Here you can download a free Simplification and Approximation questions PDF with answers for IBPS RRB PO and IBPS PO 2022 by Cracku. These are some tricky questions in the IBPS RRB PO and IBPS PO 2022 exam that you need to find the Simplification and Approximation of answers for the given questions. These questions will help you to make practice and solve the Simplification and Approximation questions in the IBPS RRB PO and IBPS PO exams. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Simplification and Approximation MCQ<\/strong> PDF for IBPS RRB PO and IBPS PO 2022 for free.<\/p>\n Download Simplification & Approximation for IBPS RRB & PO Prelims<\/a><\/p>\n Download IBPS RRB PO Previous Papers<\/a><\/p>\n Question 1:\u00a0<\/b>Simplify the following expression: a)\u00a0$\\frac{14}{75}$<\/p>\n b)\u00a0$\\frac{32}{75}$<\/p>\n c)\u00a0$-\\frac{70}{27}$<\/p>\n d)\u00a0$-\\frac{14}{15}$<\/p>\n 1)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\left(\\frac{3}{4}-\\frac{1}{4}\\div\\frac{1}{4}\u00a0 \\text{of}\u00a0 \\frac{2}{5}\\right)\\div\\left(\\frac{3}{4}\\div\\frac{2}{3}\u00a0 \\text{of}\u00a0 \\frac{3}{5}\\right)$<\/p>\n =\u00a0$\\left(\\frac{3}{4}-\\frac{1}{4}\\div\\frac{2}{20}\\right)\\div\\left(\\frac{3}{4}\\div\\frac{6}{15}\\right)$<\/p>\n =\u00a0$\\left(\\frac{3}{4}-\\frac{1}{4}\\times\\frac{20}{2}\\right)\\div\\left(\\frac{3}{4}\\times\\frac{15}{6}\\right)$<\/p>\n =\u00a0$\\left(\\frac{3}{4}-\\frac{5}{2}\\right)\\div\\left(\\frac{15}{8}\\right)$<\/p>\n =\u00a0$\\left(\\frac{3-10}{4}\\right)\\div\\left(\\frac{15}{8}\\right)$<\/p>\n =\u00a0$\\left(\\frac{-7}{4}\\right)\\div\\left(\\frac{15}{8}\\right)$<\/p>\n =\u00a0$\\left(\\frac{-7}{4}\\right)\\times\\left(\\frac{8}{15}\\right)$<\/p>\n = $-\\frac{14}{15}$<\/p>\n Hence, the correct answer is Option D<\/p>\n Question 2:\u00a0<\/b>Simplify the following expression: a)\u00a0$8\\frac{1}{2}$<\/p>\n b)\u00a0-4<\/p>\n c)\u00a0$3\\frac{23}{36}$<\/p>\n d)\u00a0$7\\frac{29}{36}$<\/p>\n 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{7}{12}\\div\\frac{1}{10}\u00a0\u00a0\\text{of}\u00a0\u00a0\\frac{2}{3}-\\frac{5}{3}\\times\\frac{9}{10}+\\frac{5}{8}\\div\\frac{3}{4}\u00a0 \\text{of}\u00a0 \\frac{2}{3}$<\/p>\n =\u00a0$\\frac{7}{12}\\div\\frac{1}{15}-\\frac{5}{3}\\times\\frac{9}{10}+\\frac{5}{8}\\div\\frac{1}{2}$<\/p>\n =\u00a0$\\frac{7}{12}\\times\\frac{15}{1}-\\frac{5}{3}\\times\\frac{9}{10}+\\frac{5}{8}\\times\\frac{2}{1}$<\/p>\n =\u00a0$\\frac{35}{4}-\\frac{3}{2}+\\frac{5}{4}$<\/p>\n =\u00a0$\\frac{35-6+5}{4}$<\/p>\n =\u00a0$\\frac{34}{4}$<\/p>\n =\u00a0$\\frac{17}{2}$<\/p>\n =\u00a0$8\\frac{1}{2}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 3:\u00a0<\/b>The value of $3\\frac{1}{5} \\div 4\\frac{1}{2}\u00a0 \\text{of}\u00a0\u00a0 5\\frac{1}{3} – \\frac{1}{8} \\div \\frac{1}{2}\u00a0 \\text{of}\u00a0\u00a0 \\frac{1}{4} + \\frac{1}{4}\\left(\\frac{1}{2} \\div \\frac{1}{8} \\times \\frac{1}{4}\\right)$ is:<\/p>\n a)\u00a0$-\\frac{37}{60}$<\/p>\n b)\u00a0$-\\frac{17}{60}$<\/p>\n c)\u00a0$\\frac{17}{60}$<\/p>\n d)\u00a0$\\frac{37}{60}$<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $3\\frac{1}{5} \\div 4\\frac{1}{2}\u00a0 \\text{of}\u00a0 5\\frac{1}{3} – \\frac{1}{8} \\div \\frac{1}{2}\u00a0 \\text{of}\u00a0 \\frac{1}{4} + \\frac{1}{4}\\left(\\frac{1}{2} \\div \\frac{1}{8} \\times \\frac{1}{4}\\right)$<\/p>\n =\u00a0$\\frac{16}{5} \\div \\frac{9}{2}\u00a0 \\text{of}\u00a0 \\frac{16}{3} – \\frac{1}{8} \\div \\frac{1}{2}\u00a0 \\text{of}\u00a0 \\frac{1}{4} + \\frac{1}{4}\\left(\\frac{1}{2} \\times \\frac{8}{1} \\times \\frac{1}{4}\\right)$<\/p>\n =\u00a0$\\frac{16}{5}\\div24-\\frac{1}{8}\\div\\frac{1}{8}+\\frac{1}{4}\\left(1\\right)$<\/p>\n =\u00a0$\\frac{16}{5\\times24}-1+\\frac{1}{4}$<\/p>\n =\u00a0$\\frac{2}{15}-1+\\frac{1}{4}$<\/p>\n =\u00a0$\\frac{8-60+15}{60}$<\/p>\n =\u00a0$-\\frac{37}{60}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 4:\u00a0<\/b>The value of $423 \\div \\left[270 \\div \\frac{3}{7} \\times 35 + \\left(17 \\div \\frac{1}{3}\\right) – \\left(8\\frac{1}{2} – \\frac{5}{2}\\right)\\right]$ is:<\/p>\n a)\u00a0$\\frac{51}{2455}$<\/p>\n b)\u00a0$\\frac{47}{2455}$<\/p>\n c)\u00a0$\\frac{43}{2455}$<\/p>\n d)\u00a0$\\frac{41}{2455}$<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $423\\div\\left[270\\div\\frac{3}{7}\\times35+\\left(17\\div\\frac{1}{3}\\right)-\\left(8\\frac{1}{2}-\\frac{5}{2}\\right)\\right]$<\/p>\n =\u00a0$423\\div\\left[270\\div\\frac{3}{7}\\times35+51-6\\right]$<\/p>\n =\u00a0$423\\div\\left[270\\times\\frac{7}{3}\\times35+51-6\\right]$<\/p>\n =\u00a0$423\\div\\left[22050+51-6\\right]$<\/p>\n =\u00a0$423\\div22095$<\/p>\n =\u00a0$\\frac{423}{22095}$<\/p>\n =\u00a0$\\frac{47}{2455}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 5:\u00a0<\/b>The value of $\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}-\\frac{1}{5}\\times\\left(\\frac{7}{8}-\\frac{5}{4}\\right)\\right]$ is:<\/p>\n a)\u00a010<\/p>\n b)\u00a00<\/p>\n c)\u00a01<\/p>\n