question_answer 1)

The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one in its simplest form. (a) \[1\frac{1}{3}+3\frac{2}{3}=1+\frac{1}{3}+3+\frac{2}{3}\] (b) \[=(1+3)+\left( \frac{1}{3}+\frac{2}{3} \right)=4+\frac{1}{3}+\frac{2}{3}\] (c) \[\frac{1}{3}+\frac{2}{3}=\frac{1+2}{3}=\frac{3}{3}=1\] (d) \[\therefore \] (e) \[4+\frac{1}{3}+\frac{2}{3}=4+1=5\] (f) \[1\frac{1}{3}+3\frac{2}{3}=5\] (g) \[4\frac{2}{3}+3\frac{1}{4}=4+\frac{2}{3}+3+\frac{1}{4}\] (h) \[=(4+3)+\left( \frac{2}{3}+\frac{1}{4} \right)=7+\frac{2}{3}+\frac{1}{4}\] (i) \[\frac{2}{3}+\frac{1}{4}=\frac{2\times 4}{3\times 4}+\frac{1\times 3}{4\times 3}\] (j) \[\because \] (k) \[=\frac{8}{12}+\frac{3}{12}=\frac{8+3}{12}=\frac{11}{12}\] (l) \[\therefore \] TIPS To convert the given fraction into its simplest form, divide numerator and denominator both by their HCF.

Answer:

First, we have to convert all the fractions in the simplest form. (a) We have, \[7+\frac{2}{3}+\frac{1}{4}=7+\frac{11}{12}=7\frac{11}{12}\] [\[\frac{7\times 12+11}{12}=\frac{95}{12}\]HCF of 2 and 12 = 2] (b) We have, \[4\frac{2}{3}+3\frac{1}{4}=7\frac{11}{12}\] [\[\frac{95}{12}\]HCF of 3 and 15 = 3] (c) We have, \[\frac{16}{5}-\frac{7}{5}=\frac{16-7}{5}=\frac{9}{5}\] [\[\frac{16}{5}-\frac{7}{5}=\frac{9}{5}\]HCF of 8 and 50 = 2] (d) We have, \[\frac{4}{3}-\frac{1}{2}\] [\[\therefore \]HCF of 16 and 100 = 4] (e) We have, \[\frac{4}{3}-\frac{1}{2}=\frac{4\times 2}{3\times 2}-\frac{1\times 3}{2\times 3}=\frac{8}{6}-\frac{3}{6}=\frac{8-3}{6}=\frac{5}{6}\] [\[\frac{4}{3}-\frac{1}{2}=\frac{5}{6}\]HCF of 10 and 60 = 10] (f) We have, \[\frac{2}{5}\] [\[\frac{3}{4}\] HCF of 15 and 75 = 15] (g) We have, \[=\frac{2}{5}m\] [\[=\frac{3}{4}m\]HCF of 12 and 60 = 12] (h) We have, \[\therefore \] [\[=\frac{2}{5}+\frac{3}{4}\]HCF of 16 and 96 = 16] (i) We have, \[\therefore \] [\[\frac{2}{5}+\frac{3}{4}=\frac{2\times 4}{5\times 4}+\frac{3\times 5}{4\times 5}=\frac{8}{20}+\frac{15}{20}=\frac{8+15}{20}=\frac{23}{20}\]HCF of 12 and 75 = 3] (j) We have, \[\frac{23}{20}m.\] [\[1\frac{2}{2}\]HCF of 12 and 72 = 12] (k) We have,\[1\frac{1}{3}\] [\[\frac{1}{4}.\]HCF of 3 and 18 = 3] (l) We have, \[=\frac{10}{20}\] [\[=\frac{25}{50}\]HCF of 4 and 25 = 1] Now, on grouping into three groups of equivalent fractions with the help of simplest form of fractions, we get (i) \[=\frac{40}{80}\] \[\frac{10}{20}=\frac{10\div 10}{20\div 10}=\frac{1}{2}\] (ii) \[\because \] \[\frac{25}{50}=\frac{25\div 25}{50\div 25}=\frac{1}{2}\] (iii) \[\because \] \[\frac{40}{80}=\frac{40\div 40}{80\div 40}=\frac{1}{2}\]