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#### Question 1:

Compare the fractions:

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

We have the following:

(i)

By cross multiplication, we get:
5 $×$ 12 = 60 and 7 $×$ 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 $×$ 15 = 75 and 9 $×$ 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 $×$ 16 = 176 and 12 $×$ 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) $\frac{4}{5},\frac{7}{10},\frac{11}{15},\frac{17}{20}$

(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.

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

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

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

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

(ii) The given fractions are:

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.

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

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

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

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

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

#### Question 3:

Arrange the following fractions in descending order:

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

We have the following:
(i) The given fractions are

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.
$\frac{3}{4}=\frac{3×6}{4×6}=\frac{18}{24}$

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

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

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

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

(ii) The given fractions are
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.
$\frac{2}{3}=\frac{2×10}{3×10}=\frac{20}{30}$

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

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

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

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 .

#### 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?

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 $×$ 5 = 10  and 4 $×$ 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) $8\frac{3}{4}+10\frac{2}{5}$

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

(ii) $\frac{8}{9}+\frac{7}{12}$

[∵ LCM of 9 and 12 = 36]

= $\frac{32+21}{36}$

= $\frac{53}{36}=1\frac{17}{36}$

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

[∵ LCM of 6 and 8 = 24]

= $\frac{20+21}{24}$

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

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

[∵ LCM of 12, 16 and 24 = 48]

= $\frac{28+33+18}{48}$

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

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

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

[∵ LCM of 5, 10 and 15 = 30]

= $\frac{114+69+32}{30}$

=

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

= $\frac{35}{4}+\frac{52}{5}$

[∵ 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) $2\frac{5}{9}-1\frac{7}{15}$

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

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

[∵ LCM of 6 and 4 = 12]

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

= $\frac{1}{12}$

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

= $\frac{16}{5}-\frac{7}{10}$

=           [∵ 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) $8\frac{5}{6}-3\frac{3}{8}+1\frac{7}{12}$

(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?

Total weight of fruits bought by Aneeta =
Now, we have:

[∵ LCM of 2 and 4 = 4]

Hence, the total weight of the fruits purchased by Aneeta is .

#### Question 9:

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

We have:

Perimeter of the rectangle ABCD = AB + BC + CD +DA
=
=
=           [∵ LCM of 2 and 4 = 4]
=

Hence, the perimeter of ABCD is .

#### 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?

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

#### Question 11:

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

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 $8\frac{2}{5}$?

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?

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?

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?

First we have to compare the fractions: .
By cross multiplication, we have:
2 $×$ 9 = 18 and 5 $×$ 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?

First, we have to compare the cost of the pen and the pencil.
Cost of the pen = Rs

Cost of the pencil = Rs
Now, we have to compare fractions
By cross multiplication, we get:

83 $×$ 4 = 332 and 19 $×$ 5 = 95

However, 332 > 95

$\frac{83}{5}>\frac{19}{4}$

So, the cost of pen is more than that of the pencil.
Now,

=      [∵ LCM of 4 and 5 = 20]

=

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

#### Question 1:

Find the product:

(i) $\frac{3}{5}×\frac{7}{11}$
(ii) $\frac{5}{8}×\frac{4}{7}$
(iii) $\frac{4}{9}×\frac{15}{16}$
(iv) $\frac{2}{5}×15$
(v) $\frac{8}{15}×20$
(vi) $\frac{5}{8}×1000$
(vii) $3\frac{1}{8}×16$
(viii) $2\frac{4}{15}×12$
(ix) $3\frac{6}{7}×4\frac{2}{3}$
(x) $9\frac{1}{2}×1\frac{9}{19}$
(xi) $4\frac{1}{8}×2\frac{10}{11}$
(xii) $5\frac{5}{6}×1\frac{5}{7}$

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

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

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

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

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

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

(vii) $3\frac{1}{8}×16=\frac{25}{8}×\frac{16}{1}=\frac{25×16}{8×1}=\frac{25×2}{1×1}=50\phantom{\rule{0ex}{0ex}}$

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

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

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

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

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

#### Question 2:

Simplify:

(i) $\frac{2}{3}×\frac{5}{44}×\frac{33}{35}$
(ii) $\frac{12}{25}×\frac{15}{28}×\frac{35}{36}$
(iii) $\frac{10}{27}×\frac{28}{65}×\frac{39}{56}$
(iv) $1\frac{4}{7}×1\frac{13}{22}×1\frac{1}{15}$
(v) $2\frac{2}{17}×7\frac{2}{9}×1\frac{33}{52}$
(vi) $3\frac{1}{16}×7\frac{3}{7}×1\frac{25}{39}$