d)\u00a05<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}-\\frac{1}{5}\\times\\left(\\frac{7}{8}-\\frac{5}{4}\\right)\\right]=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}-\\frac{1}{5}\\times\\left(\\frac{7-10}{8}\\right)\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}-\\frac{1}{5}\\times\\left(\\frac{-3}{8}\\right)\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}-\\left(\\frac{-3}{40}\\right)\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{4}{5}+\\frac{3}{40}\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{32+3}{40}\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{1}{5}\\left[\\frac{35}{40}\\right]$<\/p>\n $=\\frac{33}{40}+\\frac{7}{40}$<\/p>\n $=\\frac{40}{40}$<\/p>\n $=1$<\/p>\n Hence, the correct answer is Option C<\/p>\n Question 6:\u00a0<\/b>If the numerator of a fraction is increased by 15% and denominator is decreased by 20%, then the fraction, so obtained, is $\\frac{17}{65}$ What is the original fraction?<\/p>\n a)\u00a0$\\frac{272}{1495}$<\/p>\n b)\u00a0$\\frac{267}{1495}$<\/p>\n c)\u00a0$\\frac{278}{1495}$<\/p>\n d)\u00a0$\\frac{281}{1495}$<\/p>\n 6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the numerator of the fraction = $x$<\/p>\n Denominator of the fraction = $y$<\/p>\n Numerator when increased by 15% =\u00a0$\\frac{115}{100}x$<\/p>\n Denominator when decreased by 20% = $\\frac{80}{100}y$<\/p>\n Given, new fraction = $\\frac{17}{65}$<\/p>\n $=$> \u00a0$\\frac{\\frac{115}{100}x}{\\frac{80}{100}y}=\\frac{17}{65}$<\/p>\n $=$> \u00a0$\\frac{115x}{80y}=\\frac{17}{65}$<\/p>\n $=$> \u00a0$\\frac{x}{y}=\\frac{17\\times80}{65\\times115}$<\/p>\n $=$> \u00a0$\\frac{x}{y}=\\frac{272}{1495}$<\/p>\n $\\therefore$The original fraction $=\\frac{x}{y}=\\frac{272}{1495}$<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 7:\u00a0<\/b>The value of $\\frac{1}{\\left(9 – 4\\sqrt{5}\\right)^2} + \\frac{1}{\\left(9 + 4\\sqrt{5}\\right)^2}$ is:<\/p>\n a)\u00a0322<\/p>\n b)\u00a0424<\/p>\n c)\u00a0246<\/p>\n d)\u00a0286<\/p>\n 7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{1}{\\left(9 – 4\\sqrt{5}\\right)^2} + \\frac{1}{\\left(9 + 4\\sqrt{5}\\right)^2}=\\frac{1}{81+80-72\\sqrt{5}}+\\frac{1}{81+80+72\\sqrt{5}}$<\/p>\n $=\\frac{1}{161-72\\sqrt{5}}+\\frac{1}{161+72\\sqrt{5}}$<\/p>\n $=\\frac{161+72\\sqrt{5}+161-72\\sqrt{5}}{161^2-\\left(72\\sqrt{5}\\right)^2}$<\/p>\n $=\\frac{322}{25921-25920}$<\/p>\n $=322$<\/p>\n Question 8:\u00a0<\/b>The value of $\\frac{5\\frac{1}{2} \\div 3\\frac{2}{3} of \\frac{1}{4} + \\left(5\\frac{1}{9} – 7\\frac{7}{8} \\div 9\\frac{9}{20}\\right) \\times \\frac{9}{11}}{5 \\div 5 of \\frac{1}{10} – 10 \\times 10 \\div 20}$ is:<\/p>\n a)\u00a0$1\\frac{4}{5}$<\/p>\n b)\u00a0$1\\frac{9}{10}$<\/p>\n