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

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

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

1. Which of the following are proper fractions?

        `frac\{1}{2}`  ,  `frac\{3}{5}` ,  `frac\{10}{7}` , `frac\{7}{4}` , 2 ,  `frac\{15}{8}` ,  `frac\{16}{16}` ,  `frac\{10}{11}` , `frac\{23}{10}` 

 Solution: 

    A fraction whose denominator is greater than its numerator is called Proper Fraction. 

    Therefore,  `frac\{1}{2}`  ,  `frac\{3}{5}` , and `frac\{10}{11}` are  Proper Fraction. 

2. Which of the following are improper fractions?

        `frac\{3}{2}`  ,  `frac\{5}{6}` ,  `frac\{9}{4}` ,  `frac\{8}{8}` , 3 ,  `frac\{27}{16}` ,  `frac\{23}{31}` ,  `frac\{19}{18}` , `frac\{10}{13}` , `frac\{26}{26}`

 Solution: 

    A fraction whose denominator is smaller than its numerator is called Improper Fraction. 

    Therefore,  `frac\{3}{2}`, `frac\{9}{4}` ,  `frac\{8}{8}`, 3 ,  `frac\{27}{16}`,  `frac\{19}{18}` and `frac\{26}{26}` are Improper Fraction. 

3. Write six improper fractions with denominator 5.

 Solution: 

    Six improper fractions with denominator 5 are

    `frac\{6}{5}` , `frac\{7}{5}` , `frac\{8}{5}` , `frac\{9}{5}` , `frac\{10}{5}` and `frac\{11}{5}`

4. Write six improper fractions with numerator 13. 

 Solution: 

    Six improper fractions with denominator 13 are

    `frac\{13}{10}` , `frac\{13}{9}` , `frac\{13}{8}` , `frac\{13}{7}` , `frac\{13}{6}` and `frac\{13}{5}`

5. Convert each of the following into an improper fraction:

    (i) 5`frac\{5}{7}`     

    (ii) 9`frac\{3}{8}`

    (iii) 6`frac\{3}{10}`         

    (iv) 3`frac\{5}{11}`

    (v) 10`frac\{9}{14}`

    (vi) 12`frac\{7}{15}`

    (vii) 8`frac\{8}{13}`

    (viii) 51`frac\{2}{3}`

 Solution: 

    (i) 5`frac\{5}{7}` = `frac\{5 × 7 + 5}{7}` = `frac\{35 + 5}{7}` = `frac\{40}{7}`

    (ii) 9`frac\{3}{8}` = `frac\{9 × 8 + 3}{8}` = `frac\{72 + 3}{8}` = `frac\{75}{8}`

    (iii) 6`frac\{3}{10}` = `frac\{6 × 10 + 3}{10}` = `frac\{60 + 3}{10}` = `frac\{63}{10}`

    (iv) 3`frac\{5}{11}` = `frac\{3 × 11 + 5}{11}` = `frac\{33 + 5}{11}` = `frac\{38}{11}`

    (v) 10`frac\{9}{14}` = `frac\{10 × 14 + 9}{14}` = `frac\{140 + 9}{14}` = `frac\{149}{14}`

    (vi) 12`frac\{7}{15}` = `frac\{12 × 15 + 7}{15}` = `frac\{180 + 7}{15}` = `frac\{187}{15}`

    (vii) 8`frac\{8}{13}` = `frac\{8 × 13 + 8}{7}` = `frac\{104 + 8}{13}` = `frac\{112}{13}`

    (viii) 51`frac\{2}{3}` = `frac\{51 × 3 + 2}{3}` = `frac\{153 + 2}{3}` = `frac\{155}{3}`

6. Convert each of the following into a mixed fraction:

    (i) `frac\{17}{5}`         

    (ii) `frac\{62}{7}`         

    (iii) `frac\{101}{8}`         

    (iv) `frac\{95}{13}`

    (v) `frac\{81}{11}`       

    (vi) `frac\{87}{16}`         

    (vii) `frac\{103}{12}`     

    (viii) `frac\{117}{20}`

 Solution: 

    (i) On dividing 17 by 5, we get:

        Quotient = 3

        Remainder = 2

       ∴ `frac\{17}{5}`  =  3 + `frac\{2}{5}`  = 3`frac\{2}{5}`  

    (ii) On dividing 62 by 7, we get:

        Quotient = 8

        Remainder = 6

       ∴  `frac\{62}{7}`  =  8 + `frac\{6}{7}`  = 8`frac\{6}{7}`  

    (iii) On dividing 101 by 8, we get:

        Quotient = 12

        Remainder = 5

       ∴  `frac\{101}{8}`  =  12 + `frac\{5}{8}`  = 12`frac\{5}{8}`  

    (iv) On dividing 95 by 13, we get:

        Quotient = 7

        Remainder = 4

       ∴  `frac\{95}{13}`  =  7 + `frac\{4}{13}`  = 7`frac\{4}{13}`  

    (v) On dividing 81 by 11, we get:

        Quotient = 7

        Remainder = 4

       ∴  `frac\{81}{11}`  =  7 + `frac\{4}{11}`  = 7`frac\{4}{11}`  

    (vi) On dividing 87 by 16, we get:

        Quotient = 5

        Remainder = 7

       ∴  `frac\{87}{16}`  =  5 + `frac\{7}{16}`  = 5`frac\{7}{16}`  

    (vii) On dividing 103 by 12, we get:

        Quotient = 8

        Remainder = 7

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

    (viii) On dividing 117 by 20, we get:

        Quotient = 5

        Remainder = 17

       ∴ `frac\{117}{20}`  =  5 + `frac\{17}{20}`  = 5`frac\{17}{20}`  

7. Fill up the blanks with `>` `<` or `=`.

    (i) `frac\{1}{2}` ▭ 1         

    (ii) `frac\{3}{4}` ▭ 1         

    (iii) 1 ▭ `frac\{6}{7}`         

    (iv)  `frac\{6}{6}` ▭ 1

    (v)  `frac\{3016}{3016}` ▭ 1

    (vi)  `frac\{11}{5}` ▭ 1

 Solution: 

    (i) `frac\{1}{2}`  <  1                    

    (ii) `frac\{3}{4}`  <  1                 

    (iii) 1  >  `frac\{6}{7}`            

    (iv)  `frac\{6}{6}`  =  1                 

    (v)  `frac\{3016}{3016}`  =  1             

    (vi)  `frac\{11}{5}`  >  1 

8. Draw number lines and locate the following points:

        (i)  `frac\{1}{4}` , `frac\{1}{2}` , `frac\{3}{4}` , `frac\{4}{4}` ,        

        (ii) `frac\{1}{8}` , `frac\{2}{8}` , `frac\{3}{8}` , `frac\{5}{8}` ,`frac\{7}{8}`     

        (iii) `frac\{2}{5}` , `frac\{3}{5}` , `frac\{4}{5}` , `frac\{8}{5}`

 Solution: