5}{8}\times \frac{29}{3}=\frac{145}{24}=6\frac{1}{24}$
• Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed $\frac{2}{5}$ of the water. Pratap consumed the remaining water. (i) How much water did Vidya drink? (ii) What fraction of the total quantity of water did Pratap drink? Sol. (i) Water consumed by Vidya $=\frac{2}{5}$ of 5 litres $=\frac{2}{5}\times 5=2$ litres (ii) Water consumed by Pratap $=1-\frac{2}{5}=\frac{3}{5}$ of the total water.

EXERCISE 2.3

• Find: (i) $\frac{1}{4}$ of (a) $\frac{1}{4}$ (b) $\frac{3}{5}$ (c) $\frac{4}{3}$ (ii) $\frac{1}{7}$ of (a) $\frac{2}{9}$ (b) $\frac{6}{5}$ (c) $\frac{3}{10}$ Sol. (i) (a) $\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$ (b) $\frac{1}{4}\times \frac{3}{5}=\frac{3}{20}$ (c) $\frac{1}{4}\times \frac{4}{3}=\frac{1}{3}$ (ii) (a) $\frac{1}{7}\times \frac{2}{9}=\frac{2}{63}$ (b) $\frac{1}{7}\times \frac{6}{5}=\frac{6}{35}$ (c) $\frac{1}{7}\times \frac{3}{10}=\frac{3}{70}$
• Multiply and reduce to lowest from (if possible): (i) $\frac{2}{3}\times 2\frac{2}{3}$ (ii) $\frac{2}{7}\times \frac{7}{9}$ (iii) $\frac{3}{8}\times \frac{6}{4}$ (iv) $\frac{9}{5}\times \frac{3}{5}$ (v) $\frac{1}{3}\times \frac{15}{8}$ (vi) $\frac{11}{2}\times \frac{3}{10}$ (vii) $\frac{4}{5}\times \frac{12}{7}$ Sol. (i) $\frac{2}{3}\times 2\frac{2}{3}=\frac{2}{3}\times \frac{8}{3}=\frac{16}{9}=1\frac{7}{9}$ (ii) $\frac{2}{7}\times \frac{7}{9}=\frac{2}{9}$ (iii) $\frac{3}{8}\times \frac{6}{4}=\frac{9}{16}$ (iv) $\frac{9}{5}\times \frac{3}{5}=\frac{27}{25}=1\frac{2}{25}$ (v) $\frac{1}{3}\times \frac{15}{8}=\frac{5}{8}$ (vi) $\frac{11}{2}\times \frac{3}{10}=\frac{33}{20}=1\frac{13}{20}$ (vii) $\frac{4}{5}\times \frac{12}{7}=\frac{48}{35}=1\frac{13}{35}$
• Multiply the following fractions : (i) $\frac{2}{5}\times 5\frac{1}{4}$ (ii) $6\frac{2}{5}\times \frac{7}{9}$ (iii) $\frac{3}{2}\times 5\frac{1}{3}$ (iv) $\frac{5}{6}\times 3\frac{3}{7}$ (v) $3\frac{2}{5}\times \frac{4}{7}$ (vi) $2\frac{3}{5}\times 3$ (vii) $3\frac{4}{7}\times \frac{3}{5}$ Sol. (i) $\frac{2}{5}\times 5\frac{1}{4}=\frac{2}{5}\times \frac{21}{4}=\frac{21}{10}$ This is an improper fraction, and it can be written as a mixed fraction as $2\frac{1}{10}$. (ii) $6\frac{2}{5}\times \frac{7}{9}=\frac{32}{5}\times \frac{7}{9}=\frac{224}{45}$ This is an improper fraction, and it can be written as a mixed fraction as $4\frac{44}{45}$. (iii) $\frac{3}{2}\times 5\frac{1}{3}=\frac{3}{2}\times \frac{16}{3}=8$ This is a whole number. (iv) $\frac{5}{6}\times 2\frac{3}{7}=\frac{5}{6}\times \frac{17}{7}=\frac{85}{42}$ This is an improper fraction, and it can be written as a mixed fraction as $2\frac{1}{42}$. (v) $3\frac{2}{5}\times \frac{4}{7}=\frac{17}{5}\times \frac{4}{7}=\frac{68}{35}$ This is an improper fraction, and it can be written as a mixed fraction as $1\frac{33}{35}$. (vi) $2\frac{3}{5}\times 3=\frac{13}{5}\times 3=\frac{39}{5}$ This is an improper fraction, and it can be written as a mixed fraction as $7\frac{4}{5}$. (vii) $3\frac{4}{7}\times \frac{3}{5}=\frac{25}{7}\times \frac{3}{5}=\frac{15}{7}$ This is an improper fraction and it can be written as a mixed fraction as $2\frac{1}{7}$.
• Which is greater : (i) $\frac{2}{7}$ of $\frac{3}{4}$ or $\frac{3}{5}$ of $\frac{5}{8}$ (ii) $\frac{1}{2}$ of $\frac{6}{7}$ or $\frac{2}{3}$ of $\frac{3}{7}$ Sol. (i) $\frac{2}{7}\times \frac{3}{4}=\frac{3}{14}\frac{3}{5}\times \frac{5}{8}=\frac{3}{8}$ Converting these fractions into like fractions, $\frac{3}{14}=\frac{3\times 4}{14\times 4}=\frac{12}{56}\frac{3}{8}=\frac{3\times 7}{8\times 7}=\frac{21}{56}$ Since $\frac{21}{56}>\frac{12}{56},\therefore \frac{3}{8}>\frac{3}{14}$ Therefore, $\frac{3}{5}$ of $\frac{5}{8}$ is greater. (ii) $\frac{1}{2}\times \frac{6}{7}=\frac{3}{7}\frac{2}{3}\times \frac{3}{7}=\frac{2}{7}$ Since $3>2$, $\therefore \frac{3}{7}>\frac{2}{7}$ Therefore, $\frac{1}{2}$ of $\frac{6}{7}$ is greater.
• Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is $\frac{3}{4}\mathrm{m}$. Find the distance between the first and the last sapling? $\begin{array}{cccc}\text{ 1st }& \text{ 2nd }& \text{ 3rd }& \text{ 4th }\\ \text{ Sol. Sapling }& \text{ Sapling }& \text{ Sapling }& \text{ Sapling }\end{array}$

