Chapter 2: Fractions and Decimals

Exercise 2.1

Question 1

What fraction of each of the following figure is shaded part?

Some integers are marked on the following number line. Integers, ML Aggarwal Understanding Mathematics Solutions ICSE Class 7.

Answer

(i) Total parts = 8

Shaded parts = 2

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{2}{8}$ .

Hence, fraction = $\dfrac{1}{4}$ .

(ii) Total parts = 10

Shaded parts = 3

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{3}{10}$ .

Hence, fraction = $\dfrac{3}{10}$ .

(iii) Total parts = 12

Shaded parts = 5

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{5}{12}$ .

Hence, fraction = $\dfrac{5}{12}$ .

(iv) Total parts = 13

Shaded parts = 7

Fraction = $\dfrac{\text{Shaded parts}}{\text{Total parts}} = \dfrac{7}{13}$ .

Hence, fraction = $\dfrac{7}{13}$ .

Question 2

What fraction of an hour is 35 minutes?

Answer

We know, 1 hour = 60 minutes.

Fraction = $\dfrac{35 \text{ minutes}}{60 \text{ minutes}}$

$\Rightarrow \dfrac{35}{60}\\[1em] \Rightarrow \dfrac{35 \div 5}{60 \div 5}\\[1em] \Rightarrow \dfrac{7}{12}$

Hence, the required fraction = $\dfrac{7}{12}$ .

Question 3

Convert the following fractions into improper fractions:

(i) $2\dfrac{7}{9}$

(ii) $5\dfrac{4}{11}$

Answer

We use the rule: improper fraction = $\dfrac{(\text{natural number} \times \text{denominator}) + \text{numerator}}{\text{denominator}}$ .

(i) $2\dfrac{7}{9}$

$\Rightarrow 2\dfrac{7}{9} = \dfrac{2 \times 9 + 7}{9}\\[1em] = \dfrac{18 + 7}{9}\\[1em] = \dfrac{25}{9}.$

Hence, the required fraction is $2\dfrac{7}{9} = \dfrac{25}{9}$ .

(ii) $5\dfrac{4}{11}$

$\Rightarrow 5\dfrac{4}{11} = \dfrac{5 \times 11 + 4}{11}\\[1em] = \dfrac{55 + 4}{11}\\[1em] = \dfrac{59}{11}.$

Hence, the required fraction is $5\dfrac{4}{11} = \dfrac{59}{11}$ .

Question 4

Convert the following fractions into mixed fractions:

(i) $\dfrac{73}{8}$

(ii) $\dfrac{94}{13}$

Answer

(i) $\dfrac{73}{8}$

Dividing 73 by 8, we get quotient = 9 and remainder = 1.

$\Rightarrow \dfrac{73}{8} = 9\dfrac{1}{8}.$

Hence, $\dfrac{73}{8} = 9\dfrac{1}{8}$ .

(ii) $\dfrac{94}{13}$

Dividing 94 by 13, we get quotient = 7 and remainder = 3.

$\Rightarrow \dfrac{94}{13} = 7\dfrac{3}{13}.$

Hence, $\dfrac{94}{13} = 7\dfrac{3}{13}$ .

Question 5

Fill in the missing numbers in the following equivalent fractions:

(i) $\dfrac{3}{7} = \dfrac{...}{35}$

(ii) $\dfrac{5}{...} = \dfrac{30}{18}$

(iii) $\dfrac{...}{9} = \dfrac{56}{72}$

Answer

(i) $\dfrac{3}{7} = \dfrac{...}{35}$

To get 35 from 7, we have to multiply 7 by 5.

So, multiply both numerator and denominator by 5.

$\Rightarrow \dfrac{3}{7} = \dfrac{3 \times 5}{7 \times 5}$

= $\dfrac{15}{35}$ .

Hence, the missing number is 15.

(ii) $\dfrac{5}{...} = \dfrac{30}{18}$

To get 5 from 30, we have to divide 30 by 6.

So, divide both numerator and denominator by 6.

$\Rightarrow \dfrac{30}{18} = \dfrac{30 \div 6}{18 \div 6}$

= $\dfrac{5}{3}.$

Hence, the missing number is 3.

(iii) $\dfrac{...}{9} = \dfrac{56}{72}$

To get 9 from 72, we have to divide 72 by 8.

So, divide both numerator and denominator by 8.

$\Rightarrow \dfrac{56}{72} = \dfrac{56 \div 8}{72 \div 8}$

= $\dfrac{7}{9}.$

Hence, the missing number is 7.

Question 6

Reduce the following fractions to their simplest form:

(i) $\dfrac{48}{72}$

(ii) $\dfrac{276}{115}$

(iii) $\dfrac{72}{336}$

Answer

(i) $\dfrac{48}{72}$

By prime factorisation,

$\Rightarrow \dfrac{48}{72} = \dfrac{2 \times 2 \times 2 \times 2 \times 3}{2 \times 2 \times 2 \times 3 \times 3}\\[1em] = \dfrac{2}{3}.$

Hence, $\dfrac{48}{72}$ in simplest form is $\dfrac{2}{3}$ .

(ii) $\dfrac{276}{115}$

By prime factorisation,

$\Rightarrow \dfrac{276}{115} = \dfrac{2 \times 2 \times 3 \times 23}{5 \times 23}\\[1em] = \dfrac{2 \times 2 \times 3}{5}\\[1em] = \dfrac{12}{5}.$

Hence, $\dfrac{276}{115}$ in simplest form is $\dfrac{12}{5}$ .

(iii) $\dfrac{72}{336}$

By prime factorisation,

$\Rightarrow \dfrac{72}{336} = \dfrac{2 \times 2 \times 2 \times 3 \times 3}{2 \times 2 \times 2 \times 2 \times 3 \times 7}\\[1em] = \dfrac{3}{2 \times 7}\\[1em] = \dfrac{3}{14}.$

Hence, $\dfrac{72}{336}$ in simplest form is $\dfrac{3}{14}$ .

Question 7

Convert the following fractions into equivalent like fractions:

(i) $\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}$

(ii) $\dfrac{7}{25}, \dfrac{9}{10}, \dfrac{19}{40}$

Answer

(i) $\dfrac{3}{4}, \dfrac{5}{6}, \dfrac{7}{8}$

L.C.M. of 4, 6 and 8 is :

$\begin{array}{l|r} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 4, 6 and 8 = 2 x 2 x 2 x 3 = 24.

$\Rightarrow \dfrac{3}{4} = \dfrac{3 \times 6}{4 \times 6} = \dfrac{18}{24}\\[1em] \Rightarrow \dfrac{5}{6} = \dfrac{5 \times 4}{6 \times 4} = \dfrac{20}{24}\\[1em] \Rightarrow \dfrac{7}{8} = \dfrac{7 \times 3}{8 \times 3} = \dfrac{21}{24}$

Hence, the equivalent like fractions are $\dfrac{18}{24}, \dfrac{20}{24}$ and $\dfrac{21}{24}$ .

(ii) $\dfrac{7}{25}, \dfrac{9}{10}, \dfrac{19}{40}$

L.C.M. of 25, 10 and 40 is :

$\begin{array}{l|r} 2 & 25, 10, 40 \\ \hline 2 & 25, 5, 20 \\ \hline 2 & 25, 5, 10 \\ \hline 5 & 25, 5, 5 \\ \hline 5 & 5, 1, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 25, 10 and 40 = 2 x 2 x 2 x 5 x 5 = 200.

$\Rightarrow \dfrac{7}{25} = \dfrac{7 \times 8}{25 \times 8} = \dfrac{56}{200}\\[1em] \Rightarrow \dfrac{9}{10} = \dfrac{9 \times 20}{10 \times 20} = \dfrac{180}{200}\\[1em] \Rightarrow \dfrac{19}{40} = \dfrac{19 \times 5}{40 \times 5} = \dfrac{95}{200}$

Hence, the equivalent like fractions are $\dfrac{56}{200}, \dfrac{180}{200}$ and $\dfrac{95}{200}$ .

Question 8

Arrange the given fractions in descending order:

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

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

Answer

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

LCM of 9, 3 and 21:

$\begin{array}{l|r} 3 & 9, 3, 21 \\ \hline 3 & 3, 1, 7 \\ \hline 7 & 1, 1, 7 \\ \hline & 1, 1, 1 \end{array}$

LCM of 9, 3 and 21 = 3 x 3 x 7 = 63.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{2}{9} = \dfrac{2 \times 7}{9 \times 7} = \dfrac{14}{63}\\[1em] \Rightarrow \dfrac{2}{3} = \dfrac{2 \times 21}{3 \times 21} = \dfrac{42}{63}\\[1em] \Rightarrow \dfrac{8}{21} = \dfrac{8 \times 3}{21 \times 3} = \dfrac{24}{63}$

As 42 > 24 > 14,

$\dfrac{42}{63} \gt \dfrac{24}{63} \gt \dfrac{14}{63} \Rightarrow \dfrac{2}{3} \gt \dfrac{8}{21} \gt \dfrac{2}{9}$ .

Hence, the given fractions in descending order are $\dfrac{2}{3}, \dfrac{8}{21}, \dfrac{2}{9}$ .

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

LCM of 5, 7 and 10 :

$\begin{array}{l|r} 2 & 5, 7, 10 \\ \hline 5 & 5, 7, 5 \\ \hline 7 & 1, 7, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 5, 7 and 10 = 2 x 5 x 7 = 70.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{1}{5} = \dfrac{1 \times 14}{5 \times 14} = \dfrac{14}{70}\\[1em] \Rightarrow \dfrac{3}{7} = \dfrac{3 \times 10}{7 \times 10} = \dfrac{30}{70}\\[1em] \Rightarrow \dfrac{7}{10} = \dfrac{7 \times 7}{10 \times 7} = \dfrac{49}{70}$

As 49 > 30 > 14,

$\dfrac{49}{70} \gt \dfrac{30}{70} \gt \dfrac{14}{70} \Rightarrow \dfrac{7}{10} \gt \dfrac{3}{7} \gt \dfrac{1}{5}$ .

Hence, the given fractions in descending order are $\dfrac{7}{10}, \dfrac{3}{7}, \dfrac{1}{5}$ .

Question 9

Arrange the given fractions in ascending order:

(i) $\dfrac{5}{7}, \dfrac{3}{8}, \dfrac{9}{14}, \dfrac{20}{21}$

(ii) $\dfrac{13}{18}, \dfrac{8}{15}, \dfrac{17}{24}, \dfrac{7}{12}$

Answer

(i) $\dfrac{5}{7}, \dfrac{3}{8}, \dfrac{9}{14}, \dfrac{20}{21}$

LCM of 7, 8, 14 and 21:

$\begin{array}{l|r} 2 & 7, 8, 14, 21 \\ \hline 2 & 7, 4, 7, 21 \\ \hline 2 & 7, 2, 7, 21 \\ \hline 3 & 7, 1, 7, 21 \\ \hline 7 & 7, 1, 7, 7 \\ \hline & 1, 1, 1, 1 \end{array}$

LCM of 7, 8, 14 and 21 = 2 x 2 x 2 x 3 x 7 = 168.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{5}{7} = \dfrac{5 \times 24}{7 \times 24} = \dfrac{120}{168}\\[1em] \Rightarrow \dfrac{3}{8} = \dfrac{3 \times 21}{8 \times 21} = \dfrac{63}{168}\\[1em] \Rightarrow \dfrac{9}{14} = \dfrac{9 \times 12}{14 \times 12} = \dfrac{108}{168}\\[1em] \Rightarrow \dfrac{20}{21} = \dfrac{20 \times 8}{21 \times 8} = \dfrac{160}{168}$

As 63 < 108 < 120 < 160,

$\dfrac{63}{168} \lt \dfrac{108}{168} \lt \dfrac{120}{168} \lt \dfrac{160}{168} \Rightarrow \dfrac{3}{8} \lt \dfrac{9}{14} \lt \dfrac{5}{7} \lt \dfrac{20}{21}$ .

Hence, the given fractions in ascending order are $\dfrac{3}{8}, \dfrac{9}{14}, \dfrac{5}{7}, \dfrac{20}{21}$ .

(ii) $\dfrac{13}{18}, \dfrac{8}{15}, \dfrac{17}{24}, \dfrac{7}{12}$

LCM of 18, 15, 24 and 12:

$\begin{array}{l|r} 2 & 18, 15, 24, 12 \\ \hline 2 & 9, 15, 12, 6 \\ \hline 2 & 9, 15, 6, 3 \\ \hline 3 & 9, 15, 3, 3 \\ \hline 3 & 3, 5, 1, 1 \\ \hline 5 & 1, 5, 1, 1 \\ \hline & 1, 1, 1, 1 \end{array}$

LCM of 18, 15, 24 and 12 = 2 x 2 x 2 x 3 x 3 x 5 = 360.

Write the given fractions as equivalent like fractions.

$\Rightarrow \dfrac{13}{18} = \dfrac{13 \times 20}{18 \times 20} = \dfrac{260}{360}\\[1em] \Rightarrow \dfrac{8}{15} = \dfrac{8 \times 24}{15 \times 24} = \dfrac{192}{360}\\[1em] \Rightarrow \dfrac{17}{24} = \dfrac{17 \times 15}{24 \times 15} = \dfrac{255}{360}\\[1em] \Rightarrow \dfrac{7}{12} = \dfrac{7 \times 30}{12 \times 30} = \dfrac{210}{360}$

As 192 < 210 < 255 < 260,

$\dfrac{192}{360} \lt \dfrac{210}{360} \lt \dfrac{255}{360} \lt \dfrac{260}{360} \Rightarrow \dfrac{8}{15} \lt \dfrac{7}{12} \lt \dfrac{17}{24} \lt \dfrac{13}{18}$ .

Hence, the given fractions in ascending order are $\dfrac{8}{15}, \dfrac{7}{12}, \dfrac{17}{24}, \dfrac{13}{18}$ .

Exercise 2.2

Question 1

Evaluate the following:

(i) $\dfrac{4}{3} + \dfrac{7}{8}$

(ii) $8\dfrac{1}{2} - 3\dfrac{5}{8}$

(iii) $\dfrac{5}{12} + \dfrac{1}{18} - \dfrac{2}{9}$

Answer

(i) $\dfrac{4}{3} + \dfrac{7}{8}$

LCM of 3 and 8

$\begin{array}{l|r} 2 & 3, 8 \\ \hline 2 & 3, 4 \\ \hline 2 & 3, 2 \\ \hline 3 & 3, 1 \\ \hline & 1, 1 \end{array}$

LCM of 3 and 8 = 2 x 2 x 2 x 3 = 24.

$\Rightarrow \dfrac{4 \times 8}{3 \times 8} + \dfrac{7 \times 3}{8 \times 3}\\[1em] \Rightarrow \dfrac{32}{24} + \dfrac{21}{24}\\[1em] \Rightarrow \dfrac{32 + 21}{24}\\[1em] \Rightarrow \dfrac{53}{24}\\[1em] \Rightarrow 2\dfrac{5}{24}$

Hence, $\dfrac{4}{3} + \dfrac{7}{8} = 2\dfrac{5}{24}$ .

(ii) $8\dfrac{1}{2} - 3\dfrac{5}{8}$

LCM of 2 and 8:

$\begin{array}{l|r} 2 & 2, 8 \\ \hline 2 & 2, 4 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$

LCM of 2 and 8 = 2 x 2 x 2 = 8.

$\Rightarrow 8\dfrac{1}{2} - 3\dfrac{5}{8} \\[1em] \Rightarrow \dfrac{17}{2} - \dfrac{29}{8}\\[1em] \Rightarrow \dfrac{17 \times 4}{2 \times 4} - \dfrac{29}{8}\\[1em] \Rightarrow \dfrac{68}{8} - \dfrac{29}{8}\\[1em] \Rightarrow \dfrac{68 - 29}{8}\\[1em] \Rightarrow \dfrac{39}{8}\\[1em] \Rightarrow 4\dfrac{7}{8}$

Hence, $8\dfrac{1}{2} - 3\dfrac{5}{8} = 4\dfrac{7}{8}$ .

