Rs Aggarwal 2017 for Class 7 Math Chapter 2 - Fractions

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Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 2 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among Class 7 students for Math Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

Page No 21:

Answer:

We have the following:

(i) $\frac{5}{8} and \frac{7}{12}$

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

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

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

Page No 21:

Answer:

(i) The given fractions are $\frac{3}{4}, \frac{5}{6}, \frac{7}{9} and \frac{11}{12}$ .

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 $\frac{3}{4}, \frac{7}{9}, \frac{5}{6} and \frac{11}{12} .$

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

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 $\frac{7}{10}, \frac{11}{15}, \frac{4}{5} and \frac{17}{20} .$

Page No 22:

Answer:

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

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 $\frac{7}{8}, \frac{3}{4}, \frac{17}{24} and \frac{7}{12} .$

(ii) The given fractions are $\frac{2}{3}, \frac{3}{5}, \frac{7}{10} and \frac{8}{15} .$
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 $\frac{7}{10}, \frac{2}{3}, \frac{3}{5} and \frac{8}{15}$ .

Page No 22:

Answer:

We will compare the given fractions $\frac{2}{7} and \frac{4}{5}$ 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.

Page No 22:

Answer:

(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]

      = $\frac{175 + 208}{20}$

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

Page No 22:

Answer:

(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]

       = $\frac{49}{45} = 1 \frac{4}{45}$

Page No 22:

Answer:

(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]

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

Page No 22:

Answer:

Total weight of fruits bought by Aneeta = $(3 \frac{3}{4} + 4 \frac{1}{2}) kg$
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 $8 \frac{1}{4} kg$ .

Page No 22:

Answer:

We have:

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

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

Page No 22:

Answer:

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

Page No 22:

Answer:

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 $10 \frac{2}{5}$ .

Page No 22:

Answer:

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 $1 \frac{2}{15}$ .

Page No 22:

Answer:

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

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

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

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

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

Page No 22:

Answer:

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

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

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

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

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

Page No 22:

Answer:

First we have to compare the fractions: $\frac{2}{3} and \frac{5}{9}$ .
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 $\frac{5}{9}$ .

Page No 22:

Answer:

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

Cost of the pencil = Rs $4 \frac{3}{4} = Rs \frac{19}{4}$
Now, we have to compare fractions $\frac{83}{5} and \frac{19}{4} .$
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, $Rs (\frac{83}{5} - \frac{19}{4})$

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

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

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

Page No 26:

Answer:

(i) $\frac{3}{5} \times \frac{7}{11} = \frac{3 \times 7}{5 \times 11} = \frac{21}{55}$

(ii) $\frac{5}{8} \times \frac{4}{7} = \frac{5 \times 4}{8 \times 7} = \frac{5 \times 1}{2 \times 7} = \frac{5}{14}$

(iii) $\frac{4}{9} \times \frac{15}{16} = \frac{4 \times 15}{9 \times 16} = \frac{1 \times 5}{3 \times 4} = \frac{5}{12}$

(iv) $\frac{2}{5} \times 15 = \frac{2}{5} \times \frac{15}{1} = \frac{2 \times 15}{5 \times 1} = \frac{2 \times 3}{1 \times 1} = 6$

(v) $\frac{8}{15} \times 20 = \frac{8}{15} \times \frac{20}{1} = \frac{8 \times 20}{15 \times 1} = \frac{8 \times 4}{3 \times 1} = \frac{32}{3} = 10 \frac{2}{3}$

(vi) $\frac{5}{8} \times 1000 = \frac{5}{8} \times \frac{1000}{1} = \frac{5 \times 1000}{8 \times 1} = \frac{5 \times 125}{1 \times 1} = 625$

(vii) $3 \frac{1}{8} \times 16 = \frac{25}{8} \times \frac{16}{1} = \frac{25 \times 16}{8 \times 1} = \frac{25 \times 2}{1 \times 1} = 50$

(viii) $2 \frac{4}{15} \times 12 = \frac{34}{15} \times \frac{12}{1} = \frac{34 \times 12}{15 \times 1} = \frac{34 \times 4}{5 \times 1} = \frac{136}{5} = 27 \frac{1}{5}$

(ix) $3 \frac{6}{7} \times 4 \frac{2}{3} = \frac{27}{7} \times \frac{14}{3} = \frac{27 \times 14}{7 \times 3} = \frac{9 \times 2}{1 \times 1} = 18$

(x) $9 \frac{1}{2} \times 1 \frac{9}{19} = \frac{19}{2} \times \frac{28}{19} = \frac{19 \times 28}{2 \times 19} = \frac{1 \times 14}{1 \times 1} = 14$

(xi) $4 \frac{1}{8} \times 2 \frac{10}{11} = \frac{33}{8} \times \frac{32}{11} = \frac{33 \times 32}{8 \times 11} = \frac{3 \times 4}{1 \times 1} = 12$

(xii) $5 \frac{5}{6} \times 1 \frac{5}{7} = \frac{35}{6} \times \frac{12}{7} = \frac{35 \times 12}{6 \times 7} = \frac{5 \times 2}{1 \times 1} = 10$

Page No 26:

Answer:

We have the following:

(i) $\frac{2}{3} \times \frac{5}{44} \times \frac{33}{35} = \frac{2 \times 5 \times 33}{3 \times 44 \times 35} = \frac{1 \times 1 \times 11}{1 \times 22 \times 7} = \frac{1 \times 1 \times 1}{1 \times 2 \times 7} = \frac{1}{14}$