c)\u00a0$3\\frac{4}{5}$<\/p>\n d)\u00a0$9\\frac{1}{2}$<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{5\\frac{1}{2} \\div 3\\frac{2}{3} of \\frac{1}{4} + \\left(5\\frac{1}{9} – 7\\frac{7}{8} \\div 9\\frac{9}{20}\\right) \\times \\frac{9}{11}}{5 \\div 5 of \\frac{1}{10} – 10 \\times 10 \\div 20}$<\/p>\n = $\\frac{\\frac{11}{2} \\div \\frac{11}{3} of \\frac{1}{4} + \\left(\\frac{46}{9} – \\frac{63}{8} \\div \\frac{189}{20}\\right) \\times \\frac{9}{11}}{5 \\div 5 of \\frac{1}{10} – 10 \\times 10 \\div 20}$<\/p>\n = $\\frac{\\frac{11}{2} \\div \\frac{11}{12} + \\left(\\frac{46}{9} – \\frac{63}{8} \\div \\frac{189}{20}\\right) \\times \\frac{9}{11}}{5 \\div \\frac{1}{2} – 10 \\times 10 \\div 20}$<\/p>\n = $\\frac{ 6 + \\left(\\frac{46}{9} – \\frac{5}{6}\\right) \\times \\frac{9}{11}}{10 – 10 \\times 0.5}$<\/p>\n = $\\frac{ 6 + \\left( \\frac{77}{18}\\right) \\times \\frac{9}{11}}{5}$<\/p>\n = $\\frac{ 6 + ( \\frac{7}{2})}{5}$<\/p>\n =$\\frac{19}{10} = 1 \\frac{9}{10}$<\/p>\n Question 9:\u00a0<\/b>The value ok $\\frac{\\left(1\\frac{1}{9} \\times 1 \\frac{1}{20} \\div \\frac{21}{38} – \\frac{1}{3}\\right) \\div \\left(2\\frac{4}{9} \\div 1\\frac{7}{15} of \\frac{3}{5}\\right)}{\\frac{1}{5} of \\frac{1}{5} \\div \\frac{1}{125} – \\frac{1}{25} \\div \\frac{1}{5} of \\frac{1}{5}}$ lies between ………<\/p>\n a)\u00a00.1 and 0.15<\/p>\n b)\u00a00.2 and 0.25<\/p>\n c)\u00a00.15 and 0.2<\/p>\n d)\u00a00.25 and 0.3<\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{\\left(1\\frac{1}{9} \\times 1 \\frac{1}{20} \\div \\frac{21}{38} – \\frac{1}{3}\\right) \\div \\left(2\\frac{4}{9} \\div 1\\frac{7}{15} of \\frac{3}{5}\\right)}{\\frac{1}{5} of \\frac{1}{5} \\div \\frac{1}{125} – \\frac{1}{25} \\div \\frac{1}{5} of \\frac{1}{5}}$<\/p>\n =$\\frac{\\left(\\frac{10}{9} \\times\u00a0 \\frac{21}{20} \\div \\frac{21}{38} – \\frac{1}{3}\\right) \\div \\left(\\frac{22}{9} \\div \\frac{22}{25} \\right)}{\\frac{1}{25}\u00a0 \\div \\frac{1}{125} – \\frac{1}{25} \\div \\frac{1}{25}}$<\/p>\n =$\\frac{\\left(\\frac{10}{9} \\times \\frac{19}{10} – \\frac{1}{3}\\right) \\div \\left(\\frac{25}{9} \\right)}{5 – 1}$<\/p>\n =$\\frac{\\left(\\frac{19}{9} – \\frac{1}{3}\\right) \\div \\left(\\frac{25}{9} \\right)}{4}$<\/p>\n =$\\frac{\\left(\\frac{16}{9} \\right) \\times\u00a0 \\left(\\frac{9}{25} \\right)}{4}$<\/p>\n = $\\frac{4}{25}$ = 0.16<\/p>\n Question 10:\u00a0<\/b>In an office, $\\frac{5}{8}$ of the total number of employees are males and the rest are females. $\\frac{2}{5}$ of the number of males are non technical workers while $\\frac{2}{3}$ of the number of females are technical workers, What fraction of the total number of employees are technical workers?