RS Aggarwal Class 6 Maths Chapter 5- Fractions Test Paper-5

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

RS Aggarwal Class 6 Math Solution Chapter 5- Fractions  Test Paper-5 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 

Test Paper-5

A.1.    Define a fraction. Give five example of fractions.

 Solution: 

    A fraction is defined as a number representing a part of a whole, where the whole may         be a single object or a group of objects.

    Examples: `frac\{2}{7}` , `frac{7}{5}` , `frac\{11}{3}` , `frac\{14}{6}` , `frac\{4}{9}`

    2. What fraction of an hour is 35 minutes?

 Solution: 

    1 hour = 60 minutes

    ∴ Fraction for 35 minutes = `frac\{cancel 35^7}{cancel 60^12}` = `frac\{7}{12}`

    So, `frac\{7}{12}` part of an hour is equal to 35 minutes.

    3. Find the equivalent fraction of `frac\{5}{8}` with denominator 56.

 Solution: 

    To make denominator 56, we multiply both numerator and denominator by 7

    ∴  `frac\{5}{8}` =  `frac\{5 × 7}{8 × 7}` =  `frac\{35}{56}`

    So, the required equivalent fraction is `frac\{35}{56}`

    4. Represent 2`frac\{3}{5}` on the number line.

 Solution: 

    Let OA = AB = BC = 1 unit

    ∴ OB = 2 units and OC = 3 units

    Divide BC into 5 equal parts and take 3 parts out to reach point P.

    Clearly, point P represents the number  2`frac\{3}{5}`.

    5. Find the sum of 2`frac\{4}{5}` + 1`frac\{3}{10}` + 3`frac\{1}{15}`

 Solution: 

        2`frac\{4}{5}` + 1`frac\{3}{10}` + 3`frac\{1}{15}`

    = `frac\{14}{5}` + `frac\{13}{10}` + `frac\{46}{15}`

        L.C.M of 5, 10, 15 = 30

    = `frac\{84 + 39 + 92}{30}`

    = `frac\{cancel 215^43}{cancel 30^6}`

    = `frac\{43}{6}` = 7`frac\{1}{6}`

    6. The cost of a pen is Rs. 16`frac\{2}{3}` and that of a pencil is Rs. 4`frac\{1}{6}` Which cost more and             by how much?

 Solution: 

    Cost of a pen =  Rs. 16`frac\{2}{3}` =  Rs. `frac\{50}{3}` =  Rs. `frac\{50 × 2}{3 × 2}` = Rs. `frac\{100}{6}`

    Cost of pencil = Rs. 4`frac\{1}{6}` = Rs. `frac\{25}{6}` 

    Rs. `frac\{100}{6}` > Rs. `frac\{25}{6}` 

    So, the cost of pen is more than pencil.

    Now we find difference between their cost.

     Rs. `(frac\{50}{3}` −  `frac\{25}{6})` 

    = Rs. `(frac\{100 −  25}{6})`

    = Rs. `frac\{cancel 75^25}{cancel 6^2}`

    = Rs. `frac\{25}{2}`

    = Rs. 12`frac\{1}{2}`

    Hence, the cost of a pen is Rs. 12`frac\{1}{2}` more than the cost of a pencil.

    

    7. Of `frac\{3}{4}` and `frac\{5}{7}`, which is greater and by how much?

 Solution: 

    First we compare `frac\{3}{4}` and `frac\{5}{7}`

    By cross multiply

    3 × 7 = 21 and 5 × 4 = 20

    So, 21 > 20

    ∴ `frac\{3}{4}` > `frac\{5}{7}`

    Now, we find their difference. 

        `frac\{3}{4}` − `frac\{5}{7}`

    = `frac\{21 − 20}{28}`

    = `frac\{1}{28}`

    Hence, `frac\{3}{4}` is greater than `frac\{5}{7}` by `frac\{1}{28}`.