We have the following:

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

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

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

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

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

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

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

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

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

#### Question 3:

Find:

(i) $\frac{1}{3}$ of 24
(ii) $\frac{3}{4}$ of 32
(iii) $\frac{5}{9}$ of 45
(iv) $\frac{7}{50}$ of 1000
(v) $\frac{3}{20}$ of 1020
(vi) $\frac{5}{11}$ of Rs 220
(vii) $\frac{4}{9}$ of 54 metres
(viii) $\frac{6}{7}$ of 35 litres
(ix) $\frac{1}{6}$ of an hour
(x) of an year
(xi) $\frac{7}{20}$  of a kg
(xii) $\frac{9}{20}$ of a metre
(xiii) $\frac{7}{8}$ of a day
(xiv) $\frac{3}{7}$ of a week
(xv) $\frac{7}{50}$ of a litre

We have the following:

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

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

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

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

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

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

(vii) $\frac{4}{9}$of 54 m = $\left(\frac{4}{9}×54\right)\mathrm{m}$ = (4 $×$ 6) m = 24 m

(viii) $\frac{6}{7}$ of 35 L = $\left(\frac{6}{7}×35\right)\mathrm{L}$ = (6 $×$ 5) L = 30 L

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

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

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

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

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

(xiv) $\frac{3}{7}$ of a week = $\frac{3}{7}$ of 7 days = $\left(7×\frac{3}{7}\right)$ days = 3 days

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

#### Question 4:

Apples are sold at Rs $18\frac{2}{5}$ per kg. What is the cost of $3\frac{3}{4}$ kg of apples?

Cost of 1kg of apples =
∴ Cost of of apples =
=

Hence, the cost of of apples is Rs 69.

#### Question 5:

Cloth is being sold at Rs $42\frac{1}{2}$ per metre. What is the cost of $5\frac{3}{5}$ metres of this cloth?

Cost of 1 m of cloth =
∴ Cost of of cloth = Rs $\left(\frac{85}{2}×5\frac{3}{5}\right)\phantom{\rule{0ex}{0ex}}$
= Rs
Hence, the cost of of cloth is Rs 238.

#### Question 6:

A car covers a certain distance at a uniform speed of $66\frac{2}{3}$ km per hour. How much distance will it cover in 9 hours?

Distance covered by the car in 1 h =
Distance covered by the car in 9 h  =
=

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

#### Question 7:

One tin holds $12\frac{3}{4}$ litres of oil. How many litres of oil can 26 such tins hold?

Capacity of 1 tin =
∴ Capacity of 26 such tins =
=

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

#### Question 8:

For a particular show in a circus, each ticket costs Rs $35\frac{1}{2}$. If 308 tickets are sold for the  show, how much amount has been collected?

Cost of 1 ticket = Rs $35\frac{1}{2}$= Rs $\frac{71}{2}$
∴ Cost of 308 tickets = Rs

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

#### Question 9:

Nine boards are stacked on the top of each other. The thickness of each board is $3\frac{2}{3}$ cm. How high is the stack?

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

Hence, the height of the stack is 33 cm.

#### Question 10:

Rohit takes $4\frac{4}{5}$ minutes to make complete round of a circular park. How much time will he take to make 15 rounds?

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 =$\left(15×\frac{24}{5}\right)$ min
= (3 $×$ 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.

#### Question 11:

Amit weighs 35 kg. His sister Kavita's weight is $\frac{3}{5}$ of Amit's weight. How much does Kavita weigh?

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

#### Question 12:

There are 42 students in a class and $\frac{5}{7}$ of the students are boys. How many girls are there in the class?

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

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

Hence, there are 12 girls in the class.

#### Question 13:

Sapna earns Rs 12000 per month. She spends $\frac{7}{8}$ of her income and deposits rest of the money in a bank. How much money does she deposit in the bank each month?

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

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

#### Question 14:

Each side of a square field is $4\frac{2}{3}$ m. Find its area.

Side of the square field =
∴ Area of the square = (side)2
=
=

Hence, the area of the square field is .

#### Question 15:

Find the area of a rectangular park which is $41\frac{2}{3}$ m long and $18\frac{3}{5}$ m broad.

