Unit: 2: Fraction and Decimals

# Question: 1

Solve:

(i)             $2-\frac{3}{5}$

(ii)          $4+\frac{7}{8}$

(iii)      $\frac{3}{5}+\frac{2}{7}$

(iv)       $\frac{9}{11}-\frac{4}{15}$

(v)          $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}$

(vi)       $2\frac{2}{3}+3\frac{1}{2}$

(vii)    $8\frac{1}{2}-3\frac{5}{8}$

## Solution

(i)             $2-\frac{3}{5}=\frac{10-3}{5}=\frac{7}{5}=1\frac{2}{5}$

(ii)          $4+\frac{7}{8}=\frac{32+7}{8}=\frac{39}{8}=4\frac{7}{8}$

(iii)      $\frac{3}{5}+\frac{2}{7}=\frac{21+10}{35}=\frac{31}{35}$

(iv)       $\frac{9}{11}-\frac{4}{15}=\frac{135-44}{165}=\frac{91}{165}$

(v)          $\frac{7}{10}+\frac{2}{5}+\frac{3}{2}=\frac{7+4+15}{10}=\frac{26}{10}=\frac{13}{5}=2\frac{3}{5}$

(vi)       $2\frac{2}{3}+3\frac{1}{2}=\frac{8}{3}+\frac{7}{2}=\frac{16+21}{6}=\frac{37}{6}=6\frac{1}{6}$

(vii)    $8\frac{1}{2}-3\frac{5}{8}=\frac{17}{2}-\frac{29}{8}=\frac{68-29}{8}=\frac{39}{8}=4\frac{7}{8}$

# Question: 2

Arrange the following in descending order:

(i)             $\frac{2}{9},\frac{2}{3},\frac{8}{21}$

(ii)          $\frac{1}{5},\frac{3}{7},\frac{7}{10}$

## Solution

(i)             $\frac{2}{9},\text{\hspace{0.17em}}\frac{2}{3},\text{\hspace{0.17em}}\frac{8}{21}$
Convert these fractions into like fractions.

$\frac{2}{9}=\frac{14}{63},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{2}{3}=\frac{42}{63},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{8}{21}=\frac{24}{63}$
Now arrange these fractions in descending order.

$\frac{42}{63}>\frac{24}{63}>\frac{14}{63}$

Therefore, $\frac{2}{3}>\frac{8}{21}>\frac{2}{9}$

(ii)          $\frac{1}{5},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{7},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{7}{10}$

[Converting into like fractions]

$\frac{49}{70}>\frac{30}{70}>\frac{14}{70}$
[Arranging in descending order]

Therefore,  $\frac{7}{10}>\frac{3}{7}>\frac{1}{5}$

# Question: 3

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?

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

## Solution

Sum of first row

$=\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{[Given]}$

Sum of second row

$=\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}+\frac{15}{11}$

Sum of third row

$=\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}+\frac{15}{11}$

Sum of first column

$=\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}+\frac{15}{11}$

Sum of second column $=\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}+\frac{15}{11}$

Sum of third column

$=\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}+\frac{15}{11}$

Sum of first diagonal (left to right) $=\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}+\frac{15}{11}$

Sum of second diagonal (left to right) $=\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2+5+8}{11}+\frac{15}{11}$

Since the sum of fractions in each row, in each column and along the diagonals are same, therefore it is a magic square.

# Question: 4

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

## Solution

Given:

The sheet of paper is in rectangular form.

Length of sheet  and

Perimeter of rectangle

$=2\left(12\frac{1}{2}+10\frac{2}{3}\right)$

$=2\left(\frac{25}{2}+\frac{32}{3}\right)$

$=2\left(\frac{25×3×32×2}{6}\right)$

$=2\left(\frac{75+64}{6}\right)$

$=2×\frac{139}{6}$

$=\frac{139}{3}$

Thus, the perimeter of the rectangular sheet is

# Question: 5

Find the perimeters of (i) $\Delta \text{ABE}$ (ii) the rectangle BCDE in this figure. Whose perimeter is greater?

## Solution

(i)             In $\Delta \text{ABE,}$

The perimeter of $\Delta \text{ABE}$

$=\text{AB}+\text{BE}+\text{AE}$

$\begin{array}{l}=\frac{5}{2}+2\frac{3}{4}+3\frac{3}{5}\\ =\frac{5}{2}+\frac{11}{4}+\frac{18}{5}\end{array}$

Thus, the perimeter of $\Delta \text{ABE}$ is .

(ii)          In $\text{BCDE,}$

Perimeter of rectangle

$\begin{array}{l}=2\left(2\frac{3}{4}+\frac{7}{6}\right)\\ =2\left(\frac{11}{4}+\frac{7}{6}\right)\end{array}$

Thus, the perimeter of rectangle $\text{BCDE}$ is

Comparing the perimeter of triangle and that of rectangle,

Therefore, the perimeter of triangle $\text{ABE}$ is greater than that of rectangle $\text{BCDE}\text{.}$

# Question: 6

Salil wants to put a picture in a frame. The picture is  wide.

