Enroll to CMAT crash course<\/a><\/p>\n Take Free CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>The rational number, which equals the number 2.($\\overline{357}$) with recurring decimal is:<\/p>\n a)\u00a02335\/1001<\/p>\n b)\u00a02379\/997<\/p>\n c)\u00a02355\/999<\/p>\n d)\u00a00<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b>Instructions: Study the following table carefully to answer these questions.<\/p>\n Percentage of marks obtained by six students in six different subjects. Question 2:\u00a0<\/b>What is the average percentage of marks obtained by all subjects in Hindi? (approximated to two places of decimal)<\/p>\n a)\u00a077.45%<\/p>\n b)\u00a079.33%<\/p>\n c)\u00a075.52%<\/p>\n d)\u00a073.52%<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b>\u200bInstructions: Study the table carefully to answer the questions that follow: a)\u00a086.36<\/p>\n b)\u00a0101.71<\/p>\n c)\u00a0111.79<\/p>\n d)\u00a0133.33<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b>Instructions: Study the table carefully to answer the questions that follow: Question 4:\u00a0<\/b>The number of children visiting Super Market C forms what percent of the number of children visiting Supermarket F? (rounded off to two digits after decimal)<\/p>\n a)\u00a091.49<\/p>\n b)\u00a049.85<\/p>\n c)\u00a0121.71<\/p>\n d)\u00a0109.30<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>What is the percent rice in production in 2007 from 2006? (Round off to two digits afer decimal.) a)\u00a028.18%<\/p>\n b)\u00a018.18%<\/p>\n c)\u00a016.28%<\/p>\n d)\u00a026.18%<\/p>\n e)\u00a0None of these<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 6:\u00a0<\/b>What is the value of $(sin60\u00b0 + \\frac{2}{\\sqrt3}$) ?<\/p>\n a)\u00a0$(2\\sqrt{3}+1\u221a3)$<\/p>\n b)\u00a0$7\/2\\sqrt{3}$<\/p>\n c)\u00a03<\/p>\n d)\u00a0$2+\\sqrt{3}$<\/p>\n Question 7:\u00a0<\/b>Arrange the fractions $\\frac{3}{4},\\frac{5}{12},\\frac{13}{16},\\frac{16}{29},\\frac{3}{8}$ ascending order of magnitude<\/p>\n a)\u00a0$\\frac{3}{4}<\\frac{3}{8}<\\frac{13}{16}<\\frac{16}{29}<\\frac{5}{12}$<\/p>\n b)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{3}{4}<\\frac{13}{16}$<\/p>\n c)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{13}{16}<\\frac{3}{4}$<\/p>\n d)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{13}{16}<\\frac{16}{29}<\\frac{3}{4}$<\/p>\n Question 8:\u00a0<\/b>Arrange the fractions $\\frac{3}{4},\\frac{5}{12},\\frac{13}{16},\\frac{16}{29},\\frac{3}{8}$ ascending order of magnitude<\/p>\n a)\u00a0$\\frac{3}{4}<\\frac{3}{8}<\\frac{13}{16}<\\frac{16}{29}<\\frac{5}{12}$<\/p>\n b)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{3}{4}<\\frac{13}{16}$<\/p>\n c)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{13}{16}<\\frac{3}{4}$<\/p>\n d)\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{13}{16}<\\frac{16}{29}<\\frac{3}{4}$<\/p>\n Question 9:\u00a0<\/b>A ship is full of cargo containers. It drops 2\/3 of cargo containers at the first port and takes 60 more, at the second port it drops two third of the new total and takes eleven more. On arriving at the third port it is found that it have 48 cargo containers. Find the number of cargo containers in the ship at the starting?<\/p>\n a)\u00a0189<\/p>\n b)\u00a0159<\/p>\n c)\u00a0161<\/p>\n d)\u00a0153<\/p>\n Question 10:\u00a0<\/b>Out of a group of three numbers, second is thrice of the first and also twice of the third. If the sum of the three numbers is 55 then find the largest number in the group.<\/p>\n a)\u00a026<\/p>\n b)\u00a029<\/p>\n c)\u00a030<\/p>\n d)\u00a032<\/p>\n Question 11:\u00a0<\/b>If $a^{2}+b^{2}=80$ and $ab=32$, then calculate the value of $\\frac{a-b}{a+b}$<\/p>\n a)\u00a00.337<\/p>\n b)\u00a00.339<\/p>\n c)\u00a00.333<\/p>\n d)\u00a00.335<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 2.($\\overline{357}$)=2+0.($\\overline{357}$) =2+357\/999=(2\u00d7999+357)\/999=2355\/999<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Average percentage = $\\frac{88 + 92 + 76 + 83 + 65 + 72}{6}$ = 79.33<\/p>\n 3)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Marks obtained by Gravita in Brand Management = 88 % of 100 = $\\frac{75}{100}$ x 100 = 88 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Number of children visiting supermarket C = 20% of 45640 = $\\frac{20}{100}$ x 45640 = 9128 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The rice production in 2006 is 1100 and the rice production in 2007 is 1300. 