Question

Arrange fraction in ascending order 2/4,1/8,3/12,7/16​

Answer 1

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Answer 2

$GIVEN \ SET \ OF \ FRACTION :\\ \\ \frac{2}{4} , \ \frac{1}{8} , \ \frac{3}{12} , \ \frac{7}{16} \\\\\\ TO \ FIND :$$\\\\ The\ arrangement\ of \ the \ given \ set\ of\ fraction \ in \ ascending \ order. \\\\\\ SOLUTION :$$\\ If \ the \ denominator \ in\ each \ fraction \ is \ equal,$$then \ the\ fraction \ with\ the \ smallest\ numerator\ is \ the \ smallest.$

$Step \ 1 : \ The\ first \ step \ to \ equalise \ the\ denominators\ is\ to\ \ find \ their \ L.C.M.\\\\ The\ L.C.M. \ of \ 4, 8, 12 , \ and \ 16 \ is \ 48.\\\\ Step \ 2 : The \ L.C.M.\ is \ then\ divided\ by \ the\ denominator\ in\ each\ fraction,$$\\\\ \frac{48}{4} = 12\\\\ \frac{48}{8} = 6\\\\ \frac{48}{12} = 4\\\\\frac{48}{16} = 3$

$\ Step \ 3 :\ The \ quotient\ is \ multiplied \ by \ both \ numerator \ and \ denominator \ in \ each \ case,\\\\ \frac{2* 12}{4* 12} = \frac{24}{48} \\\\ \frac{1*6} {8 * 6} = \frac{6}{48} \\\\ \frac{3*4}{12 *4} = \frac{12}{48} \\\\ \frac{7*3} {16* 3} = \frac{21}{48}$$\\\\Step \ 4 : \ Comparing\ the \ numerators \ of \ all \ fraction, \ we \ find :\\\\ 6 < 12 < 21 < 24\\\\ => \frac{6}{48} < \frac{12}{48} < \frac{21}{48} < \frac{24}{48}\\\\ => \frac{1}{8} < \frac{3}{12} < \frac{7}{16} < \frac{2}{4}\\\\ Answer \ - \ The \ given \ set\ of \ fraction \ is \ arranged\ in \ the \ ascending \ order \ as \ : \\\\ \frac{1}{8} < \frac{3}{12} < \frac{7}{16} < \frac{2}{4}$