To fit in the frame the picture cannot be more than  wide. How much should the picture be trimmed?

## Solution

Given:

The width of the picture

and the width of picture frame

Therefore, the picture should be trimmed

$\begin{array}{l}=7\frac{3}{5}-7\frac{3}{10}\\ =\frac{38}{5}-\frac{73}{10}\end{array}$

Thus, the picture should be trimmed by

# Question: 7

Ritu ate $\frac{3}{5}$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?

## Solution

The part of an apple eaten by Ritu $=\frac{3}{5}$

The part of an apple eaten by Somu $=1-\frac{3}{5}=\frac{5-3}{5}=\frac{2}{5}$

Comparing the parts of apple eaten by both Ritu and Somu $\frac{3}{5}>\frac{2}{5}$

Larger share will be more by $\frac{3}{5}-\frac{2}{5}=\frac{1}{5}$ part.

Thus, Ritu’s part is $\frac{1}{5}$ more than Somu’s part.

# Question: 8

Michael finished colouring a picture in $\frac{7}{12}$ hour. Vaibhav finished colouring the same picture in $\frac{3}{4}$ hour. Who worked longer? By what fraction was it longer?

## Solution

Time taken by Michael to colour the picture

Time taken by Vaibhav to colour the picture

Converting both fractions in like fractions,

Here, $\frac{7}{12}<\frac{9}{12}$

Thus, Vaibhav worked longer time.

Vaibhav worked longer time by

Thus, Vaibhav took $\frac{1}{6}$ hour more than Michael.

# Question: 1

Which of the drawings (a) to (d) show :

(i)             $2×\frac{1}{5}$         (a)

(ii)          $2×\frac{1}{2}$         (b)

(iii)      $3×\frac{2}{3}$         (c)

(iv)       $3×\frac{1}{4}$        (d)

## Solution

(i)             – (d) Since      $2×\frac{1}{5}=\frac{1}{5}+\frac{1}{5}$

(ii)          – (b) Since      $2×\frac{1}{2}=\frac{1}{2}+\frac{1}{2}$

(iii)      – (a) Since       $3×\frac{2}{3}=\frac{2}{3}+\frac{2}{3}+\frac{2}{3}$

(iv)       (c) Since       $3×\frac{1}{4}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$

# Question: 2

Some pictures (a) to (c) are given below. Tell which of them show:

(i)             $3×\frac{1}{5}=\frac{3}{5}$          (a)

(ii)          $2×\frac{1}{3}=\frac{2}{3}$          (b)

(iii)      $3×\frac{3}{4}=2\frac{1}{4}$        (c)

## Solution

(i)             – (c)    Since    $3×\frac{1}{5}=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}$

(ii)          – (a)    Since    $2×\frac{1}{3}=\frac{1}{3}+\frac{1}{3}$

(iii)      – (b)    Since    $3×\frac{3}{4}=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}$

# Question: 3

Multiply and reduce to lowest form:

(i)             $7×\frac{3}{5}$

(ii)          $4×\frac{1}{3}$

(iii)      $2×\frac{6}{7}$

(iv)       $5×\frac{2}{9}$

(v)          $\frac{2}{3}×4$

(vi)       $\frac{5}{2}×6$

(vii)    $11×\frac{4}{7}$

(viii) $20×\frac{4}{5}$

(ix)       $13×\frac{1}{3}$

(x)          $15×\frac{3}{5}$

## Solution

(i)             $7×\frac{3}{5}=\frac{7×3}{5}=\frac{21}{5}=4\frac{1}{5}$

(ii)          $4×\frac{1}{3}=\frac{4×1}{3}=\frac{4}{3}=1\frac{1}{3}$

(iii)      $2×\frac{6}{7}=\frac{2×6}{7}=\frac{12}{7}=1\frac{5}{7}$

(iv)       $5×\frac{2}{9}=\frac{5×2}{9}=\frac{10}{9}=1\frac{1}{9}$

(v)          $\frac{2}{3}×4=\frac{2×4}{3}=\frac{8}{3}=2\frac{2}{3}$

(vi)       $\frac{5}{2}×6=5×3=15$

(vii)    $11×\frac{4}{7}=\frac{11×4}{7}=\frac{44}{7}=6\frac{2}{7}$

(viii) $20×\frac{4}{5}=4×4=16$

(ix)       $13×\frac{1}{3}=\frac{13×1}{3}=\frac{13}{3}=4\frac{1}{3}$

(x)          $15×\frac{3}{5}=3×3=9$

# Question: 4

(i)             $\frac{1}{2}$ of the circles in box

(ii)          $\frac{2}{3}$ of the triangles in box

(iii)      $\frac{3}{5}$ of the squares in box.

