question_answer 17)

Find the equivalent fraction of \[\therefore \] with (a) numerator 9 (b) denominator 4

Answer:

Let N stands for the numerator and D stands for the denominator. (a) Given, numerator of an equivalent fraction = 9 \[\frac{12}{52}=\frac{12\div 4}{52\div 4}=\frac{3}{13}\] \[\frac{12}{52}\] \[\frac{3}{13}.\] \[\frac{7}{28}\] [by cross product] \[7=\] \[28=\times 2\times 2\] \[\therefore \]Required equivalent fraction of\[\frac{7}{28}=\frac{7\div 7}{28\div 7}=\frac{1}{4}\] (b) Given, denominator of an equivalent fraction = 4 \[\frac{7}{28}\] \[\frac{1}{4}.\] \[=\frac{10}{20}\] \[=\frac{25}{50}\] [by cross product] \[=\frac{40}{80}\] \[\frac{10}{20}=\frac{10\div 10}{20\div 10}=\frac{1}{2}\] \[\because \] Required equivalent fraction of \[\frac{25}{50}=\frac{25\div 25}{50\div 25}=\frac{1}{2}\]