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

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

RS Aggarwal Class 6 Math Solution Chapter 5- Fractions  Exercise 5D 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 5D


1. Define like and unlike fractions and give five examples of each. 


Solution:


    Like fractions: Fractions having the same denominator are called like fractions.

    Examples: `frac\{4}{12}` ,   `frac\{6}{12}` ,   `frac\{3}{12}` ,   `frac\{7}{12}` ,  `frac\{8}{12}` 


    Unlike fractions: Fractions having different denominators are called unlike fractions.


    Example: `frac\{7}{10}` , `frac\{3}{5}`, `frac\{5}{9}` , `frac\{7}{12}` , `frac\{13}{24}`


2. Convert `frac\{3}{5}`, `frac\{7}{10}` , `frac\{8}{15}` and   `frac\{11}{30}`   into like fractions.


Solution:


    We know that like fractions having same denominator.


    So, We do,


    L.C.M of 5, 10, 15 and 30 = 30


    Now, we convert the given fractions into equivalent fractions with 30 as the denominator.


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


     `frac\{7}{10}` = `frac\{7 × 3}{10 × 3}` = `frac\{21}{30}`


    `frac\{8}{15}`` = `frac\{8 × 2}{15× 2}` = `frac\{16}{30}`


    `frac\{11}{30}` = `frac\{11× 1}{30× 1}` = `frac\{11}{30}`


    Hence, the required like fractions are `frac\{18}{30}`, `frac\{21}{30}`, `frac\{16}{30}`, and `frac\{11}{30}`



3. Convert  `frac\{1}{4}` ,  `frac\{5}{8}` , `frac\{7}{12}` and   `frac\{13}{24}` into like fractions. 


Solution:


    We know that like fractions having same denominator.


    So, We do,


    L.C.M of 4, 8, 12 and 24= 24


    Now, we convert the given fractions into equivalent fractions with 24as the denominator.


   `frac\{1}{4}` = `frac\{1× 6}{4 × 6}` = `frac\{6}{24}`


    `frac\{5}{8}` = `frac\{5× 3}{8× 3}` = `frac\{15}{24}`


   `frac\{7}{12}` = `frac\{7× 2}{12× 2}` = `frac\{14}{24}`


    `frac\{13}{24}` = `frac\{13× 1}{24× 1}` = `frac\{13}{24}`


    Hence, the required like fractions are `frac\{6}{24}`, `frac\{15}{24}`, `frac\{14}{24}`, and `frac\{13}{24}`



4. Fill in the place holders with the correct symbol  > or <:


    (i) `frac\{8}{9}` `frac\{5}{9}`       (ii) `frac\{9}{10}` ⬜ `frac\{7}{10}` (iii)  `frac\{3}{7}` ⬜ `frac\{6}{7}`


    (iv) `frac\{11}{15}` ⬜ `frac\{8}{15}` (v) `frac\{6}{11}` ⬜ `frac\{5}{11}`     (vi)  `frac\{11}{20}` ⬜ `frac\{17}{20}`


Solution:


     (i) `frac\{8}{9}` > `frac\{5}{9}`       (ii) `frac\{9}{10}` > `frac\{7}{10}` (iii)  `frac\{3}{7}` < `frac\{6}{7}`

    (iv) `frac\{11}{15}` > `frac\{8}{15}` (v) `frac\{6}{11}` > `frac\{5}{11}`     (vi)  `frac\{11}{20}` < `frac\{17}{20}`



5. Fill in the place holders with the correct symbol > or <: 


    (i) `frac\{3}{4}` `frac\{3}{5}`      (ii) `frac\{7}{8}` `frac\{7}{10}`     (iii)  `frac\{4}{11}` `frac\{9}{11}`


    (iv) `frac\{8}{11}` `frac\{8}{13}`  (v) `frac\{5}{12}` `frac\{5}{8}`      (vi)  `frac\{11}{14}` ⬜ `frac\{11}{15}`


Solution:


        (i) `frac\{3}{4}` > `frac\{3}{5}`      (ii) `frac\{7}{8}` > `frac\{7}{10}`      (iii)  `frac\{4}{11}` < `frac\{9}{11}`


    (iv) `frac\{8}{11}` > `frac\{8}{13}`  (v) `frac\{5}{12}` < `frac\{5}{8}`      (vi)  `frac\{11}{14}` > `frac\{11}{15}`



Compare the fractions given below:


6. `frac\{4}{5}` ,  `frac\{5}{7}`


Solution:


     `frac\{4}{5}` and  `frac\{5}{7}`


    By cross multiplying:     4 × 7 = 28 and 5 × 5 = 25     Clearly, 28 > 25


    ∴   `frac\{4}{5}` >  `frac\{5}{7}`



7. `frac\{3}{8}` , `frac\{5}{6}`         


Solution:


     `frac\{3}{8}` and  `frac\{5}{6}`


    By cross multiplying:     3 × 6 = 18 and 5 × 8 = 40     Clearly, 18 < 40


    ∴   `frac\{3}{8}` < `frac\{5}{6}`



8.   `frac\{7}{11}` , `frac\{6}{7}`


Solution:


     `frac\{7}{11}` and  `frac\{6}{7}`


    By cross multiplying:     7 × 7 = 49 and 6 × 11 = 66     Clearly, 49 < 66


    ∴   `frac\{7}{11}` < `frac\{6}{7}`



9.  `frac\{5}{6}` ,  `frac\{9}{11}`            


Solution:


     `frac\{5}{6}` and  `frac\{9}{11}`


    By cross multiplying:     5 × 11 = 55 and 9 × 6 = 54     Clearly, 55 > 54


    ∴   `frac\{5}{6}` > `frac\{9}{11}`



10. `frac\{2}{3}` ,  `frac\{4}{9}`               


Solution:


     `frac\{2}{3}` and  `frac\{4}{9}`


    By cross multiplying:     2 × 9 = 18 and 4 × 3 = 12     Clearly, 18 > 12


    ∴   `frac\{2}{3}` > `frac\{4}{9}`



11. `frac\{6}{13}` , `frac\{3}{4}`


Solution:


     `frac\{6}{13}` and  `frac\{3}{4}`


    By cross multiplying:     6 × 4 = 24 and 3 × 13 = 39     Clearly, 24 < 39


    ∴   `frac\{6}{13}` < `frac\{3}{4}`



12.  `frac\{3}{4}` , `frac\{5}{6}`        


Solution:


     `frac\{3}{4}` and  `frac\{5}{6}`


    By cross multiplying:     3 × 6 = 18 and 5 × 4 = 20     Clearly, 18 < 20


    ∴   `frac\{3}{4}` < `frac\{5}{6}`



13.  `frac\{5}{8}` ,  `frac\{7}{12}`           


Solution:


     `frac\{5}{8}` and  `frac\{7}{12}`


    By cross multiplying:     5 × 12 = 60 and 7 × 8 = 56     Clearly, 60 > 56


    ∴   `frac\{5}{8}` > `frac\{7}{12}`



14.   `frac\{4}{9}` , `frac\{5}{6}`


Solution:


     `frac\{4}{9}` and  `frac\{5}{6}`


    By cross multiplying:     4 × 6 = 24 and 5 × 9 = 45     Clearly, 24 < 45


    ∴   `frac\{4}{9}` < `frac\{5}{6}`




15.  `frac\{4}{5}` ,  `frac\{7}{10}`           


Solution:


     `frac\{4}{5}` and  `frac\{7}{10}`


    By cross multiplying:     4 × 10 = 40 and 7 × 5 = 35     Clearly, 40 > 35


    ∴   `frac\{4}{5}` > `frac\{7}{10}`



16. `frac\{7}{8}` ,  `frac\{9}{10}`          


Solution:


     `frac\{7}{8}` and  `frac\{9}{10}`


    By cross multiplying:     7 × 10 = 70 and 9 × 8 = 72     Clearly, 70 < 72


    ∴   `frac\{7}{8}` < `frac\{9}{10}`



17.  `frac\{11}{12}` ,  `frac\{13}{15}`


Solution:


     `frac\{11}{12}` and  `frac\{13}{15}`


    By cross multiplying:     11 × 15 = 165 and 13 × 12 = 156     Clearly, 165 > 156


    ∴   `frac\{11}{12}` > `frac\{13}{15}`



Arrange the following fractions in ascending order:


18.  `frac\{1}{2}` ,   `frac\{3}{4}` ,   `frac\{5}{6}` ,   and   `frac\{7}{8}` 


Solution:


    L.C.M of 2, 4, 6 and 8 = 24

    We convert each of the given fractions into an equivalent fraction with denominator 24.


    `frac\{1}{2}` = `frac\{1 × 12}{2 × 12}` = `frac\{12}{24}`


    `frac\{3}{4}` = `frac\{3 × 6}{4 × 6}` = `frac\{18}{24}`


    `frac\{5}{6}` = `frac\{5 × 4}{6 × 4}` = `frac\{20}{24}`


    `frac\{7}{8}` = `frac\{7 × 3}{8 × 3}` = `frac\{21}{24}`


    Clearly, `frac\{12}{24}` < `frac\{18}{24}` < `frac\{20}{24}`< `frac\{21}{24}`


    ∴  `frac\{1}{2}` < `frac\{3}{4}` < `frac\{5}{6}` < `frac\{7}{8}`



19.  `frac\{2}{3}` ,   `frac\{5}{6}` ,   `frac\{7}{9}` ,   and   `frac\{11}{18}`


Solution:


    L.C.M of 3, 6, 9 and 18 = 18

    We convert each of the given fractions into an equivalent fraction with denominator 18.