• From the figure, it can be observed that gaps between $1^{\text{st }}$ and last sapling $=3$. Length of 1 gap $=\frac{3}{4}\mathrm{m}$ Therefore, distance between I and IV sapling $=3\times \frac{3}{4}=\frac{9}{4}=2\frac{1}{4}\mathrm{m}$
• Lipika reads a book for $1\frac{3}{4}$ hours every day. She reads the entire book in 6 days. How many hours in all were required by her to read the book? Sol. Number of hours Lipika reads the book per day $=1\frac{3}{4}=\frac{7}{4}$ hours Number of days $=6$ Total number of hours required by her to read the book $=\frac{7}{4}\times 6=\frac{21}{2}=10\frac{1}{2}$ hours
• A car runs 16 km using 1 litre of petrol. How much distance will it cover using $2\frac{3}{4}$ litres of petrol? Sol. Number of kms a car run per litre petrol = 16 km Quantity of petrol $=2\frac{3}{4}\mathrm{L}=\frac{11}{4}\mathrm{L}$ Number of kms a car can run for $\frac{11}{4}$ litre petrol $=\frac{11}{4}\times 16=44\mathrm{km}$ It will cover 44 km distance by using $2\frac{3}{4}$ litres of petrol.
• (a) (i) Provide the number in the box such that $\frac{2}{3}\times \square =\frac{10}{30}$. (ii) The simplest from the number obtained in $\square$ is -. (b) (i) Provide the number in the box such that $\frac{3}{5}\times \square =\frac{24}{75}$ ? (ii) The simplest form of the number obtained in $\square$ is . Sol. (a) (i) As $\frac{2}{3}\times \frac{5}{10}=\frac{10}{30}$, Therefore, the number in the box $\square$, such that $\frac{2}{3}\times \square =\frac{10}{30}$ is $\frac{5}{10}$. (ii) The simplest form of $\frac{5}{10}$ is $\frac{1}{2}$. (b) (i) As $\frac{3}{5}\times \frac{8}{15}=\frac{24}{75}$, Therefore, the number in the box $\square$, such that $\frac{3}{5}\times \square =\frac{24}{75}$ is $\frac{8}{15}$. (ii) As $\frac{8}{15}$ cannot be further simplified, therefore, its simplest form is $\frac{8}{15}$