## Solution

(i)             $\frac{1}{2}$ of $12$ circles

(ii)          $\frac{2}{3}$ of 9 triangles

(iii)      $\frac{3}{5}$ of 15 squares

# Question: 5

Find:

a.    $\frac{1}{2}$ of      (i) 24     (ii) 46

b.   $\frac{2}{3}$ of     (i) 18      (ii) 27

c.    $\frac{3}{4}$ of     (i) 16      (ii) 36

d.   $\frac{4}{5}$ of     (i) 20      (ii) 35

a.    (i)

(ii)

b.   (i)

(ii)

c.    (i)

(ii)

d.   (i)

(ii)

# Question: 6

Multiply and express as a mixed fraction:

a.    $3×5\frac{1}{5}$

b.   $5×6\frac{3}{4}$

c.    $7×2\frac{1}{4}$

d.   $4×6\frac{1}{3}$

e.    $3\frac{1}{4}×6$

f.     $3\frac{2}{5}×8$

## Solution

a.    $3×5\frac{1}{5}=3×\frac{26}{5}=\frac{3×26}{5}=\frac{78}{5}=15\frac{3}{5}$

b.   $5×6\frac{3}{4}=5×\frac{27}{4}=\frac{5×27}{4}=\frac{135}{4}=33\frac{3}{4}$

c.    $7×2\frac{1}{4}=7×\frac{9}{4}=\frac{7×9}{4}=\frac{63}{4}=15\frac{3}{4}$

d.   $4×6\frac{1}{3}=4×\frac{19}{3}=\frac{4×19}{3}=\frac{76}{3}=25\frac{1}{3}$

e.    $3\frac{1}{4}×6=\frac{13}{4}×6=\frac{13×3}{2}=\frac{39}{2}=19\frac{1}{2}$

f.     $3\frac{2}{5}×8=\frac{17}{5}×8=\frac{17×8}{5}=\frac{136}{5}=27\frac{1}{5}$

# Question: 7

Find

a.    $\frac{1}{2}$ of         (i) $2\frac{3}{4}$      (ii) $4\frac{2}{9}$

b.   $\frac{5}{8}$ of (i) $3\frac{5}{6}$       (ii) $9\frac{2}{3}$

## Solution

a.    (i) $\frac{1}{2}$ of $2\frac{3}{4}=\frac{1}{2}×2\frac{3}{4}=\frac{1}{2}×\frac{11}{4}=\frac{11}{8}=1\frac{3}{8}$

(ii) $\frac{1}{2}$ of $4\frac{2}{9}=\frac{1}{2}×4\frac{2}{9}=\frac{1}{2}×\frac{38}{9}=\frac{19}{9}=2\frac{1}{9}$

b.   (i) $\frac{5}{8}$ of $3\frac{5}{6}=\frac{5}{8}×3\frac{5}{6}=\frac{5}{8}×\frac{23}{6}=\frac{115}{48}=2\frac{19}{48}$

(ii) $\frac{5}{8}$ of $9\frac{2}{3}=\frac{5}{8}×9\frac{2}{3}=\frac{5}{8}×\frac{29}{3}=\frac{145}{24}=6\frac{1}{24}$

# Question: 8

Vidya and Pratap went for a picnic. Their mother gave them a water bag that contained 5 litres of water. Vidya consumed $\frac{2}{5}$ of the water. Pratap consumed the remaining water.

(i)             How much water did Vidya drink?

(ii)          What fraction of the total quantity of water did Pratap drink?

## Solution

Given: Total quantity of water in bottle

(i)             Water consumed by Vidya

Thus, Vidya drank 2 litres water from the bottle.

(ii)          Water consumed by Pratap

$=\left(1-\frac{2}{5}\right)$ part of bottle

Pratap consumed $\frac{3}{5}$ of $5$ litres water $=\frac{3}{5}×5=3$ litres

Thus, Pratap drank $\frac{3}{5}$ part of the total quantity of water.

# Question: 1

Find:

(i)             $\frac{1}{4}$ of (a) $\frac{1}{4}$ (b) $\frac{3}{5}$        (c) $\frac{4}{3}$

(ii)          $\frac{1}{7}$ of (a) $\frac{2}{9}$ (b) $\frac{6}{5}$        (c) $\frac{3}{10}$

## Solution

(i)             (a) $\frac{1}{4}$ of $\frac{1}{4}=\frac{1}{4}×\frac{1}{4}=\frac{1×1}{4×4}=\frac{1}{16}$

(b) $\frac{1}{4}$ of $\frac{3}{5}=\frac{1}{4}×\frac{3}{5}=\frac{1×3}{4×5}=\frac{3}{20}$

(c) $\frac{1}{4}$ of $\frac{4}{3}=\frac{1}{4}×\frac{4}{3}=\frac{1×4}{4×3}=\frac{1}{3}$

(ii)          (a) $\frac{1}{7}$ of $\frac{2}{9}=\frac{1}{7}×\frac{2}{9}=\frac{1×2}{7×9}=\frac{2}{63}$

(b) $\frac{1}{7}$ of $\frac{6}{5}=\frac{1}{7}×\frac{6}{5}=\frac{1×6}{7×5}=\frac{6}{35}$

(c) $\frac{1}{7}$ of $\frac{3}{10}=\frac{1}{7}×\frac{3}{10}=\frac{1×3}{7×10}=\frac{3}{70}$