Length of the rectangular park =

∴ Its area = length $×$ breadth

= 2
= (25 $×$ 31) m = 775 m2

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

#### Question 1:

Write down the reciprocal of:

(i) $\frac{5}{8}$
(ii) 7
(iii) $\frac{1}{12}$
(iv) $12\frac{3}{5}$

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

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

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

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

#### Question 2:

Simplify:

(i)  $\frac{4}{7}÷\frac{9}{14}$
(ii) $\frac{7}{10}÷\frac{3}{5}$
(iii) $\frac{8}{9}÷16$
(iv) $9÷\frac{1}{3}$
(v) $24÷\frac{6}{7}$
(vi) $3\frac{3}{5}÷\frac{4}{5}$
(vii) $3\frac{3}{7}÷\frac{8}{21}$
(viii) $5\frac{4}{7}÷1\frac{3}{10}$
(ix) $15\frac{3}{7}÷1\frac{23}{49}$

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

= $\frac{8}{9}$

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

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

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

= $\frac{1}{18}$

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

= 27

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

= 4 $×$ 7 = 28

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

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

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

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

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

= 3  3 = 9

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

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

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

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

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

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

#### Question 3:

Divide:

(i)
(ii)
(iii)
(iv)
(v)
(vi)

(i)  $\frac{11}{24}÷\frac{7}{8}$

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

= $\frac{11}{21}$

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

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

= 5 $×$ 2 = 10

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

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

=  $\frac{5}{3}$ = $1\frac{2}{3}$

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

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

= 4 $×$ 5 = 20

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

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

= 5 $×$ 5 = 25

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

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

= 7 $×$ 4 = 28

#### Question 4:

A rope of length $13\frac{1}{2}$ m has been divided into 9 pieces of the same length. What is the length of each piece?

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

∴ Length of each piece = $\left(\frac{27}{2}÷9\right)$ m
= $\left(\frac{27}{2}×\frac{1}{9}\right)$ 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.

#### Question 5:

18 boxes of nails weigh equally and their total weight is $49\frac{1}{2}$ kg. How much does each box weigh?

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

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

#### Question 6:

By selling oranges at the rate of Rs $3\frac{3}{4}$ per orange, a man gets Rs 210. How many oranges does he sell?

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 = $\left(210÷\frac{15}{4}\right)$
= $\left(210×\frac{4}{15}\right)$      [∵ Reciprocal of $\frac{15}{4}$ = $\frac{4}{15}$]
= (14 $×$ 4) = 56

Hence, the man sold 56 oranges.

#### Question 7:

Mangoes are sold at Rs $18\frac{1}{2}$ per kg. What is the weight of mangoes available for Rs $157\frac{1}{4}$ ?

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 = $\left(\frac{629}{4}÷\frac{37}{2}\right)$ kg
=$\left(\frac{629}{4}×\frac{2}{37}\right)$ kg      [∵ Reciprocal of $\frac{37}{2}$ = $\frac{2}{37}$]
= $\left(\frac{17}{2}\right)$ 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.

#### Question 8:

Vikas can cover a distance of $20\frac{2}{3}$ km in $7\frac{3}{4}$ hours on foot. How many km per hour does he walk?

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

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

#### Question 9:

Preeti bought $8\frac{1}{2}$ kg of sugar for Rs $148\frac{3}{4}$. Find the price of sugar per kg.

Cost of $8\frac{1}{2}$ kg of sugar = Rs $148\frac{3}{4}$
∴ Cost of 1 kg of sugar = Rs $\left(148\frac{3}{4}÷8\frac{1}{2}\right)$
= Rs $\left(\frac{595}{4}÷\frac{17}{2}\right)$
= Rs $\left(\frac{595}{4}×\frac{2}{17}\right)$ = Rs $\left(\frac{35}{2}\right)$ = Rs $17\frac{1}{2}$

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

#### Question 10:

If the cost of a notebook is Rs $7\frac{3}{4}$, how many notebooks can be purchased for Rs $69\frac{3}{4}$ ?

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

#### Question 11:

At a charity show the price of each ticket was Rs $10\frac{1}{2}$. The total amount collected by a boy was Rs $283\frac{1}{2}$. How many tickets were sold by him?

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 = $\left(\frac{567}{2}÷\frac{21}{2}\right)$

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

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

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

#### Question 12:

A group of students arranged a picnic. Each student contributed Rs $61\frac{1}{2}$. The total contribution was Rs $676\frac{1}{2}$. How many students are there in the group?

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 = $\left(\frac{1353}{2}÷\frac{123}{2}\right)$

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

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

Hence, there are 11 students in the group.

