RS Aggarwal 2019,2020 solution class 7 chapter 2 Fractions Exercise 2A

Exercise 2A

Page-21

Question 1:

Compare the fractions:

(i) $\frac{5}{8}\mathrm{and}\frac{7}{12}$
(ii) $\frac{5}{9}\mathrm{and}\frac{11}{15}$
(iii) $\frac{11}{12}\mathrm{and}\frac{15}{16}$

Answer 1:

We have the following:

(i) $\frac{5}{8} \mathrm{and} \frac{7}{12}$
  
By cross multiplication, we get:
5 $\times$ 12 = 60 and 7 $\times$ 8 = 56
However, 60 > 56
∴  $\frac{5}{8}>\frac{7}{12}$

(ii) $\frac{5}{9}\mathrm{and}\frac{11}{15}$
By cross multiplication, we get:
 5 $\times$ 15 = 75 and 9 $\times$ 11 = 99
However, 75 < 99
∴  $\frac{5}{9}<\frac{11}{15}$

(iii) $\frac{11}{12}\mathrm{and}\frac{15}{16}$
By cross multiplication, we get:
11 $\times$ 16 = 176 and 12 $\times$ 15 = 180
However, 176 < 180
∴  $\frac{11}{12}<\frac{15}{16}$

Question 2:

Arrange the following fractions in ascending order:

(i) $\frac{3}{4},\frac{5}{6},\frac{7}{9},\frac{11}{12}$
(ii) $\frac{4}{5},\frac{7}{10},\frac{11}{15},\frac{17}{20}$

Answer 2:

(i) The given fractions are $\frac{3}{4},\frac{5}{6},\frac{7}{9}\mathrm{and}\frac{11}{12}$.
    
LCM of 4, 6, 9 and 12 = 36
    
Now, let us change each of the given fractions into an equivalent fraction with 72 as its denominator.
$\frac{3}{4}=\frac{3\times 9}{4\times 9}=\frac{27}{36 }$

$\frac{5}{6}=\frac{5\times 6}{6\times 6}=\frac{30}{36}$

$\frac{7}{9}=\frac{7\times 4}{9\times 4}=\frac{28}{36}$

$\frac{11}{12}=\frac{11\times 3}{12\times 3}=\frac{33}{36}$

Clearly, $\frac{27}{36}<\frac{28}{36}<\frac{30}{36}<\frac{33}{36}$

Hence,$\frac{3}{4}<\frac{7}{9}<\frac{5}{6}<\frac{11}{12}$

∴ The given fractions in ascending order are $\frac{3}{4}, \frac{7}{9}, \frac{5}{6} \mathrm{and} \frac{11}{12}.$

(ii) The given fractions are: $\frac{4}{5}, \frac{7}{10}, \frac{11}{15} \mathrm{and} \frac{17}{20}.$

LCM of 5, 10, 15 and 20 = 60

Now, let us change each of the given fractions into an equivalent fraction with 60 as its denominator.

$\frac{4}{5}=\frac{4\times 12}{5\times 12}=\frac{48}{60}$

$\frac{7}{10}=\frac{7\times 6}{10\times 6}=\frac{42}{60}$

$\frac{11}{15}=\frac{11\times 4}{15\times 4}=\frac{44}{60}$

$\frac{17}{20}=\frac{17\times 3}{20\times 3}=\frac{51}{60}$

Clearly, $\frac{42}{60}<\frac{44}{60}<\frac{48}{60}<\frac{51}{60}$

Hence,$\frac{7}{10}<\frac{11}{15}<\frac{4}{5}<\frac{17}{20}$

∴ The given fractions in ascending order are $\frac{7}{10}, \frac{11}{15}, \frac{4}{5} \mathrm{and} \frac{17}{20}.$

Page-22

Question 3:

Arrange the following fractions in descending order:

(i) $\frac{3}{4},\frac{7}{8},\frac{7}{12},\frac{17}{24}$
(ii) $\frac{2}{3},\frac{3}{5},\frac{7}{10},\frac{8}{15}$

Answer 3:

We have the following:
(i) The given fractions are $\frac{3}{4}, \frac{7}{8}, \frac{7}{12} \mathrm{and} \frac{17}{24}.$

LCM of 4,8,12 and 24 = 24

Now, let us change each of the given fractions into an equivalent fraction with 24 as its denominator.
$\frac{3}{4}=\frac{3\times 6}{4\times 6}=\frac{18}{24}$

$\frac{7}{8}=\frac{7\times 3}{8\times 3}=\frac{21}{24}$

$\frac{7}{12}=\frac{7\times 2}{12\times 2}=\frac{14}{24}$

$\frac{17}{24}=\frac{17\times 1}{24\times 1}=\frac{17}{24}$

Clearly, $\frac{21}{24}>\frac{18}{24}>\frac{17}{24}>\frac{14}{24}$

Hence, $\frac{7}{8}>\frac{3}{4}>\frac{17}{24}>\frac{7}{12}$

∴ The given fractions in descending order are $\frac{7}{8}, \frac{3}{4}, \frac{17}{24} \mathrm{and} \frac{7}{12}.$

(ii) The given fractions are $\frac{2}{3}, \frac{3}{5}, \frac{7}{10} \mathrm{and} \frac{8}{15}.$
LCM of 3,5,10 and 15 = 30
Now, let us change each of the given fractions into an equivalent fraction with 30 as its denominator.
$\frac{2}{3}=\frac{2\times 10}{3\times 10}=\frac{20}{30}$

$\frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}$

$\frac{7}{10}=\frac{7\times 3}{10\times 3}=\frac{21}{30}$

$\frac{8}{15}=\frac{8\times 2}{15\times 2}=\frac{16}{30}$

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

Hence, $\frac{7}{10}>\frac{2}{3}>\frac{3}{5}>\frac{8}{15}$

∴ The given fractions in descending order are $\frac{7}{10}, \frac{2}{3}, \frac{3}{5} \mathrm{and} \frac{8}{15}$.

Question 4:

Reenu got $\frac{2}{7}$ part of an apple while Sonal got $\frac{4}{5}$ part of it. Who got the larger part and by how much?

Answer 4:

We will compare the given fractions $\frac{2}{7}\mathrm{and}\frac{4}{5}$ in order to know who got the larger part of the apple.
We have,
By cross multiplication, we get:
2 $\times$ 5 = 10  and 4 $\times$ 7 = 28

However, 10 < 28
∴ $\frac{2}{7}<\frac{4}{5}$
Thus, Sonal got the larger part of the apple.

Now, $\frac{4}{5}-\frac{2}{7}=\frac{28-10}{35}=\frac{18}{35}$

∴ Sonal got $\frac{18}{35}$ part of the apple more than Reenu.

Question 5:

Find the sum:

(i) $\frac{5}{9}+\frac{3}{9}$
(ii) $\frac{8}{9}+\frac{7}{12}$
(iii) $\frac{5}{6}+\frac{7}{8}$
(iv) $\frac{7}{12}+\frac{11}{16}+\frac{9}{24}$
(v) $3\frac{4}{5}+2\frac{3}{10}+1\frac{1}{15}$
(vi) $8\frac{3}{4}+10\frac{2}{5}$

Answer 5:

(i) $\frac{5}{9}+\frac{3}{9}=\frac{8}{9}$

(ii) $\frac{8}{9}+\frac{7}{12}$
    
     $= \frac{32}{36} + \frac{21}{36}$            [∵ LCM of 9 and 12 = 36]

      = $\frac{32+21}{36}$          
         
      = $\frac{53}{36}=1\frac{17}{36}$

(iii) $\frac{5}{6}+\frac{7}{8}$

      $= \frac{20}{24} + \frac{21}{24}$                   [∵ LCM of 6 and 8 = 24]