6)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Expression : $sin(60^\\circ)+\\frac{2}{\\sqrt3}$<\/p>\n =\u00a0$\\frac{\\sqrt3}{2}+\\frac{2}{\\sqrt3}$<\/p>\n = $\\frac{3+4}{2\\sqrt3}$<\/p>\n = $\\frac{7}{2\\sqrt3}$<\/p>\n => Ans – (B)<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Fractions\u00a0: $\\frac{3}{4},\\frac{5}{12},\\frac{13}{16},\\frac{16}{29},\\frac{3}{8}$<\/p>\n L.C.M. of denominators (4,12,16,29,8) = 1392<\/p>\n Thus, fractions are changed to\u00a0: $\\frac{1044}{1392},$\u00a0$\\frac{580}{1392},$\u00a0$\\frac{1131}{1392},$\u00a0$\\frac{768}{1392},$\u00a0$\\frac{522}{1392}$<\/p>\n Now the denominators are equal and fraction are arranged on the basis of numerators<\/p>\n =\u00a0$\\frac{522}{1392}<$\u00a0$\\frac{580}{1392}<$\u00a0$\\frac{768}{1392}<$\u00a0$\\frac{1044}{1392}<$\u00a0$\\frac{1131}{1392}$<\/p>\n $\\equiv$\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{3}{4}<\\frac{13}{16}$<\/p>\n => Ans – (B)<\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Fractions\u00a0: $\\frac{3}{4},\\frac{5}{12},\\frac{13}{16},\\frac{16}{29},\\frac{3}{8}$<\/p>\n L.C.M. of denominators (4,12,16,29,8) = 1392<\/p>\n Thus, fractions are changed to\u00a0: $\\frac{1044}{1392},$\u00a0$\\frac{580}{1392},$\u00a0$\\frac{1131}{1392},$\u00a0$\\frac{768}{1392},$\u00a0$\\frac{522}{1392}$<\/p>\n Now the denominators are equal and fraction are arranged on the basis of numerators<\/p>\n =\u00a0$\\frac{522}{1392}<$\u00a0$\\frac{580}{1392}<$\u00a0$\\frac{768}{1392}<$\u00a0$\\frac{1044}{1392}<$\u00a0$\\frac{1131}{1392}$<\/p>\n $\\equiv$\u00a0$\\frac{3}{8}<\\frac{5}{12}<\\frac{16}{29}<\\frac{3}{4}<\\frac{13}{16}$<\/p>\n => Ans – (B)<\/p>\n 9)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Let number of cargo containers in the starting = $3x$<\/p>\n It drops 2\/3 of cargo containers at the first port and takes 60 more.<\/p>\n => Number of cargo containers when the ship left first port = $3x-(\\frac{2}{3}\\times3x)+60=(x+60)$<\/p>\n Similarly, number of cargo containers after leaving second port = on arriving at third port,<\/p>\n => $\\frac{x+60}{3}+11=48$<\/p>\n => $\\frac{x+60}{3}=48-11=37$<\/p>\n => $x+60=37\\times3=111$<\/p>\n => $x=111-60=51$<\/p>\n $\\therefore$ Number of cargo containers in the ship at the starting = $3\\times51=153$<\/p>\n => Ans – (D)<\/p>\n 10)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Let second number = $6x$<\/p>\n => First number = $2x$ and third number = $3x$<\/p>\n => Sum = $6x+2x+3x=55$<\/p>\n => $x=\\frac{55}{11}=5$<\/p>\n $\\therefore$ Largest number = $6\\times5=30$<\/p>\n => Ans – (C)<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Given\u00a0:\u00a0$a^{2}+b^{2}=80$ ———-(i) and $ab=32$<\/p>\n => $2ab=64$ ———–(ii)<\/p>\n Adding equations (i) and (ii), we get\u00a0:<\/p>\n => $a^2+b^2+2ab=80+64$<\/p>\n => $(a+b)^2=144$<\/p>\n => $(a+b)=\\sqrt{144}=12$<\/p>\n Similarly, $(a-b)=4$<\/p>\n $\\therefore$\u00a0$\\frac{a-b}{a+b}$<\/p>\n = $\\frac{4}{12}=0.333$<\/p>\n => Ans – (C)<\/p>\n
\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/05\/07\/4_iFDbeNp.png\")
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\nPercentage of Marks obtained by different students in different subjects of MBA<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/05\/07\/3_31QTgsK.png\")
\nQuestion 3:\u00a0<\/b>The marks obtained by Garvita in Brand Management is what percent of the marks obtained by Archita in the same subject? (rounded off to two digits after decimal)<\/p>\n
\nThe number of person visiting six different Super-markets and the percentage of Men, Women and Children visiting those Super markets<\/p>\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/05\/07\/1_4M6z45b.png\")
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\n![\"\"](\"https:\/\/cracku.in\/media\/uploads\/2015\/05\/27\/5.png\")
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\nMarks obtained by Archit in Brand Management = 76% of 100 = $\\frac{76}{100}$ x 100 = 76
\nPercentage of marks obtained by Gravita to Archit = $\\frac{88}{76}$ x 100 = 115.79%<\/p>\n
\nNumber of children visiting supermarket F = 14% of 59650 = $\\frac{14}{100}$ x 59650 = 8351
\nPrecentage =\u00a0$\\frac{9128}{8351}$ x 100 = 109.30%<\/p>\n
\nSo, the percentage increase is $\\frac{1300-1100}{1100}=18.18\\%$<\/p>\n