Unit: 2: Fraction and Decimals

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)    $8\frac{1}{2}-3\frac{5}{8}$

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

Arrange the following in descending order:

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

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

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

$\frac{2}{9}=\frac{14}{63},\frac{2}{3}=\frac{42}{63},\frac{8}{21}=\frac{24}{63}$   
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)          $\frac{1}{5},\frac{3}{7},\frac{7}{10}$

        $\frac{1}{5}=\frac{14}{70},\frac{3}{7}=\frac{30}{70},\frac{7}{10}=\frac{49}{70}$      
[Converting into like fractions]

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

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

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?

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

 

 

$\text{(Along the first row }\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{15}{11}).$

Sum of first row

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

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.

A rectangular sheet of paper is $12\frac{1}{2}\text{ cm}$ long and $10\frac{2}{3}\text{ cm}$ wide. Find its perimeter.

Given:

The sheet of paper is in rectangular form.

Length of sheet $=12\frac{1}{2}\text{ cm}$ and

Breadth of sheet $=10\frac{2}{3}\text{ cm}$

Perimeter of rectangle $=2\text{ (length}+\text{breadth)}$

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

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

$=2\left(\frac{25\times 3\times 32\times 2}{6}\right)$

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

$=2\times \frac{139}{6}$

$=\frac{139}{3}$

$=46\frac{1}{3}\text{ cm}\text{. }$

Thus, the perimeter of the rectangular sheet is $46\frac{1}{3}\text{ cm}\text{.}$

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

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

        $\text{AB}=\frac{5}{2}\text{ cm, BE}=2\frac{3}{4}\text{ cm, AE}=3\frac{3}{5}\text{ cm}$

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

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

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

        $\begin{array}{l}=\frac{50+55+72}{20}\\ =\frac{177}{20}\\ =8\frac{17}{20}\text{ cm}\end{array}$

Thus, the perimeter of $\Delta \text{ABE}$ is $8\frac{17}{20}\text{ cm}$.

(ii)          In $\text{BCDE,}$

        $\text{BE}=2\frac{3}{4}\text{ cm, ED}=\frac{7}{6}\text{ cm}$

        Perimeter of rectangle $=2\text{ (length}+\text{breadth)}$

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

        $\begin{array}{l}=2\left(\frac{33+14}{12}\right)\\ =\left(\frac{47}{6}\right)\\ =7\frac{5}{6}\text{ cm}\end{array}$

Thus, the perimeter of rectangle $\text{BCDE}$ is $7\frac{5}{6}\text{ cm}\text{.}$

Comparing the perimeter of triangle and that of rectangle,

        $8\frac{17}{20}\text{ cm}>7\frac{5}{6}\text{ cm}$

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

Salil wants to put a picture in a frame. The picture is $7\frac{3}{5}\text{ cm}$ wide.

To fit in the frame the picture cannot be more than $7\frac{3}{10}\text{ cm}$ wide. How much should the picture be trimmed?

Given:

The width of the picture $=7\frac{3}{5}\text{ cm}$

and the width of picture frame $=7\frac{3}{10}\text{ cm}$

Therefore, the picture should be trimmed

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

$\begin{array}{l}=\frac{76-73}{10}\\ =\frac{3}{10}\text{ cm}\end{array}$

Thus, the picture should be trimmed by $\frac{3}{10}\text{ cm}\text{.}$

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?

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.

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?

Time taken by Michael to colour the picture $=\frac{7}{12}\text{ hour}$ 

Time taken by Vaibhav to colour the picture $=\frac{3}{4}\text{ hour}$

Converting both fractions in like fractions, $\frac{7}{12}\text{ and }\frac{3}{4}=\frac{3\times 3}{4\times 3}=\frac{9}{12}$

Here, $\frac{7}{12}<\frac{9}{12}$          $\Rightarrow \frac{7}{12}<\frac{3}{4}$

Thus, Vaibhav worked longer time.