(iii) $\dfrac{5}{12} + \dfrac{1}{18} - \dfrac{2}{9}$

LCM of 12, 18 and 9 :

$\begin{array}{l|r} 2 & 12, 18, 9 \\ \hline 2 & 6, 9, 9 \\ \hline 3 & 3, 9, 9 \\ \hline 3 & 1, 3, 3 \\ \hline & 1, 1, 1 \end{array}$

LCM of 12, 18 and 9 =2 x 2 x 3 x 3 = 36.

$\Rightarrow \dfrac{5 \times 3}{12 \times 3} + \dfrac{1 \times 2}{18 \times 2} - \dfrac{2 \times 4}{9 \times 4}\\[1em] \Rightarrow \dfrac{15}{36} + \dfrac{2}{36} - \dfrac{8}{36}\\[1em] \Rightarrow \dfrac{15 + 2 - 8}{36}\\[1em] \Rightarrow \dfrac{9}{36}\\[1em] \Rightarrow \dfrac{1}{4}$

Hence, $\dfrac{5}{12} + \dfrac{1}{18} - \dfrac{2}{9} = \dfrac{1}{4}$ .

Question 2

Simplify the following:

(i) $7\dfrac{3}{4} - 3\dfrac{5}{6} + \dfrac{7}{8}$

(ii) $6\dfrac{1}{8} - 2\dfrac{1}{12} - 5\dfrac{1}{10} + 3\dfrac{7}{25}$

Answer

(i) $7\dfrac{3}{4} - 3\dfrac{5}{6} + \dfrac{7}{8}$

LCM of 4, 6 and 8 :

$\begin{array}{l|r} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$

LCM of 4, 6 and 8 = 2 x 2 x 2 x 3 = 24.

$\Rightarrow \dfrac{31}{4} - \dfrac{23}{6} + \dfrac{7}{8}\\[1em] \Rightarrow \dfrac{31 \times 6}{4 \times 6} - \dfrac{23 \times 4}{6 \times 4} + \dfrac{7 \times 3}{8 \times 3}\\[1em] \Rightarrow \dfrac{186}{24} - \dfrac{92}{24} + \dfrac{21}{24}\\[1em] \Rightarrow \dfrac{186 - 92 + 21}{24}\\[1em] \Rightarrow \dfrac{115}{24}\\[1em] \Rightarrow 4\dfrac{19}{24}$

Hence, $7\dfrac{3}{4} - 3\dfrac{5}{6} + \dfrac{7}{8} = 4\dfrac{19}{24}$ .

(ii) $6\dfrac{1}{8} - 2\dfrac{1}{12} - 5\dfrac{1}{10} + 3\dfrac{7}{25}$

LCM of 8, 12, 10 and 25 :

$\begin{array}{l|r} 2 & 8, 12, 10, 25 \\ \hline 2 & 4, 6, 5, 25 \\ \hline 2 & 2, 3, 5, 25 \\ \hline 3 & 1, 3, 5, 25 \\ \hline 5 & 1, 1, 5, 25 \\ \hline 5 & 1, 1, 1, 5 \\ \hline & 1, 1, 1, 1 \end{array}$

LCM of 8, 12, 10 and 25 = 2 x 2 x 2 x 3 x 5 x 5 = 600.

$\Rightarrow \dfrac{49}{8} - \dfrac{25}{12} - \dfrac{51}{10} + \dfrac{82}{25}\\[1em] \Rightarrow \dfrac{49 \times 75}{8 \times 75} - \dfrac{25 \times 50}{12 \times 50} - \dfrac{51 \times 60}{10 \times 60} + \dfrac{82 \times 24}{25 \times 24}\\[1em] \Rightarrow \dfrac{3675}{600} - \dfrac{1250}{600} - \dfrac{3060}{600} + \dfrac{1968}{600}\\[1em] \Rightarrow \dfrac{3675 - 1250 - 3060 + 1968}{600}\\[1em] \Rightarrow \dfrac{1333}{600}\\[1em] \Rightarrow 2\dfrac{133}{600}$

Hence, $6\dfrac{1}{8} - 2\dfrac{1}{12} - 5\dfrac{1}{10} + 3\dfrac{7}{25} = 2\dfrac{133}{600}$ .

Question 3

Aliyah studies for $5\dfrac{2}{3}$ hours daily. She devotes $2\dfrac{4}{5}$ hours of her time for science and mathematics. How much time does she devote for other subjects?

Answer

Total time Aliyah studies daily = $5\dfrac{2}{3}$ hours.

Time devoted for science and mathematics = $2\dfrac{4}{5}$ hours.

Time devoted for other subjects = $5\dfrac{2}{3} - 2\dfrac{4}{5}$

LCM of 3 and 5:

$\begin{array}{l|r} 3 & 3, 5 \\ \hline 5 & 1, 5 \\ \hline & 1, 1 \end{array}$

LCM of 3 and 5 = 3 x 5 = 15.

$\Rightarrow \dfrac{17}{3} - \dfrac{14}{5}\\[1em] \Rightarrow \dfrac{17 \times 5}{3 \times 5} - \dfrac{14 \times 3}{5 \times 3}\\[1em] \Rightarrow \dfrac{85}{15} - \dfrac{42}{15}\\[1em] \Rightarrow \dfrac{85 - 42}{15}\\[1em] \Rightarrow \dfrac{43}{15}\\[1em] \Rightarrow 2\dfrac{13}{15}$

Hence, Aliyah devotes $2\dfrac{13}{15}$ hours for other subjects.

Question 4

Ram solved $\dfrac{2}{7}$ part of an exercise while Shwetha solved $\dfrac{4}{5}$ of it. Who solved lesser part? By how much?

Answer

Part of exercise solved by Ram = $\dfrac{2}{7}$ .

Part of exercise solved by Shwetha = $\dfrac{4}{5}$ .

Compare $\dfrac{2}{7}$ and $\dfrac{4}{5}$ by cross multiplication:

2 × 5 = 10 and 7 × 4 = 28.

Since 10 < 28, therefore, $\dfrac{2}{7} \lt \dfrac{4}{5}$ .

So, Ram solved the lesser part of the exercise.

Difference = $\dfrac{4}{5} - \dfrac{2}{7}$

LCM of 5 and 7 = 35.

$\Rightarrow \dfrac{4 \times 7}{5 \times 7} - \dfrac{2 \times 5}{7 \times 5}\\[1em] \Rightarrow \dfrac{28}{35} - \dfrac{10}{35}\\[1em] \Rightarrow \dfrac{28 - 10}{35}\\[1em] \Rightarrow \dfrac{18}{35}$

Hence, Ram solved the lesser part by $\dfrac{18}{35}$ of the exercise.

Question 5

Sonali had ₹ $35\dfrac{3}{5}$ . She got ₹ $16\dfrac{1}{15}$ from her mother and spent ₹ $28\dfrac{2}{3}$ on food. How much money is left with her?

Answer

Money Sonali had = ₹ $35\dfrac{3}{5}$ .

Money she got from her mother = ₹ $16\dfrac{1}{15}$ .

Money she spent on food = ₹ $28\dfrac{2}{3}$ .

Money left = $35\dfrac{3}{5} + 16\dfrac{1}{15} - 28\dfrac{2}{3}$

LCM of 5, 15 and 3 = 15.

$\Rightarrow \dfrac{178}{5} + \dfrac{241}{15} - \dfrac{86}{3}\\[1em] \Rightarrow \dfrac{178 \times 3}{5 \times 3} + \dfrac{241}{15} - \dfrac{86 \times 5}{3 \times 5}\\[1em] \Rightarrow \dfrac{534}{15} + \dfrac{241}{15} - \dfrac{430}{15}\\[1em] \Rightarrow \dfrac{534 + 241 - 430}{15}\\[1em] \Rightarrow \dfrac{345}{15}\\[1em] \Rightarrow 23$

Hence, ₹23 is left with Sonali.

Exercise 2.3

Question 1

Evaluate the following:

(i) $7 \times \dfrac{3}{5}$

(ii) $21 \times \dfrac{3}{14}$

(iii) $3\dfrac{2}{5} \times 8$

(iv) $5 \times 6\dfrac{3}{4}$

Answer

(i) $7 \times \dfrac{3}{5}$

$\Rightarrow \dfrac{7 \times 3}{5}\\[1em] \Rightarrow \dfrac{21}{5}\\[1em] \Rightarrow 4\dfrac{1}{5}$

Hence, $7 \times \dfrac{3}{5} = 4\dfrac{1}{5}$ .

(ii) $21 \times \dfrac{3}{14}$

$\Rightarrow \dfrac{21 \times 3}{14}\\[1em] \Rightarrow \dfrac{21}{14} \times 3\\[1em] \Rightarrow \dfrac{3}{2} \times 3\\[1em] \Rightarrow \dfrac{9}{2}\\[1em] \Rightarrow 4\dfrac{1}{2}$

Hence, $21 \times \dfrac{3}{14} = 4\dfrac{1}{2}$ .

(iii) $3\dfrac{2}{5} \times 8$

$\Rightarrow \dfrac{17}{5} \times 8\\[1em] \Rightarrow \dfrac{17 \times 8}{5}\\[1em] \Rightarrow \dfrac{136}{5}\\[1em] \Rightarrow 27\dfrac{1}{5}$

Hence, $3\dfrac{2}{5} \times 8 = 27\dfrac{1}{5}$ .

(iv) $5 \times 6\dfrac{3}{4}$

$\Rightarrow 5 \times \dfrac{27}{4}\\[1em] \Rightarrow \dfrac{5 \times 27}{4}\\[1em] \Rightarrow \dfrac{135}{4}\\[1em] \Rightarrow 33\dfrac{3}{4}$

Hence, $5 \times 6\dfrac{3}{4} = 33\dfrac{3}{4}$ .

Question 2

Find:

(i) $\dfrac{2}{3}$ of 18

(ii) $\dfrac{1}{2}$ of $4\dfrac{2}{9}$

(iii) $\dfrac{5}{8}$ of $9\dfrac{2}{3}$

Answer

(i) $\dfrac{2}{3}$ of 18

$\Rightarrow \dfrac{2}{3} \times 18\\[1em] \Rightarrow \dfrac{2 \times 18}{3}\\[1em] \Rightarrow 2 \times 6\\[1em] \Rightarrow 12$

Hence, $\dfrac{2}{3}$ of 18 = 12.

(ii) $\dfrac{1}{2}$ of $4\dfrac{2}{9}$

$\Rightarrow \dfrac{1}{2} \times 4\dfrac{2}{9}\\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{38}{9}\\[1em] \Rightarrow \dfrac{1 \times 38}{2 \times 9}\\[1em] \Rightarrow \dfrac{38}{18}\\[1em] \Rightarrow \dfrac{19}{9}\\[1em] \Rightarrow 2\dfrac{1}{9}$

Hence, $\dfrac{1}{2}$ of $4\dfrac{2}{9} = 2\dfrac{1}{9}$ .

(iii) $\dfrac{5}{8}$ of $9\dfrac{2}{3}$

$\Rightarrow \dfrac{5}{8} \times 9\dfrac{2}{3}\\[1em] \Rightarrow \dfrac{5}{8} \times \dfrac{29}{3}\\[1em] \Rightarrow \dfrac{5 \times 29}{8 \times 3}\\[1em] \Rightarrow \dfrac{145}{24}\\[1em] \Rightarrow 6\dfrac{1}{24}$

Hence, $\dfrac{5}{8}$ of $9\dfrac{2}{3} = 6\dfrac{1}{24}$ .

Question 3

Evaluate the following:

(i) $\dfrac{3}{7} \times \dfrac{5}{9}$

(ii) $\dfrac{2}{5} \times 5\dfrac{1}{4}$

(iii) $2\dfrac{1}{3} \times 5\dfrac{4}{21}$

(iv) $3\dfrac{1}{6} \times 7\dfrac{4}{23}$

Answer

(i) $\dfrac{3}{7} \times \dfrac{5}{9}$

$\Rightarrow \dfrac{3 \times 5}{7 \times 9}\\[1em] \Rightarrow \dfrac{3}{9} \times \dfrac{5}{7}\\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{5}{7}\\[1em] \Rightarrow \dfrac{5}{21}$

Hence, $\dfrac{3}{7} \times \dfrac{5}{9} = \dfrac{5}{21}$ .

(ii) $\dfrac{2}{5} \times 5\dfrac{1}{4}$

$\Rightarrow \dfrac{2}{5} \times \dfrac{21}{4}\\[1em] \Rightarrow \dfrac{2}{4} \times \dfrac{21}{5}\\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{21}{5}\\[1em] \Rightarrow \dfrac{21}{10}\\[1em] \Rightarrow 2\dfrac{1}{10}$

Hence, $\dfrac{2}{5} \times 5\dfrac{1}{4} = 2\dfrac{1}{10}$ .

(iii) $2\dfrac{1}{3} \times 5\dfrac{4}{21}$

$\Rightarrow \dfrac{7}{3} \times \dfrac{109}{21}\\[1em] \Rightarrow \dfrac{7}{21} \times \dfrac{109}{3}\\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{109}{3}\\[1em] \Rightarrow \dfrac{109}{9}\\[1em] \Rightarrow 12\dfrac{1}{9}$

Hence, $2\dfrac{1}{3} \times 5\dfrac{4}{21} = 12\dfrac{1}{9}$ .

(iv) $3\dfrac{1}{6} \times 7\dfrac{4}{23}$

$\Rightarrow \dfrac{19}{6} \times \dfrac{165}{23}\\[1em] \Rightarrow \dfrac{19}{23} \times \dfrac{165}{6}\\[1em] \Rightarrow \dfrac{19}{23} \times \dfrac{55}{2}\\[1em] \Rightarrow \dfrac{19 \times 55}{23 \times 2}\\[1em] \Rightarrow \dfrac{1045}{46}\\[1em] \Rightarrow 22\dfrac{33}{46}$

Hence, $3\dfrac{1}{6} \times 7\dfrac{4}{23} = 22\dfrac{33}{46}$ .

Question 4

Find the value of:

(i) $\dfrac{1}{3}$ of ₹42

(ii) $\dfrac{3}{7}$ of $4\dfrac{2}{3}$ kg

(iii) $4\dfrac{1}{2}$ times of $5\dfrac{1}{2}$ metres

Answer

(i) $\dfrac{1}{3}$ of ₹42

$\Rightarrow \dfrac{1}{3} \times 42\\[1em] \Rightarrow \dfrac{42}{3}\\[1em] \Rightarrow ₹14$

Hence, $\dfrac{1}{3}$ of ₹42 = ₹14.

(ii) $\dfrac{3}{7}$ of $4\dfrac{2}{3}$ kg

$\Rightarrow \dfrac{3}{7} \times 4\dfrac{2}{3}\\[1em] \Rightarrow \dfrac{3}{7} \times \dfrac{14}{3}\\[1em] \Rightarrow \dfrac{3}{3} \times \dfrac{14}{7}\\[1em] \Rightarrow 1 \times 2\\[1em] \Rightarrow 2 \text{ kg}$

Hence, $\dfrac{3}{7}$ of $4\dfrac{2}{3}$ kg = 2 kg.

(iii) $4\dfrac{1}{2}$ times of $5\dfrac{1}{2}$ metres

$\Rightarrow 4\dfrac{1}{2} \times 5\dfrac{1}{2}\\[1em] \Rightarrow \dfrac{9}{2} \times \dfrac{11}{2}\\[1em] \Rightarrow \dfrac{9 \times 11}{2 \times 2}\\[1em] \Rightarrow \dfrac{99}{4}\\[1em] \Rightarrow 24\dfrac{3}{4} \text{ metres}$

Hence, $4\dfrac{1}{2}$ times of $5\dfrac{1}{2}$ metres = $24\dfrac{3}{4}$ metres.

Question 5

Which is greater:

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

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

Answer

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

First, find $\dfrac{2}{7}$ of $\dfrac{3}{4}$ :

$\Rightarrow \dfrac{2}{7} \times \dfrac{3}{4}\\[1em] \Rightarrow \dfrac{2}{4} \times \dfrac{3}{7}\\[1em] \Rightarrow \dfrac{1}{2} \times \dfrac{3}{7}\\[1em] \Rightarrow \dfrac{3}{14}$

Next, find $\dfrac{3}{5}$ of $\dfrac{5}{8}$ :

$\Rightarrow \dfrac{3}{5} \times \dfrac{5}{8}\\[1em] \Rightarrow \dfrac{3}{8} \times \dfrac{5}{5}\\[1em] \Rightarrow \dfrac{3}{8} \times 1\\[1em] \Rightarrow \dfrac{3}{8}$

Now compare $\dfrac{3}{14}$ and $\dfrac{3}{8}$ .