# Question: 2

Multiply and reduce to lowest form (if possible):

(i)             $\frac{2}{3}×2\frac{2}{3}$

(ii)          $\frac{2}{7}×\frac{7}{9}$

(iii)      $\frac{3}{8}×\frac{6}{4}$

(iv)       $\frac{9}{5}×\frac{3}{5}$

(v)          $\frac{1}{3}×\frac{15}{8}$

(vi)       $\frac{11}{2}×\frac{3}{10}$

(vii)    $\frac{4}{5}×\frac{12}{7}$

## Solution

(i)             $\frac{2}{3}×2\frac{2}{3}=\frac{2}{3}×\frac{8}{3}=\frac{2×8}{3×3}=\frac{16}{9}=1\frac{7}{9}$

(ii)          $\frac{2}{7}×\frac{7}{9}=\frac{2×7}{7×9}=\frac{2}{9}$

(iii)      $\frac{3}{8}×\frac{6}{4}=\frac{3×6}{8×4}=\frac{3×3}{8×2}=\frac{9}{16}$

(iv)       $\frac{9}{5}×\frac{3}{5}=\frac{9×3}{5×5}=\frac{27}{25}=1\frac{2}{25}$

(v)          $\frac{1}{3}×\frac{15}{8}=\frac{1×15}{3×8}=\frac{1×5}{1×8}=\frac{5}{8}$

(vi)       $\frac{11}{2}×\frac{3}{10}=\frac{11×3}{2×10}=\frac{33}{20}=1\frac{13}{20}$

(vii)    $\frac{4}{5}×\frac{12}{7}=\frac{4×12}{5×7}=\frac{48}{35}=1\frac{13}{35}$

# Question: 3

For the fractions given below:

a.    Multiply and reduce the product to lowest form (if possible)

b.   Tell whether the fraction obtained is proper or improper.

c.    If the fraction obtained is improper then convert it into a mixed fraction.

(i) $\frac{2}{5}×5\frac{1}{4}$

(ii)        $6\frac{2}{5}×\frac{7}{9}$

(iii)    $\frac{3}{2}×5\frac{1}{3}$

(iv)     $\frac{5}{6}×2\frac{3}{7}$

(v)        $3\frac{2}{5}×\frac{4}{7}$

(vi)     $2\frac{3}{5}×3$

(vii)  $3\frac{4}{7}×\frac{3}{5}$

## Solution

(i)             $\frac{2}{5}×5\frac{1}{4}=\frac{2}{5}×\frac{21}{4}=\frac{2×21}{5×4}=\frac{1×21}{5×2}=\frac{21}{10}=2\frac{1}{10}$

(ii)          $6\frac{2}{5}×\frac{7}{9}=\frac{32}{5}×\frac{7}{9}=\frac{32×7}{5×9}=\frac{224}{45}=4\frac{44}{45}$

(iii)      $\frac{3}{2}×5\frac{1}{3}=\frac{3}{2}×\frac{16}{3}=\frac{48}{6}=8$

(iv)       $\frac{5}{6}×2\frac{3}{7}=\frac{5}{6}×\frac{17}{7}=\frac{85}{42}=2\frac{1}{42}$

(v)          $3\frac{2}{5}×\frac{4}{7}=\frac{17}{5}×\frac{4}{7}=\frac{68}{35}=1\frac{33}{35}$

(vi)       $2\frac{3}{5}×3=\frac{13}{5}×\frac{3}{1}=\frac{13×3}{5×1}=\frac{39}{5}=7\frac{4}{5}$

(vii)    $3\frac{4}{7}×\frac{3}{5}=\frac{25}{7}×\frac{3}{5}=\frac{5×3}{7×1}=\frac{15}{7}=2\frac{1}{7}$

# Question: 4

Which is greater:

(i)             $\frac{2}{7}$ of $\frac{3}{4}$      or     $\frac{3}{5}$ of $\frac{5}{8}$

(ii)          $\frac{1}{2}$ of $\frac{6}{7}$      or     $\frac{2}{3}$ of $\frac{3}{7}$

## Solution

(i)             Left side expression $=\frac{2}{7}$ of $\frac{3}{4}$ $=\frac{2}{7}×\frac{3}{4}=\frac{3}{14}$

Right side expression $=\frac{3}{5}$ of $\frac{5}{8}$

Since, numerators are same and denominator of second term is smaller, so it will be greater fraction.

$\therefore \frac{3}{14}<\frac{3}{8}$

Thus,  is greater.

(ii)          $\frac{1}{2}$ of $\frac{6}{7}$ or $\frac{2}{3}$ of $\frac{3}{7}$

Left side expression $=\frac{1}{2}×\frac{6}{7}=\frac{3}{7}$

Right side expression

Clearly, $\frac{3}{7}>\frac{2}{7}$

Thus,  is greater.

# Question: 5

Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is . Find the distance between the first and the last sapling.

