question_answer 15)

Find (a) \[\frac{1}{2}\] of                        (i) \[2\frac{3}{4}\]                                            (ii) \[4\frac{2}{9}\] (b) \[\frac{5}{8}\]of                        (i) \[3\frac{5}{6}\]                                            (iii) \[9\frac{2}{3}\]

Answer:

                (a) (i) \[\frac{1}{2}\]of \[2\frac{3}{4}=\frac{1}{2}\times 2\frac{3}{4}=\frac{1}{2}\times \frac{11}{4}=\frac{1\times 11}{2\times 4}\]\[=\frac{11}{8}=1\frac{3}{8}\] (ii) \[\frac{1}{2}\] of \[4\frac{2}{9}=\frac{1}{2}\] of  \[\frac{38}{9}=\frac{1}{2}\times \frac{38}{9}=\frac{1\times 38}{2\times 9}=\frac{38}{18}\] \[=\frac{38\div 2}{18\div 2}\] \[=\frac{19}{9}=2\frac{1}{9}\] (b) (i) \[\frac{5}{8}\]of \[3\frac{5}{6}=\frac{5}{8}\] of \[\frac{23}{6}=\frac{5}{8}\times \frac{23}{6}=\frac{5\times 23}{8\times 6}\] \[=\frac{115}{48}=2\frac{19}{48}\]                 (ii) \[\frac{5}{8}\]of \[9\frac{2}{3}=\frac{5}{8}\]of \[\frac{29}{3}=\frac{5}{8}\times \frac{29}{3}=\frac{5\times 29}{8\times 3}\]\[=\frac{145}{24}=6\frac{1}{24}\]