#### Question 13:

24 litres of milk was distributed equally among all the students of a hostel. If each student got $\frac{2}{5}$ litre of milk, how many students are there in the hostel?

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 = $\left(24÷\frac{2}{5}\right)$

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

= (12 $×$ 5) = 60

Hence, there are 60 students in the hostel.

#### Question 14:

A bucket contains $20\frac{1}{4}$ litres of water. A small jug has a capacity of $\frac{3}{4}$ litre. How many times the jug has to be filled with water from the bucket to get it emptied?

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 =
= $\left(\frac{81}{4}×\frac{4}{3}\right)$      [∵ Reciprocal of $\frac{3}{4}$ = $\frac{4}{3}$]
= $\left(\frac{81}{3}\right)$ = 27

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

#### Question 15:

The product of two numbers is $15\frac{5}{6}$. If one of the numbers is $6\frac{1}{3}$, find the other.

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 =

=      [∵ Reciprocal of $\frac{19}{3}$ = $\frac{3}{19}$]

=

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

#### Question 16:

By what number should $9\frac{4}{5}$ be multiplied to get 42?

Product of the two numbers = 42
One of the numbers = $9\frac{4}{5}$ = $\frac{49}{5}$
∴ The other number =
=           [∵ Reciprocal of $\frac{49}{5}$ = $\frac{5}{49}$]
=

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

#### Question 17:

By what number should $6\frac{2}{9}$ be divided to obtain $4\frac{2}{3}$?

Required number =
=
=      [ ∵ Reciprocal of $\frac{14}{3}$ = $\frac{3}{14}$]
= $\left(\frac{4}{3}\right)=1\frac{1}{3}$

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

#### Question 1:

Mark (✓) against the correct answer
Which of the following is a vulgar fraction?

(a) $\frac{3}{10}$
(b) $\frac{13}{10}$
(c) $\frac{10}{3}$
(d) none of these

(c) $\frac{10}{3}$

$\frac{10}{3}$ is a vulgar fraction, because its denominator is other than 10, 100, 1000, etc.

#### Question 2:

Mark (✓) against the correct answer
Which of the following is an improper fraction?

(a) $\frac{7}{10}$
(b) $\frac{7}{9}$
(c) $\frac{9}{7}$
(d) none of these

(c) $\frac{9}{7}$
$\frac{9}{7}$ is an improper fraction, because its numerator is greater than its denominator.

#### Question 3:

Mark (✓) against the correct answer
Which of the following is a reducible fraction?

(a) $\frac{105}{112}$
(b) $\frac{104}{121}$
(c) $\frac{77}{72}$
(d) $\frac{46}{63}$

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

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

$\frac{105÷7}{112÷7}=\frac{15}{16}$

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

#### Question 4:

Mark (✓) against the correct answer
$\frac{2}{3},\frac{4}{6},\frac{6}{9},\frac{8}{12}$ are
(a) like fractions
(b) irreducible fractions
(c) equivalent fractions
(d) none of these

(c) equivalent fractions

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

Thus, are equivalent fractions.

#### Question 5:

Mark (✓) against the correct answer
Which of the following statements is true?

(a) $\frac{9}{16}=\frac{13}{24}$
(b) $\frac{9}{16}<\frac{13}{24}$
(c) $\frac{9}{16}>\frac{13}{24}$
(d) none of these

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

#### Question 6:

Mark (✓) against the correct answer
Reciprocal of $1\frac{3}{4}$ is

(a) $1\frac{4}{3}$
(b) $4\frac{1}{3}$
(c) $3\frac{1}{4}$
(d) none of these

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

#### Question 7:

Mark (✓) against the correct answer
$\left(\frac{3}{10}+\frac{8}{15}\right)=?$

(a) $\frac{11}{10}$
(b) $\frac{11}{15}$
(c) $\frac{5}{6}$
(d) none of these

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

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

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

#### Question 8:

Mark (✓) against the correct answer
$\left(3\frac{1}{4}-2\frac{1}{3}\right)=?$

(a) $1\frac{1}{12}$
(b) $\frac{1}{12}$
(c) $1\frac{1}{11}$
(d) $\frac{11}{12}$

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

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

#### Question 9:

Mark (✓) against the correct answer
$36÷\frac{1}{4}=?$

(a) 9
(b) $\frac{1}{9}$
(c) $\frac{1}{144}$
(d) 144

(d) 144

$36÷\frac{1}{4}=36×4$   [∵ Reciprocal of $\frac{1}{4}$= 4]
= 144

#### Question 10:

Mark (✓) against the correct answer
By what number should $2\frac{3}{5}$ be multiplied to get $1\frac{6}{7}$ ?

