Here you can download CMAT 2022 – important MAH – CET Series Completion Questions PDF by Cracku. Very Important MAH – CET 2022 and These questions will help your MAH – CET preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Series Completion Questions for MAH – CET<\/a><\/p>\n Enroll to MAH-CET Crash Course<\/a><\/p>\n Question 1:\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 2:\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 Question 3:\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 4:\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 Question 5:\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 Take Free MAH-CET mock tests here<\/a><\/p>\n Question 6:\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 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 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 9:\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 Question 10:\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 11:\u00a0<\/b>What will come in the place of question mark (?) in the following series ? a)\u00a096<\/p>\n b)\u00a0106<\/p>\n c)\u00a0126<\/p>\n d)\u00a0130<\/p>\n e)\u00a0None of these<\/p>\n Question 12:\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 13:\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 Question 14:\u00a0<\/b>2, 5, 12, 27, 58, ?<\/p>\n a)\u00a0122<\/p>\n b)\u00a0121<\/p>\n c)\u00a0123<\/p>\n d)\u00a0120<\/p>\n e)\u00a0None of these<\/p>\n Question 15:\u00a0<\/b>18 19.7 16.3 23.1 9.5 ?<\/p>\n a)\u00a036.5<\/p>\n b)\u00a036.8<\/p>\n c)\u00a036.7<\/p>\n d)\u00a036.9<\/p>\n e)\u00a0None of these<\/p>\n Question 16:\u00a0<\/b>68, ?, 77, 104, 168, 293<\/p>\n a)\u00a069<\/p>\n b)\u00a070<\/p>\n c)\u00a068<\/p>\n d)\u00a074<\/p>\n e)\u00a0None of these<\/p>\n Question 17:\u00a0<\/b>In how many different ways can the numbers \u2018256974\u2019 be arranged, using each digit only once in each arrangement, such that the digits 6 and 5 are at the extreme ends in each arrangement ?<\/p>\n a)\u00a048<\/p>\n b)\u00a0720<\/p>\n c)\u00a036<\/p>\n d)\u00a0360<\/p>\n e)\u00a0None of these<\/p>\n Question 18:\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 19:\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 20:\u00a0<\/b>If $(11)^{3}$ is subtracted from $(46)^{2}$ . what will be the remainder ?<\/p>\n a)\u00a0787<\/p>\n b)\u00a0785<\/p>\n c)\u00a0781<\/p>\n d)\u00a0783<\/p>\n e)\u00a0None of these<\/p>\n Get 5 MAH-CET mocks at just Rs.299<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\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 2)\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 3)\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 4)\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 5)\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 6)\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 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(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 9)\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 10)\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 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Each number is of the form $(n^3+1)$ where $n$ is a natural number<\/p>\n $1^3+1$ = 2<\/p>\n $2^3+1$ =\u00a09<\/p>\n $3^3+1$ =\u00a028<\/p>\n $4^3+1$ =\u00a065<\/p>\n $5^3+1$ =\u00a0126<\/strong><\/p>\n => Ans – (C)<\/p>\n 12)\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 13)\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 14)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Each number is multiplied by 2 and then consecutive natural numbers are added<\/p>\n 2 $\\times 2 + 1$ = 5<\/p>\n 5\u00a0$\\times 2 + 2$ =\u00a012<\/p>\n 12\u00a0$\\times 2 + 3$ =\u00a027<\/p>\n 27\u00a0$\\times 2 + 4$ =\u00a058<\/p>\n 58\u00a0$\\times 2 + 5$ =\u00a0121<\/strong><\/p>\n 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The pattern is\u00a0:<\/p>\n 18 $+ 1.7 \\times 2^0$ = 19.7<\/p>\n 19.7\u00a0$- 1.7 \\times 2^1$ =\u00a016.3<\/p>\n 16.3\u00a0$+ 1.7 \\times 2^2$ =\u00a023.1<\/p>\n 23.1\u00a0$- 1.7 \\times 2^3$ =\u00a09.5<\/p>\n 9.5\u00a0$+ 1.7 \\times 2^4$ =\u00a036.7<\/strong><\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Cubes of consecutive natural numbers are added<\/p>\n 68 $+ 1^3$ =\u00a069<\/strong><\/p>\n 69\u00a0$+ 2^3$ =\u00a077<\/p>\n 77\u00a0$+ 3^3$ =\u00a0104<\/p>\n 104\u00a0$+ 4^3$ =\u00a0168<\/p>\n 168\u00a0$+ 5^3$ =\u00a0293<\/p>\n 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Case 1 : 6 at left end and 5 is at right end : 6 _ _ _ _ 5<\/p>\n Now, four empty places can be filled by 2,9,7 and 4 in = $4!$ ways<\/p>\n = $4 \\times 3 \\times 2 \\times 1 = 24$<\/p>\n Case 2 : 6 at right end and 5 at left end : 5 _ _ _ _ 6<\/p>\n Similarly, no. of ways = $4!$<\/p>\n = $4 \\times 3 \\times 2 \\times 1 = 24$<\/p>\n $\\therefore$ Total no. of ways = $24 + 24 = 48$<\/p>\n 18)\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 19)\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 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Here<\/p>\n $(46)^2$ = 2116<\/p>\n $(11)^3$ = 1331<\/p>\n So, 2116 – 1331 = 785<\/p>\n
\n1 4 14 45 139 422 ?<\/p>\n
\n2 5 11 23 47 95 ?<\/p>\n
\n$54.786 \\div 10.121 \\times 4.454 = ?$<\/p>\n
\n1, 5, 17, 53, 161, 485, ?<\/p>\n
\n20\u00a0 23\u00a0 30\u00a0 43\u00a0 64\u00a0 ?<\/p>\n
\n33\u00a0 16.5\u00a0 ?\u00a0 24.75\u00a0 \u00a0 49.5\u00a0 123.75<\/p>\n
\n44\u00a0 ?\u00a0 99\u00a0 \u00a0148.5 \u00a0222.75\u00a0 334.125<\/p>\n
\n121\u00a0 238\u00a0 472\u00a0 ?\u00a0 1876\u00a0 3748<\/p>\n
\n9\u00a0 10\u00a0 39\u00a0 220\u00a0 ?\u00a0 14382<\/p>\n
\n2 \u00a09 \u00a028 \u00a065 ?<\/p>\n
\n$\\frac{1}{2}$ = 0.5<\/p>\n