NCERT Solutions for Class 6 Maths Chapter 7 - Fractions

Exercise 7.1

1. Write the Fraction representing the shaded portion:

Ans: Total number of parts = 4

Shaded parts = 2

fraction of shaded parts = \[\dfrac{2}{4}\]

Ans: Total number of parts = 9

Shaded parts = 8

fraction of shaded parts = \[\dfrac{8}{9}\]

Ans: Total number of parts = 8

Shaded parts = 4

fraction of shaded parts = \[\dfrac{4}{8}\]

Ans: Total number of parts = 4

Shaded parts = 1

fraction of shaded parts = \[\dfrac{1}{4}\]

Ans: Total number of parts = 7

Shaded parts = 3

fraction of shaded parts = \[\dfrac{3}{7}\]

Ans: Total number of parts = 12

Shaded parts = 9

fraction of shaded parts = \[\dfrac{9}{{12}}\]

Ans: Total number of parts = 10

Shaded parts = 10

fraction of shaded parts = \[\dfrac{{10}}{{10}}\]

Ans: Total number of parts = 9

Shaded parts = 4

fraction of shaded parts = \[\dfrac{4}{9}\]

Ans: Total number of parts = 8

Shaded parts = 4

fraction of shaded parts = \[\dfrac{4}{8}\]

Ans: Total number of parts = 2

Shaded parts = 1

fraction of shaded parts = \[\dfrac{1}{2}\]

2. Colour the part according to the given fraction:

1. \[\mathbf{\dfrac{1}{6}}\]

Ans:

2. \[\mathbf{\dfrac{1}{4}}\]

Ans:

3. \[\mathbf{\dfrac{1}{3}}\]

Ans:

4. \[\mathbf{\dfrac{3}{4}}\]

Ans:

5. \[\mathbf{\dfrac{4}{9}}\]

Ans:

3. Identify the error, if any?

Ans: There is an error present in each of them. The figures are not equally divided into parts.

4. What fraction of a day is 8 hours? 

Ans: Total number of hours in a day = 24

Therefore, fraction of 8 hours = \[\dfrac{8}{{24}}\] or \[\dfrac{1}{3}\]

5.  What fraction of an hour is 40 minutes? 

Ans: Minutes in an hour = 60

fraction of 40 minutes = \[\dfrac{{40}}{{60}}\]or \[\dfrac{2}{3}\]

6.  Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich. 

1. How can Arya divide his sandwiches so that each person has an equal share? 

Ans: He will have to divide each sandwich into 3 equal parts

2. What part of a sandwich will each boy receive? 

Ans: Each boy will receive \[1 \times \dfrac{1}{3}\]or \[\dfrac{1}{3}\] sandwiches of each type.

7.  Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished? 

Ans: Total number of dresses = 30

Number of dresses finished by Kanchan = 20

fraction of dresses finished = \[\dfrac{{20}}{{30}}\]or \[\dfrac{2}{3}\]

8.  Write the natural numbers from 2 to 12. What fraction of them are prime numbers? 

Ans: Natural numbers from 2 to 12: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Prime numbers are: 2, 3, 5, 7, 11

Total natural numbers = 11

Total prime numbers = 5

fraction of prime numbers = \[\dfrac{5}{{11}}\]

9. Write the natural numbers from 102 to 113. What fraction of them are prime numbers? 

Ans: Natural numbers from 102 to 113: 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113

Prime numbers from 102 to 113: 103, 107, 109, 113

Natural numbers = 12

Prime numbers = 4

fraction of prime numbers = \[\dfrac{4}{{12}}\] or \[\dfrac{1}{3}\]

10. What fraction of these circles have ‘X’s in them?

Ans: Total number of circles = 8

Number of circles with X in them = 4

fraction of circles with X in them = \[\dfrac{4}{8}\] or \[\dfrac{1}{2}\]

11.Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?

Ans: Total number of CDs with Kristin = 3 + 5

                                                       = 8

Total number of CDs she bought = 3

CDs she received as gift = 5

fraction of CDs she bought = \[\dfrac{3}{8}\]

fraction of CDs she received as gift = \[\dfrac{5}{8}\]

Exercise 7.2

1.Draw number lines and locate the points on them:

1. \[\mathbf{\dfrac{1}{2},\dfrac{1}{4},\dfrac{3}{4},\dfrac{4}{4}}\]

Ans:

2. \[\mathbf{\dfrac{1}{8},\dfrac{2}{8},\dfrac{3}{8},\dfrac{7}{8}}\]

Ans:

3. \[\mathbf{\dfrac{2}{5},\dfrac{3}{5},\dfrac{8}{5},\dfrac{4}{5}}\]

Ans:

2. Express the following fractions as mixed fractions:

1. \[\mathbf{\dfrac{{20}}{3}}\]

Ans: We will divide 20 by 3 to get the answer

$\quad \,6$

$3{\overline{)\,\,{20}\quad\quad}}$

$\quad\underline{\,-18\,\,}  $

$\quad{\,\, 2 \,\,}$

\[\therefore \dfrac{{20}}{3} = 6\dfrac{2}{3}\]

2. \[\mathbf{\dfrac{{11}}{5}}\]

Ans: We will divide 11 by 5 to get the answer

$\quad \,2$

$5{\overline{)\,\,{11}\quad\quad}}$

$\quad\underline{\,-10\,\,}  $

$\quad{\,\, 1 \,\,}$

\[\therefore \dfrac{{11}}{5} = 2\dfrac{1}{5}\]

3. \[\mathbf{\dfrac{{17}}{7}}\]

Ans: We will divide 17 by 7 to get the answer

$\quad \,2$

$7{\overline{)\,\,{17}\quad\quad}}$

$\quad\underline{\,-14\,\,}  $

$\quad{\,\, 3 \,\,}$

\[\therefore \dfrac{{17}}{7} = 2\dfrac{3}{7}\]

4. \[\mathbf{\dfrac{{28}}{5}}\]

Ans: We will divide 28 by 5 to get the answer

$\quad \,5$

$5{\overline{)\,\,{28}\quad\quad}}$

$\quad\underline{\,-25\,\,}  $

$\quad{\,\, 3 \,\,}$

\[\therefore \dfrac{{28}}{5} = 5\dfrac{3}{5}\]

5. \[\mathbf{\dfrac{{19}}{6}}\]

Ans: We will divide 19 by 6 to get the answer

$\quad \,3$

$6{\overline{)\,\,{19}\quad\quad}}$

$\quad\underline{\,-18\,\,}  $

$\quad{\,\, 1 \,\,}$

\[\therefore \dfrac{{19}}{6} = 3\dfrac{1}{6}\]

6. \[\mathbf{\dfrac{{35}}{9}}\]

Ans: We will divide 35 by 9 to get the answer

$\quad \,3$

$9{\overline{)\,\,{35}\quad\quad}}$

$\quad\underline{\,-27\,\,}  $

$\quad{\,\, 8 \,\,}$

\[\therefore \dfrac{{35}}{9} = 3\dfrac{8}{9}\]

3. Express the following as improper fractions:

1. \[\mathbf{7\dfrac{3}{4}}\]

Ans: 

$7\dfrac{3}{4} = \dfrac{{\left( {7 \times 4} \right) + 3}}{4}\\$

$7\dfrac{3}{4} = \dfrac{{28 + 3}}{4}\\$

$7\dfrac{3}{4} = \dfrac{{31}}{4}$

2. \[\mathbf{5\dfrac{6}{7}}\]

