Here you can download CMAT 2022 – important Number Series Questions PDF by Cracku. Very Important Number Series Questions for CMAT 2022 based on asked questions in previous exam papers. These questions will help your CMAT preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Number Series Questions for cmat<\/a><\/p>\n Enroll to CMAT crash course<\/a><\/p>\n Take CMAT mock tests here<\/a><\/p>\n Download CMAT previous papers PDF<\/a><\/p>\n Question 1:\u00a0<\/b>What should come next in the number series ? a)\u00a01<\/p>\n b)\u00a02<\/p>\n c)\u00a04<\/p>\n d)\u00a08<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>What will come in place of both the question marks (?) in the following question ?$\\frac{(?)^{0.6}}{104}=\\frac{26}{(?)^{1.4}}$<\/p>\n a)\u00a058<\/p>\n b)\u00a0-48<\/p>\n c)\u00a0-56<\/p>\n d)\u00a042<\/p>\n e)\u00a0-52<\/p>\n Question 3:\u00a0<\/b>Out of the fractions $\\frac{1}{2}, \\frac{7}{8}, \\frac{3}{4}, \\frac{5}{6}$, and $\\frac{6}{7}$ what is the difference between the largest and smallest fractions ?<\/p>\n a)\u00a0$\\frac{7}{13}$<\/p>\n b)\u00a0$\\frac{3}{8}$<\/p>\n c)\u00a0$\\frac{4}{7}$<\/p>\n d)\u00a0$\\frac{1}{6}$<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>What will come in place of the question mark (?) in the following number series? a)\u00a01589<\/p>\n b)\u00a01598<\/p>\n c)\u00a01958<\/p>\n d)\u00a01985<\/p>\n e)\u00a01835<\/p>\n Question 5:\u00a0<\/b>What will come in place of thequestion mark (?) in the following number series? a)\u00a01008<\/p>\n b)\u00a0948<\/p>\n c)\u00a0944<\/p>\n d)\u00a0940<\/p>\n e)\u00a01005<\/p>\n Get 5 CMAT mocks at just Rs.299<\/a><\/p>\n Question 6:\u00a0<\/b>What will come in place of thequestion mark (?) in the following number series? a)\u00a044<\/p>\n b)\u00a055<\/p>\n c)\u00a066<\/p>\n d)\u00a033<\/p>\n e)\u00a035<\/p>\n Question 7:\u00a0<\/b>What will come in place of thequestion mark (?) in the following number series? a)\u00a018.5<\/p>\n b)\u00a016.5<\/p>\n c)\u00a08.5<\/p>\n d)\u00a08.25<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>What will come in place of the question mark (?) in the following number series? a)\u00a095<\/p>\n b)\u00a090<\/p>\n c)\u00a0100<\/p>\n d)\u00a0105<\/p>\n e)\u00a096<\/p>\n Question 9:\u00a0<\/b>What should come in place of the question mark (?) in the following number series ? a)\u00a01168<\/p>\n b)\u00a01254<\/p>\n c)\u00a01457<\/p>\n d)\u00a01372<\/p>\n e)\u00a0None of these<\/p>\n Question 10:\u00a0<\/b>What approximate value should come in place of the question mark (?) in the following question? a)\u00a084<\/p>\n b)\u00a048<\/p>\n c)\u00a0118<\/p>\n d)\u00a058<\/p>\n e)\u00a024<\/p>\n