RD Sharma 2020 2021 Solutions for Class 6 Maths Chapter 6 - Fractions

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{28}{9} \\ \mathrm{Quotient}=3, \mathrm{Remainder}=1, \mathrm{Denominator}=9 \\ \Rightarrow \frac{28}{9}=3\frac{1}{9} \\ \left(\mathrm{ii}\right) \frac{226}{15} \\ \mathrm{Quotient}=15, \mathrm{Remainder}=1, \mathrm{Denominator}=15 \\ \Rightarrow \frac{226}{15}=15\frac{1}{15} \\ \left(\mathrm{iii}\right) \frac{145}{9} \\ \mathrm{Quotient}=16, \mathrm{Remainder}=1, \mathrm{Denominator}=9 \\ \Rightarrow \frac{145}{9}=16\frac{1}{9} \\ \left(\mathrm{iv}\right) \frac{128}{5} \\ \mathrm{Quotient}=25, \mathrm{Remainder}=3, \mathrm{Denominator}=5 \\ \Rightarrow \frac{128}{5}=25\frac{3}{5} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) 7\frac{1}{4}=\frac{7\times 4+1}{4}=\frac{29}{4} \\ \left(\mathrm{ii}\right) 8\frac{5}{7}=\frac{8\times 7+5}{7}=\frac{61}{7} \\ \left(\mathrm{iii}\right) 5\frac{3}{10}=\frac{5\times 10+3}{10}=\frac{53}{10} \\ \left(\mathrm{iv}\right) 12\frac{7}{15}=\frac{12\times 15+7}{15}=\frac{187}{15} \end{gathered}$$
 

Answer:

(i)

Yes, they are equivalent.


(ii)

 
Yes, they are equivalent.

Answer:

Answer:

(i)
$$\begin{gathered} \frac{2}{7}=\frac{6}{} \\ \mathrm{As} 2\times 3=6, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \frac{2}{7}\times \frac{3}{3}=\frac{6}{21} \end{gathered}$$

(ii)
$$\begin{gathered} \frac{5}{8}=\frac{10}{} \\ \mathrm{As} 5\times 2=10, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 2. \\ \Rightarrow \frac{5}{8}\times \frac{2}{2}=\frac{10}{16} \end{gathered}$$

(iii)
$$\begin{gathered} \frac{4}{5}=\frac{}{20} \\ \mathrm{As} 5\times 4=20, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 4. \\ \Rightarrow \frac{4}{5}\times \frac{4}{4}=\frac{16}{20} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{45}{60}=\frac{15}{} \\ \mathrm{As} 45÷3=15, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \hspace{0.17em}\& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \left(\frac{{\displaystyle \frac{45}{3}}}{{\displaystyle \frac{60}{3}}}\right)=\frac{15}{20} \end{gathered}$$

(v)
$$\begin{gathered} \frac{18}{24}=\frac{}{4} \\ \mathrm{As} 24÷6=4, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 6. \\ \Rightarrow \left(\frac{{\displaystyle \frac{18}{6}}}{{\displaystyle \frac{24}{6}}}\right)=\frac{3}{4} \end{gathered}$$

Answer:

(i)
$$\begin{gathered} \frac{3}{5}=\frac{9}{} \\ \mathrm{Consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 3\times 3=9, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \frac{3}{5}\times \frac{3}{3}=\frac{9}{15} \end{gathered}$$
(ii)
$$\begin{gathered} \frac{3}{5}=\frac{}{30} \\ \mathrm{Consider} \mathrm{the} \mathrm{denominators}. \\ \mathrm{As} 5\times 6=30, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 6. \\ \Rightarrow \frac{3}{5}\times \frac{6}{6}=\frac{18}{30} \end{gathered}$$

(iii)
$$\begin{gathered} \frac{3}{5}=\frac{21}{} \\ \mathrm{Consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 3\times 7=21, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 7. \\ \Rightarrow \frac{3}{5}\times \frac{7}{7}=\frac{21}{35} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{3}{5}=\frac{}{40} \\ \mathrm{Consider} \mathrm{the} \mathrm{denominators}. \\ \mathrm{As} 5\times 8=40, \mathrm{we} \mathrm{will} \mathrm{multip}\mathrm{ly} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 8. \\ \Rightarrow \frac{3}{5}\times \frac{8}{8}=\frac{24}{40} \end{gathered}$$

Answer:

(i)
$$\begin{gathered} \frac{45}{60}=\frac{15}{\mathrm{Denominator}} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 45÷3=15, \mathrm{we} \mathrm{will} \mathrm{divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \left(\frac{{\displaystyle \frac{45}{3}}}{{\displaystyle \frac{60}{3}}}\right)=\frac{15}{20} \end{gathered}$$

(ii)
$$\begin{gathered} \frac{45}{60}=\frac{\mathrm{Numerator}}{4} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{denominators}. \\ \mathrm{As} 60÷15=4, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 15. \\ \Rightarrow \left(\frac{{\displaystyle \frac{45}{15}}}{{\displaystyle \frac{60}{15}}}\right)=\frac{3}{4} \end{gathered}$$

(iii)
$$\begin{gathered} \frac{45}{60}=\frac{\mathrm{Numerator}}{240} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{denominators}. \\ \mathrm{As} 60\times 4=240, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 4. \\ \Rightarrow \frac{45}{60}\times \frac{4}{4}=\frac{180}{240} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{45}{60}=\frac{135}{\mathrm{Denominator}} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 45\times 3=135, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \frac{45}{60}\times \frac{3}{3}=\frac{135}{180} \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Firstly}, \mathrm{we} \mathrm{will} \mathrm{reduce} \frac{35}{42} \mathrm{into} \mathrm{the} \mathrm{lowest} \mathrm{term}. \\ \mathrm{Now}, \mathrm{we} \mathrm{will} \mathrm{divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 35 \& 42. \\ \Rightarrow \frac{35÷7}{42÷7}=\frac{5}{6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{i}\right) \frac{5}{6}=\frac{15}{\mathrm{Denominator}} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 5\times 3=15, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \frac{5\times 3}{6\times 3}=\frac{15}{18} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\frac{5}{6}=\frac{\mathrm{Numerator}}{18} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{denominators}. \\ \mathrm{As} 6\times 3=18, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 3. \\ \Rightarrow \frac{5\times 3}{6\times 3}=\frac{15}{18} \end{gathered}$$


$$\begin{gathered} \left(iii\right) \frac{5}{6}=\frac{\mathrm{numerator}}{30} \\ We will consider \mathrm{the} \mathrm{denominators} , \\ \mathrm{Since},6\times 5=30,\mathrm{therefore} \mathrm{multiplying} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 5 \\ \Rightarrow \frac{5\times 5}{6\times 5} =\frac{25}{30} \end{gathered}$$


$$\begin{gathered} \left(\mathrm{iv}\right) \frac{5}{6}=\frac{30}{\mathrm{Denominator}} \\ \mathrm{We} \mathrm{will} \mathrm{consider} \mathrm{the} \mathrm{numerators}. \\ \mathrm{As} 5\times 6=30, \mathrm{we} \mathrm{will} \mathrm{multiply} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 6. \\ \Rightarrow \frac{5\times 6}{6\times 6}= \frac{30}{36} \end{gathered}$$

Answer:

(i) 
$\frac{5}{9}\times \frac{6}{6}=\frac{30}{54}$ 
So, the given fractions are equivalent.

(ii) 
$\frac{2}{7}\times \frac{8}{8}=\frac{16}{56}$
$\frac{2}{7 } \mathrm{is} \mathrm{not} \mathrm{equivalent} \mathrm{to} \frac{16}{42}.$

(iii)
 $\frac{7}{13}\times \frac{5}{5}=\frac{35}{65}$
$\frac{5}{11}\times \frac{7}{7}=\frac{35}{77}$
$\frac{7}{13} \mathrm{is} \mathrm{not} \mathrm{equivalent} \mathrm{to} \frac{5}{11}.$

(iv)
$$\begin{gathered} \frac{4}{11}\times \frac{8}{8}=\frac{32}{88} \\ \frac{4}{11} \mathrm{is} \mathrm{equivalent} \mathrm{to} \frac{32}{88}. \end{gathered}$$

(v)
$$\begin{gathered} \frac{3}{10}\times \frac{4}{4}=\frac{12}{40} \\ \frac{3}{10}\mathrm{is} \mathrm{not} \mathrm{equivalent} \mathrm{to} \frac{12}{50}. \end{gathered}$$

(vi)

$$\begin{gathered} \frac{9}{27}=\frac{1}{3} \\ \frac{25}{75}=\frac{1}{3} \\ \frac{9}{27} \mathrm{is} \mathrm{equivalent} \mathrm{to} \frac{25}{75}. \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right)\frac{250}{400} \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 250 \& 400, \mathrm{we} \mathrm{get}: \\ =\left(\frac{{\displaystyle \frac{250}{50}}}{{\displaystyle \frac{400}{50}}}\right) \\ =\frac{5}{8} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) \frac{180}{200} \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 180 \& 200, \mathrm{we} \mathrm{get}: \\ =\left(\frac{{\displaystyle \frac{180}{20}}}{{\displaystyle \frac{200}{20}}}\right) \\ =\frac{9}{10} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) \frac{660}{990} \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 660 \& 990, \mathrm{we} \mathrm{get}: \\ =\left(\frac{{\displaystyle \frac{660}{30}}}{{\displaystyle \frac{990}{30}}}\right) \\ =\frac{\left({\displaystyle \frac{22}{11}}\right)}{\left({\displaystyle \frac{33}{11}}\right)} \\ =\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\frac{180}{360} \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 180 \& 360, \mathrm{we} \mathrm{get}: \\ =\left(\frac{{\displaystyle \frac{180}{180}}}{{\displaystyle \frac{360}{180}}}\right) \\ =\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \frac{220}{550} \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 220 \& 550, \mathrm{we} \mathrm{get}: \\ =\left(\frac{{\displaystyle \frac{220}{11}}}{{\displaystyle \frac{550}{11}}}\right) \\ =\left(\frac{20}{50}\right) \\ =\left(\frac{{\displaystyle \frac{20}{10}}}{{\displaystyle \frac{50}{10}}}\right) \\ =\frac{2}{5} \end{gathered}$$

(i) - (d)
(ii) - (e)
(iii) - (a)
(iv) -(c)
(v)- (b)

Answer:

$$\begin{gathered} \frac{2}{4}=\frac{3}{6}=\frac{79}{158} \\ \frac{3}{9}=\frac{2}{6}=\frac{58}{174} \end{gathered}$$

Answer:

Total pencils Ravish had = 20
Pencils used by Ravish = 10
Fraction of pencils used by ravish = $\frac{10÷10}{20÷10}=\frac{1}{2} (\mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 10 \& 20)$

Total pencils Shikha had = 50
Pencils used by Shikha = 25
Fraction of pencils used by Shikha = $\frac{25÷25}{50÷25}=\frac{1}{2} (\mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 25 \& 50)$

Total pencils Priya had = 80
Pencils used by Priya = 40
Fraction of pencils used by Priya = $\frac{40÷40}{80÷40}$=$\frac{1}{2} (\mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCFs} \mathrm{of} 40 \& 80)$   

Yes, each of them has utilised an equal fraction of pencils.

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{40}{75} \\ \mathrm{Factors} \mathrm{of} 40 \mathrm{are} 1, 2, 4, 5, 8, 10, 20 \mathrm{and} 40. \\ \mathrm{Factors} \mathrm{of} 75 \mathrm{are} 1, 3, 5, 15 \mathrm{and} 75. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 40 \& 75 \mathrm{are} 1 \& 5. \\ \mathrm{HCF} = 5 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 5. \\ \frac{40÷5}{75÷5}=\frac{8}{15} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \frac{42}{ 28} \\ \mathrm{Factors} \mathrm{of} 42\mathrm{are}1,2,3,6,7,14,21,42 \\ \mathrm{Factors} \mathrm{of} 28 \mathrm{are}1,2,4,7,14,28 \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 42 \& 28 \mathrm{are} 1,2,7,14 \\ \mathrm{HCF}=14 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 14 \\ \frac{42÷14}{ 28÷14}=\frac{3}{2} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \frac{12}{52} \\ \mathrm{Factors} \mathrm{of} 12 \mathrm{are} 1, 2, 3, 4, 6 \mathrm{and} 12. \\ \mathrm{Factors} \mathrm{of} 52 \mathrm{are} 1, 2, 4, 13, 26 \mathrm{and} 52. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 12 \& 52 \mathrm{are} 1, 2, \mathrm{and} 4. \\ \mathrm{HCF}= 4 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 4. \\ \frac{12÷4}{52÷4}=\frac{3}{13} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iv}\right)\frac{40}{72} \\ \mathrm{Factors} \mathrm{of} 40 \mathrm{are} 1,2,4,5,8,10,20,40 \\ \mathrm{Factors} \mathrm{of} 72 \mathrm{are} 1,2,3,4,6,8,9,12,18,24,36,72 \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 40 \& 72 \mathrm{are} 1,2,4,8 \\ \mathrm{HCF}=8 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 8 \\ \frac{40÷8}{72÷8}=\frac{5}{9} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{v}\right) \frac{80}{24} \\ \mathrm{Factors} \mathrm{of} 80 \mathrm{are} 1, 2, 4, 5, 8, 10, 16, 20, 40 \mathrm{and} 80. \\ \mathrm{Factors} \mathrm{of} 24 \mathrm{are} 1, 2, 3, 4, 6, 8, 12 \mathrm{and} 24. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 80 \& 24 \mathrm{are} 1, 2, 4 \mathrm{and} 8. \\ \mathrm{HCF}=8 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 4. \\ \frac{80÷8}{24÷8}=\frac{10}{3} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{vi}\right) \frac{84}{56} \\ \mathrm{Factors} \mathrm{of} 84 \mathrm{are} 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 \mathrm{and} 84. \\ \mathrm{Factors} \mathrm{of} 56 \mathrm{are} 1, 2, 4, 7, 8, 14, 28 \mathrm{and} 56. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 84 \& 56 \mathrm{are} 1, 2, 4, 7, 14 \mathrm{and} 28. \\ \mathrm{HCF}=28 \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 28. \\ \frac{84÷28}{56÷25}=\frac{3}{2} \end{gathered}$$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{75}{80} \\ \mathrm{Factors} \mathrm{of} 75 \mathrm{are} 1, 3, 5, 15, 25 \mathrm{and} 75. \\ \mathrm{Factors} \mathrm{of} 80 \mathrm{are} 1, 2, 4, 5, 8, 10, 16, 20, 40 \mathrm{and} 80. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 75 \& 80 \mathrm{are} 1 \& 5. \\ \mathrm{HCF} \mathrm{of} 75 \& 80 \mathrm{is} 5. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} 5, \mathrm{we} \mathrm{get}: \\ \frac{75÷5}{80÷5}=\frac{15}{16} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \frac{52}{76} \\ \mathrm{Factors} \mathrm{of} 52 \mathrm{are} 1, 2, 4, 13, 26 \mathrm{and} 52. \\ \mathrm{Factors} \mathrm{of} 76 \mathrm{are} 1, 2, 4, 19, 38 \mathrm{and} 76. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 52 \& 76 \mathrm{are} 1, 2 \mathrm{and} 4. \\ \mathrm{HCF} \mathrm{of} 52 \mathrm{and} 76 \mathrm{is} 4. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} 4, \mathrm{we} \mathrm{get}: \\ \frac{52÷4}{76÷4}=\frac{13}{19} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \frac{84}{98} \\ \mathrm{Factors} \mathrm{of} 84 \mathrm{are} 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 \mathrm{and} 84. \\ \mathrm{Factors} \mathrm{of} 98 \mathrm{are} 1, 2, 7, 14, 49 \mathrm{and} 98. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 84 \mathrm{and} 98 \mathrm{are} 1, 2, 7 \mathrm{and} 14. \\ \mathrm{HCF} \mathrm{of} 84 \mathrm{and} 98 \mathrm{is} 14. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} 14, \mathrm{we} \mathrm{get}: \\ \frac{84÷14}{98÷14}=\frac{6}{7} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \frac{68}{17} \\ \mathrm{Factors} \mathrm{of} 68 \mathrm{are} 1, 2, 4, 17, 34 \mathrm{and} 68. \\ \mathrm{Factors} \mathrm{of} 17 \mathrm{are} 1 \mathrm{and} 17. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 68 \mathrm{and} 17 \mathrm{are} 1 \mathrm{and} 17. \\ \mathrm{HCF} \mathrm{of} 68 \mathrm{and} 17 \mathrm{is} 17. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} 17, \mathrm{we} \mathrm{get}: \\ \frac{68÷17}{17÷17}=\frac{4}{1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \frac{150}{50} \\ \mathrm{Factors} \mathrm{of} 150 \mathrm{are} 1, 2, 3, 5, 6, 10, 15, 25, 50 \mathrm{and} 150. \\ \mathrm{Factors} \mathrm{of} 50 \mathrm{are} 1, 2, 5, 10, 25 \mathrm{and} 50. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 150 \mathrm{and} 50 \mathrm{are} 1, 2, 5, 10, 25 \mathrm{and} 50. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} 50, \mathrm{we} \mathrm{get}: \\ \frac{150÷50}{50÷50}=\frac{3}{1} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right) \frac{162}{108} \\ \mathrm{Factors} \mathrm{of} 162 \mathrm{are} 1, 2, 3, 6, 9, 18, 27, 54, 81 \mathrm{and} 162. \\ \mathrm{Factors} \mathrm{of} 108 \mathrm{are} 1, 2, 3, 4, 6, 9, 12, 18, 27 \mathrm{and} 54. \\ \mathrm{Common} \mathrm{factors} \mathrm{of} 162 \mathrm{and} 108 \mathrm{are} 1, 2, 3, 6, 9, 18, 27 \mathrm{and} 54. \\ \mathrm{HCF} \mathrm{of} 162 \mathrm{and} 108 \mathrm{is} 54. \\ \mathrm{Dividing} \mathrm{both} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator} \mathrm{by} 54, \mathrm{we} \mathrm{get}: \\ \frac{162÷54}{108÷54}=\frac{3}{2} \end{gathered}$$

Answer:

(i) Ascending order



Descending order


(ii) Ascending order


Descending order


(iii) Ascending order


Descending order

Answer:

(i) $\frac{5}{6}>\frac{2}{6}$ because 5 > 2 and the denominator is the same.
(ii) $\frac{3}{6}>\frac{0}{6}$ because 3 > 0 and the denominator is the same.
(iii) $\frac{1}{6}<\frac{6}{6}$ because 6 > 1 and the denominator is the same.
(iv) $\frac{8}{6}>\frac{5}{6}$ because 8 > 5 and the denominator is the same.

Answer:

(i) $\frac{3}{6}<\frac{5}{6}$ because 3 < 5 and the denominator is the same.
(ii) $\frac{4}{5}>\frac{0}{5}$ because 4 > 0 and the denominator is the same.
(iii) $\frac{3}{20}<\frac{4}{20}$ because 3 < 4 and the denominator is the same.
(iv) $\frac{1}{7}<\frac{1}{4}$ because 7 > 4; if the numerator is the same, then the fraction that has smaller denominator is greater.

Answer:

(i) $\frac{6}{7}>\frac{6}{11}$ because if the numerator is the same, then the fraction with smaller denominator is greater.
(ii) $\frac{3}{7}<\frac{5}{7}$ because 3 < 5 and the denominator is the same.

(iii)
$$\begin{gathered} \frac{8}{12}=\frac{2\times 2\times 2}{2\times 2\times 3}=\frac{2}{3} \\ so \frac{2}{3}=\frac{8}{12} \end{gathered}$$

(iv)
$$\begin{gathered} \frac{1}{5}=\frac{1}{5}\times \frac{3}{3}=\frac{3}{15} \\ \frac{3}{15}<\frac{4}{15} \end{gathered}$$ (Because 3 < 4 and the denominator is the same. Therefore, $\frac{1}{5}<\frac{4}{15}$.)

(v)
$\frac{8}{3}>\frac{8}{13}$  because if the numerator is the same, then the fraction with smaller denominator is greater. 

(vi)
 $$\begin{gathered} \frac{4}{9}=\frac{4}{9}\times \frac{8}{8}=\frac{32}{72} \\ \frac{15}{8}=\frac{15}{8}\times \frac{9}{9}=\frac{135}{72} \\ \frac{32}{72}<\frac{135}{72} \end{gathered}$$ (Because 135 > 32 and the denominator is the same)

∴$\frac{4}{9}<\frac{15}{8}$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2}{12} \\ \mathrm{HCF} \mathrm{of} 2 \& 12 \mathrm{is} 2. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 2 \&12. \\ \Rightarrow \frac{2÷2}{12÷2}=\frac{1}{6} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{ii}\right) \frac{3}{15} \\ \mathrm{HCF} \mathrm{of} 3 \&15 \mathrm{is} 3. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 3 \&15. \\ \Rightarrow \frac{3÷3}{15÷3}=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \frac{8}{50} \\ \mathrm{HCF} \mathrm{of} 8 \& 50 \mathrm{is} 2. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 8 \& 50. \\ \Rightarrow \frac{8÷2}{50÷2}=\frac{4}{25} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \frac{16}{100} \\ \mathrm{HCF} \mathrm{of} 16 \&100 \mathrm{is} 4. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \&\mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 16 \& 100. \\ \Rightarrow \frac{16÷4}{100÷4}=\frac{4}{25} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \frac{10}{60} \\ \mathrm{HCF} \mathrm{of} 10 \& 60 \mathrm{is} 10. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 10 \& 60. \\ \Rightarrow \frac{10÷10}{60÷10}=\frac{1}{6} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right) \frac{15}{75} \\ \mathrm{HCF} \mathrm{of} 15 \& 75 \mathrm{is} 15. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 15 \& 75. \\ \Rightarrow \frac{15÷15}{75÷15}=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vii}\right) \frac{12}{60} \\ \mathrm{HCF} \mathrm{of} 12 \& 60 \mathrm{is} 12. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 12 \& 60. \\ \Rightarrow \frac{12÷12}{60÷12}=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{viii}\right) \frac{16}{96} \\ \mathrm{HCF} \mathrm{of} 16 \& 96 \mathrm{is} 16. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 16 \& 96. \\ \Rightarrow \frac{16÷16}{96÷16}=\frac{1}{6} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ix}\right) \frac{12}{75} \\ \mathrm{HCF} \mathrm{of} 12 \& 75 \mathrm{is} 3. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 12 \& 75. \\ \Rightarrow \frac{12÷3}{75÷3}=\frac{4}{25} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{x}\right) \frac{12}{72} \\ \mathrm{HCF} \mathrm{of} 12 \& 72 \mathrm{is} 12. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 12 \& 72. \\ \Rightarrow \frac{12÷12}{72÷12}=\frac{1}{6} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xi}\right) \frac{3}{18} \\ \mathrm{HCF} \mathrm{of} 3 \& 18 \mathrm{is} 3. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 3 \& 18. \\ \Rightarrow \frac{3÷3}{18÷3}=\frac{1}{6} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{xii}\right) \frac{4}{25} \\ \mathrm{HCF} \mathrm{of} 4 \& 25 \mathrm{is} 1. \\ \mathrm{Divide} \mathrm{both} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 4 \& 25. \\ \Rightarrow \frac{4÷1}{25÷1}=\frac{4}{25} \end{gathered}$$

$\mathrm{Three} \mathrm{groups} \mathrm{of} \mathrm{equal} \mathrm{fractions}:\frac{2}{12},\frac{10}{60},\frac{16}{96},\frac{12}{72},\frac{3}{18}; \frac{3}{15},\frac{15}{75},\frac{12}{60}; \frac{8}{50},\frac{16}{100},\frac{12}{75},\frac{4}{25}$

Answer:

Total pages in the book = 100
Fraction of the book read by Isha = $\frac{25÷25}{100÷25}=\frac{1}{4} (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 25 \& 100)$
Fraction of the book read by Nagma = $\frac{1}{2}$
Now, compare $\frac{1}{4} \& \frac{1}{2}$.
LCM of 4 & 2 is 4.

Convert each fraction into equivalent fraction with 4 as its denominator.

$$\begin{gathered} \frac{1\times 1}{4\times 1} \&\frac{1\times 2}{2\times 2} \\ \frac{1}{4 } \&\frac{2}{4} \\ \frac{1}{4}<\frac{2}{4} \end{gathered}$$
Therefore, Isha read less.




 

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \frac{2}{9},\frac{7}{9},\frac{3}{9},\frac{4}{9},\frac{1}{9},\frac{6}{9},\frac{5}{9} \\ \mathrm{When} \mathrm{denominators} \mathrm{are} \mathrm{the} \mathrm{same} \& \mathrm{numerators} \mathrm{are} \mathrm{different}, \mathrm{then} \mathrm{the} \mathrm{fraction} \mathrm{with} \mathrm{greater} \mathrm{numerator} \mathrm{has} \mathrm{a} \mathrm{larger} \mathrm{value}. \\ \left(\mathrm{ii}\right) \frac{7}{8},\frac{7}{25},\frac{7}{11},\frac{7}{18},\frac{7}{10} \\ \mathrm{When} \mathrm{numerators} \mathrm{are} \mathrm{the} \mathrm{same} \& \mathrm{denominators} \mathrm{are} \mathrm{different}, \mathrm{then} \mathrm{the} \mathrm{fraction} \mathrm{with} \mathrm{smaller} \mathrm{denominator} \mathrm{has} \mathrm{a} \mathrm{smaller} \mathrm{value}. \\ \frac{7}{25}<\frac{7}{18}<\frac{7}{11}<\frac{7}{10}<\frac{7}{8} \\ \left(\mathrm{iii}\right) \frac{37}{47},\frac{37}{50},\frac{37}{100},\frac{37}{1000},\frac{37}{85},\frac{37}{41} \\ \mathrm{When} \mathrm{numerators} \mathrm{are} \mathrm{the} \mathrm{same} \& \mathrm{denominators} \mathrm{are} \mathrm{different}, \mathrm{then} \mathrm{the} \mathrm{fraction} \mathrm{with} \mathrm{greater} \mathrm{denominator} \mathrm{has} \mathrm{a} \mathrm{smaller} \mathrm{value}. \\ \frac{37}{1000}<\frac{37}{100}<\frac{37}{85}<\frac{37}{50}<\frac{37}{47}<\frac{37}{41} \\ \left(\mathrm{iv}\right) \frac{3}{5},\frac{1}{5},\frac{4}{5},\frac{2}{5} \\ \mathrm{When} \mathrm{denominators} \mathrm{are} \mathrm{the} \mathrm{same} \& \mathrm{numerators} \mathrm{are} \mathrm{different}, \mathrm{then} \mathrm{the} \mathrm{fraction} \mathrm{with} \mathrm{greater} \mathrm{numerator} \mathrm{has} \mathrm{a} \mathrm{larger} \mathrm{value}. \\ \frac{1}{5}<\frac{2}{5}<\frac{3}{5}<\frac{4}{5} \end{gathered}$$

(v)
$$\begin{gathered} \mathrm{LCM} \mathrm{of} 2, 4 \mathrm{and} 5 \mathrm{is} 20. \\ \frac{2}{5}=\frac{2}{5}\times \frac{4}{4}=\frac{8}{20} \\ \frac{3}{4}=\frac{3}{4}\times \frac{5}{5}=\frac{15}{20} \\ \frac{1}{2}=\frac{1}{2}\times \frac{10}{10}=\frac{10}{20} \\ \frac{3}{5}=\frac{3}{5}\times \frac{4}{4}=\frac{12}{20} \end{gathered}$$

$\frac{2}{5}<\frac{1}{2}<\frac{3}{5}<\frac{3}{4}$
(vi)$$\begin{gathered} \frac{3}{8},\frac{3}{12},\frac{3}{6},\frac{3}{4} \\ \mathrm{When} \mathrm{numerators} \mathrm{are} \mathrm{the} \mathrm{same} \& \mathrm{denominators} \mathrm{are} \mathrm{different}, \mathrm{then} \mathrm{the} \mathrm{fraction} \mathrm{with} \mathrm{smaller} \mathrm{denominator} \mathrm{has} \mathrm{a} \mathrm{greater} \mathrm{value}. \\ \frac{3}{12}<\frac{3}{8}<\frac{3}{6}<\frac{3}{4} \end{gathered}$$
(vii)
$$\begin{gathered} \frac{4÷2}{6÷2}=\frac{2}{3} (\mathrm{Dividing} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 4 \& 6) \\ \frac{6}{12}=\frac{1}{2} (\mathrm{Dividing} \mathrm{the} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{the} \mathrm{HCF} \mathrm{of} 6 \& 12) \\ \mathrm{LCM} \mathrm{of} 2, 8, 3 \mathrm{and} 16 \mathrm{is} 48. \\ \frac{4}{6}=\frac{2}{3}\times \frac{16}{16}=\frac{32}{48} \\ \frac{6}{12}=\frac{1}{2}\times \frac{24}{24}=\frac{24}{48} \\ \frac{3}{8}=\frac{3}{8}\times \frac{6}{6}=\frac{18}{48} \\ \frac{5}{16}=\frac{5}{16}\times \frac{3}{3}=\frac{15}{48} \end{gathered}$$

When denominators are the same & numerators are different, then the fraction with greater numerator has a greater value.

$\frac{5}{16}<\frac{3}{8}<\frac{6}{12}<\frac{4}{6}$

Answer:

When numerators are the same and denominators are different, then the fraction with greater denominator has a smaller value.

When denominators are the same and numerators are different, then the fraction with greater numerator has a larger value.

(i)$\frac{8}{5}>\frac{8}{9}>\frac{8}{13}>\frac{8}{17}$
(ii) $\mathrm{LCM} \mathrm{of} 9,12, 3 \mathrm{and} 15 \mathrm{is} 180.$
 $\frac{5}{9}\times \frac{20}{20}=\frac{100}{180}$
$\frac{3}{12}\times \frac{15}{15}=\frac{45}{180}$
$\frac{1}{3}\times \frac{60}{60}=\frac{60}{180}$
$\frac{4}{15}\times \frac{12}{12}=\frac{48}{180}$
$\frac{5}{9}>\frac{1}{3}>\frac{4}{15}>\frac{3}{12}$
(iii)$\frac{9}{14}>\frac{13}{28}>\frac{11}{35}>\frac{2}{7}$

Answer:

Two fractions are equal when:

Numerator of the first fraction $\times$ Denominator of the second fraction = Numerator of the second fraction $\times$ Denominator of the first fraction
(i)
$$\begin{gathered} 5\times 5=25 \\ 4\times 9=36 \\ \mathrm{So}, \\ 5\times 5\ne 4\times 9 \\ \frac{5}{9}\mathrm{is} \mathrm{not} \mathrm{equal} \mathrm{to} \frac{4}{5}. \end{gathered}$$
(ii)
$$\begin{gathered} 9\times 9=81 \\ 5\times 16=80 \\ \mathrm{So}, \\ 9\times 9\ne 5\times 16 \\ \frac{9}{16} \mathrm{is} \mathrm{not} \mathrm{equal} \mathrm{to} \frac{5}{9}. \end{gathered}$$

(iii)
$$\begin{gathered} 4\times 20=80 \\ 16\times 5=80 \\ \mathrm{So}, \\ 4\times 20=16\times 5 \\ \frac{4}{5} \mathrm{is} \mathrm{equal} \mathrm{to} \frac{16}{20}. \end{gathered}$$
(iv)
$$\begin{gathered} 1\times 30=30 \\ 4\times 15=60 \\ \mathrm{So}, \\ 1\times 30\ne 4\times 15 \\ \frac{1}{15} \mathrm{is} \mathrm{not} \mathrm{equal} \mathrm{to} \frac{4}{30}. \end{gathered}$$

Answer:

(i) The given fractions are $\frac{1}{5} \& \frac{2}{5}$

$$\begin{gathered} \frac{1}{5}+\frac{2}{5}=\frac{3}{5} \\ \frac{1+2}{5}=\frac{3}{5} \\ \frac{3}{5}=\frac{3}{5} \end{gathered}$$

(ii) The given fractions are $\frac{3}{6} \& \frac{2}{6}$

$$\begin{gathered} \frac{3}{6}+\frac{2}{6}=\frac{5}{6} \\ \frac{3+2}{6}=\frac{5}{6} \\ \frac{5}{6}=\frac{5}{6} \end{gathered}$$

Answer:

(i)$$\begin{gathered} \frac{5}{12}+\frac{1}{12}=\frac{5+1}{12} \\ =\frac{6÷6}{12÷6}=\frac{1}{2} (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$

(ii)$$\begin{gathered} \frac{3}{15}+\frac{7}{15}=\frac{3+7}{15} \\ =\frac{10÷5}{15÷5}=\frac{2}{3} (\mathrm{Dividing} \mathrm{numerator} \hspace{0.17em}\& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$

(iii)$$\begin{gathered} \frac{3}{22}+\frac{7}{22}=\frac{3+7}{22} \\ =\frac{10÷2}{22÷2}=\frac{5}{11} (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$

(iv)$\frac{1}{4}+\frac{0}{4}=\frac{1+0}{4}=\frac{1}{4}$

(v)$\frac{4}{13}+\frac{2}{13}+\frac{1}{13}=\frac{4+2+1}{13}=\frac{7}{13}$

(vi)$$\begin{gathered} \frac{0}{15}+\frac{2}{15}+\frac{1}{15}=\frac{0+2+1}{15} \\ =\frac{3÷3}{15÷3}=\frac{1}{5} (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$

(vii)$\frac{7}{31}-\frac{4}{31}+\frac{9}{31}=\frac{7-4+9}{31}=\frac{12}{31}$

(viii)$3\frac{2}{7}+\frac{1}{7}-2\frac{3}{7}=\frac{23}{7}+\frac{1}{7}-\frac{17}{7}=\frac{23+1-17}{7}=\frac{7}{7}=1$
(ix)$$\begin{gathered} 2\frac{1}{3}-1\frac{2}{3}+4\frac{1}{3}=\frac{7}{3}-\frac{5}{3}+\frac{13}{3} \\ =\frac{7-5+13}{3}=\frac{15÷3}{3÷3}=5 (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$

(x)$1-\frac{2}{3}+\frac{7}{3}=\frac{3-2+7}{3}=\frac{8}{3}$

(xi)$\frac{16}{7}-\frac{5}{7}+\frac{9}{7}=\frac{16-5+9}{7}=\frac{20}{7}$

Answer:

Shikha painted $\frac{1}{5}$ of the wall space.
Ravish painted $\frac{3}{5}$ of the wall space.
Wall space painted by both of them together  = $\frac{1}{5}+\frac{3}{5}=\frac{1+3}{5}=\frac{4}{5}$
Unpainted part of the room = $1-\frac{4}{5}=\frac{5-4}{5}=\frac{1}{5}$

Answer:

 Quantity of sugar bought by Ramesh = $2\frac{1}{2}\mathrm{kg}=\frac{(2\times 2)+1}{2}=\frac{5}{2}\mathrm{kg}$
 Quantity of sugar bought by Rohit = $3\frac{1}{2}\mathrm{kg}=\frac{(3\times 2)+1}{2}=\frac{7}{2}\mathrm{kg}$
Total amount of sugar bought by them =

$$\begin{gathered} \frac{5}{2}+\frac{7}{2}=\frac{5+7}{2} \\ =\frac{12÷2}{2÷2}=6 \mathrm{kg} (\mathrm{Dividing} \mathrm{numerator} \& \mathrm{denominator} \mathrm{by} \mathrm{their} \mathrm{HCF}) \end{gathered}$$