Ans: 

$5\dfrac{6}{7} = \dfrac{{\left( {5 \times 7} \right) + 6}}{7}\\$

$5\dfrac{6}{7} = \dfrac{{35 + 6}}{7}\\$

$5\dfrac{6}{7} = \dfrac{{41}}{7}$

3. \[\mathbf{2\dfrac{5}{6}}\]

Ans: 

$2\dfrac{5}{6} = \dfrac{{\left( {2 \times 6} \right) + 5}}{6}\\$

$2\dfrac{5}{6} = \dfrac{{12 + 5}}{6}\\$

$2\dfrac{5}{6} = \dfrac{{17}}{6}$

4. \[\mathbf{10\dfrac{3}{5}}\]

Ans: 

$10\dfrac{3}{5} = \dfrac{{\left( {10 \times 5} \right) + 3}}{5}$

$10\dfrac{3}{5} = \dfrac{{50 + 3}}{5}\\$

$10\dfrac{3}{5} = \dfrac{{53}}{5}$

5. \[\mathbf{9\dfrac{3}{7}}\]

Ans: 

$9\dfrac{3}{7} = \dfrac{{\left( {9 \times 7} \right) + 3}}{7}\\$

$9\dfrac{3}{7} = \dfrac{{63 + 3}}{7}\\$

$9\dfrac{3}{7} = \dfrac{{66}}{7}$

6. \[\mathbf{8\dfrac{4}{9}}\]

Ans: 

$8\dfrac{4}{9} = \dfrac{{\left( {8 \times 9} \right) + 4}}{9}\\$

$8\dfrac{4}{9} = \dfrac{{72 + 4}}{9}\\$

$8\dfrac{4}{9} = \dfrac{{76}}{9}$

Exercise 7.3

1. Write the fractions. Are all these fractions equivalent?

Ans: \[\dfrac{1}{2},\dfrac{2}{4},\dfrac{3}{6},\dfrac{4}{8}\]. All of these fractions are equivalent.

Ans: \[\dfrac{4}{{12}},\dfrac{3}{9},\dfrac{2}{6},\dfrac{1}{3},\dfrac{6}{{15}}\]. These fractions are not equivalent.

2.  Write the fraction and pair up the equivalent fractions to each row:

(image will be uploaded soon)

Ans: The corresponding matches are

(a) - (ii), (b) – (iv), (c) – (i), (d) – (v), (e) – (iii)

3. Replace \[\mathbf{\boxed{}}\] in each of the following by the correct number:

1. $\mathbf{\dfrac{2}{7} = \dfrac{8}{{\boxed{}}}}$

Ans:

$\dfrac{2}{7} = \dfrac{{2 \times 4}}{{7 \times 4}} $

$\dfrac{2}{7} = \dfrac{8}{{\boxed{28}}} \\ $

2. $\mathbf{\dfrac{5}{8} = \dfrac{{10}}{{\boxed{}}}}$

Ans:

$\dfrac{5}{8} = \dfrac{{5 \times 2}}{{8 \times 2}}  \\$

$\dfrac{5}{8} = \dfrac{{10}}{{\boxed{16}}}  \\ $

3. $\mathbf{\dfrac{3}{5} = \dfrac{{\boxed{}}}{{20}}}$

Ans: 

$\dfrac{3}{5} = \dfrac{{3 \times 4}}{{5 \times 4}}  \\$

$\dfrac{3}{5} = \dfrac{{\boxed{12}}}{{20}}  \\ $

4. $\mathbf{\dfrac{{45}}{{60}} = \dfrac{{15}}{{\boxed{}}}}$

Ans: 

$\dfrac{{45}}{{60}} = \dfrac{{45 \div 3}}{{60 \div 3}}  \\$

$\dfrac{{45}}{{60}} = \dfrac{{15}}{{\boxed{20}}}  \\ $

5. $\mathbf{\dfrac{{18}}{{24}} = \dfrac{{\boxed{}}}{4}}$

Ans:

$\dfrac{{18}}{{24}} = \dfrac{{18 \div 6}}{{24 \div 6}}  \\$

$\dfrac{{18}}{{24}} = \dfrac{{\boxed3}}{4}  \\ $

4. Find the equivalent fraction of \[\dfrac{3}{5}\] having:

1. Denominator 20

Ans: 

$\dfrac{3}{5} = \dfrac{{3 \times 4}}{{5 \times 4}}\\$

$\dfrac{3}{5} = \dfrac{{12}}{{20}}$

2. Numerator 9

Ans: 

$\dfrac{3}{5} = \dfrac{{3 \times 3}}{{5 \times 3}}\\$

$\dfrac{3}{5} = \dfrac{9}{{15}}$

3. Denominator 30

Ans:

$\dfrac{3}{5} = \dfrac{{3 \times 6}}{{5 \times 6}}\\$

$\dfrac{3}{5} = \dfrac{{18}}{{30}}$

4. Numerator 27

Ans: 

$\dfrac{3}{5} = \dfrac{{3 \times 9}}{{5 \times 9}}\\$

$\dfrac{3}{5} = \dfrac{{27}}{{45}}$

5. Find the equivalent fraction of \[\dfrac{{36}}{{48}}\]with:

1. Numerator 9

Ans:

$\dfrac{{36}}{{48}} = \dfrac{{36 \div 4}}{{48 \div 4}}\\$

$\dfrac{{36}}{{48}} = \dfrac{9}{{12}}$

2. Denominator 4

Ans: 

$\dfrac{{36}}{{48}} = \dfrac{{36 \div 12}}{{48 \div 12}}\\$

$\dfrac{{36}}{{48}} = \dfrac{3}{4}$

6. Check whether the given fraction is equivalent:

1. \[\mathbf{\dfrac{5}{9},\dfrac{{30}}{{54}}}\]

Ans: 

$\dfrac{5}{9} = \dfrac{{5 \times 6}}{{9 \times 6}}\\$

$\dfrac{5}{9} = \dfrac{{30}}{{54}}$

Hence, both \[\dfrac{5}{9}\] and \[\dfrac{{30}}{{54}}\]are equivalent.

2. \[\mathbf{\dfrac{3}{{10}},\dfrac{{12}}{{50}}}\]

Ans: 

$\dfrac{3}{{10}} = \dfrac{{3 \times 5}}{{10 \times 5}}\\$

$\dfrac{3}{{10}} = \dfrac{{15}}{{50}}$

Hence, \[\dfrac{3}{{10}}\]and \[\dfrac{{12}}{{50}}\] are not equivalent.

3. \[\mathbf{\dfrac{7}{{13}},\dfrac{5}{{11}}}\]

Ans: 

$\dfrac{7}{{13}} = \dfrac{{7 \times 11}}{{13 \times 11}}\\$

$\dfrac{7}{{13}} = \dfrac{{77}}{{143}}$

And, \[\dfrac{5}{{13}} = \dfrac{{5 \times 11}}{{13 \times 11}}\]

\[\dfrac{5}{{13}} = \dfrac{{55}}{{143}}\]

Hence, \[\dfrac{7}{{13}}\]and \[\dfrac{5}{{13}}\]are not equivalent.