(ii) $\frac{12}{25} \times \frac{15}{28} \times \frac{35}{36} = \frac{1 \times 3 \times 5}{5 \times 4 \times 3} = \frac{1 \times 1 \times 1}{1 \times 4 \times 1} = \frac{1}{4}$

(iii) $\frac{10}{27} \times \frac{28}{65} \times \frac{39}{56} = \frac{10 \times 1 \times 3}{27 \times 5 \times 2} = \frac{1 \times 1 \times 3}{27 \times 1 \times 1} = \frac{3}{27} = \frac{1}{9}$

(iv) $1 \frac{4}{7} \times 1 \frac{13}{22} \times 1 \frac{1}{15}$

      = $\frac{11}{7} \times \frac{35}{22} \times \frac{16}{15} = \frac{11 \times 35 \times 16}{7 \times 22 \times 15} = \frac{1 \times 5 \times 16}{1 \times 2 \times 15} = \frac{1 \times 1 \times 8}{1 \times 1 \times 3} = \frac{8}{3} = 2 \frac{2}{3}$

(v) $2 \frac{2}{17} \times 7 \frac{2}{9} \times 1 \frac{33}{52}$

     = $\frac{36}{17} \times \frac{65}{9} \times \frac{85}{52} = \frac{36 \times 65 \times 85}{17 \times 9 \times 52} = \frac{4 \times 5 \times 5}{1 \times 1 \times 4} = \frac{1 \times 5 \times 5}{1 \times 1 \times 1} = 25$

(vi) $3 \frac{1}{16} \times 7 \frac{3}{7} \times 1 \frac{25}{39}$

      = $\frac{49}{16} \times \frac{52}{7} \times \frac{64}{39} = \frac{7 \times 4 \times 4}{1 \times 1 \times 3} = \frac{112}{3} = 37 \frac{1}{3}$

Page No 26:

Answer:

We have the following:

(i) $\frac{1}{3}$ of 24 = $24 \times \frac{1}{3} = \frac{24}{1} \times \frac{1}{3} = \frac{24 \times 1}{1 \times 3} = 8$

(ii) $\frac{3}{4}$ of 32 = $32 \times \frac{3}{4} = \frac{32}{1} \times \frac{3}{4} = \frac{32 \times 3}{1 \times 4} = \frac{8 \times 3}{1 \times 1} = 24$

(iii) $\frac{5}{9}$ of 45 = $45 \times \frac{5}{9} = \frac{45}{1} \times \frac{5}{9} = \frac{45 \times 5}{1 \times 9} = \frac{5 \times 5}{1 \times 1} = 25$

(iv) $\frac{7}{50}$ of 1000 = $1000 \times \frac{7}{50} = \frac{1000}{1} \times \frac{7}{50} = \frac{20 \times 7}{1 \times 1} = 140$

(v) $\frac{3}{20}$ of 1020 = $1020 \times \frac{3}{20} = \frac{1020}{1} \times \frac{3}{20} = \frac{51 \times 3}{1 \times 1} = 153$

(vi) $\frac{5}{11}$ of Rs 220 = Rs $(220 \times \frac{5}{11})$ = Rs (20 $\times$ 5 ) = Rs 100

(vii) $\frac{4}{9}$ of 54 m = $(\frac{4}{9} \times 54) m$ = (4 $\times$ 6) m = 24 m

(viii) $\frac{6}{7}$ of 35 L = $(\frac{6}{7} \times 35) L$ = (6 $\times$ 5) L = 30 L

(ix) $\frac{1}{6}$ of 1 h = $\frac{1}{6}$ of 60 min = $(60 \times \frac{1}{6})$ min = 10 min

(x) $\frac{5}{6}$ of an year = $\frac{5}{6}$ of 12 months = $(12 \times \frac{5}{6})$ months = (2 $\times$ 5) months = 10 months

(xi) $\frac{7}{20}$ of a kg = $\frac{7}{20}$ of 1000 g = $(1000 \times \frac{7}{20})$ g = (50 $\times$ 7) gm = 350 g

(xii) $\frac{9}{20}$ of 1 m = $\frac{9}{20}$ of 100 cm = $(100 \times \frac{9}{20})$ cm = (5 $\times$ 9) cm = 45 cm

(xiii) $\frac{7}{8}$ of a day = $\frac{7}{8}$ of 24 h = $(24 \times \frac{7}{8})$ h = (3 $\times$ 7) = 21 h

(xiv) $\frac{3}{7}$ of a week = $\frac{3}{7}$ of 7 days = $(7 \times \frac{3}{7})$ days = 3 days

(xv) $\frac{7}{50}$ of 1 L = $\frac{7}{50}$ of 1000 ml = $(1000 \times \frac{7}{50})$ ml = (20 $\times$ 7) ml = 140 ml

Page No 26:

Answer:

Cost of 1kg of apples = $Rs 18 \frac{2}{5} = Rs \frac{92}{5}$
∴ Cost of $3 \frac{3}{4} kg$ of apples = $Rs (\frac{92}{5} \times 3 \frac{3}{4})$
= $Rs (\frac{92}{5} \times \frac{15}{4}) = Rs (\frac{23 \times 3}{1 \times 1}) = Rs 69$

Hence, the cost of $3 \frac{3}{4} kg$ of apples is Rs 69.

Page No 26:

Answer:

Cost of 1 m of cloth = $Rs 42 \frac{1}{2} = Rs \frac{85}{2}$
∴ Cost of $5 \frac{3}{5} m$ of cloth = Rs $(\frac{85}{2} \times 5 \frac{3}{5})$
= Rs $(\frac{85}{2} \times \frac{28}{5}) = Rs (\frac{85 \times 28}{2 \times 5}) = Rs (17 \times 14) = Rs 238$
Hence, the cost of $5 \frac{3}{5} m$ of cloth is Rs 238.