    8. Convert the fractions `frac\{1}{2}` , `frac\{2}{3}` , `frac\{4}{9}` and `frac\{5}{6}` into like fractions.

 Solution: 

    The given fractions are `frac\{1}{2}` , `frac\{2}{3}` , `frac\{4}{9}` and `frac\{5}{6}`

    L.C.M. of 2, 3, 9 and 6 = 18

    Now, we have:

    `frac\{1}{2}` = `frac\{1 × 9}{2 × 9}` = `frac\{9}{18}` 

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

    `frac\{4}{9}` = `frac\{4 × 2}{9 × 2}` = `frac\{8}{18}`

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

    Hence, `frac\{9}{18}` , `frac\{12}{18}` , `frac\{8}{18}` and `frac\{15}{18}` are like fractions.

    9. Find the equivalent fraction of `frac\{3}{5}` having denominator 30.

 Solution: 

    To make denominator 30, we multiply both numerator and denominator by 6

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

    So, the required equivalent fraction is `frac\{18}{30}`

    10. Reduce `frac\{84}{98}` to the simplest form.

 Solution: 

    The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84.

    The factors of 98 are 1, 2, 7, 14, 49, 98.

    The common factors of 84 and 98 are 1, 2, 7, 14.

    The H.C.F. of 84 and 98 is 14.

    Dividing both the numerator and the denominator by the H.C.F.

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

B Mark (✓) against the correct answer in each of the following: 

    11. `frac\{24}{11}` is an example of 

        (a) a proper fraction.

        (b) an improper fraction.

        (c) a mixed fraction.

        (d) none of these.

 Solution: The correct option is (b) an improper fraction 

        In an improper fraction, the numerator is greater than the denominator.

    12. `frac\{3}{8}` is an example of 

        (a) a proper fraction.

        (b) an improper fraction.

        (c) a mixed fraction.

        (d) none of these.

 Solution: The correct option is (a) proper fraction 

        In a proper fraction, the numerator is less than the denominator.

    13. `frac\{3}{8}` and `frac\{5}{12}` on comparison give

        (a) `frac\{3}{8}` > `frac\{5}{12}`

        (b) `frac\{3}{8}` < `frac\{5}{12}`

        (c) `frac\{3}{8}` = `frac\{5}{12}`

        (d) none of these

 Solution: The correct option is  (b) `frac\{3}{8}` < `frac\{5}{12}` 

        Considering `frac\{3}{8}` and `frac\{5}{12}` 

        On cross multiplying, we get:

        3 × 12 = 36 and 8 × 5 = 40

        Clearly, 36 < 40

        ∴ `frac\{3}{8}`  < `frac\{5}{12}`

    14. The largest of the fraction `frac\{2}{3}`, `frac\{5}{9}`, `frac\{1}{2}` and `frac\{7}{12}` is 

        (a) `frac\{2}{3}`

        (b) `frac\{5}{9}`

        (c) `frac\{7}{12}`

        (d) `frac\{1}{2}`

 Solution: The correct option is (a) `frac\{2}{3}` 

        Explanation:

        L.C.M. of 3, 9, 2 and 12 = 36

        Now, we have:

        `frac\{2}{3}` = `frac\{2 × 12}{3 × 12}`  = `frac\{24}{36}` 

        `frac\{5}{9}` = `frac\{5 × 4}{9 × 4}` = `frac\{20}{36}`

        `frac\{1}{2}` = `frac\{1 × 18}{2 × 18}` = `frac\{18}{36}` 

        `frac\{7}{12}` = `frac\{7 × 3}{12 × 3}` = `frac\{21}{36}`

        Hence, `frac\{24}{36}` = `frac\{2}{3}` is the largest fraction.

    15. 3`frac\{3}{4}` − 1`frac\{1}{2}` = ?

        (a) 2`frac\{1}{2}`

        (b) 2`frac\{1}{4}`

        (c) 1`frac\{1}{2}`

        (d) 1`frac\{1}{4}`

 Solution: The correct option is  (b) 2`frac\{1}{4}` 

            3`frac\{3}{4}` − 1`frac\{1}{2}`

        = `frac\{15}{4}` − `frac\{3}{2}`

        = `frac\{15 − 6}{4}`

        = `frac\{9}{4}`

        = 2`frac\{1}{4}`

    16. Which of the following are like fractions?

        (a) `frac\{2}{3}`, `frac\{3}{4}`, `frac\{4}{5}`, `frac\{5}{6}`

        (b) `frac\{2}{5}`, `frac\{2}{7}`, `frac\{2}{9}`, `frac\{2}{11}`

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

        (d) none of these

 Solution: The correct option is  (c) `frac\{1}{8}`, `frac\{3}{8}`, `frac\{5}{8}`, `frac\{7}{8}` 

        Like fractions have same the denominator.

