question_answer 37)

Solve (a)\[\frac{0}{1}\] (b)\[\frac{1}{1}\] (c)\[\frac{1}{1}\] (d)\[\frac{1}{1}\] (e)\[\frac{0}{1}\] (f)\[\frac{0}{2}\] (g)\[\frac{1}{6}\frac{1}{3}\] (h)\[\frac{3}{4}\frac{2}{6}\] (i)\[\frac{2}{3}\frac{2}{4}\] (j)\[\frac{6}{6}\frac{3}{3}\] (k)\[\frac{5}{6}\frac{5}{5}\] (l)\[\frac{7}{12}\] (m)\[\frac{9}{10},\frac{5}{8}\] (n)\[\frac{3}{10},\frac{18}{5}\] TIPS To subtract unlike fractions, firstly convert them into equivalent fractions with same denominator which is LCM of denominators of given fractions and then subtract smaller numerator from greater numerator.

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, \[=\frac{20}{30}=\frac{2}{3}\]