7. Reduce the following fractions to simplest form:

1. \[\mathbf{\dfrac{{48}}{{60}}}\]

Ans: 

$\dfrac{{48}}{{60}} = \dfrac{{48/12}}{{60/12}}\\$

$\dfrac{{48}}{{60}} = \dfrac{4}{5}$

2. \[\mathbf{\dfrac{{150}}{{60}}}\]

Ans: 

$\dfrac{{150}}{{60}} = \dfrac{{150/30}}{{60/30}}\\$

$\dfrac{{150}}{{60}} = \dfrac{5}{2}$

3. \[\mathbf{\dfrac{{84}}{{98}}}\]

Ans: 

$\dfrac{{84}}{{98}} = \dfrac{{84/14}}{{98/14}}\\$

$\dfrac{{84}}{{98}} = \dfrac{6}{7}$

4. \[\mathbf{\dfrac{{12}}{{52}}}\]

Ans: 

$\dfrac{{12}}{{52}} = \dfrac{{12/4}}{{52/4}}\\$

$\dfrac{{12}}{{52}} = \dfrac{3}{{13}}$

5. \[\mathbf{\dfrac{7}{{28}}}\]

Ans: 

$\dfrac{7}{{28}} = \dfrac{{7/7}}{{28/7}}\\$

$\dfrac{7}{{28}} = \dfrac{1}{4}$

8. Ramesh had 20 pencils, Sheelu had 50 pencils and Jamaal had 80 pencils. After 4 months, Ramesh used up 10 pencils, Sheelu used up 25 pencils and Jamaal used up 40 pencils. What fraction did each use up? Check is each has used up an equal fraction of her/his pencils?

Ans: Pencils with Ramesh = 20

Pencils used by Ramesh = 10

fraction of used by pencils by Ramesh = \[\dfrac{{10}}{{20}}\]or \[\dfrac{1}{2}\]

Pencils with Sheelu = 50

Pencils used by Sheelu = 25

fraction of pencils used by Sheelu = \[\dfrac{{25}}{{50}}\] or \[\dfrac{1}{2}\]

Pencils with Jamaal = 80

Pencils used by Jamaal = 40

fraction of pencils used by Jamaal = \[\dfrac{{40}}{{80}}\]or \[\dfrac{1}{2}\]

Yes, they all have used the same fraction of pencils.

9. Match the equivalent fractions and write two more for each:

1. \[\mathbf{\dfrac{{250}}{{400}}}\]

Ans: 

$\dfrac{{250}}{{400}} = \dfrac{{250/50}}{{400/50}}\\$

$\dfrac{{250}}{{400}} = \dfrac{5}{8},\dfrac{{10}}{{16}},\dfrac{{15}}{{24}}$

2. \[\mathbf{\dfrac{{180}}{{200}}}\]

Ans: 

$\dfrac{{180}}{{200}} = \dfrac{{180/20}}{{200/20}}\\$

$\dfrac{{180}}{{200}} = \dfrac{9}{{10}},\dfrac{{18}}{{20}},\dfrac{{27}}{{30}}$

3. \[\mathbf{\dfrac{{660}}{{990}}}\]

Ans: 

$\dfrac{{660}}{{990}} = \dfrac{{660/330}}{{990/330}}\\$

$\dfrac{{660}}{{990}} = \dfrac{2}{3},\dfrac{4}{6},\dfrac{6}{9}$

4. \[\mathbf{\dfrac{{180}}{{360}}}\]

Ans: 

$\dfrac{{180}}{{360}} = \dfrac{{180/180}}{{360/180}}\\$

$\dfrac{{180}}{{360}} = \dfrac{1}{2},\dfrac{2}{4},\dfrac{3}{6}$

5. \[\mathbf{\dfrac{{220}}{{550}}}\]

Ans: 

$\dfrac{{220}}{{550}} = \dfrac{{220/110}}{{550/110}}\\$

$\dfrac{{220}}{{550}} = \dfrac{2}{5},\dfrac{4}{{10}},\dfrac{6}{{15}}$

1. $\mathbf{\dfrac{2}{3}}$

2. $\mathbf{\dfrac{2}{5}}$

3. $\mathbf{\dfrac{1}{2}}$

4. $\mathbf{\dfrac{5}{8}}$

5. $\mathbf{\dfrac{9}{{10}}}$

Exercise 7.4

1.  Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<’, ‘>’, ‘=’ between the fractions:

Ans: Shaded fractions are \[\dfrac{3}{8},\dfrac{6}{8},\dfrac{4}{8},\dfrac{1}{8}\]

Ascending order: \[\dfrac{1}{8} < \dfrac{3}{8} < \dfrac{4}{8} < \dfrac{6}{8}\]

Descending order: \[\dfrac{6}{8} > \dfrac{4}{8} > \dfrac{3}{8} > \dfrac{1}{8}\]

Ans: Shaded fractions are \[\dfrac{8}{9},\dfrac{4}{9},\dfrac{3}{9},\dfrac{6}{9}\]

Ascending order: \[\dfrac{3}{9} < \dfrac{4}{9} < \dfrac{6}{9} < \dfrac{8}{9}\]

Descending order: \[\dfrac{8}{9} > \dfrac{6}{9} > \dfrac{4}{9} > \dfrac{3}{9}\]

3. Show \[\dfrac{2}{6},\dfrac{4}{6},\dfrac{8}{6}\] and \[\dfrac{6}{6}\] on the number line. Put appropriate signs between the factions given:

$\dfrac{5}{6}\boxed{}\dfrac{2}{6},\dfrac{3}{6}\boxed{}0,\dfrac{1}{6}\boxed{}\dfrac{6}{6},\dfrac{8}{6}\boxed{}\dfrac{5}{6}$

Ans: Number line

\[\dfrac{5}{6}\boxed > \dfrac{2}{6}\], \[\dfrac{1}{6}\boxed < \dfrac{6}{6}\], \[\dfrac{3}{6}\boxed > \dfrac{0}{6}\], \[\dfrac{8}{6}\boxed > \dfrac{5}{6}\]

2. Compare the fractions and put an appropriate sign:

1. $\mathbf{\dfrac{3}{6}\boxed{}\dfrac{5}{6}}$

Ans: \[\dfrac{3}{6}\boxed < \dfrac{5}{6}\]

2. $\mathbf{\dfrac{1}{7}\boxed{}\dfrac{1}{4}}$

Ans: \[\dfrac{1}{7}\boxed < \dfrac{1}{4}\]

3. $\mathbf{\dfrac{4}{5}\boxed{}\dfrac{5}{5}}$

Ans: \[\dfrac{4}{5}\boxed < \dfrac{5}{5}\]

4. $\mathbf{\dfrac{3}{5}\boxed{}\dfrac{3}{7}}$

Ans: \[\dfrac{3}{5}\boxed < \dfrac{3}{7}\]

3. Make five more each pair and put appropriate signs.

Ans:

1. \[\dfrac{9}{{10}}\boxed > \dfrac{6}{{10}}\]

2. \[\dfrac{1}{3}\boxed > \dfrac{1}{6}\]

3. \[\dfrac{1}{8}\boxed < \dfrac{1}{5}\]

4. \[\dfrac{7}{8}\boxed < \dfrac{{11}}{8}\]

5. \[\dfrac{{11}}{{13}}\boxed > \dfrac{9}{{13}}\]

4. Look at the figures and write ‘<’ or ‘>’ between the given pairs of fractions:

1. $\mathbf{\dfrac{1}{6}\boxed{}\dfrac{1}{3}}$

Ans: \[\dfrac{1}{6}\boxed < \dfrac{1}{3}\]

2. $\mathbf{\dfrac{3}{4}\boxed{}\dfrac{2}{6}}$

Ans: \[\dfrac{3}{4}\boxed > \dfrac{2}{6}\]

3. $\mathbf{\dfrac{2}{3}\boxed{}\dfrac{2}{4}}$

Ans: \[\dfrac{2}{3}\boxed > \dfrac{2}{4}\]

4. $\mathbf{\dfrac{6}{6}\boxed{}\dfrac{3}{3}}$

Ans: \[\dfrac{6}{6}\boxed = \dfrac{3}{3}\]

5. $\mathbf{\dfrac{5}{6}\boxed{}\dfrac{5}{5}}$

Ans: \[\dfrac{5}{6}\boxed < \dfrac{5}{5}\]

Make five more such problems and solve them with your friends.

