Here you can download a free Number Series questions PDF with answers for MAH MBA CET 2022 by Cracku. These are some tricky questions in the MAH MBA CET 2022 exam that you need to find the Number Series of answers for the given questions. These questions will help you to make practice and solve the Number Series questions in the MAH MBA CET exam. Utilize this best PDF practice set<\/strong> which is included answers in detail. Click on the below link to download the Number Series MCQ<\/strong> PDF for MBA-CET 2022 for free.<\/p>\n Download Number Series Questions for MAH-CET<\/a><\/p>\n Enroll to MAH-CET 2022 Crash Course<\/a><\/p>\n Question 1:\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 1)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/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 Instructions<\/b><\/p>\n What will come in place of the question mark (?) in the following number series ?<\/p>\n Question 2:\u00a0<\/b>If $\\frac{7x}{12}-12=23+\\frac{3x}{8}$, find the value of x<\/p>\n a)\u00a0164<\/p>\n b)\u00a0168<\/p>\n c)\u00a0165<\/p>\n d)\u00a0180<\/p>\n e)\u00a0140<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0$\\frac{7x}{12}-12=23+\\frac{3x}{8}$<\/p>\n => $\\frac{7x}{12} – \\frac{3x}{8} = 23 + 12$<\/p>\n => $\\frac{5x}{24} = 35$<\/p>\n => $x = 7 * 24 = 168$<\/p>\n Question 3:\u00a0<\/b>40, 20, 30, 75, 262.5, ?<\/p>\n a)\u00a01181.25<\/p>\n b)\u00a01182.25<\/p>\n c)\u00a01281.25<\/p>\n d)\u00a01161.25<\/p>\n e)\u00a01811.25<\/p>\n 3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 40 * $\\frac{1}{2}$ = 20<\/p>\n 20\u00a0* $\\frac{3}{2}$ =\u00a030<\/p>\n 30\u00a0* $\\frac{5}{2}$ =\u00a075<\/p>\n 75\u00a0* $\\frac{7}{2}$ =\u00a0262.5<\/p>\n 262.5\u00a0* $\\frac{9}{2}$ =\u00a01181.25<\/strong><\/p>\n Question 4:\u00a0<\/b>If $\\frac{7x}{12}-12=23+\\frac{3x}{8}$, find the value of x<\/p>\n a)\u00a0164<\/p>\n b)\u00a0168<\/p>\n c)\u00a0165<\/p>\n d)\u00a0180<\/p>\n e)\u00a0140<\/p>\n 4)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n Expression\u00a0:\u00a0$\\frac{7x}{12}-12=23+\\frac{3x}{8}$<\/p>\n => $\\frac{7x}{12} – \\frac{3x}{8} = 23 + 12$<\/p>\n => $\\frac{5x}{24} = 35$<\/p>\n => $x = 7 * 24 = 168$<\/p>\n Question 5:\u00a0<\/b>400, 800, 880, 896, 899.2, ?<\/p>\n a)\u00a0889.48<\/p>\n b)\u00a0898.48<\/p>\n c)\u00a0899.64<\/p>\n d)\u00a0899.84<\/p>\n e)\u00a0None of these<\/p>\n 5)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 400 + 400 = 800<\/p>\n 800 + 400 * $\\frac{1}{5}$ = 800 + 80 = 880<\/p>\n 880 + 80 *\u00a0$\\frac{1}{5}$ =\u00a0880 + 16 = 896<\/p>\n 896 + 16 *\u00a0$\\frac{1}{5}$ =\u00a0896 + 3.2 = 899.2<\/p>\n 899.2 + 3.2 *\u00a0$\\frac{1}{5}$ =\u00a0899.2 + 0.64 =\u00a0899.84<\/strong><\/p>\n Take Free MAH-CET mock tests here<\/a><\/strong><\/span><\/p>\n Enrol to 5 MAH CET Latest Mocks For Just Rs. 299<\/a><\/strong><\/span><\/p>\n Question 6:\u00a0<\/b>12, 13, 21, 85, 597, ?<\/p>\n a)\u00a04683<\/p>\n b)\u00a04663<\/p>\n c)\u00a04693<\/p>\n d)\u00a04863<\/p>\n e)\u00a0None of these<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 12 + $1^3$ = 13<\/p>\n 13 +\u00a0$2^3$ =\u00a021<\/p>\n 21 +\u00a0$4^3$ =\u00a085<\/p>\n 85 +\u00a0$8^3$ =\u00a0597<\/p>\n 597 +\u00a0$16^3$ =\u00a04693<\/strong><\/p>\n Question 7:\u00a0<\/b>20, 38, 74, 146, 290, ?