<\/p>\n a)\u00a0$\\frac{5}{8}$<\/p>\n b)\u00a0$\\frac{2}{5}$<\/p>\n c)\u00a0$\\frac{1}{2}$<\/p>\n d)\u00a0$\\frac{3}{8}$<\/p>\n 10)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let the total number of employees be 8.<\/p>\n Total number of males employee = 8 $\\times\u00a0\\frac{5}{8}$ = 5<\/p>\n Total number of females employee = 8 – 5 = 3<\/p>\n Non technical males workers = 5 $\\times\u00a0\\frac{2}{5}$ = 2<\/p>\n Technical males workers = 5 – 2 = 3<\/p>\n Technical females workers = 3 $\\times\u00a0\\frac{2}{3}$ = 2<\/p>\n total number of technical worker = 3 + 2 = 5<\/p>\n Fraction of the total number of technical workers = $\\frac{total number of technical workers}{total number of employee} = \\frac{5}{8}$<\/p>\n Question 11:\u00a0<\/b>a, b and c are three fractions such that a < b < c. If c is divided by a, the result is $\\frac{9}{2}$, which exceeds b by $\\frac{23}{6}$. The sum of a, b and c is $\\frac{19}{12}$ , What is the value of (2a + b – c)?<\/p>\n a)\u00a0$\\frac{1}{2}$<\/p>\n b)\u00a0$\\frac{1}{3}$<\/p>\n c)\u00a0$\\frac{1}{12}$<\/p>\n d)\u00a0$\\frac{1}{4}$<\/p>\n 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{c}{a} = \\frac{9}{2}$<\/p>\n $c =\u00a0\\frac{9a}{2}$<\/p>\n b +\u00a0$\\frac{23}{6} = \\frac{9}{2}$<\/p>\n b =\u00a0$\\frac{9}{2} –\u00a0\\frac{23}{6} = \\frac{2}{3}$<\/p>\n a + b + c = 19\/12<\/p>\n $a +\u00a0 \\frac{2}{3} +\u00a0\\frac{9a}{2} =\u00a0\\frac{19}{2}$<\/p>\n $\\frac{11a}{2} =\u00a0\\frac{19}{2} –\u00a0 \\frac{2}{3}$<\/p>\n $\\frac{11a}{2} =\u00a0\\frac{11}{12}$<\/p>\n a = $\\frac{1}{6}$<\/p>\n $c = \\frac{9}{2} \\times \\frac{1}{6} = \\frac{3}{4}$<\/p>\n 2a + b – c =\u00a0$\\frac{2}{6} +\u00a0\\frac{2}{3} –\u00a0\\frac{3}{4} = \\frac{3}{12} = \\frac{1}{4}$<\/p>\n Question 12:\u00a0<\/b>Three fractions $x, y$ and $z$\u00a0 are such that $x > y > z$.When small of them divided by the greatest, the result is $\\frac{9}{16}$, which exceeds $y$ by 0.0625.If $x + y\u00a0+ z\u00a0= 1 \\frac{13}{24}$,then the value of $x + z$ is<\/p>\n a)\u00a0$\\frac{7}{8}$<\/p>\n b)\u00a01<\/p>\n c)\u00a0$\\frac{25}{24}$<\/p>\n d)\u00a0$\\frac{7}{6}$<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n $\\frac{z}{x} = \\frac{9}{16}$ Question 13:\u00a0<\/b>Two-third of the number of employees of a companyare males andthe rest are females. If $\\frac{3}{8}$ of the male employees and $\\frac{2}{5}$ of the female employees are temporary employees