These are fractions with same numerator 3.

In fractions with same numerator, the fraction with smaller denominator is greater.

Since 8 < 14, therefore, $\dfrac{3}{8} \gt \dfrac{3}{14}$ .

Hence, $\dfrac{3}{5}$ of $\dfrac{5}{8}$ is greater.

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

First, find $\dfrac{1}{2}$ of $\dfrac{6}{7}$ :

$\Rightarrow \dfrac{1}{2} \times \dfrac{6}{7}\\[1em] \Rightarrow \dfrac{1 \times 6}{2 \times 7}\\[1em] \Rightarrow \dfrac{6}{14}\\[1em] \Rightarrow \dfrac{3}{7}$

Next, find $\dfrac{2}{3}$ of $\dfrac{3}{7}$ :

$\Rightarrow \dfrac{2}{3} \times \dfrac{3}{7}\\[1em] \Rightarrow \dfrac{2}{7} \times \dfrac{3}{3}\\[1em] \Rightarrow \dfrac{2}{7} \times 1\\[1em] \Rightarrow \dfrac{2}{7}$

Now compare $\dfrac{3}{7}$ and $\dfrac{2}{7}$ .

These are like fractions (same denominator 7).

In like fractions, the fraction with greater numerator is greater.

Since 3 > 2, therefore, $\dfrac{3}{7} \gt \dfrac{2}{7}$ .

Hence, $\dfrac{1}{2}$ of $\dfrac{6}{7}$ is greater.

Question 6

If 1 metre cloth costs ₹ $131\dfrac{3}{4}$ , find the cost of $5\dfrac{1}{2}$ metres cloth.

Answer

Cost of 1 metre cloth = ₹ $131\dfrac{3}{4}$ .

Cost of $5\dfrac{1}{2}$ metres cloth = $131\dfrac{3}{4} \times 5\dfrac{1}{2}$

$\Rightarrow \dfrac{527}{4} \times \dfrac{11}{2}\\[1em] \Rightarrow \dfrac{527 \times 11}{4 \times 2}\\[1em] \Rightarrow \dfrac{5797}{8}\\[1em] \Rightarrow 724\dfrac{5}{8}$

Hence, the cost of $5\dfrac{1}{2}$ metres cloth = ₹ $724\dfrac{5}{8}$ .

Question 7

If the speed of a car is $105\dfrac{1}{5}$ km/h, find the distance covered by it in $3\dfrac{3}{5}$ hours.

Answer

Speed of the car = $105\dfrac{1}{5}$ km/h.

Time taken = $3\dfrac{3}{5}$ hours.

Distance covered = Speed × Time = $105\dfrac{1}{5} \times 3\dfrac{3}{5}$

$\Rightarrow \dfrac{526}{5} \times \dfrac{18}{5}\\[1em] \Rightarrow \dfrac{526 \times 18}{5 \times 5}\\[1em] \Rightarrow \dfrac{9468}{25}\\[1em] \Rightarrow 378\dfrac{18}{25}$

Hence, the distance covered by the car = $378\dfrac{18}{25}$ km.

Question 8

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

Answer

Distance covered in 1 litre of petrol = 16 km.

Distance covered in $2\dfrac{3}{4}$ litres of petrol = $16 \times 2\dfrac{3}{4}$

$\Rightarrow 16 \times \dfrac{11}{4}\\[1em] \Rightarrow \dfrac{16 \times 11}{4}\\[1em] \Rightarrow 4 \times 11\\[1em] \Rightarrow 44 \text{ km}$

Hence, the car will cover 44 km in $2\dfrac{3}{4}$ litres of petrol.

Question 9

Sushant reads $\dfrac{1}{3}$ part of a book in 1 hour. How much part of the book will he read in $2\dfrac{1}{5}$ hours?

Answer

Part of the book read in 1 hour = $\dfrac{1}{3}$ .

Part of the book read in $2\dfrac{1}{5}$ hours = $\dfrac{1}{3} \times 2\dfrac{1}{5}$

$\Rightarrow \dfrac{1}{3} \times \dfrac{11}{5}\\[1em] \Rightarrow \dfrac{1 \times 11}{3 \times 5}\\[1em] \Rightarrow \dfrac{11}{15}$

Hence, Sushant will read $\dfrac{11}{15}$ part of the book in $2\dfrac{1}{5}$ hours.

Question 10

An ornament is made of gold and copper and weighs 52 grams. If $\dfrac{2}{13}$ of its part is copper, find the weight of pure gold in it.

Answer

Total weight of the ornament = 52 grams.

Fraction of copper = $\dfrac{2}{13}$ .

Weight of copper = $\dfrac{2}{13} \times 52$

$\Rightarrow \dfrac{2 \times 52}{13}\\[1em] \Rightarrow 2 \times 4\\[1em] \Rightarrow 8 \text{ grams}$

Weight of pure gold = Total weight − Weight of copper

⇒ 52 - 8

⇒ 44 grams

Hence, the weight of pure gold in the ornament = 44 grams.

Question 11

In a class of 40 students, $\dfrac{1}{5}$ of the total number of students like to study English and $\dfrac{2}{5}$ of the total number of students like to study Mathematics and the remaining like to study Science.

(i) How many students like to study English?

(ii) How many students like to study Mathematics?

(iii) What fraction of the total number of students like to study Science?

Answer

Total number of students = 40.

(i) Number of students who like to study English = $\dfrac{1}{5} \times 40$

$\Rightarrow \dfrac{40}{5}$

⇒ 8

Hence, 8 students like to study English.

(ii) Number of students who like to study Mathematics = $\dfrac{2}{5} \times 40$

$\Rightarrow \dfrac{2 \times 40}{5}\\[1em] \Rightarrow 2 \times 8\\[1em] \Rightarrow 16$

Hence, 16 students like to study Mathematics.

(iii) Fraction of students who like English and Mathematics = $\dfrac{1}{5} + \dfrac{2}{5} = \dfrac{1 + 2}{5} = \dfrac{3}{5}$ .

Fraction of students who like to study Science = $1 - \dfrac{3}{5}$

$\Rightarrow \dfrac{5}{5} - \dfrac{3}{5}\\[1em] \Rightarrow \dfrac{5 - 3}{5}\\[1em] \Rightarrow \dfrac{2}{5}$

Hence, $\dfrac{2}{5}$ of the total number of students like to study Science.

Question 12

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

(i) perimeter

(ii) area

Answer

Length of the sheet = $12\dfrac{1}{2}$ cm = $\dfrac{25}{2}$ cm.

Width of the sheet = $10\dfrac{2}{3}$ cm = $\dfrac{32}{3}$ cm.

(i) Perimeter = 2 × (Length + Width)

$\Rightarrow 2 \times \left(\dfrac{25}{2} + \dfrac{32}{3}\right)\\[1em] \Rightarrow 2 \times \left(\dfrac{25 \times 3}{2 \times 3} + \dfrac{32 \times 2}{3 \times 2}\right)\\[1em] \Rightarrow 2 \times \left(\dfrac{75}{6} + \dfrac{64}{6}\right)\\[1em] \Rightarrow 2 \times \dfrac{139}{6}\\[1em] \Rightarrow \dfrac{139}{3}\\[1em] \Rightarrow 46\dfrac{1}{3} \text{ cm}$

Hence, the perimeter of the sheet = $46\dfrac{1}{3}$ cm.

(ii) Area = Length × Width

$\Rightarrow \dfrac{25}{2} \times \dfrac{32}{3}\\[1em] \Rightarrow \dfrac{25 \times 32}{2 \times 3}\\[1em] \Rightarrow \dfrac{25 \times 16}{3}\\[1em] \Rightarrow \dfrac{400}{3}\\[1em] \Rightarrow 133\dfrac{1}{3} \text{ sq. cm}$

Hence, the area of the sheet = $133\dfrac{1}{3}$ sq. cm.

Question 13

In a school, $\dfrac{25}{54}$ of the students are girls and the rest are boys. If the number of boys is 2030, find the number of girls.

Answer

Fraction of girls = $\dfrac{25}{54}$ .

Fraction of boys = $1 - \dfrac{25}{54}$

$\Rightarrow \dfrac{54}{54} - \dfrac{25}{54}\\[1em] \Rightarrow \dfrac{54 - 25}{54}\\[1em] \Rightarrow \dfrac{29}{54}$

Let the total number of students be x.

Given, number of boys = 2030.

$\Rightarrow \dfrac{29}{54} \times x = 2030\\[1em] \Rightarrow x = 2030 \times \dfrac{54}{29}\\[1em] \Rightarrow x = 70 \times 54\\[1em] \Rightarrow x = 3780$

Number of girls = $\dfrac{25}{54} \times 3780$

⇒ 25 × 70

⇒ 1750

Hence, the number of girls in the school = 1750.

Question 14

In an orchard, $\dfrac{1}{5}$ are orange trees, $\dfrac{3}{13}$ are mango trees and the rest are banana trees. If the banana trees are 148 in number, find the total number of trees in the orchard.

Answer

Fraction of orange trees = $\dfrac{1}{5}$ .

Fraction of mango trees = $\dfrac{3}{13}$ .

Fraction of banana trees = $1 - \left(\dfrac{1}{5} + \dfrac{3}{13}\right)$

LCM of 5 and 13 = 65.

$\Rightarrow 1 - \left(\dfrac{1 \times 13}{5 \times 13} + \dfrac{3 \times 5}{13 \times 5}\right)\\[1em] \Rightarrow 1 - \left(\dfrac{13}{65} + \dfrac{15}{65}\right)\\[1em] \Rightarrow 1 - \dfrac{28}{65}\\[1em] \Rightarrow \dfrac{65}{65} - \dfrac{28}{65}\\[1em] \Rightarrow \dfrac{37}{65}$

Let the total number of trees be x.

Given, number of banana trees = 148.

$\Rightarrow \dfrac{37}{65} \times x = 148\\[1em] \Rightarrow x = 148 \times \dfrac{65}{37}\\[1em] \Rightarrow x = 4 \times 65\\[1em] \Rightarrow x = 260$

Hence, the total number of trees in the orchard = 260.

Exercise 2.4

Question 1

Find the reciprocal of each of the following:

(i) $\dfrac{3}{7}$

(ii) $\dfrac{13}{9}$

(iii) 8

Answer

The reciprocal of a fraction $\dfrac{a}{b}$ is $\dfrac{b}{a}$ .

(i) $\dfrac{3}{7}$

The reciprocal of $\dfrac{3}{7}$ is $\dfrac{7}{3}$ .

Hence, the reciprocal of $\dfrac{3}{7}$ is $\dfrac{7}{3}$ .

(ii) $\dfrac{13}{9}$

The reciprocal of $\dfrac{13}{9}$ is $\dfrac{9}{13}$ .

Hence, the reciprocal of $\dfrac{13}{9}$ is $\dfrac{9}{13}$ .

(iii) 8

We can write 8 as $\dfrac{8}{1}$ .

The reciprocal of $\dfrac{8}{1}$ is $\dfrac{1}{8}$ .

Hence, the reciprocal of 8 is $\dfrac{1}{8}$ .

Question 2

Evaluate the following:

(i) $14 \div \dfrac{5}{6}$

(ii) $5 \div 3\dfrac{4}{7}$

(iii) $3\dfrac{1}{5} \div 1\dfrac{2}{3}$

(iv) $2\dfrac{5}{8} \div 1\dfrac{1}{6}$

Answer

(i) $14 \div \dfrac{5}{6}$

$\Rightarrow \dfrac{14}{1} \times \dfrac{6}{5}\\[1em] \Rightarrow \dfrac{14 \times 6}{1 \times 5}\\[1em] \Rightarrow \dfrac{84}{5}\\[1em] \Rightarrow 16\dfrac{4}{5}$

Hence, $14 \div \dfrac{5}{6} = 16\dfrac{4}{5}$ .

(ii) $5 \div 3\dfrac{4}{7}$

$\Rightarrow 5 \div \dfrac{25}{7}\\[1em] \Rightarrow \dfrac{5}{1} \times \dfrac{7}{25}\\[1em] \Rightarrow \dfrac{5}{25} \times \dfrac{7}{1}\\[1em] \Rightarrow \dfrac{1}{5} \times 7\\[1em] \Rightarrow \dfrac{7}{5}\\[1em] \Rightarrow 1\dfrac{2}{5}$

Hence, $5 \div 3\dfrac{4}{7} = 1\dfrac{2}{5}$ .

(iii) $3\dfrac{1}{5} \div 1\dfrac{2}{3}$

$\Rightarrow \dfrac{16}{5} \div \dfrac{5}{3}\\[1em] \Rightarrow \dfrac{16}{5} \times \dfrac{3}{5}\\[1em] \Rightarrow \dfrac{16 \times 3}{5 \times 5}\\[1em] \Rightarrow \dfrac{48}{25}\\[1em] \Rightarrow 1\dfrac{23}{25}$

Hence, $3\dfrac{1}{5} \div 1\dfrac{2}{3} = 1\dfrac{23}{25}$ .

(iv) $2\dfrac{5}{8} \div 1\dfrac{1}{6}$

$\Rightarrow \dfrac{21}{8} \div \dfrac{7}{6}\\[1em] \Rightarrow \dfrac{21}{8} \times \dfrac{6}{7}\\[1em] \Rightarrow \dfrac{21}{7} \times \dfrac{6}{8}\\[1em] \Rightarrow 3 \times \dfrac{3}{4}\\[1em] \Rightarrow \dfrac{9}{4}\\[1em] \Rightarrow 2\dfrac{1}{4}$

Hence, $2\dfrac{5}{8} \div 1\dfrac{1}{6} = 2\dfrac{1}{4}$ .

Question 3

How many pieces each $5\dfrac{1}{6}$ metres long can be cut from a cloth $77\dfrac{1}{2}$ metres long?

Answer

Total length of cloth = $77\dfrac{1}{2}$ metres.

Length of each piece = $5\dfrac{1}{6}$ metres.

Number of pieces = $77\dfrac{1}{2} \div 5\dfrac{1}{6}$

$\Rightarrow \dfrac{155}{2} \div \dfrac{31}{6}\\[1em] \Rightarrow \dfrac{155}{2} \times \dfrac{6}{31}\\[1em] \Rightarrow \dfrac{155}{31} \times \dfrac{6}{2}\\[1em] \Rightarrow 5 \times 3\\[1em] \Rightarrow 15$

Hence, 15 pieces can be cut from the cloth.

Question 4

By what number should $4\dfrac{7}{8}$ be multiplied to get $87\dfrac{3}{4}$ ?

Answer

Let the required number be x.

$\Rightarrow 4\dfrac{7}{8} \times x = 87\dfrac{3}{4}\\[1em] \Rightarrow x = 87\dfrac{3}{4} \div 4\dfrac{7}{8}\\[1em] \Rightarrow x = \dfrac{351}{4} \div \dfrac{39}{8}\\[1em] \Rightarrow x = \dfrac{351}{4} \times \dfrac{8}{39}\\[1em] \Rightarrow x = \dfrac{351}{39} \times \dfrac{8}{4}\\[1em] \Rightarrow x = 9 \times 2\\[1em] \Rightarrow x = 18$

Hence, the required number is 18.

Question 5

In a hostel's mess, each student gets $\dfrac{1}{3}$ litre of milk every day. If the total consumption of the milk is $57\dfrac{2}{3}$ litres per day, how many students are there in the hostel?

Answer

Milk consumed by each student = $\dfrac{1}{3}$ litre.

Total consumption of milk = $57\dfrac{2}{3}$ litres.

Number of students = $57\dfrac{2}{3} \div \dfrac{1}{3}$

$\Rightarrow \dfrac{173}{3} \div \dfrac{1}{3}\\[1em] \Rightarrow \dfrac{173}{3} \times \dfrac{3}{1}\\[1em] \Rightarrow \dfrac{173 \times 3}{3}\\[1em] \Rightarrow 173$

Hence, there are 173 students in the hostel.

Question 6

The cost of $5\dfrac{1}{4}$ kg oranges is ₹336. What is the rate of oranges per kg?

Answer

Cost of $5\dfrac{1}{4}$ kg oranges = ₹336.