Five more problems

1. \[\mathbf{\dfrac{1}{2}\boxed{}\dfrac{3}{6}}\]

2. \[\mathbf{\dfrac{2}{3}\boxed{}\dfrac{3}{5}}\]

3. \[\mathbf{\dfrac{3}{4}\boxed{}\dfrac{4}{6}}\]

4. \[\mathbf{\dfrac{5}{6}\boxed{}\dfrac{2}{2}}\]

5. \[\mathbf{\dfrac{0}{1}\boxed{}\dfrac{0}{6}}\]

Solutions

1. \[\dfrac{1}{2}\boxed = \dfrac{3}{6}\]

2. \[\dfrac{2}{3}\boxed > \dfrac{3}{5}\]

3. \[\dfrac{3}{4}\boxed > \dfrac{4}{6}\]

4. \[\dfrac{5}{6}\boxed < \dfrac{2}{2}\]

5. \[\dfrac{0}{1}\boxed = \dfrac{0}{6}\]

5. How quickly can you do this? Fill appropriate sign (<, =, >):

1. $\mathbf{\dfrac{1}{2}\boxed{}\dfrac{1}{5}}$

Ans: \[\dfrac{1}{2}\boxed > \dfrac{1}{5}\]

2. $\mathbf{\dfrac{2}{4}\boxed{}\dfrac{3}{6}}$

Ans: \[\dfrac{2}{4}\boxed = \dfrac{3}{6}\]

3. $\mathbf{\dfrac{3}{5}\boxed{}\dfrac{2}{3}}$

Ans: \[\dfrac{3}{5}\boxed < \dfrac{2}{3}\]

4. $\mathbf{\dfrac{3}{4}\boxed{}\dfrac{2}{8}}$

Ans: \[\dfrac{3}{4}\boxed > \dfrac{2}{8}\]

5. $\mathbf{\dfrac{3}{5}\boxed{}\dfrac{6}{5}}$

Ans: \[\dfrac{3}{5}\boxed < \dfrac{6}{5}\]

6. $\mathbf{\dfrac{7}{9}\boxed{}\dfrac{3}{9}}$

Ans: \[\dfrac{7}{9}\boxed > \dfrac{3}{9}\]

7. $\mathbf{\dfrac{1}{4}\boxed{}\dfrac{2}{8}}$

Ans: \[\dfrac{1}{4}\boxed > \dfrac{2}{8}\]

8. $\mathbf{\dfrac{6}{{10}}\boxed{}\dfrac{4}{5}}$

Ans: \[\dfrac{6}{{10}}\boxed < \dfrac{4}{5}\]

9. $\mathbf{\dfrac{3}{4}\boxed{}\dfrac{7}{8}}$

Ans: \[\dfrac{3}{4}\boxed < \dfrac{7}{8}\]

10. $\mathbf{\dfrac{6}{{10}}\boxed{}\dfrac{4}{5}}$

Ans: \[\dfrac{6}{{10}}\boxed < \dfrac{4}{5}\]

11. $\mathbf{\dfrac{5}{{7}}\boxed{}\dfrac{15}{21}}$

Ans: \[\dfrac{5}{7}\boxed = \dfrac{{15}}{{21}}\]

6. The following fractions represent just three different numbers. Separate them groups of equivalent fractions, by changing each one to its simplest form:

1. \[\mathbf{\dfrac{2}{{12}}}\]

Ans: \[\dfrac{2}{{12}} = \dfrac{1}{6}\]

2. \[\mathbf{\dfrac{3}{{15}}}\]

Ans: \[\dfrac{3}{{15}} = \dfrac{1}{5}\]

3. \[\mathbf{\dfrac{8}{{50}}}\]

Ans: \[\dfrac{8}{{50}} = \dfrac{4}{{25}}\]

4. \[\mathbf{\dfrac{{16}}{{100}}}\]

Ans: \[\dfrac{{16}}{{100}} = \dfrac{4}{{25}}\]

5. \[\mathbf{\dfrac{{10}}{{60}}}\]

Ans: \[\dfrac{{10}}{{60}} = \dfrac{1}{6}\]

6. \[\mathbf{\dfrac{{15}}{{75}}}\]

Ans: \[\dfrac{{15}}{{75}} = \dfrac{1}{5}\]

7. \[\mathbf{\dfrac{{12}}{{60}}}\]

Ans: \[\dfrac{{12}}{{60}} = \dfrac{1}{5}\]

8. \[\mathbf{\dfrac{{16}}{{96}}}\]

Ans: \[\dfrac{{16}}{{96}} = \dfrac{1}{6}\]

9. \[\mathbf{\dfrac{{12}}{{75}}}\]

Ans: \[\dfrac{{12}}{{75}} = \dfrac{4}{{25}}\]

10. \[\mathbf{\dfrac{{12}}{{72}}}\]

Ans: \[\dfrac{{12}}{{72}} = \dfrac{1}{6}\]

11. \[\mathbf{\dfrac{3}{{18}}}\]

Ans: \[\dfrac{3}{{18}} = \dfrac{1}{6}\]

12. \[\mathbf{\dfrac{4}{{25}}}\]

Ans: \[\dfrac{4}{{25}} = \dfrac{4}{{25}}\]

I group: $\dfrac{1}{5}$[(b), (f), (g)]

II group: $\dfrac{1}{6}$[(a), (e), (h), (j), (k)]

III group: $\dfrac{4}{{25}}$[(c), (d), (i), (l)]

7. Find answers to the following. Write and indicate how you solved them:

1. Is \[\mathbf{\dfrac{5}{9}}\]equal to \[\mathbf{\dfrac{4}{5}}\]?

Ans: We have \[\dfrac{5}{9}\]and \[\dfrac{4}{5}\]

\[\dfrac{{5 \times 5}}{{9 \times 5}} = \dfrac{{25}}{{45}}\]

And,

\[\dfrac{{4 \times 9}}{{5 \times 9}} = \dfrac{{36}}{{45}}\]

Since, \[\dfrac{{25}}{{45}}\]and \[\dfrac{{36}}{{45}}\]are not equal. Hence, \[\dfrac{5}{9}\]and \[\dfrac{4}{5}\] are not equal.

2. Is \[\mathbf{\dfrac{9}{{16}}}\]equal to \[\mathbf{\dfrac{5}{9}}\]?

Ans: We have \[\dfrac{9}{{16}}\]and \[\dfrac{5}{9}\]

\[\dfrac{{9 \times 9}}{{16 \times 9}} = \dfrac{{81}}{{144}}\]

And,

\[\dfrac{{5 \times 16}}{{9 \times 16}} = \dfrac{{80}}{{144}}\]

Since, \[\dfrac{{81}}{{144}}\]and \[\dfrac{{80}}{{144}}\]are not equal. Hence, \[\dfrac{9}{{16}}\]and \[\dfrac{5}{9}\] are not equal.