and the total number of permanent employees is 740. then $\\frac{7}{15}$ of the total number of employees exceeds the number of temporary female employees by:<\/p>\n a)\u00a0400<\/p>\n b)\u00a0340<\/p>\n c)\u00a0308<\/p>\n d)\u00a0320<\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n let the total employees be x. Question 14:\u00a0<\/b>If $\\surd{33} = 5.745$, then the value of the following is approximately: a)\u00a06.32<\/p>\n b)\u00a02.035<\/p>\n c)\u00a01<\/p>\n d)\u00a00.5223<\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Given,\u00a0$\\sqrt{33}=5.745$<\/p>\n $=$>\u00a0$\\sqrt{\\frac{3}{11}}=\\sqrt{\\frac{3}{11}\\times\\frac{11}{11}}=\\ \\frac{\\sqrt{33}}{11}=\\ \\frac{5.745}{11}=0.5223$<\/p>\n Question 15:\u00a0<\/b>Ali had \u20b9 320. He spent $\\frac{3}{4}$ of it to buy a watch. Of the remainder, he used $\\frac{1}{8}$ of it to buy a pen. How much money is left?<\/p>\n a)\u00a070<\/p>\n b)\u00a0120<\/p>\n c)\u00a090<\/p>\n d)\u00a0100<\/p>\n 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n Money spent to buy a watch =\u00a0$\\frac{3}{4}\\times320$ = \u20b9 240<\/p>\n Remaining Amount = 320 – 240 =\u00a0\u20b9 80<\/p>\n Money spent to buy a pen =\u00a0$\\frac{1}{8}\\times80$ =\u00a0\u20b9 10<\/p>\n $\\therefore\\ $Money left with Ali after buying a watch and a pen = 320 – 240 -10 =\u00a0\u20b9 70<\/p>\n Hence, the correct answer is Option A<\/p>\n Question 16:\u00a0<\/b>Which one of the following is true?<\/p>\n a)\u00a0$0 > \\frac{7}{17} > \\frac{3}{7} > \\frac{3}{5}$<\/p>\n b)\u00a0$0.5 < \\frac{2}{3} < \\frac{3}{4} < \\left(\\frac{16}{25}\\right)^{0.5}$<\/p>\n c)\u00a0$\\frac{7}{24} > \\frac{1}{3} > \\frac{3}{8} > \\frac{5}{12}$<\/p>\n d)\u00a0$\\frac{1}{2} > \\frac{2}{3} > \\frac{3}{4} > \\frac{4}{5}$<\/p>\n 16)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Option A<\/u><\/u><\/u><\/p>\n $0>\\frac{7}{17}>\\frac{3}{7}>\\frac{3}{5}$<\/p>\n $=$> $0>\\frac{7}{17}\\times\\frac{35}{35}\\ >\\frac{3}{7}\\times\\frac{85}{85}>\\frac{3}{5}\\times\\frac{119}{119}$ $=$> $0>\\frac{245}{525}\\ >\\frac{255}{525}>\\frac{357}{525}$<\/p>\n Option B<\/u><\/p>\n $0.5<\\frac{2}{3}<\\frac{3}{4}<\\left(\\frac{16}{25}\\right)^{0.5}$<\/p>\n $=$> $\\frac{1}{2}<\\frac{2}{3}<\\frac{3}{4}<\\frac{4}{5}$ $=$> $\\frac{30}{60}<\\frac{40}{60}<\\frac{45}{60}<\\frac{48}{60}$ Option C<\/u><\/p>\n $\\frac{7}{24}>\\frac{1}{3}>\\frac{3}{8}>\\frac{5}{12}$<\/p>\n $=$> $\\frac{7}{24}>\\frac{8}{24}>\\frac{9}{24}>\\frac{10}{24}$<\/p>\n Option D<\/u><\/p>\n $\\frac{1}{2}>\\frac{2}{3}>\\frac{3}{4}>\\frac{4}{5}$<\/p>\n $=$> $\\frac{30}{60}>\\frac{40}{60}>\\frac{45}{60}>\\frac{48}{60}$<\/p>\n Hence, the correct answer is Option B<\/p>\n Question 17:\u00a0<\/b>In an office, there are 216 tables and 264 chairs. If $\\frac{1}{6}$ of the tables and $\\frac{1}{4}$ of the chairs are broken then how many people can work in the office if each person requires one table and one chair?