RS Aggarwal Class 6 Maths Chapter 5- Fractions Exercise 5C

 RS Aggarwal 2021-2022 for Class 6 Maths Chapter 5- Fractions

RS Aggarwal Class 6 Math Solution Chapter 5- Fractions  Exercise 5C is available here. RS Aggarwal Class 6 Math Solutions are solved by expert teachers in step by step, which help the students to understand easily.  RS Aggarwal textbooks are responsible for a strong foundation in Maths. These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students’ understanding.

 Rs Aggarwal Class 6 Math Solution Chapter 5- Fractions 

Exercise 5C

1. Write five fractions equivalent to each of the following:

    (i) `frac\{2}{3}`             (ii) `frac\{4}{5}`                 (iii) `frac\{5}{8}`             (iv) `frac\{7}{10}`

    (v) `frac\{3}{7}`             (vi) `frac\{5}{11}`             (vii) `frac\{7}{9}`             (viii) `frac\{5}{12}`

Solution: 

    (i) `frac\{2}{3}`  =  `frac\{2 × 2}{3 × 2 }`  =  `frac\{2 × 3}{3 × 3 }`  =  `frac\{2 × 4}{3 × 4 }`  =  `frac\{2 × 5}{3 × 5 }`  =  `frac\{2 × 6}{3 × 6 }`

    ∴ `frac\{2}{3}`  =  `frac\{4}{6}`  =  `frac\{6}{9}`  =   `frac\{8}{12}`  =  `frac\{10}{15}`  =  `frac\{12}{18}` 

    (ii)  `frac\{4}{5}`  =  `frac\{4 × 2}{5 × 2 }`  =  `frac\{4 × 3}{5 × 3 }`  =  `frac\{4 × 4}{5 × 4 }`  =  `frac\{4 × 5}{5 × 5 }`  =  `frac\{4 × 6}{5 × 6 }`

    ∴ `frac\{4}{5}`  =  `frac\{8}{10}`  =  `frac\{12}{15}`  =   `frac\{16}{20}`  =  `frac\{20}{25}`  =  `frac\{24}{30}` 

    (iii)  `frac\{5}{8}`  =  `frac\{5 × 2}{8 × 2 }`  =  `frac\{5 × 3}{8 × 3 }`  =  `frac\{5 × 4}{8 × 4 }`  =  `frac\{5 × 5}{8 × 5 }`  =  `frac\{5 × 6}{8 × 6 }`

    ∴ `frac\{5}{8}`  =  `frac\{10}{16}`  =  `frac\{15}{24}`  =   `frac\{20}{32}`  =  `frac\{25}{40}`  =  `frac\{30}{48}` 

    (iv)  `frac\{7}{10}`  =  `frac\{7 × 2}{10 × 2 }`  =  `frac\{7 × 3}{10 × 3 }`  =  `frac\{7 × 4}{10 × 4 }`  =  `frac\{7 × 5}{10 × 5 }`  =  `frac\{7 × 6}{10 × 6 }`

    ∴ `frac\{7}{10}`  =  `frac\{14}{20}`  =  `frac\{21}{30}`  =   `frac\{28}{40}`  =  `frac\{35}{50}`  =  `frac\{42}{60}` 

    (v)  `frac\{3}{7}`  =  `frac\{3 × 2}{7 × 2 }`  =  `frac\{3 × 3}{7 × 3 }`  =  `frac\{3 × 4}{7 × 4 }`  =  `frac\{3 × 5}{7 × 5 }`  =  `frac\{3 × 6}{7 × 6 }`

    ∴ `frac\{3}{7}`  =  `frac\{6}{14}`  =  `frac\{9}{21}`  =   `frac\{12}{28}`  =  `frac\{15}{35}`  =  `frac\{18}{42}` 

    (vi)  `frac\{5}{11}`  =  `frac\{5 × 2}{11 × 2 }`  =  `frac\{5 × 3}{11 × 3 }`  =  `frac\{5 × 4}{11 × 4 }`  =  `frac\{5 × 5}{11 × 5 }`  =  `frac\{5 × 6}{11 × 6 }`

    ∴ `frac\{5}{11}`  =  `frac\{10}{22}`  =  `frac\{15}{33}`  =   `frac\{20}{44}`  =  `frac\{25}{55}`  =  `frac\{30}{66}` 

