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

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

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

Find the sum: 

1. `frac\{5}{8}` + `frac\{1}{8}`

Solution: 

    `frac\{5}{8}` + `frac\{1}{8}`

    = `frac\{5 + 1}{8}`

    = `frac\{cancel 6^3}{cancel 8^4}`

    = `frac\{3}{4}`

2. `frac\{4}{9}` + `frac\{8}{9}`

Solution: 

     `frac\{4}{9}` + `frac\{8}{9}`

     = `frac\{4 + 8}{9}`

    = `frac\{cancel 12^4}{cancel 9^3}`

    = `frac\{4}{3}`

3. 1`frac\{3}{5}` + 2`frac\{4}{5}`

Solution: 

    1`frac\{3}{5}` + 2`frac\{4}{5}`

    = `frac\{8}{5}` + `frac\{14}{5}`    ( Changing into Improper Fraction)

    = `frac\{8 + 14}{5}` 

    = `frac\{22}{5}`    = 4`frac\{2}{5}`

4. `frac\{2}{9}` + `frac\{5}{6}`

Solution: 

     `frac\{2}{9}` + `frac\{5}{6}`

        L.C.M of 9 and 6 = 18

     = `frac\{4 + 15}{18}`

    = `frac\{19}{18}`    = 1`frac\{1}{18}`

5. `frac\{7}{12}` + `frac\{9}{16}`

Solution: 

     `frac\{7}{12}` + `frac\{9}{16}`

        L.C.M of 12 and 16 = 48

     = `frac\{28 + 27}{48}`

    = `frac\{55}{48}`    = 1`frac\{7}{48}`

6. `frac\{4}{15}` + `frac\{17}{20}`

Solution: 

      `frac\{4}{15}` + `frac\{17}{20}`

        L.C.M of 15 and 20 = 60

     = `frac\{16 + 51}{60}`

    = `frac\{67}{60}`    = 1`frac\{7}{60}`

7. 2`frac\{3}{4}` + 5`frac\{5}{6}`

Solution: 

    2`frac\{3}{4}` + 5`frac\{5}{6}`

    = `frac\{11}{4}` + `frac\{35}{6}`         ( Changing into Improper Fraction)

       L.C.M of 4 and 6 = 12

    = `frac\{33 + 70}{12}`

    = `frac\{103}{12}`    = 8`frac\{7}{12}`

8. 3`frac\{1}{8}` + 1`frac\{5}{12}`

Solution: 

     3`frac\{1}{8}` + 1`frac\{5}{12}`

    = `frac\{25}{8}` + `frac\{17}{12}`         ( Changing into Improper Fraction)

       L.C.M of 8 and 12 = 24

    = `frac\{75 + 34}{24}`

    = `frac\{109}{24}`    = 4`frac\{13}{24}`

9. 2`frac\{7}{10}` + 3`frac\{8}{15}`

Solution: 

      2`frac\{7}{10}` + 3`frac\{8}{15}`

    = `frac\{27}{10}` + `frac\{53}{15}`         ( Changing into Improper Fraction)

       L.C.M of 10 and 15 = 30

    = `frac\{81 + 106}{30}`

    = `frac\{187}{30}`    = 6`frac\{7}{30}`

10. 3`frac\{2}{3}` + 1`frac\{5}{6}` + 2

Solution: 

      3`frac\{2}{3}` + 1`frac\{5}{6}` + 2

    = `frac\{11}{3}` + `frac\{11}{6}` + `frac\{2}{1}`        ( Changing into Improper Fraction)

       L.C.M of 3, 6 and 1 = 6

    = `frac\{22 + 11 + 12}{6}`

    = `frac\{cancel 45^15}{cancel 6^2}`

    = `frac\{15}{2}`    = 7`frac\{1}{2}`

11. 3 + 1`frac\{4}{15}` + 1`frac\{3}{20}`

Solution: 

      3 + 1`frac\{4}{15}` + 1`frac\{3}{20}`

    = `frac\{3}{1}` + `frac\{19}{15}` + `frac\{23}{20}`        ( Changing into Improper Fraction)

       L.C.M of 1, 15 and 20 = 60

    = `frac\{180+ 76 + 69}{60}`

    = `frac\{cancel 325^65}{cancel 60^12}`

    = `frac\{65}{12}`    = 5`frac\{5}{12}`

12. 3`frac\{1}{3}` + 4`frac\{1}{4}` + 6`frac\{1}{6}`

Solution:  

       3`frac\{1}{3}` + 4`frac\{1}{4}` + 6`frac\{1}{6}`

    = `frac\{10}{3}` + `frac\{17}{4}` + `frac\{37}{6}`        ( Changing into Improper Fraction)