(a) $1\frac{5}{7}$
(b) $\frac{5}{7}$
(c) $1\frac{1}{7}$
(d) $\frac{1}{7}$

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

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

= $\frac{13}{7}÷\frac{13}{5}$

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

= $\frac{5}{7}$

#### Question 11:

Mark (✓) against the correct answer
By what number should $1\frac{1}{2}$ be divided to get $\frac{2}{3}$ ?

(a) $2\frac{2}{3}$
(b) $1\frac{2}{3}$
(c) $\frac{4}{9}$
(d) $2\frac{1}{4}$

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

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

= $\frac{3}{2}÷\frac{2}{3}$

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

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

#### Question 12:

Mark (✓) against the correct answer
$1\frac{3}{5}÷\frac{2}{3}=?$

(a) $1\frac{1}{15}$
(b) $1\frac{9}{10}$
(c) $2\frac{2}{5}$
(d) none of these

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

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

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

= $\left(\frac{4×3}{5}\right)=\frac{12}{5}=2\frac{2}{5}$

#### Question 13:

Mark (✓) against the correct answer
$2\frac{1}{5}÷1\frac{1}{5}=?$

(a) 1
(b) 2
(c) $1\frac{1}{5}$
(d) $1\frac{5}{6}$

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

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

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

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

#### Question 14:

Mark (✓) against the correct answer
The reciprocal of $1\frac{2}{3}$ is
(a) $3\frac{1}{2}$
(b) $2\frac{1}{3}$
(c) $1\frac{1}{3}$
(d) $\frac{3}{5}$

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

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

#### Question 15:

Mark (✓) against the correct answer
Which one of the following is the correct statement?

(a) $\frac{2}{3}<\frac{3}{5}<\frac{14}{15}$
(b) $\frac{3}{5}<\frac{2}{3}<\frac{14}{15}$
(c) $\frac{14}{15}<\frac{3}{5}<\frac{2}{3}$
(d) none of these

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

The given fractions are

LCM of 5, 3 and 15 = 15

Now, we have:

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

Clearly,

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

#### Question 16:

Mark (✓) against the correct answer
A car runs 16 km using 1 litre of petrol. How much distance will it cover in $2\frac{3}{4}$ litres of petrol?
(a) 24 km
(b) 36 km
(c) 44 km
(d) $32\frac{3}{4}$ km

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

= $\left(16×\frac{11}{4}\right)$ km

= (4 $×$ 11) km = 44 km

#### Question 17:

Mark (✓) against the correct answer
Lalit reads a book for $1\frac{3}{4}$ hours evrey day and reads the entire book in 6 days. How many hours does he take to read the entire book?

(a) $10\frac{1}{2}$ hours
(b) $9\frac{1}{2}$ hours
(c) $7\frac{1}{2}$ hours
(d) $11\frac{1}{2}$ hours

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

= $\left(6×\frac{7}{4}\right)$ h

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

#### Question 1:

Define:

(i) Fractions
(ii) Vulgar fractions
(iii) Improper fractions

Give two examples of each.

(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 $\frac{41}{35}$

#### Question 2:

What should be added to $6\frac{3}{5}$ to get 15?

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 $8\frac{2}{5}$.

#### Question 3:

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

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]

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

#### Question 4:

Find:

(i) $\frac{12}{25}$ of a litre
(ii) $\frac{5}{8}$ of a kilogram
(iii) $\frac{3}{5}$ of an hour

We have:

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

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

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

#### Question 5:

Milk is sold at Rs $37\frac{3}{4}$ per litre. Find the cost of $6\frac{2}{5}$ litres milk.

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

#### Question 6:

The cost of $5\frac{1}{4}$ kg of mangoes is Rs 189. At what rate per kg are the mangoes being sold?

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

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

#### Question 7:

Simplify:

(i) $1\frac{3}{4}×2\frac{2}{5}×3\frac{4}{7}$
(ii) $5\frac{5}{9}÷3\frac{1}{3}$

We have:

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

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

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

(ii) $5\frac{5}{9}÷3\frac{1}{3}$

= $\frac{50}{9}÷\frac{10}{3}$

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

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

#### Question 8:

By what number should $6\frac{2}{9}$ be divided to obtain $4\frac{2}{3}$?