Download CMAT Previous Papers PDF<\/a><\/p>\n Question 11:\u00a0<\/b>What should come in place of the question mark (?) in the following number series? a)\u00a0168<\/p>\n b)\u00a0154<\/p>\n c)\u00a0191<\/p>\n d)\u00a0172<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\u00a0<\/b>What should come in place of the question mark (?) in the following number series? a)\u00a01268<\/p>\n b)\u00a01234<\/p>\n c)\u00a01272<\/p>\n d)\u00a01216<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n What should come in place of question mark (?) in the following number series ?<\/p>\n Question 13:\u00a0<\/b>31, 35, 44, 60, 85, ?<\/p>\n a)\u00a0121<\/p>\n b)\u00a0111<\/p>\n c)\u00a0109<\/p>\n d)\u00a097<\/p>\n e)\u00a0None of these<\/p>\n Question 14:\u00a0<\/b>9, 49, 201, 1009, ?, 20209, 80841<\/p>\n a)\u00a04054<\/p>\n b)\u00a04049<\/p>\n c)\u00a04050<\/p>\n d)\u00a04041<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>1, 121, 441, 961, 1681, ?<\/p>\n a)\u00a02701<\/p>\n b)\u00a02511<\/p>\n c)\u00a02611<\/p>\n d)\u00a02801<\/p>\n e)\u00a0None of these<\/p>\n Question 16:\u00a0<\/b>668, 656, 632, 584, ?, 296<\/p>\n a)\u00a0392<\/p>\n b)\u00a0438<\/p>\n c)\u00a0488<\/p>\n d)\u00a0536<\/p>\n e)\u00a0None of these<\/p>\n Question 17:\u00a0<\/b>36, 20, ?, 8, 6, 5<\/p>\n a)\u00a010<\/p>\n b)\u00a012<\/p>\n c)\u00a014<\/p>\n d)\u00a016<\/p>\n e)\u00a0None of these<\/p>\n Question 18:\u00a0<\/b>What is the least number to be added to 2530 to make it a perfect square ?<\/p>\n a)\u00a050<\/p>\n b)\u00a065<\/p>\n c)\u00a075<\/p>\n d)\u00a080<\/p>\n e)\u00a0None of these<\/p>\n Question 19:\u00a0<\/b>What would be the compound interest accrued on an amount of Rs. 9,000 at the rate of 11 p.c.p.a. in two years ?<\/p>\n a)\u00a0Rs. 2089.90<\/p>\n b)\u00a0Rs. 2140.90<\/p>\n c)\u00a0Rs. 2068.50<\/p>\n d)\u00a0Rs. 2085.50<\/p>\n e)\u00a0None of these<\/p>\n Question 20:\u00a0<\/b>16 8 12 30 ? 472.5<\/p>\n a)\u00a0104<\/p>\n b)\u00a0103<\/p>\n c)\u00a0106<\/p>\n d)\u00a0105<\/p>\n e)\u00a0None of these<\/p>\n Take CMAT Mock Tests<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The series can be shown as below<\/p>\n 1 8 3 6 5 4 7 2 9 After this, the series has to repeat itself and hence, the next term will be 1<\/p>\n 2)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $\\frac{(x)^{0.6}}{104}=\\frac{26}{(x)^{1.4}}$<\/p>\n ${(x)^{0.6}} * {(x)^{01.4}}$ = 104*26<\/p>\n ${(x)^{2}}$ = 104*26<\/p>\n x = \u00b1<\/span><\/span>52<\/span><\/p>\n 3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Given values are , $\\frac{7}{8}$ = 0.87<\/p>\n $\\frac{3}{4}$ = 0.75<\/p>\n $\\frac{5}{6}$ = 0.83<\/p>\n $\\frac{6}{7}$ = 0.86<\/p>\n \u2234 