    (vii)  `frac\{7}{9}`  =  `frac\{7 × 2}{9 × 2 }`  =  `frac\{7 × 3}{9 × 3 }`  =  `frac\{7 × 4}{9 × 4 }`  =  `frac\{7 × 5}{9 × 5 }`  =  `frac\{7 × 6}{9 × 6 }`

    ∴ `frac\{7}{9}`  =  `frac\{14}{18}`  =  `frac\{21}{27}`  =   `frac\{28}{36}`  =  `frac\{35}{45}`  =  `frac\{42}{54}` 

    (viii)   `frac\{5}{12}`  =  `frac\{5 × 2}{12 × 2 }`  =  `frac\{5 × 3}{12 × 3 }`  =  `frac\{5 × 4}{12 × 4 }`  =  `frac\{5 × 5}{12 × 5 }`  =  `frac\{5 × 6}{12 × 6 }`

    ∴ `frac\{5}{12}`  =  `frac\{10}{24}`  =  `frac\{15}{36}`  =   `frac\{20}{48}`  =  `frac\{25}{60}`  =  `frac\{30}{72}` 

2. Which of the following are the pairs of equivalent fractions?

    (i) `frac\{5}{6}` and `frac\{20}{24}`             (ii) `frac\{3}{8}` and `frac\{15}{40}`                (iii) `frac\{4}{7}` and  `frac\{16}{21}`             

    (iv) `frac\{2}{9}` and `frac\{14}{63}`                (v) `frac\{1}{3}` and `frac\{9}{24}`              (vi) `frac\{2}{3}` and `frac\{33}{22}`

Solution: 

   (i) `frac\{5}{6}` and `frac\{20}{24}`

        First we do cross multiply

        5 × 24 = 20 × 6

           120  =  120

        Both products are equal. 

        So, `frac\{5}{6}` and `frac\{20}{24}` are equivalent fractions

     (ii) `frac\{3}{8}` and `frac\{15}{40}`    

        First we do cross multiply

        3 × 40 = 15 × 8

           120  =  120

        Both products are equal. 

        So, `frac\{3}{8}` and `frac\{15}{40}` are equivalent fractions

     (iii) `frac\{4}{7}` and  `frac\{16}{21}`

        First we do cross multiply

        4 × 21 = 16 × 7

             84  ≠ 112

        Both products are not equal. 

        So, `frac\{4}{7}` and  `frac\{16}{21}` are not equivalent fractions

     (iv) `frac\{2}{9}` and `frac\{14}{63}`

        First we do cross multiply

        2 × 63 = 14 × 9

             126  = 126

        Both products are  equal. 

        So,  `frac\{2}{9}` and `frac\{14}{63}` are  equivalent fractions

    (v) `frac\{1}{3}` and `frac\{9}{24}`

        First we do cross multiply

        1 × 24 = 9 × 3

             24  ≠  27

        Both products are  not equal. 

        So,  `frac\{1}{3}` and `frac\{9}{24}` are not equivalent fractions

    (vi) `frac\{2}{3}` and `frac\{33}{22}`

        First we do cross multiply

        2 × 22 =  33 × 3

             44  ≠  99

        Both products are  not equal. 

        So,  `frac\{2}{3}` and `frac\{33}{22}` are not equivalent fractions

3. Find the equivalent fraction of `frac\{3}{5}` having

    (i) denominator 30                                             (ii) numerator 24

 Solution:  

    (i) In  `frac\{3}{5}`, denominator is 5, But we have to make it 30

    Therefore, we multiply both numerator and denominator by 6, Then obtained fraction        will be equivalent to  `frac\{3}{5}` and having denominator 30

    ∴  `frac\{3}{5}` =   `frac\{3 × 6}{5 × 6}` = `frac\{18}{30}`

    (ii) In `frac\{3}{5}`, numerator is 3, But we have to make it 24

    Therefore, we multiply both numerator and denominator by 8, Then obtained fraction           will be equivalent to `frac\{3}{5}` and having numerator 24

    ∴  `frac\{3}{5}` =   `frac\{3 × 8}{5 × 8}` = `frac\{24}{40}`

4. Find the equivalent fraction of `frac\{5}{9}` having

    (i) denominator 54                                             (ii) numerator 35

Solution: 

     (i) In  `frac\{5}{9}`, denominator is 9, But we have to make it 54

    Therefore, we multiply both numerator and denominator by 6, Then obtained fraction        will be equivalent to  `frac\{5}{9}` and having denominator 54