Vaibhav worked longer time by

$\begin{array}{l}\frac{3}{4}-\frac{7}{12}\\ =\frac{9-7}{12}\\ =\frac{2}{12}\\ =\frac{1}{6}\text{ hour}\text{.}\end{array}$

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

 

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

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

 

 

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

 

 

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

 

 

(iv)       $3\times \frac{1}{4}$        (d) 

 

 

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

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

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

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

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

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

 

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

 

 

 

 

 

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

 

 

 

 

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

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

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

Multiply and reduce to lowest form:

(i)             $7\times \frac{3}{5}$

(ii)          $4\times \frac{1}{3}$

(iii)      $2\times \frac{6}{7}$

(iv)       $5\times \frac{2}{9}$

(v)          $\frac{2}{3}\times 4$

(vi)       $\frac{5}{2}\times 6$

(vii)    $11\times \frac{4}{7}$

(viii) $20\times \frac{4}{5}$

(ix)       $13\times \frac{1}{3}$

(x)          $15\times \frac{3}{5}$

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

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

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

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

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

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

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

(viii) $20\times \frac{4}{5}=4\times 4=16$

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

(x)          $15\times \frac{3}{5}=3\times 3=9$

Shade:

(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.                 

 

 

 

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

        $=\frac{1}{2}\times 12=6\text{ circles}$                         

 

 

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

        $=\frac{2}{3}\times 9=2\times 3=6\text{ triangles}$             

 

 

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

        $=\frac{3}{5}\times 153\times 3=9\text{ squares}$            

 

 

 

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) $\frac{1}{2}\text{ of }24=12$ 

(ii) $\frac{1}{2}\text{ of }46=23$ 

b.   (i) $\frac{2}{3}\text{ of }18=\frac{2}{3}\times 18=2\times 6=12$ 

(ii) $\frac{2}{3}\text{ of }27=\frac{2}{3}\times 27=2\times 9=18$

c.    (i) $\frac{3}{4}\text{ of }16=\frac{3}{4}\times 16=3\times 4=12$

(ii) $\frac{3}{4}\text{ of }36=\frac{3}{4}\times 36=3\times 9=27$ 

d.   (i) $\frac{4}{5}\text{ of }20=\frac{4}{5}\times 20=4\times 4=16$

(ii) $\frac{4}{5}\text{ of }35=\frac{4}{5}\times 35=4\times 7=28$

Multiply and express as a mixed fraction:

a.    $3\times 5\frac{1}{5}$

b.   $5\times 6\frac{3}{4}$

c.    $7\times 2\frac{1}{4}$

d.   $4\times 6\frac{1}{3}$

e.    $3\frac{1}{4}\times 6$

f.     $3\frac{2}{5}\times 8$

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

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

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

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

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

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

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) $9\frac{2}{3}$

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

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

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

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

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?

Given: Total quantity of water in bottle $=5\text{ litres}$

(i)             Water consumed by Vidya $=\frac{2}{5}\text{ of }5\text{ litres}=\frac{2}{5}\times 5=2\text{ litres}$

Thus, Vidya drank 2 litres water from the bottle.

(ii)          Water consumed by Pratap

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

$=\frac{5-2}{5}=\frac{3}{5}\text{ part of bottle}$

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

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

 

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) $\frac{3}{10}$

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

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

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

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

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

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

Multiply and reduce to lowest form (if possible):

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

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

(iii)      $\frac{3}{8}\times \frac{6}{4}$

(iv)       $\frac{9}{5}\times \frac{3}{5}$

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

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

(vii)    $\frac{4}{5}\times \frac{12}{7}$

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

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

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

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

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

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

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

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) $\frac{2}{5}\times 5\frac{1}{4}$ 

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

(iii)    $\frac{3}{2}\times 5\frac{1}{3}$

(iv)     $\frac{5}{6}\times 2\frac{3}{7}$

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

(vi)     $2\frac{3}{5}\times 3$

(vii)  $3\frac{4}{7}\times \frac{3}{5}$

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

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

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

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

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

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

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

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 $\frac{3}{7}$

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

Right side expression $=\frac{3}{5}$ of $\frac{5}{8}$$=\frac{3}{5}\times \frac{5}{8}=\frac{3}{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, $\frac{3}{5}\text{ of }\frac{5}{8}$ is greater.

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

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

Right side expression $=\frac{2}{3}\times \frac{3}{7}=\frac{2}{7}$

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

        Thus, $\frac{1}{2}\text{ of }\frac{6}{7}$ is greater.

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

The distance between two adjacent saplings $=\frac{3}{4}\text{ m}$

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 $=3\times \frac{3}{4}=\frac{9}{4}\text{ m}=2\frac{1}{4}\text{ m}$

Thus, the distance between the first and the last sapling is $2\frac{1}{4}\text{ m}\text{.}$ 

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?

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 $=1\frac{3}{4}\times 6=\frac{7}{4}\times 6=\frac{21}{2}=10\frac{1}{2}$

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

A car runs $16\text{ km}$ using $1$ litre of petrol. How much distance will it cover using $2\frac{3}{4}$ litres of petrol.