## Solution

The distance between two adjacent saplings

Saili planted $4$ saplings in a row, then the gaps between the first and the fourth sapling   $=3$

Therefore, The distance between the first and the last sapling

Thus, the distance between the first and the last sapling is

# Question: 6

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

## Solution

Time Lipika devotes every day to read a book $=1\frac{3}{4}$ hours.

She took $6$ days to read a book.

Now, no of hours taken by her to read the entire book $=1\frac{3}{4}×6=\frac{7}{4}×6=\frac{21}{2}=10\frac{1}{2}$

Thus, $10$.5 hours were required by her to read the entire book.

# Question: 7

A car runs  using $1$ litre of petrol. How much distance will it cover using $2\frac{3}{4}$ litres of petrol.

## Solution

In $1$ litre of pertrol, car covers the distance

In $2\frac{3}{4}$ litres of petrol, car covers the distance $=2\frac{3}{4}$ of

Thus, the car will cover .

# Question: 8

a.    (i) Provide the number in the box $\overline{)}$, such that $\frac{2}{3}×\overline{)}=\frac{10}{30}.$

(ii) The simplest form of the number obtained in $\overline{)}$ is _______.

b.   (i) Provide the number in the box $\overline{)}$, such that $\frac{3}{5}×\overline{)}=\frac{24}{75}?$

(ii) The simplest form of the number obtained in $\overline{)}$ is _____.

## Solution

a.    (i) $\frac{2}{3}×\overline{)\frac{5}{10}}=\frac{10}{30}$

(ii) The simplest form of

b.   (i) $\frac{3}{5}×\overline{)\frac{8}{15}}=\frac{24}{75}$

(ii) The simplest form of

# Question: 1

Find:

(i)             $12÷\frac{3}{4}$

(ii)          $14÷\frac{5}{6}$

(iii)      $8÷\frac{7}{3}$

(iv)       $4÷\frac{8}{3}$

(v)          $3÷2\frac{1}{3}$

(vi)       $5÷3\frac{4}{7}$

## Solution

(i)             $12÷\frac{3}{4}$$=12×\frac{4}{3}$

$=16$

(ii)          $14÷\frac{5}{6}$$=14×\frac{6}{5}$

$=\frac{84}{5}$

$=16\frac{4}{5}$

(iii)      $8÷\frac{7}{3}$$=8×\frac{3}{7}$

$=\frac{24}{7}$

$=3\frac{3}{7}$

(iv)       $4÷\frac{8}{3}$$=4×\frac{3}{8}$

$=\frac{3}{2}$

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

(v)          $3÷2\frac{1}{3}$$=3÷\frac{7}{3}$

$=3×\frac{3}{7}$

$=\frac{9}{7}$

$=1\frac{2}{7}$

(vi)       $5÷3\frac{4}{7}$$=5÷\frac{25}{7}$

$=5×\frac{7}{25}$

$=\frac{7}{5}$

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

# Question: 2

Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.

(i)             $\frac{3}{4}$

(ii)          $\frac{5}{8}$

(iii)      $\frac{9}{7}$

(iv)       $\frac{6}{5}$

(v)          $\frac{12}{7}$

(vi)       $\frac{1}{8}$

(vii)    $\frac{1}{11}$

## Solution

(i)             Reciprocal of   Improper fraction

(ii)          Reciprocal of   Improper fraction

(iii)      Reciprocal of   Proper fraction

(iv)       Reciprocal of   Proper fraction

(v)          Reciprocal of   Proper fraction

(vi)       Reciprocal of   Whole number

(vii)    Reciprocal of  Whole number

# Question: 3

Find:

(i)             $\frac{7}{3}÷2$

(ii)           $\frac{4}{9}÷5$

(iii)       $\frac{6}{13}÷7$

(iv)        $4\frac{1}{3}÷3$

(v)           $3\frac{1}{2}÷4$

(vi)        $4\frac{3}{7}÷7$

## Solution

(i)             $\frac{7}{3}÷2=\frac{7}{3}×\frac{1}{2}$

$\begin{array}{l}=\frac{7×1}{3×2}\\ =\frac{7}{6}\\ =1\frac{1}{6}\end{array}$

(ii)          $\frac{4}{9}÷5=\frac{4}{9}×\frac{1}{5}$

$\begin{array}{l}=\frac{4×1}{9×5}\\ =\frac{4}{45}\end{array}$

(iii)      $\frac{6}{13}÷7=\frac{6}{13}×\frac{1}{7}$

$\begin{array}{l}=\frac{6×1}{13×7}\\ =\frac{6}{91}\end{array}$

(iv)       $4\frac{1}{3}÷3=\frac{13}{3}÷3$

$\begin{array}{l}=\frac{13}{3}×\frac{1}{3}\\ =\frac{13}{9}\\ =1\frac{4}{9}\end{array}$

(v)          $3\frac{1}{2}÷4=\frac{7}{2}÷4$

$\begin{array}{l}=\frac{7}{2}×\frac{1}{4}\\ =\frac{7}{8}\end{array}$

(vi)       $4\frac{3}{7}÷7=\frac{31}{7}÷7$

$\begin{array}{l}=\frac{31}{7}×\frac{1}{7}\\ =\frac{31}{49}\end{array}$

# Question: 4

Find:

(i)             $\frac{2}{5}÷\frac{1}{2}$

(ii)          $\frac{4}{9}÷\frac{2}{3}$

(iii)      $\frac{3}{7}÷\frac{8}{7}$

(iv)       $2\frac{1}{3}÷\frac{3}{5}$

(v)          $3\frac{1}{2}÷\frac{8}{3}$

(vi)       $\frac{2}{5}÷1\frac{1}{2}$

(vii)    $3\frac{1}{5}÷1\frac{2}{3}$

(viii)$2\frac{1}{5}÷1\frac{1}{5}$

## Solution

(i)             $\frac{2}{5}÷\frac{1}{2}=\frac{2}{5}×\frac{2}{1}=\frac{2×2}{5×1}=\frac{4}{5}$

(ii)          $\frac{4}{9}÷\frac{2}{3}=\frac{4}{9}×\frac{3}{2}=\frac{2}{3}$

(iii)      $\frac{3}{7}÷\frac{8}{7}=\frac{3}{7}×\frac{7}{8}=\frac{3}{8}$

(iv)       $2\frac{1}{3}÷\frac{3}{5}=\frac{7}{3}÷\frac{3}{5}=\frac{7}{3}×\frac{5}{3}=\frac{35}{9}=3\frac{8}{9}$

(v)          $3\frac{1}{2}÷\frac{8}{3}=\frac{7}{2}÷\frac{8}{3}=\frac{7}{2}×\frac{3}{8}=\frac{7×3}{2×8}=\frac{21}{16}=1\frac{5}{16}$

(vi)       $\frac{2}{5}÷1\frac{1}{2}=\frac{2}{5}÷\frac{3}{2}=\frac{2}{5}×\frac{2}{3}=\frac{2×2}{5×3}=\frac{4}{15}$

(vii)    $3\frac{1}{5}÷1\frac{2}{3}=\frac{16}{5}÷\frac{5}{3}=\frac{16}{5}×\frac{3}{5}=\frac{16×3}{5×5}=\frac{48}{25}=1\frac{23}{25}$

(viii)$2\frac{1}{5}÷1\frac{1}{5}=\frac{11}{5}÷\frac{6}{5}=\frac{11}{5}×\frac{5}{6}=\frac{11}{6}=1\frac{5}{6}$

# Question: 1

Which is greater?

(i)

(ii)

(iii)

(iv)

(v)

(vi)

## Solution

(i)             $0.5>0.05$

(ii)          $0.7>0.5$

(iii)      $7>0.7$

(iv)       $1.37<1.49$

(v)          $2.03<2.30$

(vi)       $0.8<0.88.$

# Question: 2

Express as rupees using decimals:

(i)             7 paise

(ii)          7 rupees 7 paise

(iii)      77 rupees 77 paise

(iv)       50 paise

(v)          235 paise.

## Solution

(i)

(ii)

$\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7+\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.07\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}7.07\end{array}$

(iii)

$\begin{array}{l}=\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}77+\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.77\\ =\text{\hspace{0.17em}}\text{Rs}\text{\hspace{0.17em}}\text{\hspace{0.17em}}77.77\end{array}$

(iv)

(v)

# Question: 3

(i)             Express  in metre and kilometre.

(ii)          Express  in cm, m and km.

## Solution

(i)             Express  in metre and kilometre.

Now,

(ii)          Express  in cm, m and km.

Now,

Again,

Express in kg:

(i)

(ii)

(iii)

(iv)

We know that,

(i)

(ii)

(iii)

(iv)

# Question: 5

Write the following decimal numbers in the expanded form:

(i)             $20.03$

(ii)          $2.03$

(iii)      $200.03$

(iv)       $2.034$

## Solution

(i)             $20.03=2×10+0×1+0×\frac{1}{10}+3×\frac{1}{100}$

(ii)          $2.03=2×1+0×\frac{1}{10}+3×\frac{1}{100}$

(iii)      $200.03=2×100+0×10+0×1+0×\frac{1}{10}+3×\frac{1}{100}$

(iv)       $2.034=2×1+0×\frac{1}{10}+3×\frac{1}{100}+4×\frac{1}{1000}$

# Question: 6

Write the place value of $2$ in the following decimal numbers:

(i)             $2.56$

(ii)          $21.37$

(iii)      $10.25$

(iv)       $9.42$

(v)          $63.352.$

## Solution

(i)             Place value of $2$ in

(ii)          Place value of $2$ in

(iii)      Place value of $2$ in $10.25=2×\frac{1}{10}=2\text{\hspace{0.17em}}\text{tenths}$

(iv)       Place value of $2$ in

(v)          Place value of $2$ in

# Question: 7

Dinesh went from place A to place B and from there to place C. A is  from B and B is  from C. Ayub went from place A to place D and from there to place C. D is  from A and C is  from D. Who travelled more and by how much?