3. Is \[\mathbf{\dfrac{4}{5}}\]equal to \[\mathbf{\dfrac{{16}}{{20}}}\]?

Ans: We have \[\dfrac{4}{5}\]and \[\dfrac{{16}}{{20}}\]

\[\dfrac{{4 \times 20}}{{5 \times 20}} = \dfrac{{80}}{{100}}\]

And,

\[\dfrac{{16 \times 5}}{{20 \times 5}} = \dfrac{{80}}{{100}}\]

Since, \[\dfrac{{80}}{{100}}\]and \[\dfrac{{80}}{{100}}\]are equal. Hence, \[\dfrac{4}{5}\]and \[\dfrac{{16}}{{20}}\] are equal.

4. Is \[\mathbf{\dfrac{1}{{15}}}\]equal to \[\mathbf{\dfrac{4}{{30}}}\]?

Ans: We have \[\dfrac{1}{{15}}\]and \[\dfrac{4}{{30}}\]

\[\dfrac{{1 \times 2}}{{15 \times 2}} = \dfrac{2}{{30}}\]

And,

\[\dfrac{{4 \times 1}}{{30 \times 1}} = \dfrac{4}{{30}}\]

Since, \[\dfrac{2}{{30}}\]and \[\dfrac{4}{{30}}\]are not equal. Hence, \[\dfrac{1}{{15}}\]and \[\dfrac{4}{{30}}\] are not equal.

8. Ila read 25 pages of a book containing 100 pages. Lalita read \[\dfrac{2}{5}\] of the same book. Who read less?

Ans: Total pages in the book = 100

Pages read by Ila = 25

fraction of pages read by Ila = \[\dfrac{{25}}{{100}}\]or \[\dfrac{1}{4}\]

fraction of pages read by Lalita = \[\dfrac{2}{5}\]

\[\dfrac{1}{4} < \dfrac{2}{5}\]

Hence, Ila read less pages.

9. Rafiq exercised for \[\dfrac{3}{6}\] of an hour, while Rohit exercised for \[\dfrac{3}{4}\] of an hour. Who exercised for a longer time?

Ans: fraction of an hour exercised by Rafiq = \[\dfrac{3}{6}\]

fraction of hour exercised by Rohit = \[\dfrac{3}{4}\]

\[\dfrac{3}{6} < \dfrac{3}{4}\]

Hence, Rohit exercised for longer time than Rafiq.

10. In a class A of 25 students, 20 passed in first class; in another class B of 30 students, 24 passed in first class. In which class was a greater fraction of students getting first class?

Ans: Total students in class A = 25

Number of students passed in class A = 20

fraction of students passed in class A = \[\dfrac{{20}}{{25}}\]or \[\dfrac{4}{5}\]

Total students in class B = 30

Number of students passed in class B = 24

fraction of students passed in class B = \[\dfrac{{24}}{{30}}\]or \[\dfrac{4}{5}\]

Therefore, each class has same fraction of passed students.

Exercise 7.5

1. Write the fractions appropriately as additions or subtractions

Ans: The fraction form is 

$\dfrac{1}{5} + \dfrac{2}{5} = \dfrac{{1 + 2}}{5}  \\$

$\dfrac{1}{5} + \dfrac{2}{5} = \dfrac{3}{5}  \\ $

Ans: The fraction form is

$\dfrac{5}{5} - \dfrac{3}{5} = \dfrac{{5 - 3}}{5}  \\$

$\dfrac{5}{5} - \dfrac{3}{5} = \dfrac{2}{5}  \\ $

Ans: The fraction form is

$\dfrac{2}{6} + \dfrac{3}{6} = \dfrac{{2 + 3}}{6}  \\$

$\dfrac{2}{6} + \dfrac{3}{6} = \dfrac{5}{6}  \\$

2. Solve 

1. \[\mathbf{\dfrac{1}{{18}} + \dfrac{1}{{18}}}\]

Ans: We are given \[\dfrac{1}{{18}} + \dfrac{1}{{18}}\]

$\dfrac{1}{{18}} + \dfrac{1}{{18}} = \dfrac{2}{{18}}\\$

$= \dfrac{1}{9}$

2. \[\mathbf{\dfrac{8}{{15}} + \dfrac{3}{{15}}}\]

Ans: We are given \[\dfrac{8}{{15}} + \dfrac{3}{{15}}\]

$ \dfrac{8}{{15}} + \dfrac{3}{{15}} = \dfrac{{8 + 3}}{{18}}\\$

3. \[\mathbf{\dfrac{7}{7} - \dfrac{5}{7}}\]

Ans: We are given \[\dfrac{7}{7} - \dfrac{5}{7}\]

$\dfrac{7}{7} - \dfrac{5}{7} = \dfrac{{7 - 5}}{7}\\$

$= \dfrac{2}{7}$

4. \[\mathbf{\dfrac{1}{{22}} + \dfrac{{21}}{{22}}}\]

Ans: We are given \[\dfrac{1}{{22}} + \dfrac{{21}}{{22}}\]

$\dfrac{1}{{22}} + \dfrac{{21}}{{22}} = \dfrac{{1 + 21}}{{22}}\\$

$= \dfrac{23}{22}$

5. \[\mathbf{\dfrac{{12}}{{15}} - \dfrac{7}{{15}}}\]

Ans: We are given \[\dfrac{{12}}{{15}} - \dfrac{7}{{15}}\]

$\dfrac{{12}}{{15}} - \dfrac{7}{{15}} = \dfrac{{12 - 7}}{{15}}\\$

$= \dfrac{5}{{15}}$

6. \[\mathbf{\dfrac{5}{8} + \dfrac{3}{8}}\]

Ans: We are given \[\dfrac{5}{8} + \dfrac{3}{8}\]

$\dfrac{5}{8} + \dfrac{3}{8} = \dfrac{{5 + 3}}{8}\\$

 $= \dfrac{8}{8}\\$

 $= 1$

7. \[\mathbf{1 - \dfrac{2}{3},\left( {1 = \dfrac{3}{3}} \right)}\]

Ans: We are given \[1 - \dfrac{2}{3}\]

$1 - \dfrac{2}{3} = \dfrac{3}{3} - \dfrac{2}{3}\\$

$= \dfrac{{3 - 2}}{3}\\$

$= \dfrac{1}{3}$

8. \[\mathbf{\dfrac{1}{4} + \dfrac{0}{4}}\]

Ans: We are given \[\dfrac{1}{4} + \dfrac{0}{4}\]

$\dfrac{1}{4} + \dfrac{0}{4} = \dfrac{{1 + 0}}{4}\\$

$= \dfrac{1}{4}$

9. \[\mathbf{3 - \dfrac{{12}}{5}}\]

Ans: We are given \[3 - \dfrac{{12}}{5}\]

$3 - \dfrac{{12}}{5} = \dfrac{{15}}{5} - \dfrac{{12}}{5}\\$

$= \dfrac{{15 - 12}}{5}\\$

$= \dfrac{3}{5}$

3. Shubham painted \[\dfrac{2}{3}\]of the wall space in his room. His sister Madhavi helped and painted \[\dfrac{1}{3}\] of the wall space. How much did they paint together?