EXERCISE 2.4

• Find (i) $12÷\frac{3}{4}$ (ii) $14÷\frac{5}{6}$ (iii) $8÷\frac{7}{3}$ (iv) $4÷\frac{8}{3}$ (v) $3÷2\frac{1}{3}$ (vi) $5÷3\frac{4}{7}$ Sol. (i) $12÷\frac{3}{4}=12\times \frac{4}{3}=16$ (ii) $14÷\frac{5}{6}=14\times \frac{6}{5}=\frac{84}{5}$ (iii) $8÷\frac{7}{3}=8\times \frac{3}{7}=\frac{24}{7}$ (iv) $4÷\frac{8}{3}=4\times \frac{3}{8}=\frac{3}{2}$ (v) $3÷2\frac{1}{3}=3÷\frac{7}{3}=3\times \frac{3}{7}=\frac{9}{7}$ (vi) $5÷3\frac{4}{7}=5÷\frac{25}{7}=5\times \frac{7}{25}=\frac{7}{5}$
• Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers. (i) $\frac{3}{7}$ (ii) $\frac{5}{8}$ (iii) $\frac{9}{7}$ (iv) $\frac{6}{5}$ (v) $\frac{12}{7}$ (vi) $\frac{1}{8}$ (vii) $\frac{1}{11}$ Sol. A proper fraction is the fraction which has its denominator greater than its numerator while improper fraction is the fraction which has its numerator greater than its denominator. Whole numbers are a collection of all positive integers including 0 . (i) Reciprocal of $\frac{3}{7}=\frac{7}{3}$ Therefore, it is an improper fraction. (ii) Reciprocal of $\frac{5}{8}=\frac{8}{5}$ Therefore, it is an improper fraction. (iii) Reciprocal of $\frac{9}{7}=\frac{7}{9}$ Therefore, it is a proper fraction. (iv) Reciprocal of $\frac{6}{5}=\frac{5}{6}$ Therefore, it is a proper fraction. (v) Reciprocal of $\frac{12}{7}=\frac{7}{12}$ Therefore, it is a proper fraction. (vi) Reciprocal of $\frac{1}{8}=\frac{8}{1}$ Therefore, it is a whole number. (vii) Reciprocal of $\frac{1}{11}=\frac{11}{1}$ Therefore, it is a whole number.
• Find (i) $\frac{7}{3}÷2$ (ii) $\frac{4}{9}÷5$ (iii) $\frac{6}{13}÷7$ (iv) $4\frac{1}{3}÷3$ (v) $3\frac{1}{2}÷4$ (vi) $4\frac{3}{7}÷7$ Sol. (i) $\frac{7}{3}÷2=\frac{7}{3}\times \frac{1}{2}=\frac{7}{6}$ (ii) $\frac{4}{9}÷5=\frac{4}{9}\times \frac{1}{5}=\frac{4}{45}$ (iii) $\frac{6}{13}÷7=\frac{6}{13}\times \frac{1}{7}=\frac{6}{91}$ (iv) $4\frac{1}{3}÷3=\frac{13}{3}÷3=\frac{13}{3}\times \frac{1}{3}=\frac{13}{9}$ (v) $3\frac{1}{2}÷4=\frac{7}{2}÷4=\frac{7}{2}\times \frac{1}{4}=\frac{7}{8}$ (vi) $4\frac{3}{7}÷7=\frac{31}{7}\times \frac{1}{7}=\frac{31}{49}$
• Find (i) $\frac{2}{5}÷\frac{1}{2}$ (ii) $\frac{4}{9}÷\frac{2}{3}$ (iii) $\frac{3}{7}÷\frac{8}{7}$ (iv) $2\frac{1}{3}÷\frac{3}{5}$ (v) $3\frac{1}{2}÷\frac{8}{3}$ (vi) $\frac{2}{5}÷1\frac{1}{2}$ (vii) $3\frac{1}{5}÷1\frac{2}{3}$ (viii) $2\frac{1}{5}÷1\frac{1}{5}$ Sol. (i) $\frac{2}{5}÷\frac{1}{2}=\frac{2}{5}\times 2=\frac{4}{5}$ (ii) $\frac{4}{9}÷\frac{2}{3}=\frac{4}{9}\times \frac{3}{2}=\frac{2}{3}$ (iii) $\frac{3}{7}÷\frac{8}{7}=\frac{3}{7}\times \frac{7}{8}=\frac{3}{8}$ (iv) $2\frac{1}{3}÷\frac{3}{5}=\frac{7}{3}÷\frac{3}{5}=\frac{7}{3}\times \frac{5}{3}=\frac{35}{9}$ (v) $3\frac{1}{2}÷\frac{8}{3}=\frac{7}{2}÷\frac{8}{3}=\frac{7}{2}\times \frac{3}{8}=\frac{21}{16}$ (vi) $\frac{2}{5}÷1\frac{1}{2}=\frac{2}{5}÷\frac{3}{2}=\frac{2}{5}\times \frac{2}{3}=\frac{4}{15}$ (vii) $3\frac{1}{5}÷1\frac{2}{3}=\frac{16}{5}÷\frac{5}{3}=\frac{16}{5}\times \frac{3}{5}=\frac{48}{25}$ (viii) $2\frac{1}{5}÷1\frac{1}{5}=\frac{11}{5}÷\frac{6}{5}=\frac{11}{5}\times \frac{5}{6}=\frac{11}{6}$

4.0NCERT Solutions Class 6 Maths Chapter 7 Subtopics

Before understanding the NCERT solutions, it is important to know the subtopics covered under the chapter fractions. This section explains the different subtopics covered in Chapter 7, "Fractions." In the NCERT solutions, you will learn about:

• Fractional Units and Equal Shares
• Fractional Units as Parts of a Whole
• Measuring Using Fractional Units
• Marking Fraction Lengths on the Number Line
• Mixed Fractions
• Equivalent Fractions
• Addition and Subtraction of Fractions
• A Pinch of History