    ∴  `frac\{5}{9}` =   `frac\{5 × 6}{9 × 6}` = `frac\{30}{54}`

    (ii) In `frac\{5}{9}`, numerator is 5, But we have to make it 35

    Therefore, we multiply both numerator and denominator by 7, Then obtained fraction           will be equivalent to `frac\{5}{9}` and having numerator 35

    ∴  `frac\{5}{9}` =   `frac\{5 × 7}{9 × 7}` = `frac\{35}{63}`

5. Find the equivalent fraction of `frac\{6}{11}` having

    (i) denominator 77                                             (ii) numerator 60

 Solution: 

    (i) In `frac\{6}{11}`, denominator is 11, But we have to make it 77

    Therefore, we multiply both numerator and denominator by 7, Then obtained fraction        will be equivalent to `frac\{6}{11}` and having denominator 77

    ∴  `frac\{6}{11}` =   `frac\{6 × 7}{11 × 7}` = `frac\{42}{77}`

    (ii) In `frac\{6}{11}`, numerator is 6, But we have to make it 60

    Therefore, we multiply both numerator and denominator by 10, Then obtained fraction           will be equivalent to `frac\{6}{11}` and having numerator 60

    ∴  `frac\{6}{11}` =   `frac\{6 × 10}{11 × 10}` = `frac\{60}{110}`

6. Find the equivalent fraction of `frac\{24}{30}` having numerator 4.

 Solution: 

    In `frac\{24}{30}`, numerator is 24, But we have to make it 4

    Therefore, we divide both numerator and denominator by 6, Then obtained fraction             will be equivalent to `frac\{24}{30}` and having numerator 4

    ∴   `frac\{24}{30}` =   `frac\{24 ÷ 6}{30 ÷ 6}` = `frac\{4}{5}`

7. Find the equivalent fraction of `frac\{36}{48}` with

    (i) numerator 9                                             (ii) denominator 4

 Solution: 

    (i) In `frac\{36}{48}` , numerator is 36, But we have to make it 9

    Therefore, we divide both numerator and denominator by 4, Then obtained fraction             will be equivalent to `frac\{36}{48}`  and having numerator 9

    ∴   `frac\{36}{48}`  =   `frac\{36 ÷ 4}{48 ÷ 4}` = `frac\{9}{12}`

    (ii)  In `frac\{36}{48}` , denominator is 48, But we have to make it 4

    Therefore, we divide both numerator and denominator by 12, Then obtained fraction             will be equivalent to `frac\{36}{48}`  and having denominator 4

    ∴   `frac\{36}{48}`  =   `frac\{36 ÷ 12}{48 ÷ 12}` = `frac\{3}{4}`

8. Find the equivalent fraction of  `frac\{56}{70}` with

    (i) numerator 4                                             (ii)  denominator 10

 Solution: 

     (i) In `frac\{56}{70}` , numerator is 56, But we have to make it 4

    Therefore, we divide both numerator and denominator by 14, Then obtained fraction             will be equivalent to `frac\{56}{70}`  and having numerator 4

    ∴   `frac\{56}{70}`  =   `frac\{56 ÷ 14}{70 ÷ 14}` = `frac\{4}{5}`

     (ii)  In `frac\{56}{70}` , denominator is 70, But we have to make it 10

    Therefore, we divide both numerator and denominator by 7, Then obtained fraction             will be equivalent to `frac\{56}{70}`  and having denominator 10

    ∴   `frac\{56}{70}` =   `frac\{56 ÷ 7}{70 ÷ 7}` = `frac\{8}{10}`

9. Reduce each of the following fractions into its simplest form:

    (i) `frac\{9}{15}`         (ii) `frac\{48}{60}`         (iii) `frac\{84}{98}`         (iv) `frac\{150}{60}`         (v) `frac\{72}{90}`

 Solution: 

    (i) `frac\{9}{15}`

    First, we will find the H.C.F of 9 and 15

    Factor of 9 = 1, 3, 9

    Factor of 15 = 1, 3, 5, 15

    Common Factor = 1, 3

    H.C.F = 3

    Now, we will divide numerator and denominator by 3 to get simplest form of `frac\{9}{15}`

    Hence, `frac\{9}{15}` = `frac\{9 ÷ 3}{15 ÷ 3 }` = `frac\{3}{5}`

    (ii) `frac\{48}{60}`

    First, we will find the H.C.F of 48 and 60

    Factor of  48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

    Factor of  60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

    Common Factor = 1, 2, 3, 4, 6, 12,

    H.C.F = 12

    Now, we will divide numerator and denominator by 12 to get simplest form of `frac\{48}{60}`

