\begin{array}{l}
\text {Q1. Compare the following fractions by using the symbol } \gt \text { or } \lt \text { or } = \\\text {(i) } \quad \frac{7}{9} \text { and } \frac{8}{13} \\
\text {(ii) } \quad \frac{11}{9} \text { and } \frac{5}{9} \\
\text {(iii) } \quad \frac{37}{41} \text { and } \frac{19}{30} \\
\text {(iv) } \quad \frac{17}{15} \text { and } \frac{119}{105}
\end{array}\begin{array}{l}
\text {Sol. } \\\text {(i) } \quad \frac{7}{9} \text { and } \frac{8}{13} \\
\text {LCM of the denominators 9 and 13 is 117.} \\
\text {Now, convert the given fractions to equivalent fractions with denominator as 117.}
\text {We have, } \\\frac{7}{9} = \frac{7 \times 13}{9 \times 13} = \frac{91}{117} \\
\frac{8}{13} = \frac{8 \times 9}{13 \times 9} = \frac{72}{117} \\
\text {We know that } 91 \gt 72 \\
\Rightarrow \quad \frac{91}{117} \gt \frac{72}{117} \\
\Rightarrow \quad \frac{7}{9} \gt \frac{8}{13} \\ \\\text {(ii) } \quad \frac{11}{9} \text { and } \frac{5}{9} \\\text {The given fractions have equal denominators.} \\\text {We know that } 11 \gt 5 \\
\Rightarrow \quad \frac{11}{9} \gt \frac{5}{9} \\ \\\text {(iii) } \quad \frac{37}{41} \text { and } \frac{19}{30} \\\text {LCM of the denominators 41 and 30 is 1230.} \\
\text {Now, convert the given fractions to equivalent fractions with denominator as 1230.} \\
\text {We have, } \\\frac{37}{41} = \frac{37 \times 30}{41 \times 30} = \frac{1110}{1230} \\
\frac{19}{30} = \frac{19 \times 41}{30 \times 41} = \frac{779}{1230} \\\text {We know that } 1110 \gt 779 \\
\Rightarrow \quad \frac{1110}{1230} \gt \frac{779}{1230} \\
\Rightarrow \quad \frac{37}{41} \gt \frac{19}{30} \\ \\\text {(iv) } \quad \frac{17}{15} \text { and } \frac{119}{105} \\\text {LCM of the denominators 15 and 105 is 105.} \\
\text {Now, convert the given fractions to equivalent fractions with denominator as 105.}
\text {We have, } \\\frac{17}{15}= \frac{17 \times 7}{15 \times 7} = \frac{119}{105} \\\Rightarrow \quad \frac{17}{15} = \frac{119}{105} \\\end{array}\begin{array}{l}
\text {Q2. Arrange the following fractions in ascending order: } \\
\text {(i) } \quad \frac{3}{8}, \frac{5}{6}, \frac{6}{8}, \frac{2}{4}, \frac{1}{3} \\
\text {(ii) } \quad \frac{4}{6}, \frac{3}{8}, \frac{6}{12}, \frac{5}{16}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{3}{8}, \frac{5}{6}, \frac{6}{8}, \frac{2}{4}, \frac{1}{3} \\
\end{array}\begin{array}{c|c}
2 & 8 \quad 6 \quad 4 \quad 3 \\
\hline
2 & 4 \quad 3 \quad 2 \quad 3 \\
\hline
2 & 2 \quad 3 \quad 1 \quad 3 \\
\hline
3 & 1 \quad 3 \quad 1 \quad 3 \\
\hline
& 1 \quad 1 \quad 1 \quad 1 \\
\end{array}\begin{array}{l}
\therefore \text {LCM of denominators 8,6,4,3 is } 2 \times 2 \times 2 \times 3 = 24 \\\text {Now, convert the given fractions to equivalent fractions with denominator as 24.}
\text {We have, } \\\frac{3}{8} = \frac{3 \times 3}{8 \times 3} =\frac{9}{24} \\\frac{5}{6}= \frac{5 \times 4}{6 \times 4}=\frac{20}{24} \\
\frac{6}{8}=\frac{6 \times 3}{8 \times 3}= \frac{18}{24} \\
\frac{2}{4}= \frac{2 \times 6}{4 \times 6} = \frac{12}{24} \\
\frac{1}{3}= \frac{1 \times 8}{3 \times 8}=\frac{8}{24} \\\text {We know that } 8 \lt 9 \lt 12 \lt 18 \lt 20 \\
\Rightarrow \quad \frac{8}{24} \lt \frac{9}{24} \lt \frac{12}{24} \lt \frac{18}{24} \lt \frac{20}{24} \\
\Rightarrow \quad \frac{1}{3} \lt \frac{3}{8} \lt \frac{2}{4} \lt \frac{6}{8} \lt \frac{5}{6} \\ \\
\end{array}\begin{array}{l}
\text {(ii) } \quad \frac{4}{6}, \frac{3}{8}, \frac{6}{12}, \frac{5}{16}
\end{array}\begin{array}{c|c}
2 & 6 \quad 8 \quad 12 \quad 16 \\
\hline
2 & 3 \quad 4 \quad 6 \quad 8 \\
\hline
2 & 3 \quad 2 \quad 3 \quad 4 \\
\hline
2 & 3 \quad 1 \quad 3 \quad 2 \\
\hline
3 & 3 \quad 1 \quad 3 \quad 1 \\
\hline
& 1 \quad 1 \quad 1 \quad 1 \\
\end{array}\begin{array}{l}
\therefore \text {LCM of denominators 6,8,12 and 16 is } 2 \times 2 \times 2 \times 2 \times 3 = 48 \\\text {Now, convert the given fractions to equivalent fractions with denominator as 48.}
\text {We have, } \\\frac{4}{6} = \frac{4 \times 8}{6 \times 8} =\frac{32}{48} \\\frac{3}{8}= \frac{3 \times 6}{8 \times 6}=\frac{18}{48} \\
\frac{6}{12}=\frac{6 \times 4}{12 \times 4}= \frac{24}{48} \\
\frac{5}{16}= \frac{5 \times 4}{16 \times 4} = \frac{20}{48} \\\text {We know that } 18 \lt 20 \lt 24 \lt 32 \\
\Rightarrow \quad \frac{18}{48} \lt \frac{20}{48} \lt \frac{24}{48} \lt \frac{32}{48} \\
\Rightarrow \quad \frac{3}{8} \lt \frac{5}{16} \lt \frac{6}{12} \lt \frac{4}{6} \\ \\\end{array}\begin{array}{l}
\text {Q3. Arrange the following fractions in descending order: } \\
\text {(i) } \quad \frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20} \\
\text {(ii) } \quad \frac{2}{7}, \frac{11}{35}, \frac{9}{14}, \frac{13}{28}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{4}{5}, \frac{7}{10}, \frac{11}{15}, \frac{17}{20} \\\therefore \text {LCM of denominators 5,10,15 and 20 is 60.} \\\text {Now, convert the given fractions to equivalent fractions with denominator as 60.}
\text {We have, } \\\frac{4}{5} = \frac{4 \times 12}{5\times 12} =\frac{48}{60} \\
\frac{7}{10} = \frac{7 \times 6}{10\times 6} =\frac{42}{60} \\
\frac{11}{15} = \frac{11 \times 4}{15\times 4} =\frac{44}{60} \\
\frac{17}{20} = \frac{17 \times 3}{20\times 3} =\frac{54}{60} \\\text {We know that } 54 \gt 48 \gt 44 \gt 42\\
\Rightarrow \quad \frac{54}{60} \gt \frac{48}{60} \gt \frac{44}{60} \gt \frac{42}{60} \\
\Rightarrow \quad \frac{17}{20} \gt \frac{4}{5} \gt \frac{11}{15} \gt \frac{7}{10} \\ \\\end{array}\begin{array}{l}
\text {(ii) } \quad \frac{2}{7}, \frac{11}{35}, \frac{9}{14}, \frac{13}{28} \\\therefore \text {LCM of denominators 7,35,14 and 28 is 140.} \\\text {Now, convert the given fractions to equivalent fractions with denominator as 140.}
\text {We have, } \\\frac{2}{7} = \frac{2 \times 20}{7\times 20} =\frac{40}{140} \\\frac{11}{35} = \frac{11 \times 4}{ 35 \times 4} =\frac{44}{140} \\\frac{9}{14} = \frac{9 \times 10}{ 14 \times 10} =\frac{90}{140} \\\frac{13}{28} = \frac{14 \times 5}{ 28 \times 5} =\frac{70}{140} \\\text {We know that } 90 \gt 70 \gt 44 \gt 40\\
\Rightarrow \quad \frac{90}{140} \gt \frac{70}{140} \gt \frac{44}{140} \gt \frac{40}{140} \\
\Rightarrow \quad \frac{9}{14} \gt \frac{13}{28} \gt \frac{11}{35} \gt \frac{2}{7} \\ \\\end{array}\begin{array}{l}
\text {Q4. Write five equivalent fractions of } \frac{3}{5}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {We can get equivalent fractions by multiplying numerator and denominator by same same i.e } \\
\frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{3 \times 3}{5 \times 3} = \frac{3 \times 4}{5 \times 4} =\frac{3 \times 5}{5 \times 5} = \frac{3 \times 6}{5 \times 6} \\ \\\Rightarrow \quad \frac{3}{5} = \frac{6}{10} = \frac{9}{15} = \frac{12}{20} =\frac{15}{2} = \frac{18}{30} \\ \\\end{array}\begin{array}{l}
\text {Q5. Find the sum: } \\
\text {(i) } \quad \frac{5}{8}+\frac{3}{10} \\
\text {(ii) } \quad 4 \frac{3}{4}+9 \frac{2}{5} \\
\text {(iii) } \quad \frac{5}{6}+3+\frac{3}{4} \\
\text {(iv) } \quad 2 \frac{3}{5}+4 \frac{7}{10}+2 \frac{4}{15}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{5}{8}+\frac{3}{10} \\
\text {LCM of 8 and 10 is 40.}
\therefore \frac{5}{8}+\frac{3}{10} = \frac{5 \times 5}{8 \times 5}+\frac{3 \times 4}{10 \times 4} \\
= \frac{25}{40}+\frac{12}{40} \\
= \frac{25+12}{40} \\
= \frac{37}{40} \\ \\\text {(ii) } \quad 4 \frac{3}{4}+9 \frac{2}{5} \\
= 4 + \frac{3}{4} + 9 + \frac{2}{5} \\
\text {LCM of 4 and 5 is 20.} \\
\therefore 4 + \frac{3}{4} + 9 + \frac{2}{5} \\
= 4 + 9 + \frac{3 \times 5}{4 \times 5} + \frac{2 \times 4}{5 \times 4} \\
= 13 + \frac{15+8}{20} \\
= 13 + \frac{23}{20} \\
= 13 + 1 \frac{3}{20} \\
= 14 \frac{3}{20} \\ \\\text {(iii) } \quad \frac{5}{6}+3+\frac{3}{4} \\\text {LCM of 6 and 4 is 12.} \\
\therefore \frac{5}{6}+3+\frac{3}{4} = 3 + \frac{5 \times 2}{6 \times 2} +\frac{3 \times 3}{4 \times 3 } \\
=3 + \frac{10+9}{12} \\
=3 + \frac{19}{12} \\
=3 + 1\frac{7}{12} \\
=4\frac{7}{12} \\ \\\text {(iv) } \quad 2 \frac{3}{5}+4 \frac{7}{10}+2 \frac{4}{15} \\
\text {LCM of 5, 10 and 15 is 30.} \\
\therefore 2 + \frac{3}{5}+4 + \frac{7}{10}+2 + \frac{4}{15} \\
= 2 + 4 + 2 + \frac{3 \times 6 }{5 \times 6} + \frac{7 \times 3}{10 \times 3} + \frac{4 \times 2}{15 \times 2} \\
= 8 + \frac{18 +21 +8}{30} \\
= 8 + \frac{47}{30} \\
= 8 + 1\frac{17}{30} \\
= 9\frac{17}{30} \\
\end{array}\begin{array}{l}
\text {Q6. Find the difference of } \\
\text {(i) } \quad \frac{13}{24} \text { and } \frac{7}{16} \\
\text {(ii) } \quad \quad 6 \text { and } \frac{23}{3} \\
\text {(iii) } \quad \frac{21}{25} \text { and } \frac{18}{20} \\
\text {(iv) } \quad 3 \frac{3}{10} \text { and } 2 \frac{7}{15}\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{13}{24} – \frac{7}{16} \\
\text {LCM of 24 and 16 is 48.} \\
\therefore \frac {13 \times 2 – 7 \times 3}{48} \\
= \frac {26 – 21}{48} \\
= \frac {5}{48} \\ \\\text {(ii) } \quad \quad 6 – \frac{23}{3} \\
\text {LCM of 1 and 3 is 3.} \\
\therefore \frac {6 \times 3 – 23}{3} = \frac {18-23}{3} = \frac {-5}{3} \\ \\\text {(iii) } \quad \frac{21}{25} – \frac{18}{20} \\
\text {LCM of 25 and 20 is 100.} \\
\therefore \frac {21 \times 4 – 18 \times 5}{100} \\
= \frac {84 – 90}{100} \\
= \frac {-6}{100} \\
= \frac {-3}{50} \\ \\\text {(iv) } \quad 3 \frac{3}{10} – 2 \frac{7}{15} \\
\text {LCM of 10 and 15 is 30.} \\
\therefore 3 \frac{3}{10} – 2 \frac{7}{15} = \frac{33}{10} – \frac{37}{15} \\
= \frac{33 \times 3 – 37 \times 2}{30} \\
= \frac{99 – 74}{30} \\
= \frac{25}{30} = \frac{5}{6} \\
\end{array}\begin{array}{l}
\text {Q7. Find the difference: }
\text {(i) } \quad \frac{6}{7}-\frac{9}{11} \\
\text {(ii) } \quad 8-\frac{5}{9} \\
\text {(iii) } \quad 9-5 \frac{2}{3} \\
\text {(iv) } \quad 4 \frac{3}{10}-1 \frac{2}{15}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{6}{7}-\frac{9}{11} \\
\text {LCM of 7 and 11 is 77.} \\
\therefore \frac{6}{7}-\frac{9}{11} \\
=\frac{6 \times 11- 9 \times 7}{77} \\
=\frac{66- 63}{77} \\
=\frac{3}{77} \\ \\\text {(ii) } \quad 8-\frac{5}{9} \\
\text {LCM of 1 and 9 is 9.} \\
\therefore 8-\frac{5}{9} = \frac{8 times 9 -5}{9} \\
= \frac{72 -5}{9} \\
= \frac{67}{9} \\ \\\text {(iii) } \quad 9-5 \frac{2}{3} \\
= 9-5 – \frac{2}{3} \\
= 4 – \frac{2}{3} \\
\text {LCM of 1 and 3 is 3.} \\
\therefore 4 – \frac{2}{3} =\frac{4 \times 3 – 2}{3} \\
=\frac{12 – 2}{3} \\
=\frac{10}{3} \\ \\\text {(iv) } \quad 4 \frac{3}{10}-1 \frac{2}{15} \\
=4 + \frac{3}{10} -1 – \frac{2}{15} \\
=3 + \frac{3}{10} – \frac{2}{15} \\
\text {LCM of 10 and 15 is 30.} \\
\therefore 3 + \frac{3}{10} – \frac{2}{15} \\= \frac{3 \times 30 + 3 \times 3 – 2 \times 2}{30} \\
= \frac{90 + 9 – 4}{30} \\
= \frac{95}{30} \\
= \frac{19}{6} \\ \\\end{array}\begin{array}{l}
\text {Q8. Simplify: }
\text {(i) } \quad \frac{2}{3}+\frac{1}{6}-\frac{2}{9} \\
\text {(ii) } \quad 12-3 \frac{1}{2} \\
\text {(iii) } \quad 7 \frac{5}{6}-4 \frac{3}{8}+2 \frac{7}{12}
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {(i) } \quad \frac{2}{3}+\frac{1}{6}-\frac{2}{9} \\\text {LCM of 3 , 6 and 9 is 18.} \\
\therefore \frac{2}{3}+\frac{1}{6}-\frac{2}{9} \\
= \frac{2 \times 6 + 1 \times 3 – 2 \times 2}{18} \\
= \frac{12 + 3 – 4}{18} \\
= \frac{11}{18} \\ \\\text {(ii) } \quad 12-3 \frac{1}{2} \\
\text {LCM of 1 and 2 is 2.} \\
\therefore 12-3 \frac{1}{2} = 12- \frac{7}{2} \\
= \frac{12 \times 2 – 7}{2} \\
= \frac{24 – 7}{2} \\
= \frac{17}{2} \\ \\\text {(iii) } \quad 7 \frac{5}{6}-4 \frac{3}{8}+2 \frac{7}{12} \\
= 7 + \frac{5}{6}-4 – \frac{3}{8}+2 + \frac{7}{12} \\
= 5 + \frac{5}{6} – \frac{3}{8} + \frac{7}{12} \\
\text {LCM of 1,6,8 and 12 is 24.} \\
\therefore 5 + \frac{5}{6} – \frac{3}{8} + \frac{7}{12} \\
= \frac{5 \times 24 + 5 \times 4- 3 \times 3 +7 \times 2}{24} \\
= \frac{120 + 20 – 9 +14}{24} \\
= \frac{145}{24} \\
\end{array}\begin{array}{l}
\text {Q9. What should be added to } 5 \frac{3}{7} \text { to get 12 ? }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Let’s assume that x be the required fraction.} \\
\text {According to question, we have } \\
x+5 \frac{3}{7}=12 \\
\Rightarrow \quad x+\frac{38}{7}=12 \\
\Rightarrow \quad x=12-\frac{38}{7} \\
\Rightarrow \quad x=\frac{12 \times 7 – 38 }{7} \\
\Rightarrow \quad x= \frac{46}{7} \\
\text {Hence, } \frac{46}{7} \text { is the fraction which is added to } 5 \frac{3}{7} \text { to get 12.}
\end{array}\begin{array}{l}
\text {Q10. What should be added to } 5 \frac{4}{15} \text { to get } 12 \frac{3}{5} ?
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Let’s assume that x be the required fraction.} \\
\text {According to question, we have } \\x+5 \frac{4}{15}=12 \frac{3}{5}
\Rightarrow \quad x+\frac{15 \times 5+4}{15}=\frac{12 \times 5+3}{5} \\
\Rightarrow \quad x=\frac{63}{5}-\frac{79}{15} \\
\text {LCM of 5 and 15 is 15 } \\
\Rightarrow \quad x=\frac{63 \times 3 – 79}{15} \\ \Rightarrow \quad x = \frac{110}{15} =\frac{22}{3} \\\text {Hence, } \frac{22}{3} \text { is the fraction which is added to } 5 \frac{4}{15} \text { to get } 12 \frac{3}{5}
\end{array}\begin{array}{l}
\text {Q11. Suman studies for } 5 \frac{2}{3} \text { hours daily. She devotes } 2 \frac{4}{5}
\text { hours of her time for Science and Mathematics.} \\
\text {How much time does she devote for other subjects? }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Total hours of study } = 5 \frac{2}{3} = \frac{17}{3}\\
\text {Study hours invested for maths and science } = 2 \frac{4}{5} = \frac{14}{5}\\
\text {Let’s assume that suman devoted x hrs for other subjects. } \\
\therefore x = \frac{17}{3} – \frac{14}{5} \\
\text {LCM of 3 and 5 is 15 } \\
\Rightarrow \quad x = \frac{17 \times 5 – 14 \times 3}{15} \\
\Rightarrow \quad x = \frac{85 – 42}{15} \\
\Rightarrow \quad x = \frac{43}{15} \\
\Rightarrow \quad x = 2 \frac{13}{15} \\\text {Hence, suman devoted } 2 \frac{13}{15} \text { hours for other subjects.}
\end{array}\begin{array}{l}
\text {Q12. A piece of wire is of length } 12 \frac{3}{4} \text { m. If it is cut into two pieces in such a way } \\ \text { that the length of one piece is } 5 \frac{1}{4} \text { m, what is the length of the other piece? }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {Let’s assume that length of other piece of wire be x meters. } \\
\therefore x + 5 \frac{1}{4} = 12 \frac{3}{4} \\
\Rightarrow \quad x + \frac{21}{4} = \frac{51}{4} \\
\Rightarrow \quad x = \frac{51}{4} – \frac{21}{4} \\\Rightarrow \quad x = \frac{51 – 21}{4} \\
\Rightarrow \quad x = \frac{30}{4} \\
\Rightarrow \quad x = \frac{15}{2} \\
\Rightarrow \quad x = 7 \frac{1}{2} \\\text {Hence, length of other piece of wire is } 7 \frac{1}{2} \text { meters.}
\end{array}\begin{array}{l}
\text {Q13. A rectangular sheet of paper is } 12 \frac{1}{2} \text { cm long and } 10 \frac{2}{3} \text { cm wide.} \\
\text {Find its perimeter. }
\end{array}\begin{array}{l}
\text {Sol. } \\
\text {We know that perimeter of rectangle = 2 ( length + breadth) } \\
\therefore \text {Perimeter } = 2 \times (12 \frac{1}{2}+10 \frac{2}{3}) \\
= 2 \times (\frac{25}{2}+\frac{32}{3}) \\
=2 \times (\frac{25 \times 3+ 32 \times 2}{6} ) \\
=2 \times (\frac{139}{6}) \\
=\frac{139}{3}\text {Hence, perimeter of rectangular sheet is } \frac{139}{3} \text { cms.}
\end{array}\begin{array}{l}
\text {Q14. In a “magic square”, the sum of the numbers in each row, in each column and } \\
\text { along the diagonal is the same. Is this a magic square?} \\
\end{array}
\begin{array}{|c|c|c|}
\hline
\frac{4}{11} & \frac{9}{11} & \frac{2}{11} \\
\hline
\frac{3}{11} & \frac{5}{11} & \frac{7}{11} \\
\hline
\frac{8}{11} & \frac{1}{11} & \frac{6}{11} \\
\hline
\end{array}\begin{array}{l}
\text {Sum along rows} \\
\frac{4}{11}+\frac{9}{11}+\frac{2}{11}=\frac{4+9+2}{11}=\frac{15}{11} \\
\frac{3}{11}+\frac{5}{11}+\frac{7}{11}=\frac{3+5+7}{11}=\frac{15}{11} \\
\frac{8}{11}+\frac{1}{11}+\frac{6}{11}=\frac{8+1+6}{11}=\frac{15}{11} \\\text {Sum along columns. } \\
\frac{4}{11}+\frac{3}{11}+\frac{8}{11}=\frac{4+3+8}{11}=\frac{15}{11} \\
\frac{9}{11}+\frac{5}{11}+\frac{1}{11}=\frac{9+5+1}{11}=\frac{15}{11} \\
\frac{2}{11}+\frac{7}{11}+\frac{6}{11}=\frac{2+7+6}{11}=\frac{15}{11} \\
\text {Sum along diagonals } \\
\frac{4}{11}+\frac{5}{11}+\frac{6}{11}=\frac{4+5+6}{11}=\frac{15}{11} \\
\frac{2}{11}+\frac{5}{11}+\frac{8}{11}=\frac{2=5+8}{11}=\frac{15}{11} \\
\text {Since, all the sums along rows, columns and diagonals are equal, it is a magic square.}
\end{array}\begin{array}{l}
\text {Q15. The cost of Mathematics book is Rs. } 25 \frac{3}{4} \text { and that of Science book is Rs. } 20 \frac{1}{2}. \\
\text {Which costs more and by how much? }
\end{array}\begin{array}{l}
\text {Sol. }\\
\text {The cost of Mathematics book is Rs. } =25 \frac{3}{4} = \frac {103}{4} \\
\text { The cost of Science book is Rs. }= 20 \frac{1}{2} =\frac{41}{2} \\
\text {LCM of denominators 2 and 4 is 4.} \\
\frac{41}{2} =\frac{41 \times 2}{2 \times 2} = \frac{82}{4} \\
\text {We know that } 103 \gt 82 \\
\Rightarrow \quad \frac {103}{4} \gt \frac {82}{4} \\
\Rightarrow \quad 25 \frac{3}{4} \gt 20 \frac{1}{2} \\
\text {Hence, cost of mathematices book is more.} \\\text {Difference in costing of math and science book}=\frac {103}{4} – \frac {82}{4} \\
=\frac {103-82}{4} \\
=\frac {21}{4} \\
= 5\frac {1}{4} \\\text {Mathematices books cost more by Rs. } 5 \frac {1}{4}\end{array}\begin{array}{l}
\text {Q16. Provide the number in box }\square \text { and also give its simplest form in each of the following: } \\
\text {(i) } \quad \frac{2}{3} \times \square=\frac{10}{30} \\
\text {(ii) } \quad \frac{3}{5} \times \square=\frac{24}{75}
\end{array}\begin{array}{l}
\text {Lets assume the required be x. } \\
\text {(i) } \quad \frac{2}{3} \times x=\frac{10}{30} \\
\Rightarrow \quad \frac{2}{3} \times x=\frac{10}{30} \\
\Rightarrow \quad x=\frac{10 }{30} \times \frac{3}{2} \\
\Rightarrow \quad x=\frac{1}{2} \\ \\\text {(ii) } \quad \frac{3}{5} \times x=\frac{24}{75} \\
\Rightarrow \quad \frac{3}{5} \times x=\frac{24}{75} \\
\Rightarrow \quad x=\frac{24}{75} \times \frac{5}{3} \\
\Rightarrow \quad x=\frac{8}{15} \\
\end{array}
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