Rs Aggarwal 2019 2020 Solutions for Class 7 Maths Chapter 2 Fractions are provided here with simple step-by-step explanations. These solutions for Fractions are extremely popular among Class 7 students for Maths Fractions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 2020 Book of Class 7 Maths Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 2020 Solutions. All Rs Aggarwal 2019 2020 Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### 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:

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

#### Answer:

(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) $\frac{2}{3},\frac{3}{5},\frac{7}{10},\frac{8}{15}$

#### Answer:

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?

#### Answer:

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

#### Answer:

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

#### Answer:

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

#### 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]

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

#### 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:

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.

#### Answer:

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?

#### Answer:

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?

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

#### Question 12:

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

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

#### 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:

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:

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:

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?

#### Answer:

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) $5\frac{5}{6}×1\frac{5}{7}$

#### Answer:

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

#### Answer:

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

#### Answer:

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 $48\frac{4}{5}$ per kg. What is the cost of $3\frac{3}{4}$ kg of apples?

#### Answer:

Cost of 1 kg apple = ₹ $48\frac{4}{5}=\frac{244}{5}$

Cost of 3$\frac{3}{4}$kg apples = 3$\frac{3}{4}$ $×$

Hence, the cost of $3\frac{3}{4}$ apples is ₹ 183.

#### 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?

#### Answer:

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?

#### Answer:

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?

#### Answer:

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?

#### Answer:

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?

#### Answer:

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?

#### 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 =$\left(15×\frac{24}{5}\right)$ 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?

#### Answer:

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?

#### Answer:

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 24000 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?

#### Answer:

Total monthly income =  ₹ 24000
Monthly expenditure = $\frac{7}{8}$of  ₹ 24000
=24000$×\frac{7}{8}$ = ₹ 21000
Monthly savings = 24000−21000 = ₹ 3000

#### Question 14:

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

#### Answer:

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.

#### Answer:

Length of the rectangular park =

Its breadth =

∴ 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) $12\frac{3}{5}$

#### Answer:

(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) $15\frac{3}{7}÷1\frac{23}{49}$

#### Answer:

(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 $\frac{72}{49}$ = $\frac{49}{72}$]

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

Divide:

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

#### Answer:

(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?

#### Answer:

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?

#### Answer:

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 $6\frac{3}{4}$ per orange, a man gets Rs 378. How many oranges does he sell?

#### Answer:

Selling price of an orange = Rs $6\frac{3}{4}$ = Rs$\frac{27}{4}$
Total money received after selling oranges = Rs 378
Total no. of oranges = $\frac{378}{27}×4=56$
Hence, total no. of oranges = 56

#### Question 7:

Mangoes are sold at Rs $43\frac{1}{2}$ per kg. What is the weight of mangoes available for Rs $326\frac{1}{4}?$

#### Answer:

Selling price of 1kg mango = ₹ $43\frac{1}{2}$ = ₹ $\frac{87}{2}$

Weight of mangoes at ₹ $\frac{1305}{4}=\frac{1305}{4}×\frac{2}{87}=\frac{435}{58}=\frac{15}{2}=7\frac{1}{2}$
Hence, the weight of mangoes= $7\frac{1}{2}$ 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?

#### Answer:

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 $242\frac{1}{4}$. Find the price of sugar per kg.

#### Answer:

Cost of $\frac{17}{2}$ kg of sugar = ₹ $\frac{969}{4}\phantom{\rule{0ex}{0ex}}$
Cost of 1 kg of sugar = $\frac{\frac{969}{4}}{\frac{17}{2}}=\frac{969}{4}×\frac{2}{17}=\frac{57}{2}=28\frac{1}{2}$

Hence, the cost of sugar is Rs $28\frac{1}{2}$ per kg

#### Question 10:

If the cost of a notebook is Rs $27\frac{3}{4}$, how many notebooks can be purchased for Rs $249\frac{3}{4}?$

#### Answer:

Cost of 1 notebook = ₹ $27\frac{3}{4}$ = ₹ $\frac{111}{4}$
Number of notebooks purchased for ₹ $249\frac{3}{4}$ = $=\frac{\frac{999}{4}}{\frac{111}{4}}=\frac{999}{4}×\frac{4}{111}=9$

Hence, the number of notebooks purchased are 9.

#### Question 11:

At a charity show the price of each ticket was Rs $32\frac{1}{2}.$ The total amount collected by a boy was Rs $877\frac{1}{2}.$ How many tickets were sold by him?

#### Answer:

Total amount collected = $₹877\frac{1}{2}=₹\frac{1755}{2}$
Price of 1 ticket = ₹ $32\frac{1}{2}$ = ₹ $\frac{65}{2}$
Number of tickets sold = $\frac{\frac{1755}{2}}{\frac{65}{2}}=\frac{1755}{2}×\frac{2}{65}=27$
Hence, the number of tickets sold were 27.

#### Question 12:

A group of students arranged a picnic. Each student contributed Rs $261\frac{1}{2}$. The total contribution was Rs $2876\frac{1}{2}$. How many students are there in the group?

#### Answer:

Total contribution = ₹ $2876\frac{1}{2}=\frac{5753}{2}$
Contribution of each student = ₹ $261\frac{1}{2}$ = ₹ $\frac{523}{2}$
Number of students = $\frac{5753}{2}÷\frac{523}{2}$ = $\frac{5753}{2}×\frac{2}{523}=11$
Hence, number of students in the group are 11.

#### 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?

#### 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 = $\left(24÷\frac{2}{5}\right)$

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

#### 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 =
= $\left(\frac{81}{4}×\frac{4}{3}\right)$      [∵ 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.

#### 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 =

=      [∵ 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?

#### Answer:

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 $4\frac{2}{3}$?

#### Answer:

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

#### Answer:

(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

#### Answer:

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

#### Answer:

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

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

$\frac{105÷7}{112÷7}=\frac{15}{16}$

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

#### Answer:

(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

#### 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 $×$ 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

#### Answer:

(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

#### Answer:

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

#### Answer:

(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

#### Answer:

(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) $\frac{1}{7}$

#### Answer:

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

#### Answer:

(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

#### Answer:

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

#### Answer:

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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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

#### Answer:

(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.

#### 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 $\frac{41}{35}$

#### Question 2:

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

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

#### Question 3:

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

#### 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]

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

#### Question 4:

Find:

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

#### Answer:

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.

#### 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 $\left(\frac{151}{4}×6\frac{2}{5}\right)$
= 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?

#### Answer:

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) $5\frac{5}{9}÷3\frac{1}{3}$

#### Answer:

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 $4\frac{2}{3}$?

#### Answer:

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.

#### Answer:

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) $\frac{5}{8}$

#### Answer:

(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) $\frac{51}{85}$

#### 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 $×$ 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

#### Answer:

(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

#### Answer:

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

#### Answer:

(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) $1\frac{3}{7}$

#### Answer:

(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

#### Answer:

(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)= ......

#### 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) $13\frac{1}{2}÷8=1\frac{11}{16}$

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

(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) $16\frac{3}{4}×6\frac{2}{5}=107\frac{3}{10}.$

#### Answer:

(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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