Page No 26:

Answer:

Distance covered by the car in 1 h = $66 \frac{2}{3} km$
Distance covered by the car in 9 h = $(66 \frac{2}{3} \times 9) km$
= $(\frac{200}{3} \times 9) km = (\frac{200 \times 9}{3 \times 1}) km = (200 \times 3) km = 600 km$

Hence, the distance covered by the car in 9 h will be 600 km.

Page No 26:

Answer:

Capacity of 1 tin = $12 \frac{3}{4} L = \frac{51}{4} L$
∴ Capacity of 26 such tins = $(26 \times \frac{51}{4}) L$
= $(\frac{26}{1} \times \frac{51}{4}) L = (\frac{26 \times 51}{1 \times 4}) L = (\frac{13 \times 51}{1 \times 2}) L = (\frac{663}{2}) L = 331 \frac{1}{2} L$

Hence, 26 such tins can hold $331 \frac{1}{2}$ L of oil.

Page No 26:

Answer:

Cost of 1 ticket = Rs $35 \frac{1}{2}$ = Rs $\frac{71}{2}$
∴ Cost of 308 tickets = Rs $(\frac{71}{2} \times 308) = Rs (\frac{71}{2} \times \frac{308}{1}) = Rs (71 \times 154) = Rs 10934$

Hence, 308 tickets were sold for Rs 10,934.

Page No 26:

Answer:

Thickness of 1 board = $3 \frac{2}{3}$ cm
∴ Thickness of 9 boards = $(9 \times 3 \frac{2}{3}) cm$
= $(\frac{9}{1} \times \frac{11}{3}) cm$ = (3 $\times$ 11) cm = 33 cm

Hence, the height of the stack is 33 cm.

Page No 27:

Answer:

Time taken by Rohit to complete one round of the circular park = $4 \frac{4}{5}$ min = $\frac{24}{5}$ min

∴ Time taken to complete 15 rounds = $(15 \times \frac{24}{5})$ min
                                                                      = (3 $\times$ 24) min
                                                                      = 72 min
                                                                      = 1 h 12 min   [∵ 1 hr = 60 min]

Hence, Rohit will take 1 h 12 min to make 15 complete rounds of the circular park.

Page No 27:

Answer:

Weight of Amit = 35 kg
Weight of Kavita = $\frac{3}{5}$ of Amit's weight
= 35 kg x $\frac{3}{5}$ = $(35 \times \frac{3}{5}) kg = (7 \times 3) kg = 21 kg$
Hence, Kavita's weight is 21 kg.

Page No 27:

Answer:

Number of boys in the class = $\frac{5}{7}$ of the total no. of students
= $\frac{5}{7}$ $\times$ 42 = $(\frac{5 \times 42}{7}) = 5 \times 6 = 30$

∴ Number of girls in the class = 42 − 30 = 12

Hence, there are 12 girls in the class.

Page No 27:

Answer:

Sapna's total monthly income = Rs 12000
Monthly expenditure = $\frac{7}{8}$ of Rs 12000
= Rs $(\frac{7}{8} \times 12000)$ =Rs (7 $\times$ 1500) = Rs 10500
∴ Monthly savings = Rs 12000 − Rs 10500
= Rs 1500

Hence, Sapna deposits Rs 1500 in the bank every month.

Page No 27:

Answer:

Side of the square field = $4 \frac{2}{3} m$
∴ Area of the square = (side)²
= ${(4 \frac{2}{3} m)}^{2}$
= ${(\frac{14}{3} m)}^{2} = \frac{14}{3} m \times \frac{14}{3} m = (\frac{14 \times 14}{3 \times 3}) m^{2} = \frac{196}{9} m^{2} = 21 \frac{7}{9} m^{2}$

Hence, the area of the square field is $21 \frac{7}{9} m^{2}$ .

Page No 27:

Answer:

Length of the rectangular park = $41 \frac{2}{3} m = \frac{125}{3} m$

Its breadth = $18 \frac{3}{5} m = \frac{93}{5} m$

∴ Its area = length $\times$ breadth

= $(\frac{125}{3} \times \frac{93}{5}) m$ ²
= (25 $\times$ 31) m = 775 m²

Hence, the area of the rectangular park is 775 m².

Page No 30:

Answer:

(i) Reciprocal of $\frac{5}{8}$ = $\frac{8}{5}$             [ ∵ $\frac{5}{8} \times \frac{8}{5} = 1$ ]

(ii) Reciprocal of 7 = $\frac{1}{7}$              [ ∵ $7 \times \frac{1}{7} = 1$ ]

(iii) Reciprocal of $\frac{1}{12}$ = 12       [ ∵ $\frac{1}{12} \times 12 = 1$ ]

(iv) Reciprocal of $12 \frac{3}{5}$ = Reciprocal of $\frac{63}{5}$ = $\frac{5}{63}$       [∵ $\frac{63}{5} \times \frac{5}{63} = 1$ ]

Page No 30:

Answer:

(i) $\frac{4}{7} \div \frac{9}{14} = \frac{4}{7} \times \frac{14}{9}$               [∵ Reciprocal of $\frac{9}{14}$ = $\frac{14}{9}$ ]

    = $\frac{8}{9}$

(ii) $\frac{7}{10} \div \frac{3}{5} = \frac{7}{10} \times \frac{5}{3}$             [∵ Reciprocal of $\frac{3}{5}$ = $\frac{5}{3}$ ]

