question_answer 26)

How quickly can you do this? Fill appropriate sign. \[('<',\,\,'=',\,\,'>')\]. (a) \[=\frac{3}{21}+\frac{5}{21}=\frac{3+5}{21}=\frac{8}{21}\] (b) \[\frac{8}{21}-\frac{3}{21}=\frac{5}{21}\] (c) \[-\frac{3}{6}=\frac{3}{6}\] (d) \[\frac{3}{6}\] (e) \[\frac{3}{6}.\] (f) \[\frac{3}{6}\] (g) \[\frac{3}{6},\] (h) \[\therefore \] (i) \[=\frac{3}{6}+\frac{3}{6}=\frac{3+3}{6}=\frac{6}{6}=1\] (j) \[-\frac{3}{6}=\frac{3}{6}\] TIPS We have to use cross-product method for quicker calculation. According to this method, firstly find the products of numerator of first fraction with denominator of second fraction and denominator of first fraction with numerator of second fraction. Then, the fraction having that numerator which gives greater product will be greater.

Answer:

(a) We have,\[+\frac{5}{27}=\frac{12}{27}\] and \[\frac{5}{27}\] Here, \[\frac{12}{27}.\]and \[\frac{5}{27}\] \[\frac{12}{27}\] \[\therefore \] \[=\frac{12}{27}-\frac{5}{27}=\frac{12-5}{27}=\frac{7}{27}\] \[+\frac{5}{27}=\frac{12}{27}\] (b) We have, \[=1=\frac{7}{7}\] and \[=\frac{5}{7}\] Here, \[\therefore \]and\[=\frac{7}{7}-\frac{5}{7}=\frac{7-5}{7}=\frac{2}{7}\] \[\frac{2}{7}\] \[\frac{2}{5}\] \[\frac{3}{7}.\] \[\frac{2}{5}+\frac{3}{7}\] (c) We have, \[\therefore \]and \[\frac{2}{5}+\frac{3}{7}=\frac{2\times 7}{5\times 7}+\frac{3\times 5}{7\times 5}=\frac{14}{35}+\frac{15}{35}=\frac{14+15}{35}=\frac{29}{35}\] Here, \[\frac{5}{7}-\frac{2}{5}=\frac{11}{35}\]and \[\frac{2}{3}+\frac{1}{7}\] \[\frac{3}{10}+\frac{7}{15}\] \[\frac{4}{9}+\frac{2}{7}\] \[\frac{5}{7}+\frac{1}{3}\] \[\frac{2}{5}+\frac{1}{6}\] (d) We have, \[\frac{4}{5}+\frac{2}{3}\] and \[\frac{3}{4}-\frac{1}{3}\] Here, \[\frac{5}{6}-\frac{1}{3}\]and\[\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\] \[\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\] \[1\frac{1}{3}+3\frac{2}{3}\] \[4\frac{2}{3}+3\frac{1}{4}\] \[\frac{16}{5}-\frac{7}{5}\] (e) We have\[\frac{4}{3}-\frac{1}{2}\] Here, \[\frac{2}{3}+\frac{1}{7}\]and\[\therefore \] \[\frac{2}{3}+\frac{1}{7}=\frac{2\times 7}{3\times 7}+\frac{1\times 3}{7\times 3}+\frac{14}{21}+\frac{3}{21}=\frac{14+3}{21}=\frac{17}{21}\] \[\frac{2}{3}+\frac{1}{7}=\frac{17}{21}\] \[\frac{3}{10}+\frac{7}{15}\] \[\therefore \] (f) We have, \[\frac{7}{9}\] and \[\frac{3}{9}\] Here, \[\frac{3}{10}+\frac{7}{15}=\frac{23}{30}\] and \[\frac{4}{9}+\frac{2}{7}\] \[\therefore \] \[\frac{4}{9}+\frac{2}{7}=\frac{4\times 7}{9\times 7}+\frac{2\times 9}{7\times 9}+\frac{28}{63}+\frac{18}{63}=\frac{28+18}{63}=\frac{46}{63}\] \[\frac{4}{9}+\frac{2}{7}=\frac{46}{63}\] \[\frac{5}{7}+\frac{1}{3}\] (g) We have, \[\frac{1}{4}\] and \[\frac{2}{8}\] Here, \[\frac{5}{7}+\frac{1}{3}=\frac{5\times 3}{7\times 3}+\frac{1\times 7}{3\times 7}=\frac{15}{21}=\frac{15+7}{21}=\frac{22}{21}\]and\[\frac{5}{7}+\frac{1}{3}=\frac{22}{21}\] \[\frac{2}{5}+\frac{1}{6}\] \[\therefore \] \[\frac{2}{5}+\frac{1}{6}=\frac{2\times 6}{5\times 6}+\frac{1\times 5}{6\times 5}=\frac{12}{30}+\frac{5}{30}=\frac{12+5}{30}=\frac{17}{30}\] \[\frac{2}{5}+\frac{1}{6}=\frac{17}{30}\] (h) We have, \[\frac{6}{10}\] and \[\frac{4}{5}\] Here, \[\therefore \] and \[\frac{4}{5}+\frac{2}{3}=\frac{4\times 3}{5\times 3}+\frac{2\times 5}{3\times 5}=\frac{12}{15}+\frac{10}{15}=\frac{12+10}{15}=\frac{22}{15}\] \[\frac{4}{5}+\frac{2}{3}=\frac{22}{15}\] \[\frac{3}{4}-\frac{1}{3}\] \[\therefore \] \[\frac{3}{4}-\frac{1}{3}=\frac{3\times 3}{4\times 3}-\frac{1\times 4}{3\times 4}=\frac{9}{12}-\frac{4}{12}=\frac{9-4}{12}=\frac{5}{12}\] (i) We have, \[\frac{3}{4}\] and \[\frac{7}{8}\] Here, \[\frac{5}{6}-\frac{1}{3}\] and \[\therefore \] \[\frac{5}{6}-\frac{1}{3}=\frac{5\times 1}{6\times 1}-\frac{1\times 2}{3\times 2}=\frac{5}{6}-\frac{2}{6}=\frac{5-2}{6}=\frac{3}{6}=\frac{1}{2}\] \[\frac{5}{6}-\frac{1}{3}=\frac{1}{2}\] \[\frac{2}{3}+\frac{3}{4}+\frac{1}{2}\] \[\therefore \] (j) We have, \[\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\frac{2\times 4}{3\times 4}+\frac{3\times 3}{4\times 3}+\frac{1\times 6}{2\times 6}\] and \[=\frac{8}{12}+\frac{9}{12}+\frac{6}{12}=\frac{8+9+6}{12}=\frac{23}{12}\] Here, \[\frac{2}{3}+\frac{3}{4}+\frac{1}{2}=\frac{23}{12}\] and \[\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\] \[\therefore \] \[\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1\times 3}{2\times 3}+\frac{1\times 3}{3\times 2}+\frac{1\times 1}{6\times 1}\] \[=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{3+2+1}{6}=\frac{6}{6}=1\] \[\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\]