<\/p>\n a)\u00a0180<\/p>\n b)\u00a0186<\/p>\n c)\u00a0100<\/p>\n d)\u00a0198<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n In an office, there are 216 tables and 264 chairs. If $\\frac{1}{6}$ of the tables and $\\frac{1}{4}$ of the chairs are broken.<\/p>\n Remaining\u00a0tables =\u00a0216 of\u00a0$(1-\\frac{1}{6})$ =\u00a0216 of $\\frac{5}{6}$ = 180<\/p>\n Remaining\u00a0chairs =\u00a0264 of\u00a0$(1-\\frac{1}{4})$ =\u00a0264 of $\\frac{3}{4}$ =\u00a0198<\/p>\n In question, it is given that each person requires one table and one chair to work in the office. There are 180 tables and 198 chairs remaining. So we can say that 180 people can work in the office.<\/p>\n Question 18:\u00a0<\/b>A fraction is such that the numerator is five less than the denominator. Also four times the numerator is one more than the denominator. The fraction is:<\/p>\n a)\u00a0$\\frac{4}{7}$<\/p>\n b)\u00a0$\\frac{3}{8}$<\/p>\n c)\u00a0$\\frac{7}{12}$<\/p>\n d)\u00a0$\\frac{2}{7}$<\/p>\n 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Let’s assume the fraction is\u00a0$\\frac{P}{Q}$. P = Q-5<\/p>\n Q = P+5\u00a0 \u00a0 Eq.(i)<\/p>\n Also four times the numerator is one more than the denominator.<\/p>\n 4P = Q+1\u00a0 \u00a0\u00a0Eq.(ii)<\/p>\n Put\u00a0the value of ‘Q’ from\u00a0Eq.(i) to\u00a0Eq.(ii).<\/p>\n 4P = P+5+1<\/p>\n 4P-P = 6<\/p>\n 3P = 6<\/p>\n P = 2<\/strong><\/p>\n Put the value of ‘P’ in Eq.(i).<\/p>\n Q = P+5 = 2+5 = 7<\/strong><\/p>\n fraction =\u00a0$\\frac{P}{Q}\\ =\\ \\frac{2}{7}$<\/p>\n Question 19:\u00a0<\/b>The median of the given data $\\frac{1}{2}, \\frac{2}{7}, \\frac{3}{4}, \\frac{1}{3}, \\frac{5}{8}$ is:<\/p>\n a)\u00a0$\\frac{3}{4}$<\/p>\n b)\u00a0$\\frac{1}{3}$<\/p>\n c)\u00a0$\\frac{2}{7}$<\/p>\n d)\u00a0$\\frac{1}{2}$<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Median is the middle term when the given data is arranged in ascending order from left to right.<\/p>\n Here the given data is in fraction. So first we need to take the LCM of the\u00a0denominator.<\/p>\n LCM of (2, 7, 4, 3, 8) =\u00a0168 $\\frac{2}{7} = \\frac{48}{168}$<\/p>\n $\\frac{3}{4} = \\frac{126}{168}$<\/p>\n $\\frac{1}{3} = \\frac{56}{168}$<\/p>\n $\\frac{5}{8} = \\frac{105}{168}$<\/p>\n Now arrange the given data in ascending order.