Ans: fraction of wall painted by Shubham = \[\dfrac{2}{3}\]

fraction of wall painted by Madhavi = \[\dfrac{1}{3}\]

$\text{Total fraction of painting done by them} = \dfrac{2}{3} + \dfrac{1}{3} \\$

$= \dfrac{{2 + 1}}{3} \\$

$= \dfrac{3}{3} \\$

$= 1$

Therefore, the total wall has been painted by both of them.

4. Fill in the missing fractions:

1. \[\mathbf{\dfrac{7}{{10}} - \boxed{} = \dfrac{3}{{10}}}\]

Ans: Let the unknown fraction be x

$\dfrac{7}{{10}} - x = \dfrac{3}{{10}}\\$

$x = \dfrac{7}{{10}} - \dfrac{3}{{10}}\\$

$x = \dfrac{{7 - 3}}{{10}}\\$

$x = \dfrac{4}{{10}}$

2. \[\mathbf{\boxed{} - \dfrac{3}{{21}} = \dfrac{5}{{21}}}\]

Ans: Let the unknown fraction be x

$x - \dfrac{3}{{21}} = \dfrac{5}{{21}}\\$

$x = \dfrac{3}{{21}} + \dfrac{5}{{21}}\\$

$x = \dfrac{{3 + 5}}{{21}}\\$

$x = \dfrac{8}{{21}}$

3. \[\mathbf{\boxed{} - \dfrac{3}{6} = \dfrac{3}{6}}\]

Ans: Let the unknown fraction be x

$x - \dfrac{3}{6} = \dfrac{3}{6}\\$

$x = \dfrac{3}{6} + \dfrac{3}{6}\\$

$x = \dfrac{{3 + 3}}{6}\\$

$x = \dfrac{6}{6} \text{ or }1$

4. \[\mathbf{\boxed{} + \dfrac{5}{{27}} = \dfrac{{12}}{{27}}}\]

Ans: Let the unknown fraction x

$x + \dfrac{5}{{27}} = \dfrac{{12}}{{27}}\\$

$x = \dfrac{{12}}{{27}} - \dfrac{5}{{27}}\\$

$x = \dfrac{{12 - 5}}{{27}}\\$

$x = \dfrac{7}{{27}}$

5. Javed was given \[\dfrac{5}{7}\]of a basket of oranges. What fraction of oranges was left in the basket?

Ans: fraction of oranges given to Javed = \[\dfrac{5}{7}\]

The whole basket is represented by 1

$\text{fraction of oranges left in the basket} = 1 - \dfrac{5}{7}\\$

$= \dfrac{7}{7} - \dfrac{5}{7}\\$

$= \dfrac{{7 - 5}}{7}\\$

$= \dfrac{2}{7}$

Hence, \[\dfrac{2}{7}\] oranges are left in the basket.

Exercise 7.6

1. Solve:

1. \[\mathbf{\dfrac{2}{3} + \dfrac{1}{7}}\]

Ans: We are given \[\dfrac{2}{3} + \dfrac{1}{7}\]

$\dfrac{2}{3} + \dfrac{1}{7} = \dfrac{{2 \times 7 + 1 \times 3}}{{21}}\\$

$\dfrac{2}{3} + \dfrac{1}{7} = \dfrac{{14 + 3}}{{21}}\\$

$\dfrac{2}{3} + \dfrac{1}{7} = \dfrac{{17}}{{21}}$

2. \[\mathbf{\dfrac{3}{{10}} + \dfrac{7}{{15}}}\]

Ans: We are given \[\dfrac{3}{{10}} + \dfrac{7}{{15}}\]

$\dfrac{3}{{10}} + \dfrac{7}{{15}} = \dfrac{{3 \times 3 + 7 \times 2}}{{30}}\\$

$\dfrac{3}{{10}} + \dfrac{7}{{15}} = \dfrac{{9 + 14}}{{30}}\\$

$\dfrac{3}{{10}} + \dfrac{7}{{15}} = \dfrac{{23}}{{30}}$

3. \[\mathbf{\dfrac{4}{9} + \dfrac{2}{7}}\]

Ans: We are given \[\dfrac{4}{9} + \dfrac{2}{7}\]

$\dfrac{4}{9} + \dfrac{2}{7} = \dfrac{{4 \times 7 + 2 \times 9}}{{63}}\\$

$\dfrac{4}{9} + \dfrac{2}{7} = \dfrac{{28 + 18}}{{63}}\\$

$\dfrac{4}{9} + \dfrac{2}{7} = \dfrac{{46}}{{63}}$

4. \[\mathbf{\dfrac{5}{7} + \dfrac{1}{3}}\]

Ans: We are given \[\dfrac{5}{7} + \dfrac{1}{3}\]

$\dfrac{5}{7} + \dfrac{1}{3} = \dfrac{{5 \times 3 + 1 \times 7}}{{21}}\\$

$\dfrac{5}{7} + \dfrac{1}{3} = \dfrac{{15 + 7}}{{21}}\\$

$\dfrac{5}{7} + \dfrac{1}{3} = \dfrac{{22}}{{21}} \text{ or }1\dfrac{1}{{21}}$

5. \[\mathbf{\dfrac{2}{5} + \dfrac{1}{6}}\]

Ans: We are given \[\dfrac{2}{5} + \dfrac{1}{6}\]

$\dfrac{2}{5} + \dfrac{1}{6} = \dfrac{{2 \times 6 + 1 \times 5}}{{30}}\\$

$\dfrac{2}{5} + \dfrac{1}{6} = \dfrac{{12 + 5}}{{30}}\\$

$\dfrac{2}{5} + \dfrac{1}{6} = \dfrac{{17}}{{30}}$

6. \[\mathbf{\dfrac{4}{5} + \dfrac{2}{3}}\]

Ans: We are given \[\dfrac{4}{5} + \dfrac{2}{3}\]

$\dfrac{4}{5} + \dfrac{2}{3} = \dfrac{{4 \times 3 + 2 \times 5}}{{15}}$

$\dfrac{4}{5} + \dfrac{2}{3} = \dfrac{{12 + 10}}{{15}}$

$\dfrac{4}{5} + \dfrac{2}{3} = \dfrac{{22}}{{15}} \text{ or } 1\dfrac{7}{{15}}$

7. \[\mathbf{\dfrac{3}{4} - \dfrac{1}{3}}\]

Ans: We are given \[\dfrac{3}{4} - \dfrac{1}{3}\]

$\dfrac{3}{4} - \dfrac{1}{3} = \dfrac{{3 \times 3 - 1 \times 4}}{{12}}$

$\dfrac{3}{4} - \dfrac{1}{3} = \dfrac{{9 - 4}}{{12}}\\$

$\dfrac{3}{4} - \dfrac{1}{3} = \dfrac{5}{{12}}$

8. \[\mathbf{\dfrac{5}{6} - \dfrac{1}{3}}\]

Ans: We are given \[\dfrac{5}{6} - \dfrac{1}{3}\]

$\dfrac{5}{6} - \dfrac{1}{3} = \dfrac{{5 \times 3 - 1 \times 6}}{{18}}\\$

$\dfrac{5}{6} - \dfrac{1}{3} = \dfrac{{15 - 6}}{{18}}\\$

$\dfrac{5}{6} - \dfrac{1}{3} = \dfrac{9}{{18}} \text{ or } \dfrac{1}{2}$

9. \[\mathbf{\dfrac{2}{3} + \dfrac{3}{4} + \dfrac{1}{2}}\]

Ans: We are given \[\dfrac{2}{3} + \dfrac{3}{4} + \dfrac{1}{2}\]

