Answer:
We have,\[1\frac{1}{4}\] LCM of 5 and 7 = 35 \[\frac{1}{4}\]\[\frac{\text{Remainder}}{\text{Divisor}}\] Hence,\[=\frac{17}{4}\] (b) We have, \[\Rightarrow \] LCM of 10 and 15 = 30 \[\begin{align} & 4\overline{)17(}4 \\ & \,\,\,\,\,\frac{16}{1} \\ \end{align}\]\[=4\frac{1}{4}\] Hence, \[\frac{\text{(}Whole\times Denominator\text{)}+Numerator}{Deno\min ator}\] (c) We have, \[=4\frac{5}{6}\] LCM of 9 and 7 = 63 \[=\frac{(4\times 6)+5}{6}\]\[\frac{1}{3}\] Hence, \[\frac{2}{6}\] (d) We have,\[\frac{3}{9}\] LCM of 3 and 7 =21 \[\frac{4}{12}\]\[\frac{4}{5}\] Hence, \[\frac{28}{35}.\] (e) We have, \[\times \] LCM of 5 and 6 =30 \[=4\times 35\text{ }=140\]\[\times \] Hence, \[=5\times 28=140\] (f) We have,\[\frac{12}{8}=\frac{12\div 4}{8\div 4}=\frac{3}{2}\] LCM of 5 and 3 =15 \[\frac{3}{2}\]\[\frac{1}{14},\frac{2}{14},\frac{3}{14}\] Hence, \[\frac{1}{15},\frac{7}{27}\] (g) We have, \[\frac{11}{20}\] LCM of 4 and 3 =12 \[\frac{13}{20},\frac{13}{20}\]\[\frac{2}{3}\] Hence, \[\frac{2}{5},\frac{2}{3}\] (h) We have,\[\frac{2}{3}\] LCM of 6 and 3 = 6 \[\frac{3}{4}.\]\[\frac{8}{12}\] Hence,\[\frac{9}{12},\] (i) We have,\[\frac{9}{12}>\frac{8}{12},\] LCM of 3, 4 and 2 = 12 \[\frac{3}{4}>\frac{2}{3}\]\[\frac{2}{5}+\frac{1}{3}=\frac{2\times 3}{5\times 3}+\frac{1\times 5}{3\times 5}\] \[\because \] Hence,\[\therefore \] (j) We have,\[=\frac{\text{Number of shaded parts}}{\text{Total number of equal parts }}=\frac{2}{4}\] LCM of 2, 3 and 6 = 6 \[\therefore \] \[\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1\times 3}{2\times 3}+\frac{1\times 2}{3\times 2}+\frac{1\times 1}{6\times 1}\] \[\therefore \] Hence, \[=\frac{4}{8}\] (k) We have, \[\therefore \] \[=\frac{1}{4}\] Now, \[\therefore \] \[=\frac{3}{7}\] \[\therefore \] Hence, \[\therefore \] (l) We have, \[=\frac{10}{10}\] \[\therefore \] Now \[=\frac{4}{9}\] [\[\therefore \] LCM of 3 and 4 is 12] \[=\frac{4}{8}\] \[\therefore \]\[=\frac{1}{2}\] or \[\frac{1}{6}\] Hence, \[\frac{1}{4}\] or \[\frac{1}{3}\] (m) We have, \[\frac{3}{4}\] Hence, \[\frac{4}{9}\] (n) We have, \[\therefore \] LCM of 3 and 2 = 6 \[\therefore \] \[\therefore \] Hence, \[=\frac{20}{30}=\frac{2}{3}\]
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