     = $\frac{7}{6} = 1 \frac{1}{6}$

(iii) $\frac{8}{9} \div 16 = \frac{8}{9} \times \frac{1}{16}$               [∵ Reciprocal of 16 = $\frac{1}{16}$ ]

       = $\frac{1}{18}$

(iv) $9 \div \frac{1}{3} = 9 \times 3$                       [∵ Reciprocal of $\frac{1}{3}$ = 3]

     = 27

(v) $24 \div \frac{6}{7} = 24 \times \frac{7}{6}$                 [∵ Reciprocal of $\frac{6}{7}$ = $\frac{7}{6}$ ]

    = 4 $\times$ 7 = 28

(vi) $3 \frac{3}{5} \div \frac{4}{5} = \frac{18}{5} \div \frac{4}{5}$

     = $\frac{18}{5} \times \frac{5}{4}$             [∵ Reciprocal of $\frac{4}{5}$ = $\frac{5}{4}$ ]

      = $\frac{18}{4} = \frac{9}{2} = 4 \frac{1}{2}$

(vii) $3 \frac{3}{7} \div \frac{8}{21} = \frac{24}{7} \div \frac{8}{21}$

       = $\frac{24}{7} \times \frac{21}{8}$           [∵ Reciprocal of $\frac{8}{21}$ = $\frac{21}{8}$ ]

       = 3 3 = 9

(viii) $5 \frac{4}{7} \div 1 \frac{3}{10}$ = $\frac{39}{7} \div \frac{13}{10}$

       = $\frac{39}{7} \times \frac{10}{13}$              [∵ Reciprocal of $\frac{13}{10}$ = $\frac{10}{13}$ ]

       = $\frac{30}{7} = 4 \frac{2}{7}$

(ix) $15 \frac{3}{7} \div 1 \frac{23}{49}$ = $\frac{108}{7} \div \frac{72}{49}$

      = $\frac{108}{7} \times \frac{49}{72}$           [∵ Reciprocal of $\frac{72}{49}$ = $\frac{49}{72}$ ]

      = $\frac{9 \times 7}{1 \times 6} = \frac{3 \times 7}{1 \times 2} = \frac{21}{2} = 10 \frac{1}{2}$

Page No 30:

Answer:

(i) $\frac{11}{24} \div \frac{7}{8}$

      = $\frac{11}{24} \times \frac{8}{7}$                            [∵ Reciprocal of $\frac{7}{8}$ = $\frac{8}{7}$ ]

      = $\frac{11}{21}$

(ii) $6 \frac{7}{8} \div \frac{11}{16}$ = $\frac{55}{8} \div \frac{11}{16}$

       = $\frac{55}{8} \times \frac{16}{11}$    [∵ Reciprocal of $\frac{11}{16}$ = $\frac{16}{11}$ ]

       = 5 $\times$ 2 = 10

(iii) $5 \frac{5}{9} \div 3 \frac{1}{3}$ = $\frac{50}{9} \div \frac{10}{3}$

        = $\frac{50}{9} \times \frac{3}{10}$           [∵ Reciprocal of $\frac{10}{3}$ = $\frac{3}{10}$ ]

        = $\frac{5}{3}$ = $1 \frac{2}{3}$

(iv) $32 \div 1 \frac{3}{5}$ = $32 \div \frac{8}{5}$

      = $32 \times \frac{5}{8}$                 [∵ Reciprocal of $\frac{8}{5}$ = $\frac{5}{8}$ ]

      = 4 $\times$ 5 = 20

(v) $45 \div 1 \frac{4}{5}$ = $45 \div \frac{9}{5}$

      = $45 \times \frac{5}{9}$                [∵ Reciprocal of $\frac{9}{5}$ = $\frac{5}{9}$ ]

      = 5 $\times$ 5 = 25

(vi) $63 \div 2 \frac{1}{4}$ = $63 \div \frac{9}{4}$

      = $63 \times \frac{4}{9}$             [∵ Reciprocal of $\frac{9}{4}$ = $\frac{4}{9}$ ]

      = 7 $\times$ 4 = 28

Page No 30:

Answer:

Length of the rope = $13 \frac{1}{2}$ m = $\frac{27}{2}$ m
Number of equal pieces = 9

∴ Length of each piece = $(\frac{27}{2} \div 9)$ m
= $(\frac{27}{2} \times \frac{1}{9})$ m [∵ Reciprocal of 9 = $\frac{1}{9}$ ]
= $\frac{3}{2}$ m = $1 \frac{1}{2}$ m
Hence, the length of each piece of rope is $1 \frac{1}{2}$ m.

Page No 30:

Answer:

Weight of 18 boxes of nails = $49 \frac{1}{2}$ kg = $\frac{99}{2}$ kg
∴ Weight of 1 box = $(\frac{99}{2} \div 18)$ kg
= $(\frac{99}{2} \times \frac{1}{18})$ kg [∵ Reciprocal of 18 = $\frac{1}{18}$ ]
= $(\frac{99 \times 1}{2 \times 18})$ kg = $(\frac{11 \times 1}{2 \times 2})$ kg = $\frac{11}{4}$ kg = $2 \frac{3}{4}$ kg

Hence, the weight of each box is $2 \frac{3}{4}$ kg.

Page No 31:

Answer:

Cost of 1 orange = Rs $3 \frac{3}{4}$ = Rs $\frac{15}{4}$
Total cost of the oranges sold by the man = Rs 210

∴ Required number of oranges = $(210 \div \frac{15}{4})$
= $(210 \times \frac{4}{15})$ [∵ Reciprocal of $\frac{15}{4}$ = $\frac{4}{15}$ ]
= (14 $\times$ 4) = 56

Hence, the man sold 56 oranges.