$\dfrac{2}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{{2 \times 4 + 3 \times 3 + 1 \times 6}}{{12}}$

$\dfrac{2}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{{8 + 9 + 6}}{{12}}$

$\dfrac{2}{3} + \dfrac{3}{4} + \dfrac{1}{2} = \dfrac{{23}}{{12}} \text{ or } 1\dfrac{{11}}{{12}}$

10. \[\mathbf{\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}}\]

Ans: We are given \[\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6}\]

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{{1 \times 3 + 1 \times 2 + 1}}{6}\\$

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{{3 + 2 + 1}}{6}\\$

$\dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{6}{6} \text{ or 1}$

11. \[\mathbf{1\dfrac{1}{3} + 3\dfrac{2}{3}}\]

Ans: We are given \[\dfrac{4}{3} + \dfrac{{11}}{3}\]

$\dfrac{4}{3} + \dfrac{{11}}{3} = \dfrac{{4 + 11}}{3}\\$

$\dfrac{4}{3} + \dfrac{{11}}{3} = \dfrac{{15}}{3} \text{ or 5}$

12. \[\mathbf{4\dfrac{2}{3} + 3\dfrac{1}{4}}\]

Ans: We are given \[\dfrac{{14}}{3} + \dfrac{{13}}{4}\]

$\dfrac{{14}}{3} + \dfrac{{13}}{4} = \dfrac{{14 \times 4 + 13 \times 3}}{{12}}\\$

$\dfrac{{14}}{3} + \dfrac{{13}}{4} = \dfrac{{56 + 39}}{{12}}\\$

$\dfrac{{14}}{3} + \dfrac{{13}}{4} = \dfrac{{95}}{{12}} \text{ or }7\dfrac{{11}}{{12}}$

13. \[\mathbf{\dfrac{16}{5} - \dfrac{7}{5}}\]

Ans: We are given \[\dfrac{{16}}{5} - \dfrac{7}{5}\]

$\dfrac{{16}}{5} - \dfrac{7}{5} = \dfrac{{16 - 7}}{5}$

$\dfrac{16}{5} - \dfrac{7}{5} = \dfrac{9}{5}\; \text{or} 1\dfrac{4}{5}$

14. \[\mathbf{\dfrac{4}{3} - \dfrac{1}{2}}\]

Ans: We are given \[\dfrac{4}{3} - \dfrac{1}{2}\]

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

$\dfrac{4}{3} - \dfrac{1}{2} = \dfrac{{8 - 3}}{6}$

$\dfrac{4}{3} - \dfrac{1}{2} = \dfrac{5}{6}$

2. Sarika bought \[\dfrac{2}{5}\] meter of ribbon and Lalita \[\dfrac{3}{4}\]meter of ribbon. What is the total length of the ribbon they bought?

Ans: Length of ribbon bought by Sarita = \[\dfrac{2}{5}\]meter

Length of ribbon bought by Lalita = \[\dfrac{3}{4}\]meter

Total length of ribbon they bought = \[\dfrac{2}{5} + \dfrac{3}{4}\]

$\dfrac{2}{5} + \dfrac{3}{4} = \dfrac{{2 \times 4 + 3 \times 5}}{{20}}$

$\dfrac{2}{5} + \dfrac{3}{4} = \dfrac{{8 + 15}}{{20}}$

$\dfrac{2}{5} + \dfrac{3}{4} = \dfrac{{23}}{{20}}\text{ or }1\dfrac{3}{{20}}$

They both bought $$ 1\dfrac{3}{{20}} $$ meter of ribbon. 

3. Naina was given \[1\dfrac{1}{2}\] a piece of cake and Najma was given \[1\dfrac{1}{3}\] a piece of cake. Find the total amount of cake given to both of them.

Ans: fraction of cake given to Naina = \[1\dfrac{1}{2}\]

fraction of cake given to Najma = \[1\dfrac{1}{3}\]

Total cake given to both = \[1\dfrac{1}{2} + 1\dfrac{1}{3}\]

$1\dfrac{1}{2} + 1\dfrac{1}{3} = \dfrac{3}{2} + \dfrac{4}{3}$

$1\dfrac{1}{2} + 1\dfrac{1}{3} = \dfrac{{3 \times 3 + 4 \times 2}}{6}$

$1\dfrac{1}{2} + 1\dfrac{1}{3} = \dfrac{{9 + 8}}{6}$

$1\dfrac{1}{2} + 1\dfrac{1}{3} = \dfrac{{17}}{6} \text{ or } 2\dfrac{5}{6}$

Hence, they both got \[2\dfrac{5}{6}\]fraction of cake.

4. Fill in the boxes:

1. \[\mathbf{\boxed{} - \dfrac{5}{8} = \dfrac{1}{4}}\]

Ans: Let the unknown fraction be x

$x - \dfrac{5}{8} = \dfrac{1}{4}  $

$x = \dfrac{1}{4} + \dfrac{5}{8} $

$x = \dfrac{{1 \times 2 + 5}}{8}  $

$x = \dfrac{7}{8}  $

2. \[\mathbf{\boxed{} - \dfrac{1}{5} = \dfrac{1}{2}}\]

Ans: Let the unknown fraction be x

$x - \dfrac{1}{5} = \dfrac{1}{2}$

$x = \dfrac{1}{2} + \dfrac{1}{5}$

$x = \dfrac{{1 \times 5 + 1 \times 2}}{{10}}$

$x = \dfrac{7}{{10}}$

3. \[\mathbf{\dfrac{1}{2} - \boxed{} = \dfrac{1}{6}}\]

Ans: Let the unknown fraction be x

$\dfrac{1}{2} - x = \dfrac{1}{6}$

$x = \dfrac{1}{2} - \dfrac{1}{6}$

$x = \dfrac{{1 \times 3 - 1}}{6}$

$x = \dfrac{2}{6} \text{ or }\dfrac{1}{3}$

5. Complete the addition – subtraction box:

Ans:

Ans:

6. A piece of wire \[\dfrac{7}{8}\]meter long broke into two pieces. One piece was \[\dfrac{1}{4}\] meter long. How long is the other piece?

Ans: Total length of wire piece = \[\dfrac{7}{8}\]meter

Length of one piece = \[\dfrac{1}{4}\]meter

Let the length of the other piece be x meter

So,

$\dfrac{1}{4} + x = \dfrac{7}{8}$

$x = \dfrac{7}{8} - \dfrac{1}{4}$

$x = \dfrac{{7 - 1 \times 2}}{8}$

$x = \dfrac{5}{8}$

Hence, the other piece is \[\dfrac{5}{8}\]meter.

7. Nandini house is \[\dfrac{9}{{10}}\] km from her school. She walked some distance and then took a bus for \[\dfrac{1}{2}\] km to reach the school. How far did she walk?

Ans: Total distance to be travelled by Nandini from home to school = \[\dfrac{9}{{10}}\]km

Distance covered by bus = \[\dfrac{1}{2}\]km

Let the distance she walked be x km

$\dfrac{1}{2} + x = \dfrac{9}{{10}}\\$

$x = \dfrac{9}{{10}} - \dfrac{1}{2}\\$

$x = \dfrac{{9 - 1 \times 5}}{{10}}\\$

$x = \dfrac{4}{{10}} \text{ or }\dfrac{2}{5}$

Therefore, Nandini walked \[\dfrac{2}{5}\] km.