Required difference = $\\frac{7}{8}$ – $\\frac{1}{2}$ = (7-4)\/8 = 3\/8<\/p>\n 4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The pattern followed is\u00a0:<\/p>\n 9 $\\times 1 + 1^2$ = 10<\/p>\n 10\u00a0$\\times 3 + 3^2$ =\u00a039<\/p>\n 39\u00a0$\\times 5 + 5^2$ =\u00a0220<\/p>\n 220\u00a0$\\times 7 + 7^2$ =\u00a01589<\/strong><\/p>\n 1589\u00a0$\\times 9 + 9^2$ =\u00a014382<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Each number is multiplied by 2 and then 4 is subtracted from it.<\/p>\n 121 $\\times 2 – 4$ = 238<\/p>\n 238\u00a0$\\times 2 – 4$ =\u00a0472<\/p>\n 472\u00a0$\\times 2 – 4$ =\u00a0940<\/strong><\/p>\n 940\u00a0$\\times 2 – 4$ =\u00a01876<\/p>\n 1876\u00a0$\\times 2 – 4$ =\u00a03748<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Each number is multiplied by $\\frac{3}{2}$<\/p>\n 44 $\\times \\frac{3}{2}$ =\u00a066<\/strong><\/p>\n 66\u00a0$\\times \\frac{3}{2}$ =\u00a099<\/p>\n 99\u00a0$\\times \\frac{3}{2}$ =\u00a0148.5<\/p>\n 148.5\u00a0$\\times \\frac{3}{2}$ =\u00a0222.75<\/p>\n 222.75\u00a0$\\times \\frac{3}{2}$ =\u00a0334.125<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The pattern followed is\u00a0:<\/p>\n 33 $\\times \\frac{1}{2}$ = 16.5<\/p>\n 16.5\u00a0$\\times \\frac{2}{2}$ =\u00a016.5<\/strong><\/p>\n 16.5\u00a0$\\times \\frac{3}{2}$ =\u00a024.75<\/p>\n 24.75\u00a0$\\times \\frac{4}{2}$ =\u00a049.5<\/p>\n 49.5\u00a0$\\times \\frac{5}{2}$ =\u00a0123.75<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Numbers of the form $n^2 – (n-1)$ are added, where $n$ is an integer starting from 2<\/p>\n 23 $+ 2^2 – 1$ = 23<\/p>\n 23\u00a0$+ 3^2 – 2$ =\u00a030<\/p>\n 30\u00a0$+ 4^2 – 3$ =\u00a043<\/p>\n 43\u00a0$+ 5^2 – 4$ =\u00a064<\/p>\n 64\u00a0$+ 6^2 – 5$ =\u00a095<\/strong><\/p>\n 9)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 1 * 3 + 2 = 5<\/p>\n 5\u00a0* 3 + 2 =\u00a017<\/p>\n 17\u00a0* 3 + 2 =\u00a053<\/p>\n 53\u00a0* 3 + 2 =\u00a0161<\/p>\n 161\u00a0* 3 + 2 =\u00a0485<\/p>\n 485\u00a0* 3 + 2 =\u00a01457<\/strong><\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Expression\u00a0: $54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n = $\\frac{55}{10} \\times 4.5$<\/p>\n = $24.75 \\approx 24$<\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 2 * 2 + 1 = 5<\/p>\n 5 *\u00a02 + 1 =\u00a0\u00a011<\/p>\n 11 *\u00a02 + 1 =\u00a023<\/p>\n 23 *\u00a02 + 1 =\u00a047<\/p>\n 47 *\u00a02 + 1 =\u00a095<\/p>\n 95 *\u00a02 + 1 =\u00a0191<\/strong><\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 1\u00a0* 3 +\u00a01 = 4<\/p>\n 4\u00a0* 3 +\u00a02 = 14<\/p>\n 14\u00a0* 3 + 3 = 45<\/p>\n 45\u00a0* 3 +\u00a04 = 139<\/p>\n 139\u00a0* 3 + 5 = 422<\/p>\n 422 * 3 + 6 =\u00a01272<\/strong><\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 31 + $2^2$ = 35<\/p>\n 35 +\u00a0$3^2$ =\u00a044<\/p>\n 44 +\u00a0$4^2$ =\u00a060<\/p>\n 60 +\u00a0$5^2$ =\u00a085<\/p>\n 85 +\u00a0$6^2$ =\u00a0121<\/strong><\/p>\n 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 9 \u00a0 \u00a0 \u00a0* 5 + 4 = 49<\/p>\n 49 \u00a0 \u00a0 * 4 + 5 = 201<\/p>\n 201 \u00a0 * 5 + 4 = 1009<\/p>\n 1009 \u00a0* 4 + 5 =\u00a04041<\/strong><\/p>\n 4041 \u00a0* 5 + 4 = 20209<\/p>\n 20209 * 4 + 5 = 80841<\/p>\n 15)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n $1^2$ \u00a0= 1<\/p>\n $11^2$ =\u00a0121<\/p>\n $21^2$ =\u00a0441<\/p>\n $31^2$ =\u00a0961<\/p>\n $41^2$ =\u00a01681<\/p>\n $51^2$ =\u00a02601<\/strong><\/p>\n 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n 668 – 1 * 12 = 656<\/p>\n 656 – 2 * 12 = 632<\/p>\n 632 – 4 * 12 = 584<\/p>\n 584 – 8 * 12 =\u00a0488<\/strong><\/p>\n 488 – 16 * 12 = 296<\/p>\n 17)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The pattern here followed is\u00a0:<\/p>\n $\\frac{36}{2} + 2$ = 20<\/p>\n $\\frac{20}{2} + 2$ =\u00a012<\/strong><\/p>\n $\\frac{12}{2} + 2$ = 8<\/p>\n $\\frac{8}{2} + 2$ =\u00a06<\/p>\n $\\frac{6}{2} + 2$ =\u00a05<\/p>\n 18)\u00a0Answer\u00a0(E)<\/strong><\/p>\n We know that $50^2 = 2500$ and $51^2 = 2601$<\/p>\n $\\because$ 2500\u00a0< 2530\u00a0< 2601<\/p>\n $\\therefore$ Required number = 2601 – 2530 = 71<\/p>\n 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $C.I. = P [(1 + \\frac{R}{100})^T – 1]$<\/p>\n = $9000 [(1 + \\frac{11}{100})^2 – 1]$<\/p>\n = $9000 [(1.11)^2 – 1]$<\/p>\n = $9000 \\times (1.2321 – 1)$<\/p>\n = $9000 \\times 0.2321$ = Rs. $2,088.90$<\/p>\n 20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Odd multiples of $\\frac{1}{2}$ are multiplied<\/p>\n 16 $\\times \\frac{1}{2}$ = 8<\/p>\n 8\u00a0$\\times \\frac{3}{2}$ =\u00a012<\/p>\n 12\u00a0$\\times \\frac{5}{2}$ =\u00a030<\/p>\n 30\u00a0$\\times \\frac{7}{2}$ =\u00a0105<\/strong><\/p>\n 105\u00a0$\\times \\frac{9}{2}$ =\u00a0472.5<\/p>\n
\n1 8 3 6 5 4 7 2 9 1 8 3 6 5 4 7 2 1 8 3 6 5 4 7 1 8 3 6 5 4<\/p>\n
\n9\u00a0 10\u00a0 39\u00a0 220\u00a0 ?\u00a0 14382<\/p>\n
\n121\u00a0 238\u00a0 472\u00a0 ?\u00a0 1876\u00a0 3748<\/p>\n
\n44\u00a0 ?\u00a0 99\u00a0 \u00a0148.5 \u00a0222.75\u00a0 334.125<\/p>\n
\n33\u00a0 16.5\u00a0 ?\u00a0 24.75\u00a0 \u00a0 49.5\u00a0 123.75<\/p>\n
\n20\u00a0 23\u00a0 30\u00a0 43\u00a0 64\u00a0 ?<\/p>\n
\n1, 5, 17, 53, 161, 485, ?<\/p>\n
\n$54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n
\n2 5 11 23 47 95 ?<\/p>\n
\n1 4 14 45 139 422 ?<\/p>\n
\n1 8 3 6 5 4 7 2
\n1 8 3 6 5 4 7
\n1 8 3 6 5 4<\/p>\n
\n$\\frac{1}{2}$ = 0.5<\/p>\n