<\/p>\n $\\frac{48}{168}$,\u00a0$\\frac{56}{168}$, $\\frac{84}{168}$, $\\frac{105}{168}$, $\\frac{126}{168}$<\/p>\n So medium = 3rd term =\u00a0$\\frac{84}{168}$<\/p>\n =\u00a0$\\frac{1}{2}$<\/p>\n Question 20:\u00a0<\/b>What is the value of: $\\frac{12\u00a0 of\u00a0 3 \\div 6 + 12 \\times 2 – (2 \\times 4 – 5)}{12 \\div 3 \\times 4 + (2 \\times 4 – 5)}$?<\/p>\n a)\u00a0$\\frac{27}{22}$<\/p>\n b)\u00a0$\\frac{23}{17}$<\/p>\n c)\u00a0$\\frac{27}{19}$<\/p>\n d)\u00a0$\\frac{21}{9}$<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n =\u00a0$\\frac{12\\times\\ \\frac{3}{6}+24-(8-5)}{\\frac{12}{3}\\times4+(8-5)}$<\/p>\n = $\\frac{6+24-3}{16+3}$<\/p>\n = $\\frac{27}{19}$<\/p>\n
\n$\\left(\\frac{3}{4}-\\frac{1}{4}\\div\\frac{1}{4}\u00a0 \\text{of}\u00a0 \\frac{2}{5}\\right)\\div\\left(\\frac{3}{4}\\div\\frac{2}{3}\u00a0 \\text{of} \u00a0\\frac{3}{5}\\right)$<\/p>\n
\n$\\frac{7}{12} \\div \\frac{1}{10}\u00a0 \\text{of}\u00a0 \u00a0\\frac{2}{3} – \\frac{5}{3} \\times \\frac{9}{10} + \\frac{5}{8} \\div \\frac{3}{4}\u00a0 \\text{of}\u00a0\u00a0 \\frac{2}{3}$<\/p>\n
\n$\\frac{9}{16} = y + 0.00625$
\ny = $\\frac{1}{z}$
\nx +\u00a0y +\u00a0z = $1 \\frac{13}{24}$
\nx + $\\frac{1}{z} +\u00a0z =\u00a0\\frac{37}{24}$
\nx + z =\u00a0$\\frac{37}{24} – \\frac{12}{24} = \\frac{25}{14}$<\/p>\n
\nMale\u00a0employees = $\\frac{2x}{3}$
\nFemale employees = x –\u00a0$\\frac{2x}{3}$ =\u00a0$\\frac{x}{3}$
\nPermanent male employees = 1 –\u00a0$\\frac{3}{8}$ =\u00a0$\\frac{5}{8}$ of the male employee =\u00a0$\\frac{2x}{3} \\times\u00a0\\frac{5}{8}$ =\u00a0$\\frac{5x}{12}$
\nPermanent female employees = 1 – $\\frac{2}{5}$ = $\\frac{3}{5}$\u00a0of the male employee = $\\frac{x}{3} \\times \\frac{3}{5}$ = $\\frac{x}{5}$
\nTotal number of permanent employees = 740
\n$\\frac{5x}{12}$ +\u00a0$\\frac{x}{5}$ = 740
\n$\\frac{37x}{60}$ = 740
\nx = 740 $\\times \\frac{60}{37} = 1200$
\n$\\frac{7}{15}$ of the total number of employees =\u00a01200 $\\times\u00a0\\frac{7}{15}$ = 560
\nNumber of temporary female employees =\u00a0$\\frac{x}{3} \\times \\frac{2}{5}$ = $\\frac{2x}{15}$
\n=\u00a0$\\frac{2 \\times 1200}{15}$ = 160
\n$\\frac{7}{15}$ of the total number of employees exceeds the number of temporary female employees by\u00a0= 560 – 160 = 400<\/p>\n
\n$\\surd{\\left(\\frac{3}{11}\\right)}$<\/p>\n
\n<\/u><\/p>\n
\n<\/u><\/p>\n
\n<\/u><\/u><\/p>\n
\nA fraction is such that the numerator is five less than the denominator.<\/p>\n
\n$\\frac{1}{2} =\u00a0\\frac{84}{168}$<\/p>\n