8. Asha and Samuel have bookshelves of the same size partly filled with books. Asha’s shelf is \[\dfrac{5}{6}\] th full and Samuel’s shelf is \[\dfrac{2}{5}\]th full. Whose bookshelf is more full? By what fraction?

Ans: fraction of books in Asha’s bookshelf = \[\dfrac{5}{6}\]

fraction of books in Samuel’s bookshelf = \[\dfrac{2}{5}\]

$\dfrac{5}{6} = \dfrac{{5 \times 5}}{{6 \times 5}}$

$\dfrac{5}{6} = \dfrac{{25}}{{30}}$

And,

$\dfrac{2}{5} = \dfrac{{2 \times 6}}{{5 \times 6}}$

$\dfrac{2}{5} = \dfrac{{12}}{{30}}$

We can see that since, \[\dfrac{{25}}{{30}} > \dfrac{{12}}{{30}}\], \[\therefore \dfrac{5}{6} > \dfrac{2}{5}\]

Hence, Asha’s bookshelf is more filled.

Difference between Asha’s and Samuel’s bookshelf = \[\dfrac{5}{6} - \dfrac{2}{5}\]

$\dfrac{5}{6} - \dfrac{2}{5} = \dfrac{{5 \times 5 - 2 \times 6}}{{30}}$

$\dfrac{5}{6} - \dfrac{2}{5} = \dfrac{{25 - 12}}{{30}}$

$\dfrac{5}{6} - \dfrac{2}{5} = \dfrac{{13}}{{30}}$

9. Jaidev takes \[\dfrac{5}{6} - \dfrac{2}{5} = \dfrac{{5 \times 5 - 2 \times 6}}{{30}}\] minutes to walk across the school ground. Rahul takes \[\dfrac{7}{4}\] minutes to do same. Who takes less time and by what fraction?

Ans: Time taken by Jaidev = \[\dfrac{5}{6} - \dfrac{2}{5} = \dfrac{{5 \times 5 - 2 \times 6}}{{30}}\]minutes

Time taken by Rahul = \[\dfrac{7}{4}\]minutes

$2\dfrac{1}{5} = \dfrac{{11}}{5}$

$2\dfrac{1}{5} = \dfrac{{11 \times 4}}{{5 \times 4}}$

$2\dfrac{1}{5} = \dfrac{{44}}{{20}}$

And,

$\dfrac{7}{4} = \dfrac{{7 \times 5}}{{4 \times 5}}$

$\dfrac{7}{4} = \dfrac{{35}}{{20}}$

Since, \[\dfrac{{44}}{{20}} > \dfrac{{35}}{{20}}\], \[\therefore 2\dfrac{1}{5} > \dfrac{7}{4}\]

Difference between their times = \[\dfrac{{44}}{{20}} - \dfrac{{35}}{{20}}\]

$\dfrac{{44}}{{20}} - \dfrac{{35}}{{20}} = \dfrac{{44 - 35}}{{20}}$

$\dfrac{{44}}{{20}} - \dfrac{{35}}{{20}} = \dfrac{9}{{20}}$

Therefore, Rahul takes \[\dfrac{9}{{20}}\]minutes less time than Jaidev. 

NCERT Solutions For Class 6 Maths Chapter 7 PDF – Free Download

Class 6th Maths Chapter 7 may seem a little complicated for the students since several sub-topics deal with different types of Fractions. In order to master that, students should focus on understanding the fundamentals. 

There is a wide range of options available if you are looking for an ideal study material that can help you explore the underlying areas of this chapter. NCERT Solutions Class 6 Maths Chapter 7 worksheet with answers not only allows you to grasp the critical segments of Fraction but also helps to deal with tricky questions.

Even though several sections of this chapter may look tricky in the beginning, once a student grasps it thoroughly, these questions do not seem that difficult to solve. NCERT Solutions for Class 6 Maths Chapter 7 PDF is available on our website which presents the right approach to grab excellent scores in Maths.

Core Sections of Fraction Chapter for Maths Class 6

Since understanding the basics is your primary objective to achieve the desired outcome, you should be aware of all these core sections of Fraction. 

Topics Covered in NCERT Class 6 Maths Chapter 7 Fractions

The variety of topics that comprise the Fractions Chapter needs to be covered thoroughly to score more marks and build concepts. The topics covered in NCERT Solutions Class 6 Maths Chapter 7 Fractions are given below.

Important Points of Chapter 7 — Fractions  

• In mathematics, a fraction is a number that represents the part of a whole. The whole could be a single object or a group of objects.

• A fraction is represented in \[\frac{p}{q}\] form, where p is referred to as the numerator and q is said as the denominator. 

• In a proper fraction, the numerator is less than the denominator whereas when the numerator is greater than the denominator the fraction is said to be an improper fraction.

• A mixed fraction is written as Quotient \[\frac{Remainder}{Divisor}\] which can be written as an improper fraction as \[\frac{Quotient \times Divisor + Remainder}{Divisor}\] .

• Equivalent fractions are those fractions that represent the same part of a whole.

• To write any fraction in its simplest form, divide the numerator and denominator by their HCF.

• Like fractions are those fractions that have the same denominator, whereas the fractions that have different denominators are called, unlike fractions.

• Like fractions play a vital role in performing arithmetic operations on a set of fractions.

Benefits of Opting For NCERT Solutions for Class 6 Chapter 7

In order to get a better hold of Maths, practice is the ultimate solution that can enhance your speed skills and help you get better marks in exams. While the chapter focuses on elaborating on different vital aspects so that students understand the core concepts, NCERT Maths Book Class 6 Chapter 7 Solutions helps in understanding the various approaches that can be used to solve questions in the Fractions chapter. The solutions also give a sneak peak into important formulas that form the core of the chapter.

Few Other Advantages NCERT Solution Includes:

• Accuracy is a necessary component that can be found in all NCERT solutions since the solutions are scrutinized and crafted by subject matter experts. 

• The solutions also maintain CBSE guidelines; hence students can easily rely on the methods and patterns and apply them in future.

• Students get to explore various areas of the chapter with the help of these solutions. Along with that, they also get to learn fast-solving techniques and better methods to enhance their time-management skills.

• Since the subject is solely focused on calculations and procedures, a proper grasp of each topic is mandatory. These solutions provide the opportunity to work on your weak points which need improvement to score better.

Apart from that, NCERT Solution for Class 6 Maths Chapter 7 ensures that students do not get confused with so many different elements mentioned in the chapter. All these questions are appropriately structured so that they can understand the relevance once a topic is done. Hence, NCERT Class 6 Maths Chapter 7 solutions can be considered as a useful study material to promote conceptual clarity.

Chapter Based NCERT Solutions for Class 6 Maths

• Chapter 1 - Knowing Our Numbers

• Chapter 2 - Whole Numbers

• Chapter 3 - Playing with Numbers

• Chapter 4 - Basic Geometrical Ideas

• Chapter 5 - Understanding Elementary Shapes

• Chapter 6 - Integers

• Chapter 7 - Fractions

• Chapter 8 - Decimals

• Chapter 9 - Data Handling

• Chapter 10 - Mensuration

• Chapter 11 - Algebra

• Chapter 12 - Ratio and Proportion

• Chapter 13 - Symmetry

• Chapter 14 - Practical Geometry

Along with this, students can also view additional study materials provided at our platform for Class 6 Maths Chapter 1 Knowing Our Numbers

• NCERT Books Class 6

• CBSE Class 6 Maths Syllabus

NCERT Solution Class 6 Maths of Chapter 7 All Exercises