    `frac\{2}{3}` = `frac\{2 × 6}{3 × 6}` = `frac\{12}{18}`


    `frac\{5}{6}` = `frac\{5 × 3}{6 × 3}` = `frac\{15}{18}`


    `frac\{7}{9}` = `frac\{7 × 2}{9 × 2}` = `frac\{14}{18}`


    `frac\{11}{18}` = `frac\{11 × 1}{18 × 1}` = `frac\{11}{18}`


    Clearly, `frac\{11}{18}` < `frac\{12}{18}` < `frac\{14}{18}`< `frac\{15}{18}`


    ∴  `frac\{11}{18}` < `frac\{2}{3}` < `frac\{7}{9}` < `frac\{5}{6}`



20.   `frac\{2}{5}` ,   `frac\{7}{10}` ,   `frac\{11}{15}` ,   and   `frac\{17}{30}`


Solution:


    L.C.M of 5, 10, 15 and 30 = 30

    We convert each of the given fractions into an equivalent fraction with denominator 30.


    `frac\{2}{5}`  = `frac\{2 × 6}{5 × 6}` = `frac\{12}{30}`


    `frac\{7}{10}` = `frac\{7 × 3}{10 × 3}` = `frac\{21}{30}`


    `frac\{11}{15}` = `frac\{11 × 2}{15 × 2}` = `frac\{22}{30}`


    `frac\{17}{30}` = `frac\{17 × 1}{30 × 1}` = `frac\{17}{30}`


    Clearly, `frac\{12}{30}` < `frac\{17}{30}` < `frac\{21}{30}`< `frac\{22}{30}`


    ∴  `frac\{2}{5}` < `frac\{17}{30}` < `frac\{7}{10}` < `frac\{11}{15}`



21.   `frac\{3}{4}` ,   `frac\{7}{8}` ,   `frac\{11}{16}` ,   and   `frac\{23}{32}`


Solution:


    L.C.M of 4, 8, 16 and 32 = 32

    We convert each of the given fractions into an equivalent fraction with denominator 30.


    `frac\{3}{4}`  = `frac\{3 × 8}{4 × 8}` = `frac\{24}{32}`


    `frac\{7}{8}` = `frac\{7 × 3}{8 × 3}` = `frac\{21}{32}`


    `frac\{11}{16}` = `frac\{11 × 2}{16 × 2}` = `frac\{22}{32}`


    `frac\{23}{32}` = `frac\{23 × 1}{32 × 1}` = `frac\{23}{32}`


    Clearly, `frac\{21}{32}` < `frac\{22}{32}` < `frac\{23}{32}`< `frac\{24}{32}`


    ∴  `frac\{7}{8}` < `frac\{11}{16}` < `frac\{23}{32}` < `frac\{3}{4}`




Arrange the following fractions in descending order: 


22.  `frac\{3}{4}` ,   `frac\{5}{8}` ,   `frac\{11}{12}` ,   and   `frac\{17}{24}` 


Solution:


    L.C.M of 4, 8, 12 and 24 = 24

    We convert each of the given fractions into an equivalent fraction with denominator 24.


    `frac\{3}{4}`  = `frac\{3 × 6}{4 × 6}` = `frac\{18}{24}`


    `frac\{5}{8}` = `frac\{5 × 3}{8 × 3}` = `frac\{15}{24}`


    `frac\{11}{12}` = `frac\{11 × 2}{12 × 2}` = `frac\{22}{24}`


    `frac\{17}{24}` = `frac\{17 × 1}{24 × 1}` = `frac\{17}{24}`


    Clearly, `frac\{22}{24}` > `frac\{18}{24}` > `frac\{17}{24}`> `frac\{15}{24}`


    ∴  `frac\{11}{12}` > `frac\{3}{4}` > `frac\{17}{24}` > `frac\{5}{8}`



23.  `frac\{7}{9}` ,   `frac\{5}{12}` ,   `frac\{11}{18}` ,   and   `frac\{17}{36}`


Solution:


    L.C.M of 9, 12, 18 and 36 = 36

    We convert each of the given fractions into an equivalent fraction with denominator 36.