    Hence, `frac\{48}{60}` = `frac\{48 ÷ 12}{60 ÷ 12 }` = `frac\{4}{5}`

    (iii) `frac\{84}{98}` 

    First, we will find the H.C.F of 84 and 98

    Factor of  84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84

    Factor of  98 = 1, 2, 7, 14, 49, 98

    Common Factor = 1, 2, 7, 14

    H.C.F = 14

    Now, we will divide numerator and denominator by 14 to get simplest form of `frac\{84}{98}` 

    Hence, `frac\{84}{98}`  = `frac\{84 ÷ 14}{98 ÷ 14 }` = `frac\{6}{7}`

    (iv) `frac\{150}{60}`

    First, we will find the H.C.F of 150 and 60

    Factors of 150 = 1, 2, 3, 5, 6, 10, 15, 25, 30, 75, 150.

    Factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

    Common Factors  = 1, 2, 3, 5, 6, 10, 15, 30.

    H.C.F. = 30.

    Now, we will divide numerator and denominator by 30 to get simplest form of `frac\{150}{60}` 

    Hence, `frac\{150}{60}`  = `frac\{150 ÷ 30}{60 ÷ 30}` = `frac\{5}{2}`

     (v) `frac\{72}{90}`

    First, we will find the H.C.F of 72 and 90

    Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

    Factors of 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

    Common Factors  = 1, 2, 3, 6, 9,18.

    H.C.F. = 18.

    Now, we will divide numerator and denominator by 18 to get simplest form of  `frac\{72}{90}`

    Hence, `frac\{72}{90}`  = `frac\{72 ÷ 18}{90 ÷ 18}` = `frac\{4}{5}`

10. Show that each of the following fraction is in the simplest form: 

    (i) `frac\{8}{11}`         (ii) `frac\{9}{14}`         (iii) `frac\{25}{36}`         (iv) `frac\{8}{15}`         (v) `frac\{21}{10}`

 Solution: 

    (i) In `frac\{8}{11}` , numerator = 8 and denominator = 11

        Factors of 8 = 1, 2, 4 and 8.

        Factors of 11 = 1 and 11.

        Common factor of 8 and 11 = 1.

       Thus, H.C.F. of 8 and 11 = 1.

       Hence, `frac\{8}{11}`  is the simplest form.

    (ii) In `frac\{9}{14}` , numerator = 9 and denominator = 14

        Factors of 9 = 1, 3 and 9.

        Factors of 14= 1, 2, 7 and 14.

       Common factor of 9 and 14 = 1.

       Thus, H.C.F. of 9 and 14 = 1.

       Hence, `frac\{9}{14}` is the simplest form.

    (iii) In `frac\{25}{36}` , numerator = 25 and denominator = 36

         Factors of 25 = 1, 5 and 25.

         Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36.

         Common factor of 25 and 36 = 1.

         Thus, H.C.F. of 25 and 36 = 1.

         Hence, `frac\{25}{36}`  is the simplest form.

    (iv) In `frac\{8}{15}` , numerator = 8 and denominator = 15

          Factors of 8 = 1, 2, 4 and 8.

      Factors of 15 = 1, 3, 5 and 15.

      Common factor of 8 and 15 = 1.

     Thus, H.C.F. of 8 and 15 = 1.

     Hence, `frac\{8}{15}` is the simplest form.

(v) In `frac\{21}{10}` , numerator = 21 and denominator = 10

     Factors of 21 = 1, 3, 7 and 21.

     Factors of 10 = 1, 2, 5 and 10.

     Common factor of 21 and 10 = 1.

    Thus, H.C.F. of 21 and 10 = 1.

    Hence, `frac\{21}{10}`  is the simplest form.

11. Replace ..... by the correct number in each of the following:

    (i) `frac\{2}{7}` = `frac\{8}{⬜}`                 (ii) `frac\{3}{5}` = `frac\{⬜}{35}`

    (iii) `frac\{5}{8}` = `frac\{20}{⬜}`                     (iv) `frac\{45}{60}` = `frac\{9}{⬜}`

    (v) `frac\{40}{56}` = `frac\{⬜}{7}`                     (vi) `frac\{42}{54}` = `frac\{7}{⬜}`

Solution: 

(i) 28         

(ii) 21 

(iii) 32

(iv) 12 

(v) 5   

(vi) 9