Page No 31:

Answer:

Cost of 1 kg of mangoes = Rs $18 \frac{1}{2}$ = Rs $\frac{37}{2}$
Total cost of the required mangoes = Rs $157 \frac{1}{4}$ = Rs $\frac{629}{4}$
∴ Weight of the required mangoes = $(\frac{629}{4} \div \frac{37}{2})$ kg
= $(\frac{629}{4} \times \frac{2}{37})$ kg [∵ Reciprocal of $\frac{37}{2}$ = $\frac{2}{37}$ ]
= $(\frac{17}{2})$ kg = $8 \frac{1}{2}$ kg
Hence, the weight of the mangoes available for Rs $157 \frac{1}{4}$ is $8 \frac{1}{2}$ kg.

Page No 31:

Answer:

Distance covered by Vikas in $7 \frac{3}{4}$ h = $20 \frac{2}{3}$ km
∴ Distance covered by him in 1 h = $(20 \frac{2}{3} \div 7 \frac{3}{4})$ km
                                                   = $(\frac{62}{3} \div \frac{31}{4})$ km
                                                   = $(\frac{62}{3} \times \frac{4}{31})$ km
                                                    = $(\frac{2 \times 4}{3})$ km = $(\frac{8}{3})$ km = $2 \frac{2}{3}$ km

Hence, the distance covered by Vikas in 1 h is $2 \frac{2}{3}$ km.

Page No 31:

Answer:

Cost of $8 \frac{1}{2}$ kg of sugar = Rs $148 \frac{3}{4}$
∴ Cost of 1 kg of sugar = Rs $(148 \frac{3}{4} \div 8 \frac{1}{2})$
= Rs $(\frac{595}{4} \div \frac{17}{2})$
= Rs $(\frac{595}{4} \times \frac{2}{17})$ = Rs $(\frac{35}{2})$ = Rs $17 \frac{1}{2}$

Hence, the cost of 1 kg of sugar is Rs $17 \frac{1}{2}$ .

Page No 31:

Answer:

Cost of 1 notebook = Rs $7 \frac{3}{4}$ = Rs $\frac{31}{4}$
∴ Number of notebooks purchased for Rs $69 \frac{3}{4}$ = $(69 \frac{3}{4} \div \frac{31}{4})$
                                                                        = $(\frac{279}{4} \div \frac{31}{4})$
                                                                        = $(\frac{279}{4} \times \frac{4}{31})$     [∵ Reciprocal of $\frac{31}{4}$ = $\frac{4}{13}$ ]
                                                                        = $(\frac{279}{31})$ = 9
Hence, 9 notebooks can be purchased for Rs $69 \frac{3}{4}$ .

Page No 31:

Answer:

Cost of 1 ticket = Rs $10 \frac{1}{2}$ = Rs $\frac{21}{2}$
Total amount collected by the boy = Rs $283 \frac{1}{2}$ = Rs $\frac{567}{2}$
∴ Number of tickets sold = $(\frac{567}{2} \div \frac{21}{2})$

= $(\frac{567}{2} \times \frac{2}{21})$ [∵ Reciprocal of $\frac{21}{2}$ = $\frac{2}{21}$ ]

= $\frac{567}{21} = 27$

Hence, the boy sold 27 tickets of the charity show.

Page No 31:

Answer:

Amount contributed by 1 student = Rs $61 \frac{1}{2}$ = Rs $\frac{123}{2}$
Total amount collected = Rs $676 \frac{1}{2}$ = Rs $\frac{1353}{2}$
∴ Number of students in the group = $(\frac{1353}{2} \div \frac{123}{2})$

= $(\frac{1353}{2} \times \frac{2}{123})$ [∵ Reciprocal of $\frac{123}{2}$ = $\frac{2}{123}$ ]

= $(\frac{1353}{123}) = 11$

Hence, there are 11 students in the group.

Page No 31:

Answer:

Quantity of milk given to each student = $\frac{2}{5}$ L
Total quantity of milk distributed among all the students = 24 L

∴ Number of students = $(24 \div \frac{2}{5})$

= $(24 \times \frac{5}{2})$ [∵ Reciprocal of $\frac{2}{5}$ = $\frac{5}{2}$ ]

= (12 $\times$ 5) = 60

Hence, there are 60 students in the hostel.

Page No 31:

Answer:

Capacity of the small jug = $\frac{3}{4}$ L
Capacity of the bucket = $20 \frac{1}{4}$ L = $\frac{81}{4}$ L
∴ Required number of small jugs = $(\frac{81}{4} \div \frac{3}{4})$
= $(\frac{81}{4} \times \frac{4}{3})$ [∵ Reciprocal of $\frac{3}{4}$ = $\frac{4}{3}$ ]
= $(\frac{81}{3})$ = 27

Hence, the small jug has to be filled 27 times to empty the water from the bucket.

Page No 31:

Answer:

Product of the two numbers = $15 \frac{5}{6}$ = $\frac{95}{6}$

One of the numbers = $6 \frac{1}{3}$ = $\frac{19}{3}$

∴ The other number = $(\frac{95}{6} \div \frac{19}{3})$

= $(\frac{95}{6} \times \frac{3}{19})$ [∵ Reciprocal of $\frac{19}{3}$ = $\frac{3}{19}$ ]

= $(\frac{5}{2}) = 2 \frac{1}{2}$

Hence, the other number is $2 \frac{1}{2}$ .