       L.C.M of 3, 4 and 6 = 12

    = `frac\{40 + 51 + 74}{12}`

    = `frac\{cancel 165^55}{cancel 12^4}`

    = `frac\{55}{4}`    = 13`frac\{3}{4}`

13. `frac\{2}{3}` + 3`frac\{1}{6}` + 4`frac\{2}{9}` + 2`frac\{5}{18}`

Solution: 

     `frac\{2}{3}` + 3`frac\{1}{6}` + 4`frac\{2}{9}` + 2`frac\{5}{18}`

    =  `frac\{2}{3}` + `frac\{19}{6}` + `frac\{38}{9}` + `frac\{41}{18}`     ( Changing into Improper Fraction)

         L.C.M of 3, 6, 9,  and 18 = 18

    =  `frac\{12 + 57 + 76 + 41}{18}`

    =  `frac\{cancel 186^31}{cancel 18^3}`

    =  `frac\{31}{3}`   = 10`frac\{1}{3}`

14. 2`frac\{1}{3}` + 1`frac\{1}{4}` + 2`frac\{5}{6}` + 3`frac\{7}{12}`

Solution: 

    2`frac\{1}{3}` + 1`frac\{1}{4}` + 2`frac\{5}{6}` + 3`frac\{7}{12}`

    = `frac\{7}{3}` + `frac\{5}{4}` + `frac\{17}{6}` + `frac\{43}{12}`    ( Changing into Improper Fraction)

         L.C.M of 3, 4, 6,  and 12 = 12

    = `frac\{28 + 15 + 34 + 43}{12}`

    = `frac\{cancel 120^10}{cancel 12^1}`

    = 10

15. 2 + `frac\{3}{4}` + 1`frac\{5}{8}` + 3`frac\{7}{16}`

Solution: 

    2 + `frac\{3}{4}` + 1`frac\{5}{8}` + 3`frac\{7}{16}`

    = `frac\{2}{1}` + `frac\{3}{4}` + `frac\{13}{8}` + `frac\{55}{16}`

        L.C.M of 1, 4, 8,  and 16 = 16

    = `frac\{32 + 12 + 26 + 55}{16}`

    = `frac\{125}{16}`    = 7`frac\{13}{16}`

16. Rohit bought a pencil for Rs. 3`frac\{2}{5}` and an eraser for Rs. 2`frac\{7}{10}`? What is the total cost of both the articles?

Solution: 

    Cost of a pencil = Rs. 3`frac\{2}{5}` 

    Cost of an eraser =  Rs. 2`frac\{7}{10}`

    Total cost of both articles = 3`frac\{2}{5}` + 2`frac\{7}{10}`

                                             = `frac\{17}{5}` + `frac\{27}{10}`

                                            = `frac\{34 + 27}{10}`

                                             = `frac\{61}{10}`    = Rs. 6`frac\{1}{10}`

17. Sohini bought 4`frac\{1}{2}` m of cloth for her kurta and 2`frac\{2}{3}` m of cloth for her pyjamas. How much cloth did she purchase in all?

Solution: 

    Sohini bought cloth for kurta = 4`frac\{1}{2}` m

    Sohini bought cloth for pyjamas = 2`frac\{2}{3}` m 

    Total length of cloth she purchased = 4`frac\{1}{2}` + 2`frac\{2}{3}` 

                                                             = `frac\{9}{2}` + `frac\{8}{3}` 

                                                             = `frac\{27 + 16}{3}` 

                                                             = `frac\{43}{6}`    = 7`frac\{1}{6}` m

18. While coming back home from his school, Kishan covered 4`frac\{3}{4}` km by rickshaw and 1`frac\{1}{2}` km on foot. What is the distance of his house from the school? 

Solution: 

    Distance travelled by Rickshaw = 4`frac\{3}{4}` km

    Distance travelled on foot = 1`frac\{1}{2}` km

    Total distance covered = 4`frac\{3}{4}` + 1`frac\{1}{2}`

                         = `frac\{19}{4}` + `frac\{3}{2}`

                         = `frac\{19 + 6}{4}`

                         = `frac\{25}{4}` = 6`frac\{1}{4}` km.

19. The weight of an empty gas cylinder is 16`frac\{4}{5}` kg and it contains 14`frac\{2}{3}` kg of gas. What is the weight of the cylinder filled with gas?

Solution: 

    Weight of empty cylinder = 16`frac\{4}{5}` kg.

    Weight of gas filled in it = 14`frac\{2}{3}` kg.

    Total. weight of cylinder with gas = 16`frac\{4}{5}` + 14`frac\{2}{3}`

                                   = `frac\{84}{5}` + `frac\{44}{3}`

                                   = `frac\{252 + 220}{15}`

                                   = `frac\{472}{15}`

                                   = 31`frac\{7}{15}` kg.