## Solution

Distance travelled by Dinesh when he went from place A to place  and from place B to

Total distance covered by Dinesh $=\text{AB}+\text{BC}$

Total distance covered by Ayub $=\text{AD}+\text{DC}$

On comparing the total distance of Ayub and Dinesh,

Therefore, Ayub covered more distance by

# Question: 8

Shyama bought  apples and  mangoes. Sarala bought  oranges and  bananas. Who bought more fruits?

## Solution

Total weight of fruits bought by Shyama

Total weight of fruits bought by Sarala

On comparing the quantity of fruits,

Therefore, Sarala bought more fruits.

# Question: 9

How much less is  than ?

## Solution

We have to find the difference between  and .

Difference

Therefore,  is  less than .

# Question: 1

Find:

(i)             $0.2×6$

(ii)          $8×4.6$

(iii)      $2.71×5$

(iv)       $20.1×4$

(v)          $0.05×7$

(vi)       $211.02×4$

(vii)    $2×0.86$

## Solution

(i)             $0.2×6=1.2$

(ii)          $8×4.6=36.8$

(iii)      $2.71×5=13.55$

(iv)       $20.1×4=80.4$

(v)          $0.05×7=0.35$

(vi)       $211.02×4=844.08$

(vii)    $2×0.86=1.72$

# Question: 2

Find the area of rectangle whose length is $5.7\text{\hspace{0.17em}}\text{cm}$ and breadth is $3\text{\hspace{0.17em}}\text{cm}$.

## Solution

Given:

Length of rectangle  and Breadth of rectangle

Area of rectangle $=$ Length $x$ Breadth

Thus, the area of rectangle is .

# Question: 3

Find:

(i)             $1.3×10$

(ii)          $36.8×10$

(iii)      $153.7×10$

(iv)       $168.07×10$

(v)          $31.1×100$

(vi)       $156.1×100$

(vii)    $3.62×100$

(viii)$43.07×100$

(ix)       $0.5×10$

(x)          $0.08×10$

(xi)       $0.9×100$

(xii)    $0.03×1000$

## Solution

(i)             $1.3×10=13.0$

(ii)          $36.8×10=368.0$

(iii)      $153.7×10=1537.0$

(iv)       $168.07×10=1680.7$

(v)          $31.1×100=3110.0$

(vi)       $156.1×100=15610.0$

(vii)    $3.62×100=362.0$

(viii)$43.07×100=4307.0$

(ix)       $0.5×10=5.0$

(x)          $0.08×10=0.80$

(xi)       $0.9×100=90.0$

(xii)    $0.03×1000=30.0$

# Question: 4

A two-wheeler covers a distance of  in one litre of petrol. How much distance will it cover in  of petrol?

## Solution

$\because$      In one litre, a two-wheeler covers a distance

$\therefore$      In $10$ litres, a two-wheeler covers a distance

Thus,  distance will be covered by it in $10$ litres of petrol.

# Question: 5

Find:

(i)             $2.5×0.3$

(ii)          $0.1×51.7$

(iii)      $0.2×316.8$

(iv)       $1.3×3.1$

(v)          $0.5×0.05$

(vi)       $11.2×0.15$

(vii)    $1.07×0.02$

(viii) $10.05×1.05$

(ix)       $101.01×0.01$

(x)          $100.01×1.1$

## Solution

(i)             $2.5×0.3=0.75$

(ii)          $0.1×51.7=5.17$

(iii)      $0.2×316.8=63.36$

(iv)       $1.3×3.1=4.03$

(v)          $0.5×0.05=0.025$

(vi)       $11.2×0.15=1.680$

(vii)    $1.07×0.02=0.0214$

(viii) $10.05×1.05=10.5525$

(ix)       $101.01×0.01=1.0101$

(x)          $100.01×1.1=110.011$

# Question: 1

Find:

(i)             $0.4÷2$

(ii)          $0.35÷5$

(iii)      $2.48÷4$

(iv)       $65.4÷6$

(v)          $651.2÷4$

(vi)       $14.49÷7$

(vii)    $3.96÷4$

(viii)$0.80÷5$

## Solution

(i)             $0.4÷2=\frac{4}{10}×\frac{1}{2}=\frac{2}{10}=0.2$

(ii)          $0.35÷5=\frac{35}{100}×\frac{1}{5}=\frac{7}{100}=0.07$

(iii)      $2.48÷4=\frac{248}{100}×\frac{1}{4}=\frac{62}{100}=0.62$

(iv)       $65.4÷6=\frac{654}{10}×\frac{1}{6}=\frac{109}{10}=10.9$

(v)          $651.2÷4=\frac{6512}{10}×\frac{1}{4}=\frac{1628}{10}=162.8$

(vi)       $14.49÷7=\frac{1449}{100}×\frac{1}{7}=\frac{207}{100}=2.07$

(vii)    $3.96÷4=\frac{396}{100}×\frac{1}{4}=\frac{99}{100}=0.99$

(viii)$0.80÷5=\frac{80}{100}×\frac{1}{5}=\frac{16}{100}=0.16$

# Question: 2

Find:

(i)             $4.8÷10$

(ii)          $52.5÷10$

(iii)      $0.7÷10$

(iv)       $33.1÷10$

(v)          $272.23÷10$

(vi)       $0.56÷10$

(vii)    $3.97÷10$

## Solution

(i)             $4.8÷10=\frac{4.8}{10}=0.48$

(ii)          $52.5÷10=\frac{52.5}{10}=5.25$

(iii)      $0.7÷10=\frac{0.7}{10}=0.07$

(iv)       $33.1÷10=\frac{33.1}{10}=3.31$

(v)          $272.23÷10=\frac{272.33}{10}=27.223$

(vi)       $0.56÷10=\frac{0.56}{10}=0.056$

(vii)    $3.97÷10=\frac{3.97}{10}=0.397$

# Question: 3

Find:

(i)             $2.7÷100$

(ii)          $0.3÷100$

(iii)      $0.78÷100$

(iv)       $432.6÷100$

(v)          $23.6÷100$

(vi)       $98.53÷100$

## Solution

(i)             $2.7÷100=\frac{27}{10}×\frac{1}{100}=\frac{27}{1000}=0.027$

(ii)          $0.3÷100=\frac{3}{10}×\frac{1}{100}=\frac{3}{1000}=0.003$

(iii)      $0.78÷100=\frac{78}{100}×\frac{1}{100}=\frac{78}{10000}=0.0078$

(iv)       $432.6÷100=\frac{4326}{10}×\frac{1}{100}=\frac{4326}{1000}=4.326$

(v)          $23.6÷100=\frac{236}{10}×\frac{1}{100}=\frac{236}{1000}=0.236$

(vi)       $98.53÷100=\frac{9853}{100}×\frac{1}{100}=\frac{9853}{10000}=0.9853$

# Question: 4

Find:

(i)             $7.9÷1000$

(ii)          $26.3÷1000$

(iii)      $38.53÷1000$

(iv)       $128.9÷1000$

(v)          $0.5÷1000$

## Solution

(i)             $7.9÷1000=\frac{79}{10}×\frac{1}{1000}=\frac{79}{10000}=0.0079$

(ii)          $26.3÷1000=\frac{263}{10}×\frac{1}{1000}=\frac{263}{10000}=0.0263$

(iii)      $38.53÷1000=\frac{3853}{100}×\frac{1}{1000}=\frac{3853}{100000}=0.03853$

(iv)       $128.9÷1000=\frac{1289}{10}×\frac{1}{1000}=\frac{1289}{10000}=0.1289$

(v)          $0.5÷1000=\frac{5}{10}×\frac{1}{1000}=\frac{5}{10000}=0.0005$

# Question: 5

Find:

(i)             $7÷3.5$

(ii)          $36÷0.2$

(iii)      $3.25÷0.5$

(iv)       $30.94÷0.7$

(v)          $0.5÷0.25$

(vi)       $7.75÷0.25$

(vii)    $76.5÷0.15$

(viii)$37.8÷1.4$

(ix)       $2.73÷1.3$

## Solution

(i)             $7÷3.5=7÷\frac{35}{10}=7×\frac{10}{35}=\frac{10}{5}=2$

(ii)          $36÷0.2=36÷\frac{2}{10}=36×\frac{10}{2}=18×10=180$

(iii)      $3.25÷0.5=\frac{325}{100}÷\frac{5}{10}=\frac{325}{100}×\frac{10}{5}=\frac{65}{10}=6.5$

(iv)       $30.94÷0.7=\frac{3094}{100}÷\frac{7}{10}=\frac{3094}{100}×\frac{10}{7}=\frac{442}{10}=44.2$

(v)          $0.5÷0.25=\frac{5}{10}÷\frac{25}{100}=\frac{5}{10}×\frac{100}{25}=\frac{10}{5}=2$

(vi)       $7.75÷0.25=\frac{775}{100}÷\frac{25}{100}=\frac{775}{100}×\frac{100}{25}=31$

(vii)    $76.5÷0.15=\frac{765}{10}÷\frac{15}{100}=\frac{765}{10}×\frac{100}{15}=51×10=510$

(viii)$37.8÷1.4=\frac{378}{10}÷\frac{14}{10}=\frac{378}{10}×\frac{10}{14}=27$

(ix)       $2.73÷1.3=\frac{273}{100}÷\frac{13}{10}=\frac{273}{100}×\frac{10}{13}=\frac{21}{10}=2.1$

# Question: 6

A vehicle covers a distance of  in  of petrol. How much distance will it cover in one litre of petrol?

## Solution

$\because$      In $2.4$ litres of petrol, distance covered by the vehicle

$\therefore$      In $1$ litre of petrol, distance covered by the vehicle $=43.2÷2.4$

$=\frac{432}{10}÷\frac{24}{10}=\frac{432}{10}×\frac{10}{24}$

Thus, it covered  distance in one litre of petrol.