Page No 31:

Answer:

Product of the two numbers = 42
One of the numbers = $9 \frac{4}{5}$ = $\frac{49}{5}$
∴ The other number = $(42 \div \frac{49}{5})$
= $(42 \times \frac{5}{49})$ [∵ Reciprocal of $\frac{49}{5}$ = $\frac{5}{49}$ ]
= $(\frac{6 \times 5}{7}) = \frac{30}{7} = 4 \frac{2}{7}$

Hence, the required number is $4 \frac{2}{7}$ .

Page No 31:

Answer:

Required number = $(6 \frac{2}{9} \div 4 \frac{2}{3})$
                            = $(\frac{56}{9} \div \frac{14}{3})$
                            = $(\frac{56}{9} \times \frac{3}{14})$      [ ∵ Reciprocal of $\frac{14}{3}$ = $\frac{3}{14}$ ]
                            = $(\frac{4}{3}) = 1 \frac{1}{3}$

Hence, we have to divide $6 \frac{2}{9}$ by $1 \frac{1}{3}$ to get $4 \frac{2}{3}$ .

Page No 31:

Answer:

Page No 31:

Answer:

Page No 31:

Answer:

(a) $\frac{105}{112}$

A fraction that is reducible can be reduced by dividing both the numerator and denominator by a common factor.

$\frac{105 \div 7}{112 \div 7} = \frac{15}{16}$

Thus, $\frac{105}{112}$ is a reducible fraction.

Page No 32:

Answer:

(c) equivalent fractions

Equivalent fractions are those which are the same but look different.

Thus, $\frac{2}{3}, \frac{4}{6} = \frac{2}{3}, \frac{6}{9} = \frac{2}{3}, \frac{8}{12} = \frac{2}{3}$ are equivalent fractions.

Page No 32:

Answer:

(c) $\frac{9}{16}$ > $\frac{13}{24}$
The two fraction are $\frac{9}{16}$ and $\frac{13}{24}$ .
By cross multiplication, we have:
9 $\times$ 24 = 216 and 13 $\times$ 16 = 208
However, 216 > 208
∴ $\frac{9}{16}$ > $\frac{13}{24}$

Page No 32:

Answer:

(d) none of these
Reciprocal of $1 \frac{3}{4}$ = Reciprocal of $\frac{7}{4}$ = $\frac{4}{7}$

Page No 32:

Answer:

(c) $\frac{5}{6}$

$(\frac{3}{10} + \frac{8}{15}) = (\frac{9 + 16}{30})$ [∵ LCM of 10 and 15 = 30]

= $\frac{25}{30} = \frac{5}{6}$

Page No 32:

Answer:

(d) $\frac{11}{12}$

$(3 \frac{1}{4} - 2 \frac{1}{3})$ = $(\frac{13}{4} - \frac{7}{3})$
= $(\frac{39 - 28}{12})$ [∵ LCM of 4 and 3 = 12]
= $\frac{11}{12}$

Page No 32:

Answer:

(d) 144

$36 \div \frac{1}{4} = 36 \times 4$ [∵ Reciprocal of $\frac{1}{4}$ = 4]
= 144

Page No 32:

Answer:

(b) $\frac{5}{7}$

Required number = $1 \frac{6}{7} \div 2 \frac{3}{5}$

                             = $\frac{13}{7} \div \frac{13}{5}$

                             = $\frac{13}{7} \times \frac{5}{13}$     [∵ Reciprocal of $\frac{13}{5}$ = $\frac{5}{13}$ ]

                             = $\frac{5}{7}$

Page No 32:

Answer:

(d) $2 \frac{1}{4}$

Required number = $1 \frac{1}{2} \div \frac{2}{3}$

                             = $\frac{3}{2} \div \frac{2}{3}$

                             = $\frac{3}{2} \times \frac{3}{2}$       [∵ Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$ ]

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

Page No 32:

Answer:

(c) $2 \frac{2}{5}$

$1 \frac{3}{5} \div \frac{2}{3} = \frac{8}{5} \div \frac{2}{3}$

= $\frac{8}{5} \times \frac{3}{2}$ [∵ Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$ ]

= $(\frac{4 \times 3}{5}) = \frac{12}{5} = 2 \frac{2}{5}$

Page No 32:

Answer:

(d) $1 \frac{5}{6}$

$2 \frac{1}{5} \div 1 \frac{1}{5} = \frac{11}{5} \div \frac{6}{5}$

= $\frac{11}{5} \times \frac{5}{6}$ [∵ Reciprocal of $\frac{6}{5}$ = $\frac{5}{6}$ ]

= $\frac{11}{6} = 1 \frac{5}{6}$

Page No 32:

Answer:

(d) $\frac{3}{5}$

Reciprocal of $1 \frac{2}{3}$ = Reciprocal of $\frac{5}{3}$ = $\frac{3}{5}$

Page No 32:

Answer:

(b) $\frac{3}{5} < \frac{2}{3} < \frac{14}{15}$

The given fractions are $\frac{3}{5}, \frac{2}{3} and \frac{14}{15} .$

LCM of 5, 3 and 15 = 15

Now, we have:

$\frac{2}{3} \times \frac{5}{5} = \frac{10}{15}$ , $\frac{3}{5} \times \frac{3}{3} = \frac{9}{15}$ and $\frac{14}{15} \times \frac{1}{1} = \frac{14}{15}$

Clearly, $\frac{9}{15} < \frac{10}{15} < \frac{14}{15}$

∴ $\frac{3}{5} < \frac{2}{3} < \frac{14}{15}$

Page No 33:

Answer:

(c) 44 km
Distance covered by the car on $2 \frac{3}{4}$ L of petrol = $(16 \times 2 \frac{3}{4})$ km

= $(16 \times \frac{11}{4})$ km

= (4 $\times$ 11) km = 44 km

Page No 33:

Answer:

(a) $10 \frac{1}{2}$ hours
Time taken by Lalit to read the entire book = $(6 \times 1 \frac{3}{4})$ h

= $(6 \times \frac{7}{4})$ h

= $(\frac{21}{2})$ h = $10 \frac{1}{2}$ h

Page No 34:

Answer:

(i) A number of the form $\frac{a}{b}$ , where a and b are natural numbers, is called a natural number.
Here, a is the numerator and b is the denominator.