    17. ?  −   `frac\{8}{21}` = `frac\{8}{21}`

        (a) 0

        (b) 1

        (c) `frac\{21}{8}`

        (d) `frac\{16}{21}`

 Solution: The correct option is  (d) `frac\{16}{21}` 

        ?  −   `frac\{8}{21}` = `frac\{8}{21}`

        ? = `frac\{8}{21}` +   `frac\{8}{21}`

        ? =  `frac\{8 + 8}{21}`

        ? =  `frac\{16}{21}`

C.18. Fill in the blanks: 

        (i) 9`frac\{2}{3}` +            =  19

        (ii) 6`frac\{1}{6}` −   ? = `frac\{29}{30}`

        (iii) 7 −  5`frac\{2}{3}` =          

        (iv) `frac\{72}{90}` reduced to simplest form is                . 

        (v) `frac\{42}{54}` = `frac\{7}{☐}`

 Solution: 

       (i) Let the required number be `x`

        ∴ 9`frac\{2}{3}` + `x`  =  19

        ⇒ `x` = 19 − 9`frac\{2}{3}`

        ⇒ `x` = 19  −  `frac\{29}{3}`

        ⇒ `x` = `frac\{57 − 29}{3}`

        ⇒ `x` = `frac\{28}{3}`

        ⇒ `x` = 9`frac\{1}{3}`

    (ii) Let the required number be `x`

        ∴ 6`frac\{1}{6}` −   `x` = `frac\{29}{30}`

        ⇒ `x` = 6`frac\{1}{6}` − `frac\{29}{30}`

        ⇒ `x` = `frac\{37}{6}`  −  `frac\{29}{30}`

        ⇒ `x` = `frac\{185 − 29}{30}`

        ⇒ `x` = `frac\{156}{30}`

        ⇒ `x` = 5`frac\{cancel 6^1}{cancel 30^5}`

        ⇒ `x` = 5`frac\{1}{5}`

     (iii) Let the required number be `x`

        ∴ 7 −  5`frac\{2}{3}` = `x`

        ⇒ `x` = `frac\{7}{1}` − 5`frac\{2}{3}`

        ⇒ `x` = `frac\{7}{1}` − `frac\{17}{3}`

        ⇒ `x` = `frac\{21 − 17}{3}`

        ⇒ `x` = `frac\{4}{3}`

    (iv) H.C.F 72 and 90 = 18

        Now, 

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

       (v) `frac\{42}{54}` = `frac\{7}{☐}`

            By cross multiply

            42 × ☐ = 54 × 7

            ☐ = `frac\{54 × 7}{42}`

            ☐ = `frac\{cancel 378^9}{cancel 42}`

             ☐ = 9

D.19. Write 'T' for true and 'F' for false for each of the statements given below:

        (a) 3`frac\{1}{3}` > `frac\{33}{10}`

        (b) 8 −  1`frac\{5}{6}` = 7`frac\{1}{6}`

        (c) `frac\{1}{2}`, `frac\{1}{3}` and `frac\{1}{4}` are like fractions

        (d) `frac\{3}{5}` lies between 3 and 5.

        (e) Among `frac\{1}{2}`, `frac\{1}{3}`, `frac\{3}{4}`. `frac\{4}{3}` the largest fraction is `frac\{4}{3}`.

 Solution: 

        (a) T                     

        (b) F                   `(frac\{8}{1}` − `frac\{11}{6}` = `frac\{48 −11}{6}` = `frac\{37}{6}` = 6`frac\{1}{6})`

        (c) F                    (Because like fractions have the same denominator.)

    ​    (d) F                    (It lies between 0 and 1 as all proper fractions are less than 1.)

        (e) T                    (Because it is an improper fraction, while the others are proper fractions.)