question_answer 35)

Fill in the missing fractions. (a) \[\therefore \] (b) \[-\frac{3}{21}=\frac{5}{21}\] (c) \[\because \] (d) \[=\frac{6}{8}>\frac{4}{8}>\frac{3}{8}>\frac{1}{8}\]

Answer:

(a) We have, \[\because \] Here, missing fraction is subtracted from \[=\frac{8}{9}\] to get \[=\frac{4}{9}\]. This means on adding missing term to \[=\frac{3}{9}\], we get \[=\frac{6}{9}\]So, we subtract \[\therefore \] from\[\therefore \] to get missing fraction. \[=\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\]Missing fraction \[=\frac{8}{9}>\frac{6}{9}>\frac{4}{9}>\frac{3}{9}\]or \[\frac{2}{6},\frac{4}{6},\frac{8}{6}\] [simplest form] Thus, \[\frac{6}{6}.\] (b) We have, \[\frac{1}{6}\] Here, \[\frac{2}{6},\frac{4}{6},\frac{6}{6}\]is subtracted from missing fraction to get \[\frac{8}{6}\] So, we add \[\therefore \]to \[\frac{5}{6}\frac{2}{6},\]to get missing fraction. \[\frac{3}{6}0,\]Missing fraction \[\frac{1}{6}\frac{6}{6},\] Thus,\[\frac{8}{6}\frac{5}{6}\] (c) We have, \[\frac{3}{6}\frac{5}{6}\] Here, \[\frac{1}{7}\frac{1}{4}\]is subtracted from missing fraction to get \[\frac{4}{5}\frac{5}{5}\] So, we add to \[\frac{3}{5}\frac{3}{7}\] to\[\frac{3}{6}\frac{5}{6}\] get missing fraction. \[\therefore \]Missing fraction \[\frac{3}{6}\frac{5}{6}\] Thus, \[\frac{1}{7}\frac{1}{4}\] (d) We have, \[\therefore \] Here, \[\frac{1}{7}\frac{1}{4}\]is added to missing fraction to get \[\frac{4}{5}\frac{5}{5}\] So, we subtract\[\therefore \]from \[\frac{4}{5}\frac{5}{5}\]to get missing fraction. \[\frac{3}{5}\frac{3}{7}\]Missing fraction \[\frac{3}{7}\frac{5}{7}\] Thus, \[\frac{1}{9}\frac{1}{5}\]