$\frac{2}{3}$ is a fraction with 2 as the numerator and 3 as the denominator.

$\frac{12}{5}$ is a fraction with 12 as the numerator and 5 as the denominator.

(ii) A fraction whose denominator is a whole number other than 10, 100, 1000, etc., is called a vulgar faction.
Examples: $\frac{2}{5}$ and $\frac{4}{15}$

(iii) A fraction whose numerator is greater than or equal to its denominator is called an improper fraction.
Examples: $\frac{11}{3}$ and $\frac{41}{35}$

Page No 34:

Answer:

Required number to be added = $15 - 6 \frac{3}{5}$

                                             = $\frac{15}{1} - \frac{33}{5}$

                                              = $\frac{75 - 33}{5}$ [∵ LCM of 1 and 5 = 5]

                                              = $\frac{42}{5} = 8 \frac{2}{5}$

Hence, the required number is $8 \frac{2}{5}$ .

Page No 34:

Answer:

We have,

$9 \frac{5}{6} - 4 \frac{3}{8} + 2 \frac{7}{12}$

= $\frac{59}{6} - \frac{35}{8} + \frac{31}{12}$

= $\frac{236 - 105 + 62}{24}$ [∵ LCM of 6, 8 and 12 = 24]

= $\frac{298 - 105}{24}$ = $\frac{193}{24} = 8 \frac{1}{24}$

Page No 34:

Answer:

We have:

(i) $\frac{12}{25}$ of 1 L = $\frac{12}{25}$ of 1000 ml = $(1000 \times \frac{12}{25})$ ml = (40 $\times$ 12) ml = 480 ml

(ii) $\frac{5}{8}$ of 1 kg = $\frac{5}{8}$ of 1000 g = $(1000 \times \frac{5}{8})$ g = (125 $\times$ 5) g = 625 g

(iii) $\frac{3}{5}$ of 1 h = $\frac{3}{5}$ of 60 min = $(60 \times \frac{3}{5})$ min = (12 $\times$ 3) min = 36 min

Page No 34:

Answer:

Cost of 1 L of milk = Rs $37 \frac{3}{4}$ = Rs $\frac{151}{4}$
Cost of $6 \frac{2}{5}$ L of milk = Rs $(\frac{151}{4} \times 6 \frac{2}{5})$
= Rs $(\frac{151}{4} \times \frac{32}{5})$
= Rs $(\frac{151 \times 8}{1 \times 5})$ = Rs $\frac{1208}{5}$ = Rs $241 \frac{3}{5}$
Hence, the cost of $6 \frac{2}{5}$ L of milk is Rs $241 \frac{3}{5}$ .

Page No 34:

Answer:

Cost of $5 \frac{1}{4}$ kg of mangoes = Rs 189
Cost of 1 kg of mango = Rs $(189 \div 5 \frac{1}{4})$
                                   = Rs $(189 \div \frac{21}{4})$
                                   = Rs $(189 \times \frac{4}{21})$       [∵ Reciprocal of $\frac{21}{4}$ = $\frac{4}{21}$ ]
                                   = Rs (9 $\times$ 4) = Rs 36

Hence, the mangoes are being sold at Rs 36 per kg.

Page No 34:

Answer:

We have:

(i) $1 \frac{3}{4} \times 2 \frac{2}{5} \times 3 \frac{4}{7}$

     = $\frac{7}{4} \times \frac{12}{5} \times \frac{25}{7}$

     = $\frac{7 \times 12 \times 25}{4 \times 5 \times 7} = \frac{1 \times 3 \times 5}{1 \times 1 \times 1} = 15$

(ii) $5 \frac{5}{9} \div 3 \frac{1}{3}$

     = $\frac{50}{9} \div \frac{10}{3}$

     = $\frac{50}{9} \times \frac{3}{10}$    [∵ Reciprocal of $\frac{10}{3}$ = $\frac{3}{10}$ ]

     = $\frac{5 \times 1}{3 \times 1} = \frac{5}{3} = 1 \frac{2}{3}$

Page No 34:

Answer:

Required number = $6 \frac{2}{9} \div 4 \frac{2}{3}$

                            = $\frac{56}{9} \div \frac{14}{3}$

                            = $\frac{56}{9} \times \frac{3}{14}$     [∵ Reciprocal of $\frac{14}{3}$ = $\frac{3}{14}$ ]

                            = $\frac{4}{3}$ = $1 \frac{1}{3}$

Hence, we have to divide $6 \frac{2}{9}$ by $1 \frac{1}{3}$ to obtain $4 \frac{2}{3}$ .

Page No 34:

Answer:

Side of the square = $5 \frac{2}{3}$ m = $\frac{17}{3}$ m

Its area = (side)² = ${(\frac{17}{3} m)}^{2}$ = $(\frac{17}{3} m \times \frac{17}{3} m) = (\frac{289}{9}) m^{2} = 32 \frac{1}{9} m^{2}$

Hence, the area of the square is $32 \frac{1}{9} m^{2}$ .