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

= $\frac{56}{9}÷\frac{14}{3}$

= $\frac{56}{9}×\frac{3}{14}$    [∵ 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 $4\frac{2}{3}$.

#### Question 9:

Each side of a square is $5\frac{2}{3}$ m long. Find its area.

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

Its area = (side)2 = ${\left(\frac{17}{3}\mathrm{m}\right)}^{2}$ =

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

#### Question 10:

Mark (✓) against the correct answer
Which of the following is a vulgar fraction?

(a) $\frac{7}{10}$
(b) $\frac{19}{100}$
(c) $3\frac{3}{10}$
(d) $\frac{5}{8}$

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

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

#### Question 11:

Mark (✓) against the correct answer
Which of the following is an irreducible fraction?

(a) $\frac{105}{112}$
(b) $\frac{66}{77}$
(c) $\frac{46}{63}$
(d) $\frac{51}{85}$

(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 $×$ 23 $×$1
63 = 3 $×$ 3$×$ 21 $×$1

Clearly, the HCF of 46 and 63 is 1.

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

#### Question 12:

Mark (✓) against the correct answer
Reciprocal of $1\frac{3}{5}$ is
(a) $1\frac{5}{3}$
(b) $5\frac{1}{3}$
(c) $3\frac{1}{5}$
(d) none of these

(d) none of these

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

#### Question 13:

Mark (✓) against the correct answer
$1\frac{3}{5}÷\frac{2}{3}=?$

(a) $1\frac{9}{10}$
(b) $1\frac{1}{15}$
(c) $2\frac{2}{5}$
(d) none of these

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

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

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

= $\left(\frac{4×3}{5}\right)=\frac{12}{5}=2\frac{2}{5}$

#### Question 14:

Mark (✓) against the correct answer
Which of the following is correct?

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

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

The given fractions are and $\frac{11}{15}$.

LCM of 5, 3 and 15 = 15

Now, we have:

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

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

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

#### Question 15:

Mark (✓) against the correct answer
By what number should $1\frac{3}{4}$ be divided to get $2\frac{1}{2}$?
(a) $\frac{3}{7}$
(b) $1\frac{2}{5}$
(c) $\frac{7}{10}$
(d) $1\frac{3}{7}$

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

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

= $\frac{7}{4}÷\frac{5}{2}$

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

= $\frac{7×1}{2×5}=\frac{7}{10}$

#### Question 16:

Mark (✓) against the correct answer
A car runs 9 km using 1 litre of petrol. How much distance will it cover in $3\frac{2}{3}$ litres o petrol?
(a) 36 km
(b) 33 km
(c) $2\frac{5}{11}$ km
(d) 22 km

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

#### Question 17:

Fill in the blanks.

(i) Reciprocal of $8\frac{2}{5}$ is ...... .
(ii) $13\frac{1}{2}÷8=......$
(iii) $69\frac{3}{4}÷7\frac{3}{4}=......$
(iv) $41\frac{2}{3}×18\frac{3}{5}=......$
(v) $\frac{84}{98}$(in irreducible form)= ......

(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) $13\frac{1}{2}÷8=1\frac{11}{16}$

$13\frac{1}{2}÷8=\frac{27}{2}×\frac{1}{8}=\frac{27}{16}=1\frac{11}{16}$

(iii) $69\frac{3}{4}÷7\frac{3}{4}=9$

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

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

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

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

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

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

The HCF of 84 and 98 is 14.

$\frac{84÷14}{98÷14}=\frac{6}{7}$

#### Question 18:

Write 'T' for true and 'F' for false

(i) $\frac{9}{16}<\frac{13}{24}.$
(ii) Among $\frac{2}{5},\frac{16}{35}$ and $\frac{9}{14}$, the largest is $\frac{16}{35}.$
(iii) $\frac{11}{15}-\frac{9}{20}=\frac{17}{60}.$
(iv) $\frac{11}{25}$ of a litre = 440 mL.
(v) $16\frac{3}{4}×6\frac{2}{5}=107\frac{3}{10}.$

(i) F

By cross multiplication, we have:

9 $×$ 24 = 216 and 13 $×$ 16 = 208

However, 216 > 208

∴ $\frac{9}{16}>\frac{13}{24}$

(ii) F

The LCM of 5, 35 and 14 is 70.

Now,

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 $×$ 3 $×$ 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 = $\left(1000×\frac{11}{25}\right)$ ml = (40 $×$ 11) ml = 440 ml

(v) F

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

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