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#### Question 1:

Compare the following fractions by using the symbol > or < or =:
(i)
(ii)
(iii)
(iv)

First, we need to find the LCM of denominators in each case. After that, we will equate the denominators in order to compare the two fractions.

(i)
LCM of 9 and 13 is 117.

(ii)

(iii)

(iv)

LCM of 15 and 105 is 105.

#### Question 2:

Arrange the following fractions in ascending order:
(i) $\frac{3}{8},\frac{5}{6},\frac{6}{8}\frac{2}{4},\frac{1}{3}$
(ii) $\frac{4}{3},\frac{3}{8},\frac{6}{12},\frac{5}{16}$

(i)

(ii)

#### Question 3:

Arrange the following fractions in descending order:
(i) $\frac{4}{5},\frac{7}{10},\frac{11}{15},\frac{17}{20}$
(ii) $\frac{2}{7},\frac{11}{35},\frac{9}{14},\frac{13}{28}$

(i)

(ii)

#### Question 4:

Write five equivalent fractions of $\frac{3}{5}$.

Five equivalent fractions of $\frac{3}{5}$ are:

#### Question 5:

Find the sum:
(i) $\frac{5}{8}+\frac{3}{10}$
(ii) $4\frac{3}{4}+9\frac{2}{5}$
(iii) $\frac{5}{6}+3+\frac{3}{4}$
(iv) $2\frac{3}{5}+4\frac{7}{10}+2\frac{4}{15}$

(i)

(ii)

(iii)

(iv)

#### Question 6:

Find the difference of
(i)
(ii)
(iii)
(iv)

(i)

(ii)

(iii)

(iv)

#### Question 7:

Find the difference:
(i) $\frac{6}{7}-\frac{9}{11}$
(ii) $8-\frac{5}{9}$
(iii) $9-5\frac{2}{3}$
(iv) $4\frac{3}{10}-1\frac{2}{15}$

(i)

(ii)

(iii)

(iv)

#### Question 8:

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

(i)

(ii)

(iii)

#### Question 9:

What should be added to $5\frac{3}{7}$ to get 12?

Let x be the required fraction.

According to the question:

$x+5\frac{3}{7}=12\phantom{\rule{0ex}{0ex}}⇒x+\frac{\left(5×7\right)+3}{7}=12\phantom{\rule{0ex}{0ex}}⇒x=12-\frac{38}{7}\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left(12×7\right)-\left(38×1\right)}{7}⇔\frac{46}{7}$

#### Question 10:

What should be added to $5\frac{4}{15}$ to get $12\frac{3}{5}?$

Let x be the required fraction.

According to the question:

#### Question 11:

Suman studies for $5\frac{2}{3}$ hours daily. She devotes $2\frac{4}{5}$ hours of her time for Science and Mathematics. How much time does she devote for other subjects?

Suman studies for  $5\frac{2}{3}\mathrm{hours}\phantom{\rule{0ex}{0ex}}$ daily. Therefore, we have

She studies science and mathematics for  $2\frac{4}{5}\mathrm{hours}$. Therefore, we have
$2\frac{4}{5}\mathrm{hours}=\frac{\left(2×5\right)+4}{5}=\frac{14}{5}\mathrm{hours}$
Time devoted to other subjects = Total study time $-$ Time devoted to science and mathematics

$\frac{17}{3}-\frac{14}{5}=\frac{\left(17×5\right)-\left(14×3\right)}{15}\phantom{\rule{0ex}{0ex}}=\frac{43}{15}\mathrm{hours}$

#### Question 12:

A piece of wire is of length $12\frac{3}{4}\mathrm{m}$. If it is cut into two pieces in such a way that the length of one piece is $5\frac{1}{4}\mathrm{m},$ what is the length of the other piece?

Let the length of second piece be x.

Total length of wire = Length of one piece + Length of second piece

$12\frac{3}{4}=5\frac{1}{4}+x\phantom{\rule{0ex}{0ex}}⇒\frac{\left(12×4\right)+3}{4}=\frac{\left(5×4\right)+1}{4}+x\phantom{\rule{0ex}{0ex}}⇒x=\frac{\left(12×4\right)+3}{4}-\frac{\left(5×4\right)+1}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{51}{4}-\frac{21}{4}⇔\frac{30}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{30}{4}⇔\frac{15}{2}$

#### Question 13:

A rectangular sheet of paper is long and wide. Find its perimeter.

Perimeter of rectangle = 2(length + width)

#### Question 14:

In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic  square?

 $\frac{4}{11}$ $\frac{9}{11}$ $\frac{2}{11}$ $\frac{3}{11}$ $\frac{5}{11}$ $\frac{7}{11}$ $\frac{8}{11}$ $\frac{1}{11}$ $\frac{6}{11}$

#### Question 15:

The cost of Mathematics book is Rs $25\frac{3}{4}$ and that of Science book is Rs $20\frac{1}{2}$. Which costs more and by how much?