In $1$ litre of pertrol, car covers the distance $=16\text{ km}$

In $2\frac{3}{4}$ litres of petrol, car covers the distance $=2\frac{3}{4}$ of $16\text{ km}=\frac{11}{4}\times 16=44\text{ km}$

Thus, the car will cover $44\text{ km}$.

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

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

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

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

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

(ii) The simplest form of $\frac{5}{10}\text{ is }\frac{1}{2}.$

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

(ii) The simplest form of $\frac{8}{15}\text{ is }\frac{8}{15}.$

 

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)       $5÷3\frac{4}{7}$

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

$=16$

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

$=\frac{84}{5}$

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

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

$=\frac{24}{7}$

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

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

$=\frac{3}{2}$

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

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

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

$=\frac{9}{7}$

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

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

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

$=\frac{7}{5}$

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

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)    $\frac{1}{11}$

(i)             Reciprocal of  $\frac{3}{4}=\frac{4}{3}\stackrel{}{\to }$ Improper fraction

(ii)          Reciprocal of  $\frac{5}{8}=\frac{8}{5}\stackrel{}{\to }$ Improper fraction

(iii)      Reciprocal of  $\frac{9}{7}=\frac{7}{9}\stackrel{}{\to }$ Proper fraction

(iv)       Reciprocal of  $\frac{6}{5}=\frac{5}{6}\stackrel{}{\to }$ Proper fraction

(v)          Reciprocal of  $\frac{12}{7}=\frac{7}{12}\stackrel{}{\to }$ Proper fraction

(vi)       Reciprocal of  $\frac{1}{8}=8\stackrel{}{\to }$ Whole number

(vii)    Reciprocal of  $\frac{1}{11}=11\stackrel{}{\to }$ Whole number

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)        $4\frac{3}{7}÷7$

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

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

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

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

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

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

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

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

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

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

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

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

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)$2\frac{1}{5}÷1\frac{1}{5}$

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

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

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

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

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

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

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

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

 

Which is greater?

(i)             $0.5\text{ or }0.05$

(ii)          $0.7\text{ or }0.5$

(iii)      $7\text{ or }0.7$

(iv)       $1.37\text{ or }1.49$

(v)          $2.03\text{ or }2.30$

(vi)       $0.8\text{ or }\mathrm{0.88.}$

(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)       $0.8<\mathrm{0.88.}$

Express as rupees using decimals:

(i)             7 paise

(ii)          7 rupees 7 paise

(iii)      77 rupees 77 paise

(iv)       50 paise

(v)          235 paise.

$\because 100\text{ paise}=\text{Rs}1$

$\therefore 1\text{ paise}=\text{Rs}\frac{1}{100}$

(i)             $7\text{ paise}=\text{Rs}\frac{7}{100}=\text{Rs}0.07$

(ii)          $7\text{ rupees }7\text{ paise}=\text{Rs}7+\text{Rs}\frac{7}{100}$

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

(iii)      $77\text{ rupees }77\text{ paise}=\text{Rs}77+\text{Rs}\frac{77}{100}$

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

(iv)       $50\text{ paise}=\text{Rs}\frac{50}{100}=\text{Rs}0.50$

(v)          $235\text{ paise}=\text{Rs}\frac{235}{100}=\text{Rs}2.35$

(i)             Express $5\text{ cm}$ in metre and kilometre.

(ii)          Express $35\text{ mm}$ in cm, m and km.

(i)             Express $5\text{ cm}$ in metre and kilometre.

$\because 100\text{ cm}=1\text{meter}$

$\therefore 1\text{ cm}=\frac{1}{100}\text{meter}$

$\Rightarrow 5\text{ cm}=\frac{5}{100}=0.05\text{ meter}\text{.}$

Now,

$\because 1000\text{ meters}=1\text{kilometers}$

$\therefore 1\text{ meter}=\frac{1}{1000}\text{kilometer}$

$\Rightarrow 0.05\text{ meter}=\frac{0.05}{1000}=0.00005\text{ kilometer}$

(ii)          Express $35\text{ mm}$ in cm, m and km.