Rate of oranges per kg = $336 \div 5\dfrac{1}{4}$

$\Rightarrow 336 \div \dfrac{21}{4}\\[1em] \Rightarrow \dfrac{336}{1} \times \dfrac{4}{21}\\[1em] \Rightarrow \dfrac{336}{21} \times 4\\[1em] \Rightarrow 16 \times 4\\[1em] \Rightarrow ₹64$

Hence, the rate of oranges is ₹64 per kg.

Question 7

The length of a rectangular plot of area $68\dfrac{3}{4}$ sq. m is $12\dfrac{1}{2}$ m, find its width.

Answer

Area of the rectangular plot = $68\dfrac{3}{4}$ sq. m.

Length of the plot = $12\dfrac{1}{2}$ m.

Width = Area ÷ Length = $68\dfrac{3}{4} \div 12\dfrac{1}{2}$

$\Rightarrow \dfrac{275}{4} \div \dfrac{25}{2}\\[1em] \Rightarrow \dfrac{275}{4} \times \dfrac{2}{25}\\[1em] \Rightarrow \dfrac{275}{25} \times \dfrac{2}{4}\\[1em] \Rightarrow 11 \times \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{11}{2}\\[1em] \Rightarrow 5\dfrac{1}{2} \text{ m}$

Hence, the width of the plot = $5\dfrac{1}{2}$ m.

Question 8

If the cost of $5\dfrac{1}{2}$ kg of sugar is ₹ $206\dfrac{1}{4}$ , then find the cost of $8\dfrac{1}{4}$ kg of sugar.

Answer

Cost of $5\dfrac{1}{2}$ kg of sugar = ₹ $206\dfrac{1}{4}$ .

Cost of 1 kg of sugar = $206\dfrac{1}{4} \div 5\dfrac{1}{2}$

$\Rightarrow \dfrac{825}{4} \div \dfrac{11}{2}\\[1em] \Rightarrow \dfrac{825}{4} \times \dfrac{2}{11}\\[1em] \Rightarrow \dfrac{825}{11} \times \dfrac{2}{4}\\[1em] \Rightarrow 75 \times \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{75}{2}$

Cost of $8\dfrac{1}{4}$ kg of sugar = $\dfrac{75}{2} \times 8\dfrac{1}{4}$

$\Rightarrow \dfrac{75}{2} \times \dfrac{33}{4}\\[1em] \Rightarrow \dfrac{75 \times 33}{2 \times 4}\\[1em] \Rightarrow \dfrac{2475}{8}\\[1em] \Rightarrow 309\dfrac{3}{8}$

Hence, the cost of $8\dfrac{1}{4}$ kg of sugar = ₹ $309\dfrac{3}{8}$ .

Question 9

Vamika completed $\dfrac{2}{3}$ part of her homework in 2 hours. How much part of her homework had she completed in $1\dfrac{1}{4}$ hours?

Answer

Part of homework completed in 2 hours = $\dfrac{2}{3}$ .

Part of homework completed in 1 hour = $\dfrac{2}{3} \div 2$

$\Rightarrow \dfrac{2}{3} \times \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{2}{2} \times \dfrac{1}{3}\\[1em] \Rightarrow \dfrac{1}{3}$

Part of homework completed in $1\dfrac{1}{4}$ hours = $\dfrac{1}{3} \times 1\dfrac{1}{4}$

$\Rightarrow \dfrac{1}{3} \times \dfrac{5}{4}\\[1em] \Rightarrow \dfrac{1 \times 5}{3 \times 4}\\[1em] \Rightarrow \dfrac{5}{12}$

Hence, Vamika completed $\dfrac{5}{12}$ part of her homework in $1\dfrac{1}{4}$ hours.

Exercise 2.5

Question 1

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

(i) 2.56

(ii) 21.37

(iii) 10.25

(iv) 63.352

Answer

(i) 2.56

The digit 2 is in the ones (units) place.

Hence, the place value of 2 in 2.56 is 2.

(ii) 21.37

The digit 2 is in the tens place.

Hence, the place value of 2 in 21.37 is 20.

(iii) 10.25

The digit 2 is in the tenths place.

Hence, the place value of 2 in 10.25 is $\dfrac{2}{10}$ .

(iv) 63.352

The digit 2 is in the thousandths place.

Hence, the place value of 2 in 63.352 is $\dfrac{2}{1000}$ .

Question 2

Convert the following decimal numbers to fractions (in simplest form):

(i) 0.8

(ii) 0.225

(iii) 0.0092

(iv) 3.025

Answer

(i) 0.8

$\Rightarrow \dfrac{8}{10}\\[1em] \Rightarrow \dfrac{8 \div 2}{10 \div 2}\\[1em] \Rightarrow \dfrac{4}{5}$

Hence, 0.8 = $\dfrac{4}{5}$ .

(ii) 0.225

$\Rightarrow \dfrac{225}{1000}\\[1em] \Rightarrow \dfrac{225 \div 25}{1000 \div 25}\\[1em] \Rightarrow \dfrac{9}{40}$

Hence, 0.225 = $\dfrac{9}{40}$ .

(iii) 0.0092

$\Rightarrow \dfrac{92}{10000}\\[1em] \Rightarrow \dfrac{92 \div 4}{10000 \div 4}\\[1em] \Rightarrow \dfrac{23}{2500}$

Hence, 0.0092 = $\dfrac{23}{2500}$ .

(iv) 3.025

$\Rightarrow \dfrac{3025}{1000}\\[1em] \Rightarrow \dfrac{3025 \div 25}{1000 \div 25}\\[1em] \Rightarrow \dfrac{121}{40}$

Hence, 3.025 = $\dfrac{121}{40}$ .

Question 3

Convert the following decimals to mixed fractions:

(i) 5.05

(ii) 63.125

(iii) 17.075

(iv) 317.0006

Answer

(i) 5.05

$\Rightarrow 5 + \dfrac{5}{100}\\[1em] \Rightarrow 5 + \dfrac{1}{20}\\[1em] \Rightarrow 5\dfrac{1}{20}$

Hence, 5.05 = $5\dfrac{1}{20}$ .

(ii) 63.125

$\Rightarrow 63 + \dfrac{125}{1000}\\[1em] \Rightarrow 63 + \dfrac{1}{8}\\[1em] \Rightarrow 63\dfrac{1}{8}$

Hence, 63.125 = $63\dfrac{1}{8}$ .

(iii) 17.075

$\Rightarrow 17 + \dfrac{75}{1000}\\[1em] \Rightarrow 17 + \dfrac{3}{40}\\[1em] \Rightarrow 17\dfrac{3}{40}$

Hence, 17.075 = $17\dfrac{3}{40}$ .

(iv) 317.0006

$\Rightarrow 317 + \dfrac{6}{10000}\\[1em] \Rightarrow 317 + \dfrac{3}{5000}\\[1em] \Rightarrow 317\dfrac{3}{5000}$

Hence, 317.0006 = $317\dfrac{3}{5000}$ .

Question 4

Convert the following fractions into decimal numbers:

(i) $\dfrac{3}{5}$

(ii) $\dfrac{7}{8}$

(iii) $3\dfrac{5}{16}$

(iv) $137\dfrac{13}{625}$

Answer

(i) $\dfrac{3}{5}$

$\Rightarrow \dfrac{3 \times 2}{5 \times 2}\\[1em] \Rightarrow \dfrac{6}{10}\\[1em] \Rightarrow 0.6$

Hence, $\dfrac{3}{5}$ = 0.6.

(ii) $\dfrac{7}{8}$

$\Rightarrow \dfrac{7 \times 125}{8 \times 125}\\[1em] \Rightarrow \dfrac{875}{1000}\\[1em] \Rightarrow 0.875$

Hence, $\dfrac{7}{8}$ = 0.875.

(iii) $3\dfrac{5}{16}$

$\Rightarrow 3 + \dfrac{5}{16}\\[1em] \Rightarrow 3 + \dfrac{5 \times 625}{16 \times 625}\\[1em] \Rightarrow 3 + \dfrac{3125}{10000}\\[1em] \Rightarrow 3 + 0.3125\\[1em] \Rightarrow 3.3125$

Hence, $3\dfrac{5}{16}$ = 3.3125.

(iv) $137\dfrac{13}{625}$

$\Rightarrow 137 + \dfrac{13}{625}\\[1em] \Rightarrow 137 + \dfrac{13 \times 16}{625 \times 16}\\[1em] \Rightarrow 137 + \dfrac{208}{10000}\\[1em] \Rightarrow 137 + 0.0208\\[1em] \Rightarrow 137.0208$

Hence, $137\dfrac{13}{625}$ = 137.0208.

Question 5

Which is greater?

(i) 0.5 or 0.05

(ii) 7 or 0.7

(iii) 2.03 or 2.30

(iv) 0.8 or 0.88

Answer

(i) 0.5 or 0.05

Expressing as like decimals: 0.50 and 0.05.

Comparing, 50 > 5, so 0.50 > 0.05.

Hence, 0.5 is greater.

(ii) 7 or 0.7

Expressing as like decimals: 7.0 and 0.7.

Comparing the whole number parts, 7 > 0, so 7 > 0.7.

Hence, 7 is greater.

(iii) 2.03 or 2.30

The whole number parts are equal (2 = 2).

Comparing the tenths digits, 3 > 0, so 2.30 > 2.03.

Hence, 2.30 is greater.

(iv) 0.8 or 0.88

Expressing as like decimals: 0.80 and 0.88.

Comparing, 88 > 80, so 0.88 > 0.80.

Hence, 0.88 is greater.

Question 6

Arrange the following decimal numbers in ascending order:

(i) 38.02, 38.021, 3.802, 83.02, 38.002

(ii) 46.542, 46.452, 46.254, 46.05, 64.542, 46.0542

Answer

(i) 38.02, 38.021, 3.802, 83.02, 38.002

Expressing as like decimals: 38.020, 38.021, 3.802, 83.020, 38.002.

Comparing the numbers, we get :

3.802 < 38.002 < 38.020 < 38.021 < 83.020.

Hence, the numbers in ascending order are 3.802, 38.002, 38.02, 38.021, 83.02 .

(ii) 46.542, 46.452, 46.254, 46.05, 64.542, 46.0542

Expressing as like decimals: 46.5420, 46.4520, 46.2540, 46.0500, 64.5420, 46.0542.

Comparing the numbers, we get :

46.0500 < 46.0542 < 46.2540 < 46.4520 < 46.5420 < 64.5420.

Hence, the numbers in ascending order are 46.05, 46.0542, 46.254, 46.452, 46.542, 64.542.

Question 7

Arrange the following decimal numbers in descending order:

(i) 5.6, 0.93, 1.87, 1.9, 1.78, 0.39

(ii) 71.201, 20.1, 2.01, 3.1, 2.14, 0.652

Answer

(i) 5.6, 0.93, 1.87, 1.9, 1.78, 0.39

Expressing as like decimals: 5.60, 0.93, 1.87, 1.90, 1.78, 0.39.

Comparing the numbers, we get :

5.60 > 1.90 > 1.87 > 1.78 > 0.93 > 0.39.

Hence, the numbers in descending order are 5.6, 1.9, 1.87, 1.78, 0.93, 0.39.

(ii) 71.201, 20.1, 2.01, 3.1, 2.14, 0.652

Expressing as like decimals: 71.201, 20.100, 2.010, 3.100, 2.140, 0.652.

Comparing the numbers, we get :

71.201 > 20.100 > 3.100 > 2.140 > 2.010 > 0.652.

Hence, the numbers in descending order are 71.201, 20.1, 3.1, 2.14, 2.01, 0.652.

Question 8

Express as rupees using decimals:

(i) 7 paise

(ii) 77 rupees 77 paise

(iii) 235 paise

Answer

We know, 100 paise = ₹1, so 1 paise = ₹ $\dfrac{1}{100}$ = ₹0.01.

(i) 7 paise

$\Rightarrow ₹\dfrac{7}{100}$

⇒ ₹0.07

Hence, 7 paise = ₹0.07.

(ii) 77 rupees 77 paise

$\Rightarrow ₹77 + ₹\dfrac{77}{100}\\[1em] \Rightarrow ₹77 + ₹0.77\\[1em] \Rightarrow ₹77.77$

Hence, 77 rupees 77 paise = ₹77.77.

(iii) 235 paise

$\Rightarrow ₹\dfrac{235}{100}$

⇒ ₹2.35

Hence, 235 paise = ₹2.35.

Question 9

Express 5 cm in metre and kilometre.

Answer

We know, 100 cm = 1 m and 1000 m = 1 km.

In metre:

$\Rightarrow 5 \text{ cm} = \dfrac{5}{100} \text{ m}$

$\Rightarrow$ 0.05 m

In kilometre:

$\Rightarrow 0.05 \text{ m} = \dfrac{0.05}{1000} \text{ km}$

$\Rightarrow$ 0.00005 km

Hence, 5 cm = 0.05 m = 0.00005 km.

Question 10

Express in kg using decimals:

(i) 200 g

(ii) 3470 g

(iii) 4 kg 8 g

Answer

We know, 1000 g = 1 kg, so 1 g = $\dfrac{1}{1000}$ kg = 0.001 kg.

(i) 200 g

$\Rightarrow \dfrac{200}{1000} \text{ kg}$

⇒ 0.2 kg

Hence, 200 g = 0.2 kg.

(ii) 3470 g

$\Rightarrow \dfrac{3470}{1000} \text{ kg}$

⇒ 3.470 kg

Hence, 3470 g = 3.470 kg.

(iii) 4 kg 8 g

$\Rightarrow 4 \text{ kg} + \dfrac{8}{1000} \text{ kg}\\[1em] \Rightarrow 4 \text{ kg} + 0.008 \text{ kg}\\[1em] \Rightarrow 4.008 \text{ kg}$

Hence, 4 kg 8 g = 4.008 kg.

Question 11

Add:

(i) 5.765, 9.2, 3.08

(ii) 15.49, 8.3572, 0.903, 7.8

Answer

(i) 5.765, 9.2, 3.08

Writing the numbers as like decimals and adding:

⇒ 5.765 + 9.200 + 3.080

⇒ 18.045

Hence, the sum = 18.045.

(ii) 15.49, 8.3572, 0.903, 7.8

Writing the numbers as like decimals and adding:

⇒ 15.4900 + 8.3572 + 0.9030 + 7.8000

⇒ 32.5502

Hence, the sum = 32.5502.

Question 12

Calculate the following:

(i) 72.53 - 46.782

(ii) 18.376 - 5.43 - 8.8976

(iii) 28.5 - 9.708 - 6.234

(iv) 8.2 - 4.56 - 0.7912 + 2.67

Answer

(i) 72.53 - 46.782

Writing as like decimals and subtracting:

⇒ 72.530 - 46.782

⇒ 25.748

Hence, 72.53 - 46.782 = 25.748.

(ii) 18.376 - 5.43 - 8.8976

First add the numbers to be subtracted:

⇒ 5.4300 + 8.8976

⇒ 14.3276

Now subtract:

⇒ 18.3760 - 14.3276

⇒ 4.0484

Hence, 18.376 - 5.43 - 8.8976 = 4.0484.

(iii) 28.5 - 9.708 - 6.234

First add the numbers to be subtracted:

⇒ 9.708 + 6.234

⇒ 15.942

Now subtract:

⇒ 28.500 - 15.942

⇒ 12.558

Hence, 28.5 - 9.708 - 6.234 = 12.558.

(iv) 8.2 - 4.56 - 0.7912 + 2.67

Group the positive and negative numbers separately.

Sum of positive numbers = 8.2 + 2.67 = 10.87.

Sum of numbers to be subtracted = 4.56 + 0.7912 = 5.3512.

⇒ 10.8700 - 5.3512

⇒ 5.5188

Hence, 8.2 - 4.56 - 0.7912 + 2.67 = 5.5188.

Question 13

(i) What number added to 3.56 gives 13.016?

(ii) What number should be subtracted from 30 to get 23.709?

(iii) What is the excess of 20.4 over 9.7403?

Answer

(i) What number added to 3.56 gives 13.016?

Required number = 13.016 - 3.56

⇒ 13.016 - 3.560

⇒ 9.456

Hence, the required number is 9.456.

(ii) What number should be subtracted from 30 to get 23.709?

Required number = 30 - 23.709

⇒ 30.000 - 23.709

⇒ 6.291

Hence, the required number is 6.291.

(iii) What is the excess of 20.4 over 9.7403?

Excess = 20.4 - 9.7403

⇒ 20.4000 - 9.7403

⇒ 10.6597

Hence, the excess of 20.4 over 9.7403 is 10.6597.

Exercise 2.6

Question 1

Calculate the following:

(i) 2.7 × 4

(ii) 2.71 × 5

(iii) 2.5 × 0.3

(iv) 2.3 × 4.35

(v) 238.06 × 7.5

(vi) 0.79 × 32.4

(vii) 1.07 × 0.02

(viii) 10.05 × 1.05

Answer

(i) 2.7 × 4

Multiplying without the decimal point, 27 × 4 = 108.

The number of decimal places in 2.7 is 1, so the product has 1 decimal place.

⇒ 2.7 × 4

⇒ 10.8

Hence, 2.7 × 4 = 10.8.

(ii) 2.71 × 5

Multiplying without the decimal point, 271 × 5 = 1355.

The number of decimal places in 2.71 is 2, so the product has 2 decimal places.

⇒ 2.71 × 5

⇒ 13.55

Hence, 2.71 × 5 = 13.55.

(iii) 2.5 × 0.3

Multiplying without the decimal points, 25 × 3 = 75.

The number of decimal places in 2.5 and 0.3 together is 1 + 1 = 2, so the product has 2 decimal places.

⇒ 2.5 × 0.3

⇒ 0.75

Hence, 2.5 × 0.3 = 0.75.

(iv) 2.3 × 4.35

Multiplying without the decimal points, 23 × 435 = 10005.

The number of decimal places in 2.3 and 4.35 together is 1 + 2 = 3, so the product has 3 decimal places.

⇒ 2.3 × 4.35

⇒ 10.005

Hence, 2.3 × 4.35 = 10.005.

(v) 238.06 × 7.5

Multiplying without the decimal points, 23806 × 75 = 1785450.

The number of decimal places in 238.06 and 7.5 together is 2 + 1 = 3, so the product has 3 decimal places.

⇒ 238.06 × 7.5

⇒ 1785.45

Hence, 238.06 × 7.5 = 1785.45.

(vi) 0.79 × 32.4

Multiplying without the decimal points, 79 × 324 = 25596.

The number of decimal places in 0.79 and 32.4 together is 2 + 1 = 3, so the product has 3 decimal places.

⇒ 0.79 × 32.4

⇒ 25.596

Hence, 0.79 × 32.4 = 25.596.

(vii) 1.07 × 0.02

Multiplying without the decimal points, 107 × 2 = 214.

The number of decimal places in 1.07 and 0.02 together is 2 + 2 = 4, so the product has 4 decimal places.

⇒ 1.07 × 0.02

⇒ 0.0214

Hence, 1.07 × 0.02 = 0.0214.

(viii) 10.05 × 1.05

Multiplying without the decimal points, 1005 × 105 = 105525.

The number of decimal places in 10.05 and 1.05 together is 2 + 2 = 4, so the product has 4 decimal places.

⇒ 10.05 × 1.05

⇒ 10.5525

Hence, 10.05 × 1.05 = 10.5525.

Question 2

Calculate the following:

(i) 10.8 ÷ 4

(ii) 126.35 ÷ 7

(iii) 22.5 ÷ 1.5

(iv) 4.28 ÷ 0.02

(v) 3.645 ÷ 1.35

(vi) 0.728 ÷ 0.04

(vii) 13.06 ÷ 0.08

(viii) 58.635 ÷ 4.5

Answer

(i) 10.8 ÷ 4

$\Rightarrow \dfrac{10.8}{4}$

⇒ 2.7

Hence, 10.8 ÷ 4 = 2.7.

(ii) 126.35 ÷ 7

$\Rightarrow \dfrac{126.35}{7}$

⇒ 18.05

Hence, 126.35 ÷ 7 = 18.05.

(iii) 22.5 ÷ 1.5

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$\Rightarrow \dfrac{22.5}{1.5}\\[1em] \Rightarrow \dfrac{22.5 \times 10}{1.5 \times 10}\\[1em] \Rightarrow \dfrac{225}{15}\\[1em] \Rightarrow 15$

Hence, 22.5 ÷ 1.5 = 15.

(iv) 4.28 ÷ 0.02

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{4.28}{0.02}\\[1em] \Rightarrow \dfrac{4.28 \times 100}{0.02 \times 100}\\[1em] \Rightarrow \dfrac{428}{2}\\[1em] \Rightarrow 214$

Hence, 4.28 ÷ 0.02 = 214.

(v) 3.645 ÷ 1.35

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{3.645}{1.35}\\[1em] \Rightarrow \dfrac{3.645 \times 100}{1.35 \times 100}\\[1em] \Rightarrow \dfrac{364.5}{135}\\[1em] \Rightarrow 2.7$

Hence, 3.645 ÷ 1.35 = 2.7.

(vi) 0.728 ÷ 0.04

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{0.728}{0.04}\\[1em] \Rightarrow \dfrac{0.728 \times 100}{0.04 \times 100}\\[1em] \Rightarrow \dfrac{72.8}{4}\\[1em] \Rightarrow 18.2$

Hence, 0.728 ÷ 0.04 = 18.2.

(vii) 13.06 ÷ 0.08

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{13.06}{0.08}\\[1em] \Rightarrow \dfrac{13.06 \times 100}{0.08 \times 100}\\[1em] \Rightarrow \dfrac{1306}{8}\\[1em] \Rightarrow 163.25$

Hence, 13.06 ÷ 0.08 = 163.25.

(viii) 58.635 ÷ 4.5

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$\Rightarrow \dfrac{58.635}{4.5}\\[1em] \Rightarrow \dfrac{58.635 \times 10}{4.5 \times 10}\\[1em] \Rightarrow \dfrac{586.35}{45}\\[1em] \Rightarrow 13.03$

Hence, 58.635 ÷ 4.5 = 13.03.

Question 3

Multiply each of the following numbers by 10, 100 and 1000 (verbally):

(i) 5.9

(ii) 3.76

(iii) 0.549

Answer

While multiplying a decimal by 10, 100 or 1000, the decimal point is shifted to the right by as many places as there are zeros.

(i) 5.9

$\Rightarrow 5.9 \times 10 = 59\\[1em] \Rightarrow 5.9 \times 100 = 590\\[1em] \Rightarrow 5.9 \times 1000 = 5900$

Hence, 5.9 × 10 = 59, 5.9 × 100 = 590 and 5.9 × 1000 = 5900.

(ii) 3.76

$\Rightarrow 3.76 \times 10 = 37.6\\[1em] \Rightarrow 3.76 \times 100 = 376\\[1em] \Rightarrow 3.76 \times 1000 = 3760$

Hence, 3.76 × 10 = 37.6, 3.76 × 100 = 376 and 3.76 × 1000 = 3760.

(iii) 0.549

$\Rightarrow 0.549 \times 10 = 5.49\\[1em] \Rightarrow 0.549 \times 100 = 54.9\\[1em] \Rightarrow 0.549 \times 1000 = 549$

Hence, 0.549 × 10 = 5.49, 0.549 × 100 = 54.9 and 0.549 × 1000 = 549.

Question 4

Divide each of the following numbers by 10, 100 and 1000 (verbally):

(i) 4.8

(ii) 38.53

(iii) 128.9

Answer

While dividing a decimal by 10, 100 or 1000, the decimal point is shifted to the left by as many places as there are zeros.

(i) 4.8

$\Rightarrow 4.8 \div 10 = 0.48\\[1em] \Rightarrow 4.8 \div 100 = 0.048\\[1em] \Rightarrow 4.8 \div 1000 = 0.0048$

Hence, 4.8 ÷ 10 = 0.48, 4.8 ÷ 100 = 0.048 and 4.8 ÷ 1000 = 0.0048.

(ii) 38.53

$\Rightarrow 38.53 \div 10 = 3.853\\[1em] \Rightarrow 38.53 \div 100 = 0.3853\\[1em] \Rightarrow 38.53 \div 1000 = 0.03853$

Hence, 38.53 ÷ 10 = 3.853, 38.53 ÷ 100 = 0.3853 and 38.53 ÷ 1000 = 0.03853.

(iii) 128.9

$\Rightarrow 128.9 \div 10 = 12.89\\[1em] \Rightarrow 128.9 \div 100 = 1.289\\[1em] \Rightarrow 128.9 \div 1000 = 0.1289$

Hence, 128.9 ÷ 10 = 12.89, 128.9 ÷ 100 = 1.289 and 128.9 ÷ 1000 = 0.1289.

Question 5

Find the area of a rectangle whose length is 5.7 cm and breadth is 3.5 cm.

Answer

Length of rectangle = 5.7 cm.

Breadth of rectangle = 3.5 cm.

Area of rectangle = length × breadth = 5.7 × 3.5

Multiplying without the decimal points, 57 × 35 = 1995.

The total number of decimal places is 1 + 1 = 2.

⇒ 5.7 × 3.5

⇒ 19.95 sq. cm

Hence, the area of the rectangle = 19.95 sq. cm.

Question 6

The cost of one metre cloth is ₹38.50. Find the cost of 3.6 m cloth.

Answer

Cost of 1 metre of cloth = ₹38.50.

Cost of 3.6 m of cloth = 38.50 × 3.6

Multiplying without the decimal points, 3850 × 36 = 138600.

The total number of decimal places is 2 + 1 = 3.

⇒ 38.50 × 3.6

⇒ 138.60

Hence, the cost of 3.6 m cloth = ₹138.60.

Question 7

A two-wheeler covers a distance of 45.3 km in one litre of petrol. How much distance will it cover in 5.9 litres of petrol?

Answer

Distance covered in 1 litre of petrol = 45.3 km.

Distance covered in 5.9 litres of petrol = 45.3 × 5.9

Multiplying without the decimal points, 453 × 59 = 26727.

The total number of decimal places is 1 + 1 = 2.

⇒ 45.3 × 5.9

⇒ 267.27 km

Hence, the two-wheeler will cover 267.27 km in 5.9 litres of petrol.

Question 8

If 1 kg of pure milk contains 0.245 kg of fat. How much fat is there in 12.8 kg of milk?

Answer

Fat in 1 kg of milk = 0.245 kg.

Fat in 12.8 kg of milk = 0.245 × 12.8

Multiplying without the decimal points, 245 × 128 = 31360.

The total number of decimal places is 3 + 1 = 4.

⇒ 0.245 × 12.8

⇒ 3.136 kg

Hence, there is 3.136 kg of fat in 12.8 kg of milk.

Question 9

If ₹242.46 are to be distributed among 6 children equally, find the share of each.

Answer

Total amount = ₹242.46.

Number of children = 6.

Share of each child = $\dfrac{242.46}{6}$

$\Rightarrow \dfrac{242.46}{6}$

⇒ 40.41

Hence, the share of each child = ₹40.41.

Question 10

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

Answer

Distance covered in 2.4 litres of petrol = 43.2 km.

Distance covered in 1 litre of petrol = $\dfrac{43.2}{2.4}$

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$\Rightarrow \dfrac{43.2}{2.4}\\[1em] \Rightarrow \dfrac{43.2 \times 10}{2.4 \times 10}\\[1em] \Rightarrow \dfrac{432}{24}\\[1em] \Rightarrow 18 \text{ km}$

Hence, the vehicle will cover 18 km in one litre of petrol.

Question 11

How many ice cream cones can be filled from 8.4 litres of ice cream, if one cone can be filled with 35 millilitres of ice cream?

Answer

Total ice cream = 8.4 litres = 8.4 × 1000 = 8400 mL. (Since 1 litre = 1000 mL)

Ice cream filled in one cone = 35 mL.

Number of cones = $\dfrac{8400}{35}$

$\Rightarrow \dfrac{8400}{35}$

⇒ 240

Hence, 240 ice cream cones can be filled.

Question 12

If the product of two decimal numbers is 38.745 and one of the numbers is 2.7, find the other number.

Answer

Product of two numbers = 38.745.

One of the numbers = 2.7.

Other number = $\dfrac{38.745}{2.7}$

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$\Rightarrow \dfrac{38.745}{2.7}\\[1em] \Rightarrow \dfrac{38.745 \times 10}{2.7 \times 10}\\[1em] \Rightarrow \dfrac{387.45}{27}\\[1em] \Rightarrow 14.35$

Hence, the other number is 14.35.

Question 13

If $\dfrac{2}{3}$ of a number is 10, then what is 1.75 times of that number?

Answer

Let the number be $x$ .

$\Rightarrow \dfrac{2}{3} \times x = 10\\[1em] \Rightarrow x = 10 \times \dfrac{3}{2}\\[1em] \Rightarrow x = 15$

Now, 1.75 times of the number = 1.75 × 15

$\Rightarrow 1.75 \times 15$

$\Rightarrow$ 26.25

Hence, 1.75 times of the number = 26.25.

Exercise 2.7

Question 1

Simplify the following:

(i) $\dfrac{3}{5}$ of $1\dfrac{1}{9} + 3\dfrac{1}{2}$

(ii) $\dfrac{4}{5} \times 2\dfrac{3}{8} - 2 \times \dfrac{3}{5}$

(iii) $\left(\dfrac{4}{5} + 2\right)\left(3 - \dfrac{2}{3}\right)$

Answer

(i) $\dfrac{3}{5}$ of $1\dfrac{1}{9} + 3\dfrac{1}{2}$

Solving 'of' first,

$\Rightarrow \dfrac{3}{5} \text{ of } \dfrac{10}{9} + \dfrac{7}{2}\\[1em] \Rightarrow \dfrac{3}{5} \times \dfrac{10}{9} + \dfrac{7}{2}\\[1em] \Rightarrow \dfrac{2}{3} + \dfrac{7}{2}\\[1em] \Rightarrow \dfrac{2 \times 2}{3 \times 2} + \dfrac{7 \times 3}{2 \times 3}\\[1em] \Rightarrow \dfrac{4}{6} + \dfrac{21}{6}\\[1em] \Rightarrow \dfrac{4 + 21}{6}\\[1em] \Rightarrow \dfrac{25}{6}\\[1em] \Rightarrow 4\dfrac{1}{6}$

Hence, $\dfrac{3}{5}$ of $1\dfrac{1}{9} + 3\dfrac{1}{2} = 4\dfrac{1}{6}$ .

(ii) $\dfrac{4}{5} \times 2\dfrac{3}{8} - 2 \times \dfrac{3}{5}$

Solving multiplication first,

$\Rightarrow \dfrac{4}{5} \times \dfrac{19}{8} - 2 \times \dfrac{3}{5}\\[1em] \Rightarrow \dfrac{19}{10} - \dfrac{6}{5}\\[1em] \Rightarrow \dfrac{19}{10} - \dfrac{6 \times 2}{5 \times 2}\\[1em] \Rightarrow \dfrac{19}{10} - \dfrac{12}{10}\\[1em] \Rightarrow \dfrac{19 - 12}{10}\\[1em] \Rightarrow \dfrac{7}{10}$

Hence, $\dfrac{4}{5} \times 2\dfrac{3}{8} - 2 \times \dfrac{3}{5} = \dfrac{7}{10}$ .

(iii) $\left(\dfrac{4}{5} + 2\right)\left(3 - \dfrac{2}{3}\right)$

Solving the brackets first,

$\Rightarrow \left(\dfrac{4}{5} + \dfrac{10}{5}\right)\left(\dfrac{9}{3} - \dfrac{2}{3}\right)\\[1em] \Rightarrow \left(\dfrac{4 + 10}{5}\right)\left(\dfrac{9 - 2}{3}\right)\\[1em] \Rightarrow \dfrac{14}{5} \times \dfrac{7}{3}\\[1em] \Rightarrow \dfrac{98}{15}\\[1em] \Rightarrow 6\dfrac{8}{15}$

Hence, $\left(\dfrac{4}{5} + 2\right)\left(3 - \dfrac{2}{3}\right) = 6\dfrac{8}{15}$ .

Question 2

Simplify the following:

(i) $\left(\dfrac{1}{4} \text{ of } 2 \dfrac{2}{7}\right) \div \dfrac{3}{5}$

(ii) $\left(\dfrac{3}{7} \div \dfrac{1}{2}\right) \div \dfrac{7}{8}$

(iii) $\dfrac{5}{8} \div \dfrac{3}{4} + \dfrac{2}{5}$

Answer

(i) $\left(\dfrac{1}{4} \text{ of } 2 \dfrac{2}{7}\right) \div \dfrac{3}{5}$

Solving 'of' inside the bracket first,

$\Rightarrow \left(\dfrac{1}{4} \times \dfrac{16}{7}\right) \div \dfrac{3}{5}\\[1em] \Rightarrow \dfrac{4}{7} \div \dfrac{3}{5}\\[1em] \Rightarrow \dfrac{4}{7} \times \dfrac{5}{3}\\[1em] \Rightarrow \dfrac{20}{21}$

Hence, $\left(\dfrac{1}{4} \text{ of } 2 \dfrac{2}{7}\right) \div \dfrac{3}{5} = \dfrac{20}{21}$ .

(ii) $\left(\dfrac{3}{7} \div \dfrac{1}{2}\right) \div \dfrac{7}{8}$

Solving the bracket first,

$\Rightarrow \left(\dfrac{3}{7} \times \dfrac{2}{1}\right) \div \dfrac{7}{8}\\[1em] \Rightarrow \dfrac{6}{7} \div \dfrac{7}{8}\\[1em] \Rightarrow \dfrac{6}{7} \times \dfrac{8}{7}\\[1em] \Rightarrow \dfrac{48}{49}$

Hence, $\left(\dfrac{3}{7} \div \dfrac{1}{2}\right) \div \dfrac{7}{8} = \dfrac{48}{49}$ .

(iii) $\dfrac{5}{8} \div \dfrac{3}{4} + \dfrac{2}{5}$

Solving division first,

$\Rightarrow \dfrac{5}{8} \times \dfrac{4}{3} + \dfrac{2}{5}\\[1em] \Rightarrow \dfrac{5}{6} + \dfrac{2}{5}\\[1em] \Rightarrow \dfrac{5 \times 5}{6 \times 5} + \dfrac{2 \times 6}{5 \times 6}\\[1em] \Rightarrow \dfrac{25}{30} + \dfrac{12}{30}\\[1em] \Rightarrow \dfrac{25 + 12}{30}\\[1em] \Rightarrow \dfrac{37}{30}\\[1em] \Rightarrow 1\dfrac{7}{30}$

Hence, $\dfrac{5}{8} \div \dfrac{3}{4} + \dfrac{2}{5} = 1\dfrac{7}{30}$ .

Question 3

Simplify the following:

(i) $\left(4\dfrac{1}{2} - 2\dfrac{2}{3}\right) \div \dfrac{7}{12} + 5\dfrac{1}{2} \text{ of } 3 \dfrac{5}{6}$

(ii) $\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right) - \left[8 - \Big\lbrace 5\dfrac{1}{3} - \left(3 - 2\dfrac{1}{2}\right)\Big\rbrace\right]$

Answer

(i) $\left(4\dfrac{1}{2} - 2\dfrac{2}{3}\right) \div \dfrac{7}{12} + 5\dfrac{1}{2} \text{ of } 3 \dfrac{5}{6}$

Solving the bracket and 'of' first,

$\Rightarrow \left(\dfrac{9}{2} - \dfrac{8}{3}\right) \div \dfrac{7}{12} + \dfrac{11}{2} \times \dfrac{23}{6}\\[1em] \Rightarrow \left(\dfrac{27}{6} - \dfrac{16}{6}\right) \div \dfrac{7}{12} + \dfrac{253}{12}\\[1em] \Rightarrow \dfrac{11}{6} \div \dfrac{7}{12} + \dfrac{253}{12}\\[1em] \Rightarrow \dfrac{11}{6} \times \dfrac{12}{7} + \dfrac{253}{12}\\[1em] \Rightarrow \dfrac{22}{7} + \dfrac{253}{12}$

LCM of 7 and 12 = 84.

$\Rightarrow \dfrac{22 \times 12}{7 \times 12} + \dfrac{253 \times 7}{12 \times 7}\\[1em] \Rightarrow \dfrac{264}{84} + \dfrac{1771}{84}\\[1em] \Rightarrow \dfrac{264 + 1771}{84}\\[1em] \Rightarrow \dfrac{2035}{84}\\[1em] \Rightarrow 24\dfrac{19}{84}$

Hence, $\left(4\dfrac{1}{2} - 2\dfrac{2}{3}\right) \div \dfrac{7}{12} + 5\dfrac{1}{2} \text{ of } 3 \dfrac{5}{6} = 24\dfrac{19}{84}$ .

(ii) $\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right) - \left[8 - \Big\lbrace5\dfrac{1}{3} - \left(3 - 2\dfrac{1}{2}\right)\Big\rbrace\right]$

Solving the brackets first,

$\Rightarrow \left(\dfrac{3 + 2}{6}\right) \div \left(\dfrac{3 - 2}{12}\right) - \left[8 - \Big\lbrace\dfrac{16}{3} - \left(3 - \dfrac{5}{2}\right)\Big\rbrace\right]\\[1em] \Rightarrow \dfrac{5}{6} \div \dfrac{1}{12} - \left[8 - \Big\lbrace\dfrac{16}{3} - \dfrac{1}{2}\Big\rbrace\right]\\[1em] \Rightarrow \dfrac{5}{6} \times \dfrac{12}{1} - \left[8 - \Big\lbrace\dfrac{32 - 3}{6}\Big\rbrace\right]\\[1em] \Rightarrow 10 - \left[8 - \dfrac{29}{6}\right]\\[1em] \Rightarrow 10 - \left[\dfrac{48 - 29}{6}\right]\\[1em] \Rightarrow 10 - \dfrac{19}{6}\\[1em] \Rightarrow \dfrac{60 - 19}{6}\\[1em] \Rightarrow \dfrac{41}{6}\\[1em] \Rightarrow 6\dfrac{5}{6}$

Hence, $\left(\dfrac{1}{2} + \dfrac{1}{3}\right) \div \left(\dfrac{1}{4} - \dfrac{1}{6}\right) - \left[8 - \Big\lbrace5\dfrac{1}{3} - \left(3 - 2\dfrac{1}{2}\right)\Big\rbrace\right] = 6\dfrac{5}{6}$ .

Question 4

Simplify the following:

(i) $2.3 - [1.89 - {3.6 - (2.7 - \overline{0.8 - 0.03})}]$

(ii) $4.5 - \dfrac{1}{2}$ of (7.6 - 3.5) + 2.3 × 4.05

Answer

(i) $2.3 - [1.89 - {3.6 - (2.7 - \overline{0.8 - 0.03})}]$

Solving the bar (vinculum) first,

$\Rightarrow 2.3 - [1.89 - {3.6 - (2.7 - 0.77)}]\\[1em] \Rightarrow 2.3 - [1.89 - {3.6 - 1.93}]\\[1em] \Rightarrow 2.3 - [1.89 - 1.67]\\[1em] \Rightarrow 2.3 - 0.22\\[1em] \Rightarrow 2.08$

Hence, $2.3 - [1.89 - {3.6 - (2.7 - \overline{0.8 - 0.03})}] = 2.08$ .

(ii) $4.5 - \dfrac{1}{2}$ of (7.6 - 3.5) + 2.3 × 4.05

Solving the bracket first,

$\Rightarrow 4.5 - \dfrac{1}{2} \text{ of } 4.1 + 2.3 \times 4.05\\[1em] \Rightarrow 4.5 - \dfrac{1}{2} \times 4.1 + 2.3 \times 4.05\\[1em] \Rightarrow 4.5 - 2.05 + 9.315\\[1em] \Rightarrow 2.45 + 9.315\\[1em] \Rightarrow 11.765$

Hence, $4.5 - \dfrac{1}{2}$ of (7.6 - 3.5) + 2.3 × 4.05 = 11.765.

Question 5

Simplify the following:

(i) $\dfrac{2\dfrac{1}{2} + \dfrac{1}{5}}{2\dfrac{1}{2} \div \dfrac{1}{5}}$

(ii) $\dfrac{3.5 \times 0.24}{0.21} - 0.037$

Answer

(i) $\dfrac{2\dfrac{1}{2} + \dfrac{1}{5}}{2\dfrac{1}{2} \div \dfrac{1}{5}}$

Solving the numerator and denominator separately.

Numerator = $2\dfrac{1}{2} + \dfrac{1}{5} = \dfrac{5}{2} + \dfrac{1}{5} = \dfrac{25 + 2}{10} = \dfrac{27}{10}$ .

Denominator = $2\dfrac{1}{2} \div \dfrac{1}{5} = \dfrac{5}{2} \times \dfrac{5}{1} = \dfrac{25}{2}$ .

$\Rightarrow \dfrac{27}{10} \div \dfrac{25}{2}\\[1em] \Rightarrow \dfrac{27}{10} \times \dfrac{2}{25}\\[1em] \Rightarrow \dfrac{27}{125}$

Hence, $\dfrac{2\dfrac{1}{2} + \dfrac{1}{5}}{2\dfrac{1}{2} \div \dfrac{1}{5}} = \dfrac{27}{125}$ .

(ii) $\dfrac{3.5 \times 0.24}{0.21} - 0.037$

Solving the numerator first,

$\Rightarrow \dfrac{0.84}{0.21} - 0.037\\[1em] \Rightarrow \dfrac{0.84 \times 100}{0.21 \times 100} - 0.037\\[1em] \Rightarrow \dfrac{84}{21} - 0.037\\[1em] \Rightarrow 4 - 0.037\\[1em] \Rightarrow 3.963$

Hence, $\dfrac{3.5 \times 0.24}{0.21} - 0.037 = 3.963$ .

Mental Maths

Question 1

Fill in the blanks:

(i) In fractions with same numerator, the fraction with greater denominator is ....

(ii) $\dfrac{114}{138}$ reduced to simplest form is ....

(iii) $\dfrac{154}{286} = \dfrac{...}{13}$

(iv) The reciprocal of the fraction $2\dfrac{3}{8}$ is ....

(v) There are .... minutes in $\dfrac{2}{5}$ of 2 hours.

(vi) $2\dfrac{3}{7} \times 4\dfrac{2}{3} = ....$

(vii) $1\dfrac{2}{3} \div 2\dfrac{1}{5} = ....$

(viii) If the price of 7 similar pens is ₹37.80, then the price of each pen is ....

(ix) 5.4 × 2.35 = ....

(x) 0.32 ÷ 8 = ....

(xi) 45 mg = .... g

(xii) 5.06 kg = .... kg .... g

(xiii) 7.035 m = .... m .... cm ... mm

(xiv) The product of a proper fraction and an improper fraction is ...... the improper fraction.

(xv) The lowest form of the product $2\dfrac{3}{7} \times \dfrac{7}{9}$ is ......

(xvi) Ravi ate $\dfrac{2}{7}$ part of a cake while his sister Rani ate $\dfrac{4}{5}$ of the remaining. Part of the cake left is ........ .

Answer

(i) In fractions with same numerator, the fraction with greater denominator is smaller.

(ii) $\dfrac{114}{138}$

By prime factorisation,

$\Rightarrow \dfrac{114}{138} = \dfrac{2 \times 3 \times 19}{2 \times 3 \times 23}$

= $\dfrac{19}{23}.$

So, $\dfrac{114}{138}$ reduced to simplest form is $\dfrac{19}{23}$ .

(iii) $\dfrac{154}{286} = \dfrac{...}{13}$

To get 13 from 286, we have to divide 286 by 22. So, divide 154 by 22.

$\dfrac{154}{286} = \dfrac{154 \div 22}{286 \div 22} = \dfrac{7}{13}$ .

So, $\dfrac{154}{286} = \dfrac{7}{13}$ .

(iv) The reciprocal of $2\dfrac{3}{8} = \dfrac{19}{8}$ is obtained by interchanging the numerator and denominator.

So, the reciprocal of the fraction $2\dfrac{3}{8}$ is $\dfrac{8}{19}$ .

(v) 2 hours = 2 × 60 = 120 minutes.

$\dfrac{2}{5}$ of 120 = $\dfrac{2}{5} \times 120 = 48$ minutes.

So, there are 48 minutes in $\dfrac{2}{5}$ of 2 hours.

(vi) $2\dfrac{3}{7} \times 4\dfrac{2}{3} = \dfrac{17}{7} \times \dfrac{14}{3} = \dfrac{17 \times 2}{3} = \dfrac{34}{3} = 11\dfrac{1}{3}$ .

So, $2\dfrac{3}{7} \times 4\dfrac{2}{3} = 11\dfrac{1}{3}$ .

(vii) $1\dfrac{2}{3} \div 2\dfrac{1}{5} = \dfrac{5}{3} \div \dfrac{11}{5} = \dfrac{5}{3} \times \dfrac{5}{11} = \dfrac{25}{33}$ .

So, $1\dfrac{2}{3} \div 2\dfrac{1}{5} = \dfrac{25}{33}$ .

(viii) Price of each pen = $\dfrac{37.80}{7} = 5.40$ .

So, the price of each pen is ₹5.40.

(ix) Multiplying without the decimal points, 54 × 235 = 12690. The total number of decimal places is 1 + 2 = 3.

So, 5.4 × 2.35 = 12.69.

(x) 0.32 ÷ 8 = 0.04.

So, 0.32 ÷ 8 = 0.04.

(xi) Since 1 mg = $\dfrac{1}{1000}$ g, 45 mg = $\dfrac{45}{1000}$ = 0.045 g.

So, 45 mg = 0.045 g.

(xii) 5.06 kg = 5 kg + 0.06 kg = 5 kg + (0.06 × 1000) g = 5 kg + 60 g.

So, 5.06 kg = 5 kg 60 g.

(xiii) 7.035 m = 7 m + 0.035 m = 7 m + 3.5 cm = 7 m + 3 cm + 5 mm.

So, 7.035 m = 7 m 3 cm 5 mm.

(xiv) The product of a proper fraction and an improper fraction is less than the improper fraction.

(xv) $2\dfrac{3}{7} \times \dfrac{7}{9} = \dfrac{17}{7} \times \dfrac{7}{9} = \dfrac{17}{9}$ .

So, the lowest form of the product is $\dfrac{17}{9}$ .

(xvi) Part of cake eaten by Ravi = $\dfrac{2}{7}$ .

Remaining part = $1 - \dfrac{2}{7} = \dfrac{5}{7}$ .

Part eaten by Rani = $\dfrac{4}{5}$ of $\dfrac{5}{7} = \dfrac{4}{5} \times \dfrac{5}{7} = \dfrac{4}{7}$ .

Total part eaten = $\dfrac{2}{7} + \dfrac{4}{7} = \dfrac{6}{7}$ .

Part of cake left = $1 - \dfrac{6}{7} = \dfrac{1}{7}$ .

So, part of the cake left is $\dfrac{1}{7}$ .

Question 2

State whether the following statements are true (T) or false (F):

(i) The reciprocal of 1 is 0

(ii) The reciprocal of a proper fraction is a proper fraction.

(iii) The reciprocal of an improper fraction is an improper fraction.

(iv) Product of two fractions = $\dfrac{\text{product of their denominators}}{\text{product of their numerators}}$

(v) $\dfrac{3}{20}$ of 2 kg = 300 g

(vi) The multiplicative inverse of $3\dfrac{5}{7}$ is $\dfrac{7}{26}$

(vii) $\dfrac{11}{15} - \dfrac{7}{20} = \dfrac{23}{60}$

(viii) $\dfrac{2}{3}$ of 8 is same as $\dfrac{2}{3} \div 8$

(ix) The product of two proper fractions is greater than each of the two fractions.

(x) To multiply a decimal number by 10, move the decimal point to the left by one place.

(xi) To divide a decimal number by 100, move the decimal point to the left by two places.

Answer

(i) False.

Reason: The reciprocal of 1 is 1 itself, since 1 × 1 = 1.

(ii) False.

Reason: The reciprocal of a proper fraction is an improper fraction. For example, the reciprocal of $\dfrac{2}{3}$ is $\dfrac{3}{2}$ , which is improper.

(iii) False.

Reason: The reciprocal of an improper fraction is a proper fraction. For example, the reciprocal of $\dfrac{5}{2}$ is $\dfrac{2}{5}$ , which is proper.

(iv) False.

Reason: Product of two fractions = $\dfrac{\text{product of their numerators}}{\text{product of their denominators}}$ .

(v) True.

Reason: $\dfrac{3}{20}$ of 2 kg = $\dfrac{3}{20} \times 2 = \dfrac{6}{20} = 0.3$ kg = 300 g.

(vi) True.

Reason: $3\dfrac{5}{7} = \dfrac{26}{7}$ , and its multiplicative inverse is $\dfrac{7}{26}$ .

(vii) True.

Reason: LCM of 15 and 20 = 60. So, $\dfrac{11}{15} - \dfrac{7}{20} = \dfrac{44}{60} - \dfrac{21}{60} = \dfrac{23}{60}$ .

(viii) False.

Reason: $\dfrac{2}{3}$ of 8 = $\dfrac{2}{3} \times 8 = \dfrac{16}{3}$ , whereas $\dfrac{2}{3} \div 8 = \dfrac{2}{3} \times \dfrac{1}{8} = \dfrac{1}{12}$ . They are not the same.

(ix) False.

Reason: The product of two proper fractions is smaller than each of the two fractions.

(x) False.

Reason: To multiply a decimal number by 10, move the decimal point to the right by one place.

(xi) True.

Reason: To divide a decimal number by 100, the decimal point is moved to the left by two places.

Multiple Choice Questions

Question 3

$\dfrac{5}{6}$ of 480 is

400
576
480
none of these

Answer

$\dfrac{5}{6}$ of 480 = $\dfrac{5}{6} \times 480 = 5 \times 80 = 400$ .

Hence, option 1 is the correct option.

Question 4

The reciprocal of $5\dfrac{2}{3}$ is

$5\dfrac{3}{2}$
$3\dfrac{2}{5}$
$2\dfrac{3}{5}$
$\dfrac{3}{17}$

Answer

$5\dfrac{2}{3} = \dfrac{17}{3}$ .

The reciprocal of $\dfrac{17}{3}$ is $\dfrac{3}{17}$ .

Hence, option 4 is the correct option.

Question 5

The fraction $\dfrac{11}{7}$ lies between

11 and 7
1 and 2
0 and 1
2 and 3

Answer

$\dfrac{11}{7} = 1\dfrac{4}{7}$ , which is greater than 1 and less than 2.

So, $\dfrac{11}{7}$ lies between 1 and 2.

Hence, option 2 is the correct option.

Question 6

If the cost of 1 kg almonds is ₹460, then the cost of $\dfrac{2}{5}$ kg of almonds is

₹92
₹184
₹230
₹1200

Answer

Cost of 1 kg almonds = ₹460.

Cost of $\dfrac{2}{5}$ kg almonds = $\dfrac{2}{5} \times 460 = 2 \times 92 = ₹184$ .

Hence, option 2 is the correct option.

Question 7

$2\dfrac{1}{5} \div 1\dfrac{1}{5}$ is equal to

2
$1\dfrac{1}{5}$
$2\dfrac{1}{6}$
$1\dfrac{5}{6}$

Answer

$\Rightarrow 2\dfrac{1}{5} \div 1\dfrac{1}{5}\\[1em] \Rightarrow \dfrac{11}{5} \div \dfrac{6}{5}\\[1em] \Rightarrow \dfrac{11}{5} \times \dfrac{5}{6}\\[1em] \Rightarrow \dfrac{11}{6}\\[1em] \Rightarrow 1\dfrac{5}{6}$

Hence, option 4 is the correct option.

Question 8

$5\dfrac{1}{6} \div 4\dfrac{1}{2}$ is equal to

$\dfrac{31}{6}$
$\dfrac{1}{27}$
$5\dfrac{1}{27}$
$1\dfrac{4}{27}$

Answer

$\Rightarrow 5\dfrac{1}{6} \div 4\dfrac{1}{2}\\[1em] \Rightarrow \dfrac{31}{6} \div \dfrac{9}{2}\\[1em] \Rightarrow \dfrac{31}{6} \times \dfrac{2}{9}\\[1em] \Rightarrow \dfrac{31}{27}\\[1em] \Rightarrow 1\dfrac{4}{27}$

Hence, option 4 is the correct option.

Question 9

If $\dfrac{3}{4}$ of a number is 12, then the number is

Answer

Let the number be x.

$\Rightarrow \dfrac{3}{4} \times x = 12\\[1em] \Rightarrow x = 12 \times \dfrac{4}{3}\\[1em] \Rightarrow x = 16$

Hence, option 2 is the correct option.

Question 10

Ayaz bought 3 dozen eggs. He found $\dfrac{1}{9}$ of them were rotten. The number of rotten eggs were

Answer

Total eggs = 3 dozen = 3 × 12 = 36.

Number of rotten eggs = $\dfrac{1}{9}$ of 36 = $\dfrac{1}{9} \times 36 = 4$ .

Hence, option 1 is the correct option.

Question 11

Shruti reads a novel for $1\dfrac{3}{4}$ hours daily. If she reads the entire novel in 6 days, then the time she takes to read the entire novel is

$7\dfrac{1}{2}$ hours
$9\dfrac{1}{2}$ hours
$10\dfrac{1}{2}$ hours
$11\dfrac{1}{2}$ hours

Answer

Time taken daily = $1\dfrac{3}{4}$ hours.

Total time = $1\dfrac{3}{4} \times 6 = \dfrac{7}{4} \times 6 = \dfrac{42}{4} = \dfrac{21}{2} = 10\dfrac{1}{2}$ hours.

Hence, option 3 is the correct option.

Question 12

The place value of the digit 7 in the decimal number 35.0471 is

7
$\dfrac{7}{100}$
$\dfrac{1}{1000}$
$\dfrac{7}{1000}$

Answer

In 35.0471, the digit 7 is at the thousandths place.

So, its place value = $7 \times \dfrac{1}{1000} = \dfrac{7}{1000}$ .

Hence, option 4 is the correct option.

Question 13

0.002 × 0.3 is

0.6
0.06
0.006
0.0006

Answer

Multiplying without the decimal points, 2 × 3 = 6.

The total number of decimal places is 3 + 1 = 4.

So, 0.002 × 0.3 = 0.0006.

Hence, option 4 is the correct option.

Question 14

The value of the mixed fraction $5\dfrac{3}{8}$ is

5.735
5.375
5.625
5.875

Answer

$5\dfrac{3}{8} = 5 + \dfrac{3}{8} = 5 + 0.375 = 5.375$ .

Hence, option 2 is the correct option.

Question 15

0.35 ÷ 0.7 is

50
5
0.5
0.05

Answer

$\Rightarrow \dfrac{0.35}{0.7}\\[1em] \Rightarrow \dfrac{0.35 \times 10}{0.7 \times 10}\\[1em] \Rightarrow \dfrac{3.5}{7}\\[1em] \Rightarrow 0.5$

Hence, option 3 is the correct option.

Question 16

30 m 5 cm is same as

30.5 m
3.05 m
30.05 m
30.005 m

Answer

Since 1 cm = $\dfrac{1}{100}$ m, 5 cm = 0.05 m.

So, 30 m 5 cm = 30 m + 0.05 m = 30.05 m.

Hence, option 3 is the correct option.

Question 17

0.05309 × 1000 is

5.309
53.09
530.9
none of these

Answer

While multiplying by 1000, the decimal point shifts 3 places to the right.

So, 0.05309 × 1000 = 53.09.

Hence, option 2 is the correct option.

Question 18

2.305 ÷ 1000 is

0.2305
0.02305
0.002305
none of these

Answer

While dividing by 1000, the decimal point shifts 3 places to the left.

So, 2.305 ÷ 1000 = 0.002305.

Hence, option 3 is the correct option.

Question 19

If each side of a regular hexagon is 3.5 cm, then the perimeter of the hexagon is

17.5 cm
21 cm
18.5 cm
24.5 cm

Answer

A regular hexagon has 6 equal sides.

Perimeter = 6 × side = 6 × 3.5 = 21 cm.

Hence, option 2 is the correct option.

Question 20

Which of the following numbers has the smallest value?

0.0002
$\dfrac{2}{1000}$
0.02 × 0.001
$\dfrac{2}{1000} \div 0.01$

Answer

Evaluating each option:

Option 1: 0.0002

Option 2: $\dfrac{2}{1000} = 0.002$

Option 3: 0.02 × 0.001 = 0.00002

Option 4: $\dfrac{2}{1000} \div 0.01 = 0.002 \div 0.01 = 0.2$

Comparing all the values, 0.00002 is the smallest.

Hence, option 3 is the correct option.

Statement I-II Type Questions

Question 21

Statement I: The product of a proper fraction and an improper fraction is always an improper fraction.

Statement II: The product of two proper fractions is always a proper fraction.

Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both Statement I and statement II are true.
Both Statement I and statement II are false.

Answer

According to Statement I, consider the proper fraction $\dfrac{1}{2}$ and the improper fraction $\dfrac{3}{2}$ .

Their product = $\dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4}$ , which is a proper fraction.

So, the product of a proper fraction and an improper fraction is not always an improper fraction.

∴ Statement I is false.

According to Statement II, the product of two proper fractions is always smaller than each of them, and is therefore a proper fraction. For example, $\dfrac{2}{3} \times \dfrac{4}{5} = \dfrac{8}{15}$ , which is proper.

∴ Statement II is true.

Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Question 22

Statement I: We can write any improper fraction as a mixed fraction.

Statement II: All improper fractions consisting of a natural number and a proper fraction are greater than 1.

Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both Statement I and statement II are true.
Both Statement I and statement II are false.

Answer

According to Statement I, any improper fraction can be written as a mixed fraction by dividing the numerator by the denominator. For example, $\dfrac{11}{4} = 2\dfrac{3}{4}$ .

∴ Statement I is true.

According to Statement II, a mixed fraction is made up of a natural number and a proper fraction, so it is always greater than 1. For example, $2\dfrac{3}{4} = 2 + \dfrac{3}{4} > 1$ .

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Question 23

Statement I: If we divide $\dfrac{2}{5}$ by $\dfrac{5}{2}$ , we get 1

Statement II: $\dfrac{2}{5}$ and $\dfrac{5}{2}$ are reciprocals of each other.

Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both Statement I and statement II are true.
Both Statement I and statement II are false.

Answer

According to Statement I, $\dfrac{2}{5} \div \dfrac{5}{2} = \dfrac{2}{5} \times \dfrac{2}{5} = \dfrac{4}{25}$ , which is not equal to 1.

∴ Statement I is false.

According to Statement II, $\dfrac{2}{5} \times \dfrac{5}{2} = 1$ , so $\dfrac{2}{5}$ and $\dfrac{5}{2}$ are reciprocals of each other.

∴ Statement II is true.

Statement I is false but Statement II is true.

Hence, option 2 is the correct option.

Question 24

Statement I: 2.01 is greater than 1.99

Statement II: The decimal numbers having equal number of decimal points are called like decimal numbers.

Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both Statement I and statement II are true.
Both Statement I and statement II are false.

Answer

According to Statement I, comparing the whole number parts, 2 > 1, so 2.01 > 1.99.

∴ Statement I is true.

According to Statement II, decimal numbers having an equal number of decimal places are called like decimals. For example, 2.01 and 1.99 each have 2 decimal places, so they are like decimals.

∴ Statement II is true.

Both Statement I and Statement II are true.

Hence, option 3 is the correct option.

Question 25

Statement I: Aryan arranges the following number in descending order:

1.03 + 2.6, 6.75 - 4.879, 3.67 × 0.4, 10.326 ÷ 0.3

The number on the extreme right in the list is 6.75 - 4.879

Statement II: To add two decimal numbers, we add them like whole numbers by ignoring the decimal point. Then we count the number of decimal points in each number and sum it up. Say, it comes to 5. So, we add the decimal point in the result at the 5th place from the right.

Statement I is true but statement II is false.
Statement I is false but statement II is true.
Both Statement I and statement II are true.
Both Statement I and statement II are false.

Answer

According to Statement I, evaluate each number:

$\Rightarrow 1.03 + 2.6 = 3.63\\[1em] \Rightarrow 6.75 - 4.879 = 1.871\\[1em] \Rightarrow 3.67 \times 0.4 = 1.468\\[1em] \Rightarrow 10.326 \div 0.3 = 34.42$

Arranging in descending order: 34.42, 3.63, 1.871, 1.468, i.e. 10.326 ÷ 0.3, 1.03 + 2.6, 6.75 − 4.879, 3.67 × 0.4.

The number on the extreme right is 3.67 × 0.4 (= 1.468), not 6.75 − 4.879.

∴ Statement I is false.

According to Statement II, to add two decimal numbers we write them one below the other so that the decimal points are aligned in the same column, and then add like whole numbers, placing the decimal point of the sum directly below the other decimal points. The rule described (counting and summing the number of decimal places) is the rule for multiplication, not addition.

∴ Statement II is false.

Both Statement I and Statement II are false.

Hence, option 4 is the correct option.

Check Your Progress

Question 1

What fraction is 270 gram of 3 kilogram?

Answer

3 kilogram = 3 × 1000 = 3000 gram. (Since 1 kg = 1000 g)

Required fraction = $\dfrac{270}{3000}$

$\Rightarrow \dfrac{270}{3000}$

$\Rightarrow \dfrac{9}{100}$

Hence, 270 gram is $\dfrac{9}{100}$ of 3 kilogram.

Question 2

Simplify the following:

(i) $7\dfrac{1}{2} \times 2\dfrac{4}{15}$

(ii) $3\dfrac{6}{7} \times 4\dfrac{2}{3}$

(iii) $3\dfrac{3}{7} \div \dfrac{16}{21}$

(iv) $15\dfrac{3}{7} \div 1\dfrac{23}{49}$

Answer

(i) $7\dfrac{1}{2} \times 2\dfrac{4}{15}$

$\Rightarrow \dfrac{15}{2} \times \dfrac{34}{15}\\[1em] \Rightarrow \dfrac{34}{2}\\[1em] \Rightarrow 17$

Hence, $7\dfrac{1}{2} \times 2\dfrac{4}{15} = 17$ .

(ii) $3\dfrac{6}{7} \times 4\dfrac{2}{3}$

$\Rightarrow \dfrac{27}{7} \times \dfrac{14}{3}\\[1em] \Rightarrow \dfrac{27 \times 2}{3}\\[1em] \Rightarrow \dfrac{54}{3}\\[1em] \Rightarrow 18$

Hence, $3\dfrac{6}{7} \times 4\dfrac{2}{3} = 18$ .

(iii) $3\dfrac{3}{7} \div \dfrac{16}{21}$

$\Rightarrow \dfrac{24}{7} \div \dfrac{16}{21}\\[1em] \Rightarrow \dfrac{24}{7} \times \dfrac{21}{16}\\[1em] \Rightarrow \dfrac{3 \times 3}{2}\\[1em] \Rightarrow \dfrac{9}{2}\\[1em] \Rightarrow 4\dfrac{1}{2}$

Hence, $3\dfrac{3}{7} \div \dfrac{16}{21} = 4\dfrac{1}{2}$ .

(iv) $15\dfrac{3}{7} \div 1\dfrac{23}{49}$

$\Rightarrow \dfrac{108}{7} \div \dfrac{72}{49}\\[1em] \Rightarrow \dfrac{108}{7} \times \dfrac{49}{72}\\[1em] \Rightarrow \dfrac{108 \times 7}{72}\\[1em] \Rightarrow \dfrac{756}{72}\\[1em] \Rightarrow \dfrac{21}{2}\\[1em] \Rightarrow 10\dfrac{1}{2}$

Hence, $15\dfrac{3}{7} \div 1\dfrac{23}{49} = 10\dfrac{1}{2}$ .

Question 3

A shirt was marked at ₹540. It was sold at $\dfrac{3}{4}$ of the marked price. What was the sale price?

Answer

Marked price of shirt = ₹540.

Sale price = $\dfrac{3}{4}$ of 540 = $\dfrac{3}{4} \times 540$

$\Rightarrow \dfrac{3}{4} \times 540\\[1em] \Rightarrow 3 \times 135\\[1em] \Rightarrow 405$

Hence, the sale price of the shirt = ₹405.

Question 4

In a class of 56 students, $\dfrac{1}{4}$ are in blue house and $\dfrac{3}{14}$ are in yellow house. Out of the remaining, $\dfrac{1}{3}$ are in green house and the rest are in red house. Find the number of students in each house.

Answer

Total number of students = 56.

Number of students in blue house = $\dfrac{1}{4} \times 56 = 14$ .

Number of students in yellow house = $\dfrac{3}{14} \times 56 = 3 \times 4 = 12$ .

Remaining students = 56 - 14 - 12 = 30.

Number of students in green house = $\dfrac{1}{3} \times 30 = 10$ .

Number of students in red house = 30 - 10 = 20.

Hence, blue house = 14, yellow house = 12, green house = 10 and red house = 20 students.

Question 5

Rohit bought a motor cycle for ₹36000. He paid $\dfrac{1}{6}$ of the price in cash upfront and the rest in 12 equal monthly installments. Find the amount he had to pay every month.

Answer

Total price of motor cycle = ₹36000.

Amount paid in cash = $\dfrac{1}{6}$ of 36000 = $\dfrac{1}{6} \times 36000 = ₹6000$ .

Remaining amount = 36000 - 6000 = ₹30000.

Amount paid every month = $\dfrac{30000}{12}$

$\Rightarrow \dfrac{30000}{12}$

⇒ 2500

Hence, Rohit had to pay ₹2500 every month.

Question 6

I read $\dfrac{4}{9}$ of a book on one day and $\dfrac{3}{5}$ of the remaining next day. If 100 pages of the book were still left unread, how many pages did the book contain?

Answer

Let the total number of pages in the book be 1 (whole).

Part read on first day = $\dfrac{4}{9}$ .

Remaining part = $1 - \dfrac{4}{9} = \dfrac{5}{9}$ .

Part read on second day = $\dfrac{3}{5}$ of $\dfrac{5}{9} = \dfrac{3}{5} \times \dfrac{5}{9} = \dfrac{1}{3}$ .

Total part read = $\dfrac{4}{9} + \dfrac{1}{3} = \dfrac{4}{9} + \dfrac{3}{9} = \dfrac{7}{9}$ .

Part left unread = $1 - \dfrac{7}{9} = \dfrac{2}{9}$ .

Given, $\dfrac{2}{9}$ of the book = 100 pages.

$\Rightarrow \dfrac{2}{9} \times \text{Total pages} = 100\\[1em] \Rightarrow \text{Total pages} = 100 \times \dfrac{9}{2}\\[1em] \Rightarrow \text{Total pages} = 450$

Hence, the book contained 450 pages.

Question 7

Find the value of: $\dfrac{1}{4\dfrac{2}{7}} + \dfrac{1}{1\dfrac{11}{13}} + \dfrac{1}{\dfrac{5}{9}}$

Answer

Converting the mixed fractions in the denominators to improper fractions,

$4\dfrac{2}{7} = \dfrac{30}{7}$ , $1\dfrac{11}{13} = \dfrac{24}{13}$ .

$\Rightarrow \dfrac{1}{\dfrac{30}{7}} + \dfrac{1}{\dfrac{24}{13}} + \dfrac{1}{\dfrac{5}{9}}\\[1em] \Rightarrow \dfrac{7}{30} + \dfrac{13}{24} + \dfrac{9}{5}$

LCM of 30, 24 and 5 = 120.

$\Rightarrow \dfrac{7 \times 4}{30 \times 4} + \dfrac{13 \times 5}{24 \times 5} + \dfrac{9 \times 24}{5 \times 24}\\[1em] \Rightarrow \dfrac{28}{120} + \dfrac{65}{120} + \dfrac{216}{120}\\[1em] \Rightarrow \dfrac{28 + 65 + 216}{120}\\[1em] \Rightarrow \dfrac{309}{120}\\[1em] \Rightarrow \dfrac{103}{40}\\[1em] \Rightarrow 2\dfrac{23}{40}$

Hence, the value = $2\dfrac{23}{40}$ .

Question 8

Convert the following numbers to fractions (in simplest form):

(i) 0.025

(ii) 0.876

(iii) 4.3125

Answer

(i) 0.025

$\Rightarrow \dfrac{25}{1000}$

$\Rightarrow \dfrac{1}{40}$

Hence, 0.025 = $\dfrac{1}{40}$ .

(ii) 0.876

$\Rightarrow \dfrac{876}{1000}$

$\Rightarrow \dfrac{219}{250}$

Hence, 0.876 = $\dfrac{219}{250}$ .

(iii) 4.3125

$\Rightarrow \dfrac{43125}{10000}\\[1em] \Rightarrow \dfrac{69}{16}\\[1em] \Rightarrow 4\dfrac{5}{16}$

Hence, 4.3125 = $4\dfrac{5}{16}$ .

Question 9

Write the following fractions as decimals:

(i) $1\dfrac{3}{8}$

(ii) $\dfrac{47}{125}$

(iii) $2\dfrac{9}{40}$

Answer

(i) $1\dfrac{3}{8}$

$\Rightarrow \dfrac{11}{8}\\[1em] \Rightarrow \dfrac{11 \times 125}{8 \times 125}\\[1em] \Rightarrow \dfrac{1375}{1000}\\[1em] \Rightarrow 1.375$

Hence, $1\dfrac{3}{8} = 1.375$ .

(ii) $\dfrac{47}{125}$

$\Rightarrow \dfrac{47 \times 8}{125 \times 8}\\[1em] \Rightarrow \dfrac{376}{1000}\\[1em] \Rightarrow 0.376$

Hence, $\dfrac{47}{125} = 0.376$ .

(iii) $2\dfrac{9}{40}$

$\Rightarrow \dfrac{89}{40}\\[1em] \Rightarrow \dfrac{89 \times 25}{40 \times 25}\\[1em] \Rightarrow \dfrac{2225}{1000}\\[1em] \Rightarrow 2.225$

Hence, $2\dfrac{9}{40} = 2.225$ .

Question 10

By how much does the sum of 17.443 and 29.657 exceed the sum of 13.687 and 18.548?

Answer

Sum of 17.443 and 29.657 = 17.443 + 29.657 = 47.1.

Sum of 13.687 and 18.548 = 13.687 + 18.548 = 32.235.

Required excess = 47.1 - 32.235

⇒ 47.100 - 32.235

⇒ 14.865

Hence, the first sum exceeds the second by 14.865.

Question 11

Simplify the following:

(i) 4.27 × 0.036

(ii) 0.09 × 1.04

(iii) 1.32 ÷ 0.8

(iv) 0.7038 ÷ 0.34

Answer

(i) 4.27 × 0.036

Multiplying without the decimal points, 427 × 36 = 15372. The total number of decimal places is 2 + 3 = 5.

⇒ 4.27 × 0.036

⇒ 0.15372

Hence, 4.27 × 0.036 = 0.15372.

(ii) 0.09 × 1.04

Multiplying without the decimal points, 9 × 104 = 936. The total number of decimal places is 2 + 2 = 4.

⇒ 0.09 × 1.04

⇒ 0.0936

Hence, 0.09 × 1.04 = 0.0936.

(iii) 1.32 ÷ 0.8

Multiply both the dividend and the divisor by 10 to make the divisor a whole number.

$\Rightarrow \dfrac{1.32}{0.8}\\[1em] \Rightarrow \dfrac{1.32 \times 10}{0.8 \times 10}\\[1em] \Rightarrow \dfrac{13.2}{8}\\[1em] \Rightarrow 1.65$

Hence, 1.32 ÷ 0.8 = 1.65.

(iv) 0.7038 ÷ 0.34

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{0.7038}{0.34}\\[1em] \Rightarrow \dfrac{0.7038 \times 100}{0.34 \times 100}\\[1em] \Rightarrow \dfrac{70.38}{34}\\[1em] \Rightarrow 2.07$

Hence, 0.7038 ÷ 0.34 = 2.07.

Question 12

If one kg of rice costs ₹52.70, then find the cost of 12.5 kg rice.

Answer

Cost of 1 kg rice = ₹52.70.

Cost of 12.5 kg rice = 52.70 × 12.5

Multiplying without the decimal points, 5270 × 125 = 658750. The total number of decimal places is 2 + 1 = 3.

⇒ 52.70 × 12.5

⇒ 658.75

Hence, the cost of 12.5 kg rice = ₹658.75.

Question 13

A piece of cloth is 24.5 m long. How many pieces, each of length 1.75 m, can be cut from it?

Answer

Total length of cloth = 24.5 m.

Length of each piece = 1.75 m.

Number of pieces = $\dfrac{24.5}{1.75}$

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{24.5}{1.75}\\[1em] \Rightarrow \dfrac{24.5 \times 100}{1.75 \times 100}\\[1em] \Rightarrow \dfrac{2450}{175}\\[1em] \Rightarrow 14$

Hence, 14 pieces can be cut from the cloth.

Question 14

The product of two decimal numbers is 1.599 and one of them is 0.65, find the other number.

Answer

Product of two numbers = 1.599.

One of the numbers = 0.65.

Other number = $\dfrac{1.599}{0.65}$

Multiply both the dividend and the divisor by 100 to make the divisor a whole number.

$\Rightarrow \dfrac{1.599}{0.65}\\[1em] \Rightarrow \dfrac{1.599 \times 100}{0.65 \times 100}\\[1em] \Rightarrow \dfrac{159.9}{65}\\[1em] \Rightarrow 2.46$

Hence, the other number is 2.46.

Question 15

Simplify the following:

(i) $\left(\dfrac{4}{5} - \dfrac{1}{3}\right) \div 4\dfrac{1}{5} + \dfrac{2}{3}$ of $\left(5\dfrac{1}{6} - 4\dfrac{3}{8}\right)$

(ii) $1\dfrac{2}{3}$ of $\left(\dfrac{3}{8} - \dfrac{1}{12}\right) - \left[4\dfrac{2}{3} - \Big\lbrace6 - \left(2\dfrac{2}{3} - \overline{4\dfrac{1}{2} - 3\dfrac{1}{3}}\right)\Big\rbrace\right]$

Answer

(i) $\left(\dfrac{4}{5} - \dfrac{1}{3}\right) \div 4\dfrac{1}{5} + \dfrac{2}{3}$ of $\left(5\dfrac{1}{6} - 4\dfrac{3}{8}\right)$

Solving the brackets first,

$\Rightarrow \left(\dfrac{12 - 5}{15}\right) \div \dfrac{21}{5} + \dfrac{2}{3} \text{ of } \left(\dfrac{31}{6} - \dfrac{35}{8}\right)\\[1em] \Rightarrow \dfrac{7}{15} \div \dfrac{21}{5} + \dfrac{2}{3} \text{ of } \left(\dfrac{124 - 105}{24}\right)\\[1em] \Rightarrow \dfrac{7}{15} \times \dfrac{5}{21} + \dfrac{2}{3} \times \dfrac{19}{24}\\[1em] \Rightarrow \dfrac{1}{9} + \dfrac{19}{36}\\[1em] \Rightarrow \dfrac{4}{36} + \dfrac{19}{36}\\[1em] \Rightarrow \dfrac{4 + 19}{36}\\[1em] \Rightarrow \dfrac{23}{36}$

Hence, $\left(\dfrac{4}{5} - \dfrac{1}{3}\right) \div 4\dfrac{1}{5} + \dfrac{2}{3}$ of $\left(5\dfrac{1}{6} - 4\dfrac{3}{8}\right) = \dfrac{23}{36}$ .

(ii) $1\dfrac{2}{3}$ of $\left(\dfrac{3}{8} - \dfrac{1}{12}\right) - \left[4\dfrac{2}{3} - \Big\lbrace6 - \left(2\dfrac{2}{3} - \overline{4\dfrac{1}{2} - 3\dfrac{1}{3}}\right)\Big\rbrace\right]$

Solving the bar (vinculum) first,

$\Rightarrow 4\dfrac{1}{2} - 3\dfrac{1}{3}\\[1em] = \dfrac{9}{2} - \dfrac{10}{3}\\[1em] = \dfrac{27 - 20}{6} = \dfrac{7}{6}$

Now substituting and solving the innermost bracket,

$\Rightarrow 1\dfrac{2}{3} \text{ of } \left(\dfrac{3}{8} - \dfrac{1}{12}\right) - \left[4\dfrac{2}{3} - \Big\lbrace6 - \left(\dfrac{8}{3} - \dfrac{7}{6}\right)\Big\rbrace\right]\\[1em] \Rightarrow 1\dfrac{2}{3} \text{ of } \left(\dfrac{9 - 2}{24}\right) - \left[\dfrac{14}{3} - \Big\lbrace6 - \left(\dfrac{16 - 7}{6}\right)\Big\rbrace\right]\\[1em] \Rightarrow \dfrac{5}{3} \text{ of } \dfrac{7}{24} - \left[\dfrac{14}{3} - \Big\lbrace6 - \dfrac{9}{6}\Big\rbrace\right]\\[1em] \Rightarrow \dfrac{5}{3} \times \dfrac{7}{24} - \left[\dfrac{14}{3} - \Big\lbrace\dfrac{36 - 9}{6}\Big\rbrace\right]\\[1em] \Rightarrow \dfrac{35}{72} - \left[\dfrac{14}{3} - \dfrac{9}{2}\right]\\[1em] \Rightarrow \dfrac{35}{72} - \left[\dfrac{28 - 27}{6}\right]\\[1em] \Rightarrow \dfrac{35}{72} - \dfrac{1}{6}\\[1em] \Rightarrow \dfrac{35 - 12}{72}\\[1em] \Rightarrow \dfrac{23}{72}$

Hence, $1\dfrac{2}{3}$ of $\left(\dfrac{3}{8} - \dfrac{1}{12}\right) - \left[4\dfrac{2}{3} - \Big\lbrace6 - \left(2\dfrac{2}{3} - \overline{4\dfrac{1}{2} - 3\dfrac{1}{3}}\right)\Big\rbrace\right] = \dfrac{23}{72}$ .

Question 16

What is the least fraction that must be added to $\left(1\dfrac{1}{3} \div 1\dfrac{1}{2}\right) \div 1\dfrac{1}{9}$ to make the result a natural number?

Answer

First, simplify $\left(1\dfrac{1}{3} \div 1\dfrac{1}{2}\right) \div 1\dfrac{1}{9}$ .

$\Rightarrow \left(\dfrac{4}{3} \div \dfrac{3}{2}\right) \div \dfrac{10}{9}\\[1em] \Rightarrow \left(\dfrac{4}{3} \times \dfrac{2}{3}\right) \div \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{8}{9} \div \dfrac{10}{9}\\[1em] \Rightarrow \dfrac{8}{9} \times \dfrac{9}{10}\\[1em] \Rightarrow \dfrac{8}{10}\\[1em] \Rightarrow \dfrac{4}{5}$

The result is $\dfrac{4}{5}$ , which is a proper fraction. The smallest natural number greater than $\dfrac{4}{5}$ is 1.

Least fraction to be added = $1 - \dfrac{4}{5} = \dfrac{5 - 4}{5} = \dfrac{1}{5}$ .

Hence, the least fraction that must be added is $\dfrac{1}{5}$ .

Question 17

A drum of petrol is $\dfrac{3}{4}$ full. When 30 litres of oil are drawn from it, it is $\dfrac{7}{12}$ full. Find the capacity of the drum.

Answer

Let the capacity of the drum be 1 (whole).

Part of drum that is full at first = $\dfrac{3}{4}$ .

Part of drum that is full after drawing oil = $\dfrac{7}{12}$ .

Part of drum corresponding to 30 litres = $\dfrac{3}{4} - \dfrac{7}{12}$

$\Rightarrow \dfrac{3}{4} - \dfrac{7}{12}\\[1em] \Rightarrow \dfrac{9 - 7}{12}\\[1em] \Rightarrow \dfrac{2}{12}\\[1em] \Rightarrow \dfrac{1}{6}$

So, $\dfrac{1}{6}$ of the capacity = 30 litres.

$\Rightarrow \dfrac{1}{6} \times \text{Capacity} = 30\\[1em] \Rightarrow \text{Capacity} = 30 \times 6\\[1em] \Rightarrow \text{Capacity} = 180 \text{ litres}$

Hence, the capacity of the drum is 180 litres.

Chapter 2

Fractions and Decimals

Class - 7 ML Aggarwal Understanding ICSE Mathematics

Exercise 2.1

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Exercise 2.2

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Exercise 2.3

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Exercise 2.4

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Exercise 2.5

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Exercise 2.6

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Exercise 2.7

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Mental Maths

Question 1