       = $\frac{20+21}{24}$

       =$\frac{41}{24}=1\frac{17}{24}$

(iv) $\frac{7}{12}+\frac{11}{16}+\frac{9}{24}$

     $\frac{28}{48} + \frac{33}{48} + \frac{18}{48}$             [∵ LCM of 12, 16 and 24 = 48]

      = $\frac{28+33+18}{48}$

      = $\frac{79}{48}=1\frac{31}{48}$

(v) $3\frac{4}{5}+2\frac{3}{10}+1\frac{1}{15}$

     = $\frac{19}{5}+\frac{23}{10}+\frac{16}{15}$

   $= \frac{114}{30} + \frac{69}{30} + \frac{32}{30}$            [∵ LCM of 5, 10 and 15 = 30]

     = $\frac{114+69+32}{30}$

     = $\frac{215}{30}=7\frac{5}{30} = 7\frac{1}{6}$

(vi) $8\frac{3}{4}+10\frac{2}{5}$

      = $\frac{35}{4}+\frac{52}{5}$

     $= \frac{175}{20} + \frac{208}{20}$                  [∵ LCM of 4 and 5 = 20]

      = $\frac{175+208}{20}$

      = $\frac{383}{20}=19\frac{3}{20}$

Question 6:

Find the difference:

(i) $\frac{5}{7}-\frac{2}{7}$
(ii) $\frac{5}{6}-\frac{3}{4}$
(iii) $3\frac{1}{5}-\frac{7}{10}$
(iv) $7-4\frac{2}{3}$
(v) $3\frac{3}{10}-1\frac{7}{15}$
(vi) $2\frac{5}{9}-1\frac{7}{15}$

Answer 6:

(i) $\frac{5}{7}-\frac{2}{7}=\frac{5-2}{7}=\frac{3}{7}$

(ii) $\frac{5}{6}-\frac{3}{4}$

     $= \frac{10}{12} - \frac{9}{12}$                 [∵ LCM of 6 and 4 = 12]

     = $\frac{10-9}{12}$             
    
      = $\frac{1}{12}$

(iii) $3\frac{1}{5}-\frac{7}{10}$

      = $\frac{16}{5}-\frac{7}{10}$

     =  $\frac{32}{10} - \frac{7}{10}$         [∵ LCM of 5 and 10 = 10]

      =  $\frac{32-7}{10}$
          
      = $\frac{25}{10}=\frac{5}{2}=2\frac{1}{2}$

(iv) $7-4\frac{2}{3}$

       =  $\frac{7}{1}-\frac{14}{3}$

       = $\frac{21-14}{3}$        [∵ LCM of 1 and 3 = 3]
          
       = $\frac{7}{3}=2\frac{1}{3}$

(v) $3\frac{3}{10}-1\frac{7}{15}$

      = $\frac{33}{10}-\frac{22}{15}$

      = $\frac{99-44}{30}$              [∵ LCM of 10 and 15 = 30]
      
      = $\frac{55}{30}=\frac{11}{6}=1\frac{5}{6}$

(vi) $2\frac{5}{9}-1\frac{7}{15}$

       = $\frac{23}{9}-\frac{22}{15}$

       = $\frac{115-66}{45}$                [∵ LCM of 9 and 15 = 45]
 
       =$\frac{49}{45}=1\frac{4}{45}$

Question 7:

Simplify:

(i) $\frac{2}{3}+\frac{5}{6}-\frac{1}{9}$
(ii) $8-4\frac{1}{2}-2\frac{1}{4}$
(iii) $8\frac{5}{6}-3\frac{3}{8}+1\frac{7}{12}$

Answer 7:

(i) $\frac{2}{3}+\frac{5}{6}-\frac{1}{9}$

   = $\frac{12+15-2}{18}$    [∵ LCM of 3, 6 and 9 = 18]