<\/p>\n a)\u00a0576<\/p>\n b)\u00a0578<\/p>\n c)\u00a0580<\/p>\n d)\u00a0574<\/p>\n e)\u00a0None of these<\/p>\n 7)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 20 * 2 – 2 = 38<\/p>\n 38 *\u00a02 – 2 =\u00a074<\/p>\n 74 *\u00a02 – 2 =\u00a0146<\/p>\n 146 *\u00a02 – 2 =\u00a0290<\/p>\n 290 *\u00a02 – 2 =\u00a0578<\/strong><\/p>\n Question 8:\u00a0<\/b>5, 12.5, 32.5, 105, 427.5, ?<\/p>\n a)\u00a02145<\/p>\n b)\u00a02045<\/p>\n c)\u00a02245<\/p>\n d)\u00a02005<\/p>\n e)\u00a0None of these<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 5 * 1 + 7.5 = 12.5<\/p>\n 12.5 * 2\u00a0+ 7.5 = 32.5<\/p>\n 32.5 * 3\u00a0+ 7.5 =\u00a0105<\/p>\n 105 * 4\u00a0+ 7.5 =\u00a0427.5<\/p>\n 427.5 * 5\u00a0+ 7.5 =\u00a02145<\/strong><\/p>\n Instructions<\/b><\/p>\n What will come in place of the question mark (?) in the given number series?<\/p>\n Question 9:\u00a0<\/b>2, 7, 26, 101, 400, ?<\/p>\n a)\u00a01595<\/p>\n b)\u00a0980<\/p>\n c)\u00a01294<\/p>\n d)\u00a0896<\/p>\n e)\u00a01080<\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern here followed is\u00a0:<\/p>\n 2 $\\times 4 – 1$ = 7<\/p>\n 7\u00a0$\\times 4 – 2$ =\u00a026<\/p>\n 26\u00a0$\\times 4 – 3$ =\u00a0101<\/p>\n 101\u00a0$\\times 4 – 4$ =\u00a0400<\/p>\n 400\u00a0$\\times 4 – 5$ =\u00a01595<\/strong><\/p>\n Question 10:\u00a0<\/b>721, 361, 181, 91, 46, ?<\/p>\n a)\u00a022<\/p>\n b)\u00a028<\/p>\n c)\u00a018<\/p>\n d)\u00a020.5<\/p>\n e)\u00a023.5<\/p>\n 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern is\u00a0:<\/p>\n 721 $- 360$ = 361<\/p>\n 361 $ – \\frac{360}{2} = 361 – 180$ = 181<\/p>\n 181\u00a0$ – \\frac{180}{2} = 181 – 90$ = 91<\/p>\n 91\u00a0$ – \\frac{90}{2} = 91 – 45$ = 46<\/p>\n 46\u00a0$ – \\frac{45}{2} = 46 – 22.5$ =\u00a023.5<\/strong><\/p>\n Question 11:\u00a0<\/b>3, 27, 189, 945, ?, 2835<\/p>\n a)\u00a02835<\/p>\n b)\u00a03780<\/p>\n c)\u00a0945<\/p>\n d)\u00a01890<\/p>\n e)\u00a04725<\/p>\n 11)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern is\u00a0:<\/p>\n 3 $\\times 9$ = 27<\/p>\n 27\u00a0$\\times 7$ =\u00a0189<\/p>\n 189\u00a0$\\times 5$ =\u00a0945<\/p>\n 945\u00a0$\\times 3$ =\u00a02835<\/strong><\/p>\n 2835\u00a0$\\times 1$ =\u00a02835<\/p>\n Question 12:\u00a0<\/b>154, 158, 185, 201, 326, ?<\/p>\n a)\u00a0424<\/p>\n b)\u00a0354<\/p>\n c)\u00a0362<\/p>\n d)\u00a0380<\/p>\n e)\u00a0398<\/p>\n 12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern is\u00a0:<\/p>\n 154 $ + 2^2$ = 158<\/p>\n 158\u00a0$ + 3^3$ =\u00a0185<\/p>\n 185\u00a0$ + 4^2$ =\u00a0201<\/p>\n 201\u00a0$ + 5^3$ =\u00a0326<\/p>\n 326\u00a0$ + 6^2$ =\u00a0362<\/strong><\/p>\n Question 13:\u00a0<\/b>369, 380, 402, 435, ?, 534<\/p>\n a)\u00a0452<\/p>\n b)\u00a0501<\/p>\n c)\u00a0528<\/p>\n d)\u00a0479<\/p>\n e)\u00a0494<\/p>\n 13)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n Successive multiples of 11 are added<\/p>\n 369 $ + 11 \\times 1$ = 380<\/p>\n 380\u00a0$ + 11 \\times 2$ =\u00a0402<\/p>\n 402\u00a0$ + 11 \\times 3$ =\u00a0435<\/p>\n 435\u00a0$ + 11 \\times 4$ =\u00a0479<\/strong><\/p>\n 479\u00a0$ + 11 \\times 5$ =\u00a0534<\/p>\n Question 14:\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 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/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 Question 15:\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 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Solution:<\/b><\/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 Question 16:\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 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/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 Question 17:\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 17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n Solution:<\/b><\/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 Instructions<\/b><\/p>\n What will come in place of question mark (?) in the given number series ?<\/p>\n Question 18:\u00a0<\/b>63 72 97 146 227 ?<\/p>\n a)\u00a0263<\/p>\n b)\u00a0466<\/p>\n c)\u00a0348<\/p>\n d)\u00a0329<\/p>\n e)\u00a0384<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Square of odd numbers are added<\/p>\n 63 $+ 3^2$ = 72<\/p>\n 72\u00a0$+ 5^2$ =\u00a097<\/p>\n 97\u00a0$+ 7^2$ =\u00a0146<\/p>\n 146\u00a0$+ 9^2$ =\u00a0227<\/p>\n 227\u00a0$+ 11^2$ =\u00a0348<\/strong><\/p>\n Question 19:\u00a0<\/b>16 ? 32 16 4 0.5<\/p>\n a)\u00a028<\/p>\n b)\u00a020<\/p>\n c)\u00a024<\/p>\n d)\u00a032<\/p>\n e)\u00a018<\/p>\n 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern is\u00a0:<\/p>\n 16 $\\times 2$ =\u00a032<\/strong><\/p>\n 32\u00a0$\\times 1$ =\u00a032<\/p>\n 32\u00a0$\\times \\frac{1}{2}$ =\u00a016<\/p>\n 16\u00a0$\\times \\frac{1}{4}$ =\u00a04<\/p>\n 4\u00a0$\\times \\frac{1}{8}$ =\u00a00.5<\/p>\n Question 20:\u00a0<\/b>8 20 35 54 78 ?<\/p>\n a)\u00a0114<\/p>\n b)\u00a0169<\/p>\n c)\u00a0108<\/p>\n d)\u00a0123<\/p>\n e)\u00a0136<\/p>\n 20)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n The pattern is\u00a0:<\/p>\n 8 + $12$ = 20<\/p>\n 20 + $12 + 3$ = 35<\/p>\n 35 + $12 + 3 + 4$ = 54<\/p>\n 54 + $12 + 3 + 4 + 5$ = 78<\/p>\n 78 + $12 + 3 + 4 + 5 + 6$ =\u00a0108<\/strong><\/p>\n Question 21:\u00a0<\/b>16.8 16.85 16.75 16.9 16.7 ?<\/p>\n a)\u00a017.5<\/p>\n b)\u00a017.25<\/p>\n c)\u00a016.95<\/p>\n d)\u00a016.25<\/p>\n e)\u00a018.85<\/p>\n 21)\u00a0Answer\u00a0(C)<\/strong><\/p>\n Solution:<\/b><\/p>\n Multiples of 0.05 are alternatively added and subtracted<\/p>\n 16.8 $+ 0.05 \\times 1$ = 16.85<\/p>\n 16.85 $- 0.05 \\times 2$ =\u00a016.75<\/p>\n 16.75 $+ 0.05 \\times 3$ =\u00a016.9<\/p>\n 16.9 $- 0.05 \\times 4$ =\u00a016.7<\/p>\n 16.7 $+ 0.05 \\times 5$ =\u00a016.95<\/strong><\/p>\n Question 22:\u00a0<\/b>8 ? 16 48 192 960<\/p>\n a)\u00a016<\/p>\n b)\u00a010<\/p>\n c)\u00a012<\/p>\n d)\u00a014<\/p>\n e)\u00a08<\/p>\n 22)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Solution:<\/b><\/p>\n Consecutive natural numbers are added<\/p>\n 8 $\\times 1$ =\u00a08<\/strong><\/p>\n 8\u00a0$\\times 2$ =\u00a016<\/p>\n 16\u00a0$\\times 3$ =\u00a048<\/p>\n 48\u00a0$\\times 4$ =\u00a0192<\/p>\n 192\u00a0$\\times 5$ =\u00a0960<\/p>\n