$\because 10\text{ mm}=1\text{cm}$

$\therefore 1\text{ mm}=\frac{1}{10}\text{cm}$

$\Rightarrow 35\text{ mm}=\frac{35}{10}=3.5\text{ cm}$

Now,

$\because 100\text{ cm}=1\text{meter}$

$\therefore 1\text{ cm}=\frac{1}{100}\text{meter}$

$\Rightarrow 3.5\text{ cm}=\frac{3.5}{100}=0.035\text{ meter}$

Again,

$\because 1000\text{ meters}=1\text{kilometers}$

$\therefore 1\text{ meter}=\frac{1}{1000}\text{kilometer}$

$\Rightarrow 0.035\text{ meter}=\frac{0.035}{1000}=0.000035\text{ kilometer}$

Express in kg:

(i)             $200\text{ g}$

(ii)          $3470\text{ g}$

(iii)      $4\text{ kg }8\text{ g}$

(iv)       $2598\text{ mg}$

We know that,

$1000\text{ g}=1\text{ kg}$

$\Rightarrow 1\text{ g}=\frac{1}{1000}\text{ kg}$

(i)             $\begin{array}{c}200\text{ g}=\left(200\times \frac{1}{1000}\right)\text{ kg}\\ =0.2\text{ kg}\end{array}$

(ii)          $\begin{array}{c}3470\text{ g}=\left(3470\times \frac{1}{1000}\right)\text{ kg}\\ =3.470\text{ kg}\end{array}$

(iii)      $\begin{array}{c}4\text{ kg }8\text{ g}=4\text{ kg}+\left(8\times \frac{1}{1000}\right)\text{ kg}\\ =4\text{ kg}+0.008\text{ kg}\\ =4.008\text{ kg}\end{array}$

(iv)       $\begin{array}{c}2598\text{ mg}=\left(2598\times \frac{1}{1000}\right)\text{ g}\\ =\left(2598\times \frac{1}{1000}\times \frac{1}{1000}\right)\text{ kg}\\ =0.002598\text{ kg}\end{array}$

Write the following decimal numbers in the expanded form:

(i)             $20.03$

(ii)          $2.03$

(iii)      $200.03$

(iv)       $2.034$

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

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

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

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

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

(i)             $2.56$

(ii)          $21.37$

(iii)      $10.25$

(iv)       $9.42$

(v)          $\mathrm{63.352.}$

(i)             Place value of $2$ in $2.56=2\times 1=2\text{ ones}$

(ii)          Place value of $2$ in $21.37=2\times 10=2\text{ tens}$

(iii)      Place value of $2$ in $10.25=2\times \frac{1}{10}=2\text{tenths}$

(iv)       Place value of $2$ in $9.42=2\times \frac{1}{100}=2\text{ hundredths}$

(v)          Place value of $2$ in $63.352=2\times \frac{1}{1000}=2\text{ thousandths}$

Dinesh went from place A to place B and from there to place C. A is $7.5\text{ km}$ from B and B is $12.7\text{ km}$ from C. Ayub went from place A to place D and from there to place C. D is $9.3\text{ km}$ from A and C is $11.8\text{ km}$ from D. Who travelled more and by how much?

 

 

 

Distance travelled by Dinesh when he went from place A to place $\text{B}=7.5\text{ km}$ and from place B to $\text{C}=12.7\text{ km}.$

 

 

 

 

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

$=7.5+12.7=20.2\text{ km}$

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

$=9.3+11.8=21.1\text{ km}$

      On comparing the total distance of Ayub and Dinesh,

      $21.1\text{ km}>20.2\text{ km}$

Therefore, Ayub covered more distance by $21.1-20.2=0.9\text{ km}=900\text{ m}$

Shyama bought $5\text{kg }300\text{g}$ apples and $3\text{kg }250\text{g}$ mangoes. Sarala bought $4\text{kg }800\text{g}$ oranges and $4\text{kg }150\text{g}$ bananas. Who bought more fruits?

Total weight of fruits bought by Shyama $=5\text{ kg }300\text{ g}+3\text{ kg }250\text{ g}=8\text{ kg }550\text{ g}$

Total weight of fruits bought by Sarala $=4\text{ kg }800\text{ g}+4\text{ kg }150\text{ g}=8\text{ kg }950\text{ g}$

On comparing the quantity of fruits, $8\text{ kg }550\text{ g}<8\text{ kg }950\text{ g}$

Therefore, Sarala bought more fruits.

How much less is $28\text{ km}$ than $42.6\text{ km}$?

We have to find the difference between $42.6\text{ km}$ and $28\text{ km}$.

Difference $=42.6-28.0=14.6\text{ km}$

Therefore, $28\text{ km}$ is $14.6\text{ km}$ less than $42.6\text{ km}$.