  = $\frac{27-2}{18}=\frac{25}{18}=1\frac{7}{18}$

(ii) $8-4\frac{1}{2}-2\frac{1}{4}$

 = $\frac{8}{1}-\frac{9}{2}-\frac{9}{4}$

 = $\frac{32-18-9}{4}$     [∵ LCM of 1, 2 and 4 = 4]

 = $\frac{32-27}{4}=\frac{5}{4}=1\frac{1}{4}$

(iii) $8\frac{5}{6}-3\frac{3}{8}+1\frac{7}{12}$

    = $\frac{53}{6}-\frac{27}{8}+\frac{19}{12}$

    =$\frac{212-81+38}{24}$     [∵ LCM of 6, 8 and 12 = 24]

    = $\frac{250-81}{24}=\frac{169}{24}=7\frac{1}{24}$

Question 8:

Aneeta bought $3\frac{3}{4}$ kg apples and $4\frac{1}{2}$ kg guava. What is the total weight of fruits purchased by her?

Answer 8:

Total weight of fruits bought by Aneeta = $\left(3\frac{3}{4} + 4\frac{1}{2}\right) \mathrm{kg}$
Now, we have:

$3\frac{3}{4} + 4\frac{1}{2} = \frac{15}{4} + \frac{9}{2}$

                $=\frac{15 + 18}{4}$[∵ LCM of 2 and 4 = 4]

                $=\frac{15 + 18}{4}=\frac{33}{4}=8\frac{1}{4}$

Hence, the total weight of the fruits purchased by Aneeta is $8\frac{1}{4} \mathrm{kg}$.

Question 9:

A rectangular sheet of paper is $15\frac{3}{4}$ cm long and $12\frac{1}{2}$ cm wide. Find its perimeter.

Answer 9:

We have:

Perimeter of the rectangle ABCD = AB + BC + CD +DA
= $\left(15\frac{3}{4} + 12\frac{1}{2} + 15\frac{3}{4} + 12\frac{1}{2}\right) \mathrm{cm}$
= $\left(\frac{63}{4} + \frac{25}{2} + \frac{63}{4} + \frac{25}{2}\right) \mathrm{cm}$
= $\left(\frac{63 + 50 + 63 + 50}{4}\right) \mathrm{cm}$          [∵ LCM of 2 and 4 = 4]
= $\left(\frac{226}{4}\right) \mathrm{cm}=\left(\frac{113}{2}\right) \mathrm{cm}=56\frac{1}{2} \mathrm{cm}$

Hence, the perimeter of ABCD is $56\frac{1}{2} \mathrm{cm}$.

Question 10:

A picture is $7\frac{3}{5}$ cm wide. How much should it be trimmed to fit in a frame $7\frac{3}{10}$ cm wide?

Answer 10:

Actual width of the picture = $7\frac{3}{5}\mathrm{cm}=\frac{38}{5}\mathrm{cm}$
Required width of the picture = $7\frac{3}{10}\mathrm{cm}=\frac{73}{10}\mathrm{cm}$
∴ Extra width = $\left(\frac{38}{5}-\frac{73}{10}\right)\mathrm{cm}$
                       = $\left(\frac{76-73}{10}\right)\mathrm{cm}$      [∵ LCM of 5 and 10 is 10]
                       = $\frac{3}{10}\mathrm{cm}$
Hence, the width of the picture should be trimmed by $\frac{3}{10} \mathrm{cm}$.

Question 11:

What should be added to $7\frac{3}{5}$ to get 18?

Answer 11:

Required number to be added = $18-7\frac{3}{5}$

                                                =$\frac{18}{1}-\frac{38}{5}$

                                                = $\frac{90-38}{5}$             [∵ LCM of 1 and 5 = 5]
                                                = $\frac{52}{5}=10\frac{2}{5}$

Hence, the required number is $10\frac{2}{5}$.

Question 12:

What should be added to $7\frac{4}{15}$ to get $8\frac{2}{5}$?