#### Question 16:

(i) Provide the number in box and also give its simplest from in each of the following:
(i)
(ii) .

(i) $\frac{2}{3}×x=\frac{10}{30}\phantom{\rule{0ex}{0ex}}x=\frac{10}{30}×\frac{3}{2}=\frac{1}{2}$

(ii) $\frac{3}{5}×x=\frac{24}{75}\phantom{\rule{0ex}{0ex}}x=\frac{24}{75}×\frac{5}{3}=\frac{8}{15}$

#### Question 1:

Multiply:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{7}{11}×\frac{3}{5}=\frac{7×3}{11×5}\phantom{\rule{0ex}{0ex}}⇒\frac{21}{55}$

(ii)
$\frac{3}{5}×25=\frac{3×25}{5×1}\phantom{\rule{0ex}{0ex}}⇒\frac{75}{5}⇔15$

(iii)

(iv)

#### Question 2:

Find the product:
(i) $\frac{4}{7}×\frac{14}{25}$
(ii) $7\frac{1}{2}×2\frac{4}{15}$
(iii) $3\frac{6}{7}×4\frac{2}{3}$
(iv) $6\frac{11}{14}×3\frac{1}{2}$

(i)
$\frac{4}{7}×\frac{14}{25}=\frac{4×14}{7×25}\phantom{\rule{0ex}{0ex}}⇒\frac{8}{25}$

(ii)
$7\frac{1}{2}×2\frac{4}{15}⇔\frac{\left(7×2\right)+1}{2}×\frac{\left(2×15\right)+4}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{15}{2}×\frac{34}{15}⇔\frac{17}{1}$
(iii)
$3\frac{6}{7}×4\frac{2}{3}⇔\frac{\left(3×7\right)+6}{7}×\frac{\left(4×3\right)+2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{\overline{)27}}^{9}}{\overline{)7}}×\frac{{\overline{)14}}^{2}}{\overline{)3}}⇔\frac{18}{1}$
(iv)
$6\frac{11}{14}×3\frac{1}{2}⇔\frac{\left(14×6\right)+11}{14}×\frac{\left(3×2\right)+1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{95}{\overline{){14}^{2}}}×\frac{\overline{)7}}{2}⇔\frac{95}{4}$

#### Question 3:

Simplify:
(i)
(ii)
(iii)

(i)
$\frac{12}{25}×\frac{15}{28}×\frac{35}{36}⇔\frac{\overline{)12}×{\overline{)15}}^{3}×{\overline{)35}}^{5}}{{\overline{)25}}^{5}×{\overline{)28}}^{4}×{\overline{)36}}^{3}}\phantom{\rule{0ex}{0ex}}⇒\frac{\overline{)3}×\overline{)5}}{\overline{)5}×4×\overline{)3}}⇔\frac{1}{4}$

(ii)
$\frac{10}{27}×\frac{39}{56}×\frac{28}{65}⇔\frac{\overline{)10}×{\overline{)39}}^{{\overline{)3}}^{1}}×\overline{)28}}{{\overline{)27}}^{9}×{\overline{)56}}^{\overline{)2}}×{\overline{)65}}^{\overline{)5}}}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{9}$
(iii)
$2\frac{2}{17}×7\frac{2}{9}×1\frac{33}{52}⇔\frac{\left(2×17\right)+2}{17}×\frac{\left(7×9\right)+2}{9}×\frac{\left(52×1\right)+33}{52}\phantom{\rule{0ex}{0ex}}⇔\frac{\overline{){36}^{\overline{)4}}}×{\overline{)65}}^{5}×{\overline{)85}}^{5}}{\overline{)17}×\overline{)9}×{\overline{)52}}^{\overline{)4}}}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{1}$

Find:
(i)
(ii)
(iii)

#### Question 5:

Which is greater?

$\frac{2}{7}=\frac{2}{7}×\frac{2}{2}=\frac{4}{14}$

$⇒\frac{6}{14}>\frac{2}{7}$

#### Question 6:

Find:
(i) $\frac{7}{11}$ of Rs 330
(ii) $\frac{5}{9}$ of 108 metres
(iii) $\frac{3}{7}$ of 42 litres
(iv) $\frac{1}{12}$ of an hour
(v) $\frac{5}{6}$ of an year
(vi) $\frac{3}{20}$ of a kg
(vii) $\frac{7}{20}$ of a litre
(viii) $\frac{5}{6}$ of a day
(ix) $\frac{2}{7}$ of a week

#### Question 7:

Shikha plants 4 saplings in a row in her garden. The distance between two adjacent saplings is $\frac{3}{4}\mathrm{m}$. Find the distance between the first and the last sapling.

