question_answer 21)

Match the equivalent fractions and write two more for each.

Column I	Column II
(i) \[\frac{13}{24}\]	(a) \[\frac{17}{102}or\frac{12}{102}\]
(ii) \[\therefore \]	(b) \[\frac{17}{102}>\frac{12}{102}\]
(iii)\[\frac{17}{102}\]	(c) \[\frac{1}{8},\frac{5}{8},\frac{3}{8}\]
(iv) \[\frac{1}{5},\frac{11}{5},\frac{4}{5},\frac{3}{5},\frac{7}{5}\]	(d) \[\frac{1}{7},\frac{3}{7},\frac{13}{7},\frac{11}{7},\frac{7}{7}\]
(v) \[\frac{1}{8},\frac{5}{8},\frac{3}{8}\]	(e) \[\therefore \]

TIPS Firstly, reduce (i), (ii), (iii), (iv) and (v) into simplest form. Then, match the equivalent fractions. To write more equivalent fractions, multiply numerator and denominator both by same number.

Answer:

(i) We have, \[\therefore \] Now, factors of \[=\frac{5}{3}>\frac{3}{8}>\frac{1}{8}\] and factors of \[=\frac{5}{8}>\frac{3}{8}>\frac{1}{8}\] Common factors are 2, 5 and 5 HCF of 250 and \[\frac{1}{5},\frac{11}{5},\frac{4}{5},\frac{3}{5},\frac{7}{5}\] \[=1<3<4<7<11\] \[=11>7>4>3>1\] So, \[\therefore \] is equivalent to \[\frac{1}{5}<\frac{3}{5}<\frac{4}{5}<\frac{7}{5}<\frac{11}{5}\] i.e. (i) \[=\frac{11}{5}>\frac{7}{5}>\frac{4}{5}>\frac{3}{5}>\frac{1}{5}\](d) Two more equivalent fractions are \[\frac{11}{7},\frac{3}{7},\frac{13}{7},\frac{11}{7},\frac{7}{7}\] and \[\therefore \] (ii) We have \[=1<3<7<11<13\] Now, factors of \[=13>11>7>3>1\] and factors of \[\therefore \] Common factors = 2, 2 and 5 HCF of 180 and \[=\frac{1}{7}<\frac{3}{7}<\frac{7}{7}<\frac{11}{7}<\frac{13}{7}\] \[=\frac{13}{7}>\frac{11}{7}>\frac{7}{7}>\frac{3}{7}>\frac{1}{7}\] \[\frac{1}{12},\frac{1}{23},\frac{1}{5},\frac{1}{7},\frac{1}{50},\frac{1}{9},\frac{1}{17}\] So, \[\frac{3}{7},\frac{3}{11},\frac{3}{5},\frac{3}{2},\frac{3}{13},\frac{3}{4},\frac{3}{17}\] is equivalent to \[=50>23>17>12>9>7>5\] i.e. (ii) \[\therefore \](e) Two more equivalent fractions are \[=\frac{1}{50},\frac{1}{23},\frac{1}{17},\frac{1}{12},\frac{1}{9},\frac{1}{7},\frac{1}{5}\] and \[\left[ \begin{align} & \because \frac{1}{50}\text{has larger denominator, so it is } \\ & \text{smallest fraction} \\ \end{align} \right]\] (iii) We have, \[=\frac{1}{5},\frac{1}{7},\frac{1}{9},\frac{1}{12},\frac{1}{17},\frac{1}{23},\frac{1}{50}\] Now, factors of \[600=\times \times 2\times \times \times \] and factors of \[\therefore \] Common factors = 2, 3, 5 and 11 HCF of 660 and \[=\frac{3}{17},\frac{3}{13},\frac{3}{7},\frac{3}{5},\frac{3}{4},\frac{3}{2}\] \[=\frac{3}{2},\frac{3}{4},\frac{3}{5},\frac{3}{7},\frac{3}{11},\frac{3}{13},\frac{3}{17}\] \[=\frac{2}{11},\frac{2}{21},\frac{2}{9},\frac{2}{7},\frac{2}{8},\frac{2}{15}\] So, \[=21>15>11>9>8>7\]is equivalent to \[\therefore \] i.e. (iii)\[=\frac{2}{21},\frac{2}{15},\frac{2}{11},\frac{2}{9},\frac{2}{8},\frac{2}{7}\](a) Two more equivalent fractions are \[=\frac{2}{7},\frac{2}{8},\frac{2}{9},\frac{2}{11},\frac{2}{15},\frac{2}{21}\] and \[\frac{4}{5},\frac{4}{6},\frac{4}{13},\frac{4}{2},\frac{4}{9},\frac{4}{11}\] (iv) We have, \[=13>11>9>6>5>2\] Now, factors of \[\therefore \] and factors of \[=\frac{4}{13},\frac{4}{11},\frac{4}{9},\frac{4}{6},\frac{4}{5},\frac{4}{2}\] Common factors = 2, 2, 3, 3 and 5 HCF of 180 and \[=\frac{4}{2},\frac{4}{5},\frac{4}{6},\frac{4}{9},\frac{4}{11},\frac{4}{13}\] \[\frac{7}{11},\frac{7}{13},\frac{7}{5},\frac{7}{2},\frac{7}{3},\frac{7}{4}\] \[=13>11>5>4>3>2\] So, \[\therefore \]is equivalent to \[=\frac{7}{13},\frac{7}{11},\frac{7}{5},\frac{7}{4},\frac{7}{3},\frac{7}{2}\] i.e. (iv) \[=\frac{7}{2},\frac{7}{3},\frac{7}{4},\frac{7}{5},\frac{7}{11},\frac{7}{13}\] (c) Two more equivalent fractions are \['<','=''>'\] and \[\frac{2}{6},\frac{4}{6},\frac{8}{6}\] (v) We have, \[\frac{6}{6}\] Now, factors of \[\frac{5}{6}\frac{2}{6},\] and factors of \[\frac{3}{6}0,\] Common factors = 2, 5 and 11 HCF of 220 and \[\frac{1}{6}\frac{6}{6},\] \[\frac{8}{6}\frac{5}{6}\] \[=\frac{3}{8}\] So,\[=\frac{6}{8}\] is equivalent to \[=\frac{4}{8}\] i.e. (v)\[=\frac{1}{8}\](b) Two more equivalent fractions are \[\therefore \]and\[\therefore \]