 

Find:

(i)             $0.2\times 6$

(ii)          $8\times 4.6$

(iii)      $2.71\times 5$

(iv)       $20.1\times 4$

(v)          $0.05\times 7$

(vi)       $211.02\times 4$

(vii)    $2\times 0.86$

(i)             $0.2\times 6=1.2$

(ii)          $8\times 4.6=36.8$

(iii)      $2.71\times 5=13.55$

(iv)       $20.1\times 4=80.4$

(v)          $0.05\times 7=0.35$

(vi)       $211.02\times 4=844.08$

(vii)    $2\times 0.86=1.72$

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

Given:

Length of rectangle $=5.7\text{ cm}$ and Breadth of rectangle $=3\text{ cm}$

Area of rectangle $=$ Length $x$ Breadth $=5.7\times 3=17.1{\text{ cm}}^2$

Thus, the area of rectangle is $17.1{\text{ cm}}^2$.

Find:

(i)             $1.3\times 10$

(ii)          $36.8\times 10$

(iii)      $153.7\times 10$

(iv)       $168.07\times 10$

(v)          $31.1\times 100$

(vi)       $156.1\times 100$

(vii)    $3.62\times 100$

(viii)$43.07\times 100$

(ix)       $0.5\times 10$

(x)          $0.08\times 10$

(xi)       $0.9\times 100$

(xii)    $0.03\times 1000$

(i)             $1.3\times 10=13.0$

(ii)          $36.8\times 10=368.0$

(iii)      $153.7\times 10=1537.0$

(iv)       $168.07\times 10=1680.7$

(v)          $31.1\times 100=3110.0$

(vi)       $156.1\times 100=15610.0$

(vii)    $3.62\times 100=362.0$

(viii)$43.07\times 100=4307.0$

(ix)       $0.5\times 10=5.0$

(x)          $0.08\times 10=0.80$

(xi)       $0.9\times 100=90.0$

(xii)    $0.03\times 1000=30.0$

A two-wheeler covers a distance of $55.3\text{ km}$ in one litre of petrol. How much distance will it cover in $10\text{ litres}$ of petrol?

$\because$      In one litre, a two-wheeler covers a distance $=55.3\text{ km}$

$\therefore$      In $10$ litres, a two-wheeler covers a distance $=55.3\times 10=553.0\text{ km}$

Thus, $553\text{ km}$ distance will be covered by it in $10$ litres of petrol.

Find:

(i)             $2.5\times 0.3$

(ii)          $0.1\times 51.7$

(iii)      $0.2\times 316.8$

(iv)       $1.3\times 3.1$

(v)          $0.5\times 0.05$

(vi)       $11.2\times 0.15$

(vii)    $1.07\times 0.02$

(viii) $10.05\times 1.05$

(ix)       $101.01\times 0.01$

(x)          $100.01\times 1.1$

(i)             $2.5\times 0.3=0.75$

(ii)          $0.1\times 51.7=5.17$

(iii)      $0.2\times 316.8=63.36$

(iv)       $1.3\times 3.1=4.03$

(v)          $0.5\times 0.05=0.025$

(vi)       $11.2\times 0.15=1.680$

(vii)    $1.07\times 0.02=0.0214$

(viii) $10.05\times 1.05=10.5525$

(ix)       $101.01\times 0.01=1.0101$

(x)          $100.01\times 1.1=110.011$

 

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)$0.80÷5$

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

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

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

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

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

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

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

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

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)    $3.97÷10$ 

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

Find:

(i)             $2.7÷100$

(ii)          $0.3÷100$

(iii)      $0.78÷100$

(iv)       $432.6÷100$

(v)          $23.6÷100$

(vi)       $98.53÷100$

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

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

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

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

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

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

Find:

(i)             $7.9÷1000$

(ii)          $26.3÷1000$

(iii)      $38.53÷1000$

(iv)       $128.9÷1000$

(v)          $0.5÷1000$

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

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

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

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

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

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)       $2.73÷1.3$

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

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

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

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

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

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

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

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

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

A vehicle covers a distance of $43.2\text{ km}$ in $2.4\text{ litres}$ of petrol. How much distance will it cover in one litre of petrol?

$\because$      In $2.4$ litres of petrol, distance covered by the vehicle $=43.2\text{ km}$

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

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

$=18\text{ km}$

Thus, it covered $18\text{ km}$ distance in one litre of petrol.