    `frac\{7}{9}`  = `frac\{7 × 4}{9 × 4}` = `frac\{28}{36}`


    `frac\{5}{12}` = `frac\{5 × 3}{12 × 3}` = `frac\{15}{36}`


    `frac\{11}{18}` = `frac\{11 × 2}{18 × 2}` = `frac\{22}{36}`


    `frac\{17}{36}` = `frac\{17 × 1}{36 × 1}` = `frac\{17}{36}`


    Clearly, `frac\{28}{36}` > `frac\{22}{36}` > `frac\{17}{36}`> `frac\{15}{36}`


    ∴  `frac\{7}{9}` > `frac\{11}{18}` > `frac\{17}{36}` > `frac\{5}{12}`



24.  `frac\{2}{3}` ,   `frac\{3}{5}` ,   `frac\{7}{10}` ,   and   `frac\{8}{15}`


Solution:


    L.C.M of 3, 5, 10 and 15 = 30

    We convert each of the given fractions into an equivalent fraction with denominator 30.


    `frac\{2}{3}`  = `frac\{2 × 10}{3 × 10}` = `frac\{20}{30}`


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


    `frac\{7}{10}` = `frac\{7 × 3}{10 × 3}` = `frac\{21}{30}`


    `frac\{8}{15}` = `frac\{8 × 2}{15 × 2}` = `frac\{16}{30}`


    Clearly, `frac\{21}{30}` > `frac\{20}{30}` > `frac\{18}{30}` > `frac\{16}{30}`


    ∴  `frac\{7}{10}` > `frac\{2}{3}` > `frac\{3}{5}` > `frac\{8}{15}`



25.  `frac\{5}{7}` ,   `frac\{9}{14}` ,   `frac\{17}{21}` ,   and   `frac\{31}{42}`


Solution:


    L.C.M of 7, 14, 21 and 42 = 42

    We convert each of the given fractions into an equivalent fraction with denominator 42.


    `frac\{5}{7}`  = `frac\{5 × 6}{7 × 6}` = `frac\{30}{42}`


    `frac\{9}{14}` = `frac\{9 × 3}{14 × 3}` = `frac\{27}{42}`


    `frac\{17}{21}` = `frac\{17 × 2}{21 × 2}` = `frac\{34}{42}`


    `frac\{31}{42}` = `frac\{31 × 1}{42 × 1}` = `frac\{31}{42}`


    Clearly, `frac\{34}{42}` > `frac\{31}{42}` > `frac\{30}{42}` > `frac\{27}{42}`


    ∴  `frac\{17}{21}` > `frac\{31}{42}` > `frac\{5}{7}` > `frac\{9}{14}`



26.  `frac\{1}{12}` ,   `frac\{1}{23}` ,   `frac\{1}{7}` ,   `frac\{1}{9}` ,  `frac\{1}{17}` , `frac\{1}{50}` 


Solution:


    The numerators are equal     So, The fraction having small denominator is greater than the fraction having large denominator     ∴ In descending order, we can write


    Clearly, `frac\{1}{7}` > `frac\{1}{9}` > `frac\{1}{12}` > `frac\{1}{17}` > `frac\{1}{23}` > `frac\{1}{50}`



27.  `frac\{3}{7}` ,   `frac\{3}{11}` ,   `frac\{3}{5}` ,   `frac\{3}{13}` ,   `frac\{3}{4}` ,   `frac\{3}{17}`    


Solution:


    The numerators are equal     So, The fraction having small denominator is greater than the fraction having large denominator     ∴ In descending order, we can write


    Clearly, `frac\{3}{4}` > `frac\{3}{5}` > `frac\{3}{7}` > `frac\{3}{11}` > `frac\{3}{13}` > `frac\{3}{17}`



28. Lalita read 30 pages of a book containing 100 pages while Sarita read  `frac\{2}{5}` of the book. Who read more?


Solution:


    Lalita reads 30 pages out of 100 pages.

    Sarita read `frac\{2}{5}` of the same book = `frac\{2}{5}` of 100 pages = `frac\{2}{5}` × 100 = `frac\{cancel 200^40}{cancel 5}` = 40 pages

    Hence, Sarita read more pages than Lalita as 40 is greater than 30.



29. Rafiq exercised for  `frac\{2}{3}` hour, while Rohit exercised for  `frac\{3}{4}` hour. Who exercised for a longer time?


Solution:


    To know who spent more time on exercise,

    We have to compare `frac\{2}{3}` hour with `frac\{3}{4}` hour .

    On cross multiplying:     2 × 4 = 8 and 3 × 3 = 9     Clearly, 8 < 9

    ∴ `frac\{2}{3}` < `frac\{3}{4}`

    Hence, Rohit exercised for a longer time.



30. In a school 20 students out of 25 passed in VI A. while 24 out of 30 passed in VI B. Which

section gave better results?


Solution: 

    Fraction of students who passed in VI A = `frac\{20}{25}` = `frac\{20÷5}{25÷5}` = `frac\{4}{5}`

    Fraction of students who passed in VI B = `frac\{24}{30}` = `frac\{24÷6}{30÷6}` = `frac\{4}{5}`

    In both the sections, the fraction of students who passed is the same, so both the sections have the same result.






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