Distance between the first and second saplings = $\frac{3}{4}\mathrm{m}$
Distance between the first and third saplings = $2×\frac{3}{4}\mathrm{m}=\frac{3}{2}\mathrm{m}\phantom{\rule{0ex}{0ex}}$
Distance between the first and fourth saplings $3×\frac{3}{4}\mathrm{m}=\frac{9}{4}\mathrm{m}$

#### Question 8:

Ravish reads $\frac{1}{3}$ part of a book in 1 hour. How much part of the book will he read in $2\frac{1}{5}$ houurs?

#### Question 9:

Lipika reads a book for $1\frac{3}{4}$ hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?

In one day, Lipika reads for $1\frac{3}{4}\mathrm{hours}\phantom{\rule{0ex}{0ex}}$.
$1\frac{3}{4}\mathrm{hours}=\frac{\left(1×4\right)+3}{4}=\frac{7}{4}\mathrm{hours}$
Total hours required =

#### Question 10:

Find the area of a rectangular park which is long and broad.

#### Question 11:

If milk is available at Rs $17\frac{3}{4}$ per litre, find the cost of $7\frac{2}{5}$ litres of milk.

Cost of 1 litre milk =
$17\frac{3}{4}=\frac{\left(17×4\right)+3}{4}=\frac{71}{4}\phantom{\rule{0ex}{0ex}}$

#### Question 12:

Sharda can walk in one hour. How much distance will she cover in $2\frac{2}{5}$ hours?

$8\frac{1}{3}\mathrm{km}=\frac{\left(8×3\right)+1}{3}\phantom{\rule{0ex}{0ex}}⇒8\frac{1}{3}\mathrm{km}=\frac{25}{3}\mathrm{km}$

Distance covered by ​Sharda in 1 hour =  $\frac{25}{3}$km

Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ = Distance covered in 1 hour $×\frac{12}{5}$
$⇒$Distance covered by Sharda in $\frac{12}{5}\mathrm{hours}$ =

#### Question 13:

A sugar bag contains 30 kg of sugar. After consuming $\frac{2}{3}$ of it, how much sugar is left in the bag?

#### Question 14:

Each side of a square is long. Find its area.

#### Question 15:

There are 45 students in a class and $\frac{3}{5}$ of them are boys. How many girls are there in the class?

#### Question 1:

Find the reciprocal of each of the following fractions and classify them as proper, improper and whole numbers:
(i) $\frac{3}{7}$
(ii) $\frac{5}{8}$
(iii) $\frac{9}{7}$
(iv) $\frac{6}{5}$
(v) $\frac{12}{7}$
(vi) $\frac{1}{8}$

Reciprocal of a non-zero fraction
(i)

(ii)

(iii)

(iv)

(v)

(vi)

#### Question 2:

Divide:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{3}{8}÷\frac{5}{9}=\frac{3}{8}×\frac{9}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{3×9}{8×5}⇔\frac{27}{40}$

(ii)
$3\frac{1}{4}÷\frac{2}{3}=\frac{\left(3×4\right)+1}{4}×\frac{3}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{13}{4}×\frac{3}{2}⇔\frac{39}{8}=4\frac{7}{8}$

(iii)
$\frac{7}{8}÷4\frac{1}{2}=\frac{7}{8}÷\frac{\left(4×2\right)+1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{\overline{){8}^{4}}}×\frac{\overline{)2}}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{7}{4}×\frac{1}{9}⇔\frac{7}{36}$
(iv)
$6\frac{1}{4}÷2\frac{3}{5}=\frac{\left(6×4\right)+1}{4}÷\frac{\left(2×5\right)+3}{5}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{4}÷\frac{13}{5}=\frac{25}{4}×\frac{5}{13}\phantom{\rule{0ex}{0ex}}⇒\frac{125}{52}=2\frac{21}{52}$

#### Question 3:

Divide:
(i)
(ii)
(iii)
(iv)

(i)
$\frac{3}{8}÷4=\frac{3}{8}×\frac{1}{4}\phantom{\rule{0ex}{0ex}}=\frac{3}{32}$

(ii)
$\frac{9}{16}÷6=\frac{9}{16}×\frac{1}{6}=\frac{\overline{){9}^{3}}}{16×\overline{){6}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{3}{32}$

(iii)
$9÷\frac{3}{16}=\frac{9}{1}×\frac{16}{3}⇔\frac{\overline{){9}^{3}}×16}{\overline{)3}}\phantom{\rule{0ex}{0ex}}=48$

(iv)
$10÷\frac{100}{3}=\frac{10}{1}×\frac{3}{100}⇔\frac{\overline{)10}×3}{\overline{){100}^{10}}}\phantom{\rule{0ex}{0ex}}=\frac{3}{10}$