Page No 34:

Answer:

(d) $\frac{5}{8}$

$\frac{5}{8}$ is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

Page No 34:

Answer:

(c) $\frac{46}{63}$

A fraction $\frac{a}{b}$ is said to be irreducible or in its lowest terms if the HCF of a and b is 1.
46 = 2 $\times$ 23 $\times$ 1
63 = 3 $\times$ 3 $\times$ 21 $\times$ 1

Clearly, the HCF of 46 and 63 is 1.

Hence, $\frac{46}{63}$ is an irreducible fraction.

Page No 34:

Answer:

(d) none of these

Reciprocal of $1 \frac{3}{5}$ = Reciprocal of $\frac{8}{5}$ = $\frac{5}{8}$

Page No 34:

Answer:

(c) $2 \frac{2}{5}$

$1 \frac{3}{5} \div \frac{2}{3}$ = $\frac{8}{5} \div \frac{2}{3}$

= $\frac{8}{5} \times \frac{3}{2}$ [∵ Reciprocal of $\frac{2}{3}$ = $\frac{3}{2}$ ]

= $(\frac{4 \times 3}{5}) = \frac{12}{5} = 2 \frac{2}{5}$

Page No 34:

Answer:

(b) $\frac{3}{5} < \frac{2}{3} < \frac{11}{15}$

The given fractions are $\frac{2}{3}, \frac{3}{5}$ and $\frac{11}{15}$ .

LCM of 5, 3 and 15 = 15

Now, we have:

$\frac{2}{3} \times \frac{5}{5} = \frac{10}{15}$ , $\frac{3}{5} \times \frac{3}{3} = \frac{9}{15}$ and $\frac{11}{15} \times \frac{1}{1} = \frac{11}{15}$

Clearly, $\frac{9}{15} < \frac{10}{15} < \frac{11}{15}$

∴ $\frac{3}{5} < \frac{2}{3} < \frac{11}{15}$

Page No 34:

Answer:

(c) $\frac{7}{10}$

Required number = $1 \frac{3}{4} \div 2 \frac{1}{2}$

                             = $\frac{7}{4} \div \frac{5}{2}$

                             = $\frac{7}{4} \times \frac{2}{5}$    [∵ Reciprocal of $\frac{5}{2}$ = $\frac{2}{5}$ ]

                             = $\frac{7 \times 1}{2 \times 5} = \frac{7}{10}$

Page No 35:

Answer:

(b) 33 km
Distance covered by the car on $3 \frac{2}{3}$ L of petrol = $(9 \times 3 \frac{2}{3})$ km
= $(9 \times \frac{11}{3})$ km
= (3 $\times$ 11) km = 33 km

Page No 35:

Answer:

(i) The reciprocal of $8 \frac{2}{5}$ is $\frac{5}{42}$ .

Reciprocal of $8 \frac{2}{5}$ = Reciprocal of $\frac{42}{5}$ = $\frac{5}{42}$

(ii) $13 \frac{1}{2} \div 8 = 1 \frac{11}{16}$

     $13 \frac{1}{2} \div 8 = \frac{27}{2} \times \frac{1}{8} = \frac{27}{16} = 1 \frac{11}{16}$

(iii) $69 \frac{3}{4} \div 7 \frac{3}{4} = 9$

$69 \frac{3}{4} \div 7 \frac{3}{4} = \frac{279}{4} \div \frac{31}{4}$

= $\frac{279}{4} \times \frac{4}{31} = \frac{279}{31}$ = 9

(iv) $41 \frac{2}{3} \div 18 \frac{3}{5} = 775$

        $41 \frac{2}{3} \times 18 \frac{3}{5} = \frac{125}{3} \times \frac{93}{5}$

       = $\frac{125}{3} \times \frac{93}{5} = \frac{25 \times 31}{1 \times 1} = 775$

(v) $\frac{84}{98}$ (irreducible form) = $\frac{6}{7}$

The HCF of 84 and 98 is 14.

∴ $\frac{84 \div 14}{98 \div 14} = \frac{6}{7}$

Page No 35:

Answer:

(i) F

     By cross multiplication, we have:

     9 $\times$ 24 = 216 and 13 $\times$ 16 = 208

     However, 216 > 208

     ∴ $\frac{9}{16} > \frac{13}{24}$

(ii) F

      The LCM of 5, 35 and 14 is 70.

      Now, $\frac{2}{5} = \frac{2}{5} \times \frac{14}{14} = \frac{28}{70}; \frac{16}{35} = \frac{16}{35} \times \frac{2}{2} = \frac{32}{70} and \frac{9}{14} = \frac{9}{14} \times \frac{5}{5} = \frac{45}{70}$

      Clearly, $\frac{28}{70} < \frac{32}{70} < \frac{45}{70}$

      ∴ $\frac{2}{5} < \frac{16}{35} < \frac{9}{14}$

(iii) T

       The LCM of 15 and 20 = (5 $\times$ 3 $\times$ 4) = 60

        ∴ $\frac{11}{15} - \frac{9}{20} = \frac{44 - 27}{60} = \frac{17}{60}$
(iv) T

       $\frac{11}{25}$ of 1 L = $\frac{11}{25}$ of 1000 ml = $(1000 \times \frac{11}{25})$ ml = (40 $\times$ 11) ml = 440 ml

(v) F

      $16 \frac{3}{4} \times 6 \frac{2}{5}$ = $(\frac{67}{4} \times \frac{32}{5}) = (\frac{67 \times 32}{4 \times 5}) = (\frac{67 \times 8}{5}) = \frac{536}{5} = 107 \frac{1}{5}$

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