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) \[\therefore \] |
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\[\therefore \]
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