Answer 12:

Required number to be added = $8\frac{2}{5}-7\frac{4}{15}$

                                                = $\frac{42}{5}-\frac{109}{15}$

                                                = $\frac{126-109}{15}$    [∵ LCM of 5 and 15 = 15]
                                               
                                               =$\frac{17}{15}=1\frac{2}{15}$

Hence, the required number should be $1\frac{2}{15}$.

Question 13:

A piece of wire $3\frac{3}{4}$ m long broke into two pieces. One piece is $1\frac{1}{2}$ m long. How long is the other piece?

Answer 13:

Required length of other piece of wire = $\left(3\frac{3}{4}-1\frac{1}{2}\right)\mathrm{m}$

                                                  =$\left(\frac{15}{4}-\frac{3}{2}\right)\mathrm{m}$

                                                  =$\left(\frac{15-6}{4}\right)\mathrm{m}$    [∵ LCM of 4 and 2 = 4]

                                                  = $\frac{9}{4}\mathrm{m}=2\frac{1}{4}\mathrm{m}$

Hence, the length of the other piece of wire is $2\frac{1}{4}\mathrm{m}$.

Question 14:

A film show lasted of $3\frac{2}{3}$ hours. Out of this time $1\frac{1}{2}$ hours was spent on advertisements. What was the actual duration of the film?

Answer 14:

Actual duration of the film = $\left(3\frac{2}{3}-1\frac{1}{2}\right)\mathrm{hours}$

                                           = $\left(\frac{11}{3}-\frac{3}{2}\right)\mathrm{hours}$

                                           = $\left(\frac{22-9}{6}\right)\mathrm{hours}$   [∵ LCM of 3 and 2 = 6]

                                           = $\frac{13}{6}\mathrm{hours}=2\frac{1}{6}\mathrm{hours}$

Hence, the actual duration of the film was $2\frac{1}{6}\mathrm{hours}$.

Question 15:

Of $\frac{2}{3}$ and $\frac{5}{9}$ , which is greater and by how much?

Answer 15:

First we have to compare the fractions: $\frac{2}{3} \mathrm{and} \frac{5}{9}$.
By cross multiplication, we have:
2 $\times$ 9 = 18 and 5 $\times$ 3 = 15

However, 18 > 15
∴$\frac{2}{3}>\frac{5}{9}$

So, $\frac{2}{3}$ is larger than $\frac{5}{9}$.
Now, $\frac{2}{3}-\frac{5}{9}$

      = $\frac{6-5}{9}$    [∵ LCM of 3 and 9 = 9]
      =$\frac{1}{9}$
Hence, $\frac{2}{3}$ is $\frac{1}{9}$ part more than $\frac{5}{9}$.

Question 16:

The cost of a pen is Rs $16\frac{3}{5}$ and that of a pencil is Rs $4\frac{3}{4}$. Which costs more and by how much?

Answer 16:

First, we have to compare the cost of the pen and the pencil.
Cost of the pen = Rs $16\frac{3}{5} = \mathrm{Rs} \frac{83}{5}$

Cost of the pencil = Rs $4\frac{3}{4} = \mathrm{Rs} \frac{19}{4}$
Now, we have to compare fractions $\frac{83}{5} \mathrm{and} \frac{19}{4}.$
By cross multiplication, we get:

83 $\times$ 4 = 332 and 19 $\times$ 5 = 95

However, 332 > 95

∴ $\frac{83}{5}>\frac{19}{4}$

So, the cost of pen is more than that of the pencil.
Now, $\mathrm{Rs} \left(\frac{83}{5} - \frac{19}{4}\right)$

      = $\mathrm{Rs} \left(\frac{332 - 95}{20}\right)$     [∵ LCM of 4 and 5 = 20]

      = $\mathrm{Rs} \frac{237}{20} = \mathrm{Rs} 11\frac{17}{20}$

∴ The pen costs Rs $11\frac{17}{20}$ more than the pencil.