#### Question 4:

Simplify:
(i) $\frac{3}{10}÷\frac{10}{3}$
(ii) $4\frac{3}{5}÷\frac{4}{5}$
(iii) $5\frac{4}{7}÷1\frac{3}{10}$
(iv) $4÷2\frac{2}{5}$

(i)
$\frac{3}{10}÷\frac{10}{3}=\frac{3}{10}×\frac{3}{10}⇔\frac{3×3}{10×10}\phantom{\rule{0ex}{0ex}}⇒\frac{9}{100}\phantom{\rule{0ex}{0ex}}$
(ii)
$4\frac{3}{5}÷\frac{4}{5}=\frac{\left(4×5\right)+3}{5}÷\frac{4}{5}\phantom{\rule{0ex}{0ex}}⇒4\frac{3}{5}÷\frac{4}{5}=\frac{23}{5}÷\frac{4}{5}⇔\frac{23}{5}×\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{23}{4}=5\frac{3}{4}$

(iii)
$5\frac{4}{7}÷1\frac{3}{10}=\frac{\left(5×7\right)+4}{7}÷\frac{\left(1×10\right)+3}{10}\phantom{\rule{0ex}{0ex}}⇒5\frac{4}{7}÷1\frac{3}{10}=\frac{39}{7}÷\frac{13}{10}⇔\frac{\overline{){39}^{3}}}{7}×\frac{10}{\overline{)13}}\phantom{\rule{0ex}{0ex}}⇒\frac{30}{7}=4\frac{2}{7}$

(iv)
$4÷2\frac{2}{5}=\frac{4}{1}÷\frac{\left(2×5\right)+2}{5}⇔\frac{\overline{)4}}{1}×\frac{5}{\overline{){12}^{3}}}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{3}=1\frac{2}{3}$

#### Question 5:

A wire of length $12\frac{1}{2}\mathrm{m}$ is cut into 10 pieces of equal length. Find the length of each piece.

$12\frac{1}{2}\mathrm{m}=\frac{\left(12×2\right)+1}{2}\mathrm{m}\phantom{\rule{0ex}{0ex}}=\frac{25}{2}\mathrm{m}$

Length of one piece =
Length of one piece = $\frac{\overline{){25}^{5}}}{{\overline{)20}}^{4}}=\frac{5}{4}\mathrm{m}$

#### Question 6:

The length of a rectangular plot of area . What is the width of the plot?

Area of rectangle = Length of rectangle $×$ Width of rectangle

#### Question 7:

By what number should $6\frac{2}{9}$ be multiplied to get $4\frac{4}{9}$?

Let the required number be x.

According to the question:

#### Question 8:

The product of two numbers is $25\frac{5}{6}$. If one of the numbers is $6\frac{2}{3}$, find the other.

Let the required number be x.

According to the question:

$6\frac{2}{3}×x=25\frac{5}{6}\phantom{\rule{0ex}{0ex}}\frac{\left(6×3\right)+2}{3}×x=\frac{\left(25×6\right)+5}{6}\phantom{\rule{0ex}{0ex}}\frac{20}{3}×x=\frac{155}{6}\phantom{\rule{0ex}{0ex}}x=\frac{3}{20}×\frac{155}{6}=\frac{\overline{)3}×{\overline{)155}}^{31}}{\overline{){20}^{4}}×\overline{){6}^{2}}}=\frac{31}{8}\phantom{\rule{0ex}{0ex}}=3\frac{7}{8}$

#### Question 9:

The cost of of apples is Rs 400. At what rate per kg are the apples being sold?

$6\frac{1}{4}\mathrm{kg}=\frac{\left(6×4\right)+1}{4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}⇒\frac{25}{4}\mathrm{kg}$
Cost of  $\frac{25}{4}\mathrm{kg}$ of apples = Rs. 400
Cost of 1 kg of apples =

#### Question 10:

By selling oranges at the rate of Rs $5\frac{1}{4}$ per orange, a fruit-seller gets Rs 630. How many dozens of oranges does he sell?

Cost of 1 orange =
Number of oranges sold = $630÷\frac{21}{4}$

$\because$ 12 oranges = 1 dozen

$\therefore$ 120 oranges =

#### Question 11:

In mid-day meal scheme $\frac{3}{10}$ litre of milk is given to each student of a primary school. If 30 litres of milk is distributed every day in the school, how many students are there in the school?

Number of students in the school =

#### Question 12:

In a charity show Rs 6496 were collected by selling some tickets. If the price of each ticket was Rs $50\frac{3}{4}$, how many tikets were sold?

Number of tickets sold = $6496÷\frac{203}{4}$