Download TISSNET 2022 Number Series Questions pdf by Cracku. Very Important Number Series Questions for TISSNET 2022 based on asked questions in previous exam papers. These questions will help your TISSNET exam preparation. So kindly download the PDF for reference and do more practice.<\/p>\n
Download Number Series Questions for TISSNET<\/a><\/p>\n Get 5 TISSNET mocks at just Rs.299<\/a><\/p>\n Download TISSNET previous papers PDF<\/a><\/p>\n Question 1:\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 2:\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 3:\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 4:\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 5:\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 Download TISSNET Previous Papers PDF<\/a><\/p>\n Question 6:\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 7:\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 8:\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 9:\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 10:\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 11:\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 12:\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 13:\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 14:\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 Question 15:\u00a0<\/b>9, 11, 16, 33, 98, ?<\/p>\n a)\u00a0350<\/p>\n b)\u00a0355<\/p>\n c)\u00a0360<\/p>\n d)\u00a0365<\/p>\n e)\u00a0370<\/p>\n Take TISSNET Mock Tests<\/a><\/p>\n Question 16:\u00a0<\/b>65, 70, 63, 74, 61, ?<\/p>\n a)\u00a078<\/p>\n b)\u00a058<\/p>\n c)\u00a072<\/p>\n d)\u00a046<\/p>\n e)\u00a068<\/p>\n Question 17:\u00a0<\/b>13, 14, 30, 93, ?, 1885<\/p>\n a)\u00a0358<\/p>\n b)\u00a0336<\/p>\n c)\u00a0364<\/p>\n d)\u00a0376<\/p>\n e)\u00a0380<\/p>\n Question 18:\u00a0<\/b>8, 14, 25, 46, 82, ?<\/p>\n a)\u00a0132<\/p>\n b)\u00a0130<\/p>\n c)\u00a0138<\/p>\n d)\u00a0128<\/p>\n e)\u00a0142<\/p>\n Question 19:\u00a0<\/b>14, 8, 7, 11.5, 22, ?<\/p>\n a)\u00a054<\/p>\n b)\u00a064<\/p>\n c)\u00a062<\/p>\n d)\u00a058<\/p>\n e)\u00a056<\/p>\n Question 20:\u00a0<\/b>$\\frac{1}{8}\\times121+\\frac{1}{5}\\times76-?=25$<\/p>\n a)\u00a05<\/p>\n b)\u00a045<\/p>\n c)\u00a015<\/p>\n d)\u00a035<\/p>\n e)\u00a065<\/p>\n Question 21:\u00a0<\/b>$\\sqrt{628}\\times17.996\\div15.04=?$<\/p>\n a)\u00a030<\/p>\n b)\u00a010<\/p>\n c)\u00a05<\/p>\n d)\u00a020<\/p>\n e)\u00a025<\/p>\n Question 22:\u00a0<\/b>80.04% of 150.16 + 60.02% of 50.07 = ?<\/p>\n a)\u00a0150<\/p>\n b)\u00a0125<\/p>\n c)\u00a0210<\/p>\n d)\u00a0175<\/p>\n e)\u00a0213<\/p>\n Question 23:\u00a0<\/b>$17.98^{2} + 4.05 \\times 90.11 \\div 4.98 = ?$<\/p>\n a)\u00a0396<\/p>\n b)\u00a0336<\/p>\n c)\u00a0242<\/p>\n d)\u00a0423<\/p>\n e)\u00a0816<\/p>\n Question 24:\u00a0<\/b>90.05 + 281 \u00f7 4 -151.06 = $\\sqrt[3]{?}$<\/p>\n a)\u00a027<\/p>\n b)\u00a0343<\/p>\n c)\u00a0216<\/p>\n d)\u00a0729<\/p>\n e)\u00a0176<\/p>\n Question 25:\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 26:\u00a0<\/b>26, 12, 11, 15.5, 30, ?<\/p>\n a)\u00a072<\/p>\n b)\u00a068<\/p>\n c)\u00a074<\/p>\n d)\u00a082<\/p>\n e)\u00a078<\/p>\n Question 27:\u00a0<\/b>8, 9.4, 12.2, 17.8, 29, ?<\/p>\n a)\u00a053.6<\/p>\n b)\u00a051.4<\/p>\n c)\u00a052.1<\/p>\n d)\u00a048.6<\/p>\n e)\u00a049.8<\/p>\n Question 28:\u00a0<\/b>17, 16, 30, 87, 344, ?<\/p>\n a)\u00a01735<\/p>\n b)\u00a01760<\/p>\n c)\u00a01660<\/p>\n d)\u00a01685<\/p>\n e)\u00a01715<\/p>\n Question 29:\u00a0<\/b>13, 13, 20, 37.5, 83, ?<\/p>\n a)\u00a0233<\/p>\n b)\u00a0216<\/p>\n c)\u00a0234<\/p>\n d)\u00a0235<\/p>\n e)\u00a0239<\/p>\n Question 30:\u00a0<\/b>29, 31, 37, 49, 69, ?<\/p>\n a)\u00a0108<\/p>\n b)\u00a099<\/p>\n c)\u00a094<\/p>\n d)\u00a0103<\/p>\n e)\u00a088<\/p>\n Enroll to CAT 2022 Complete Course<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\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 2)\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 3)\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 4)\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 5)\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 6)\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 7)\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 8)\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 9)\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 10)\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 11)\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 12)\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 13)\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 14)\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 15)\u00a0Answer\u00a0(B)<\/strong><\/p>\n Number of the form $(2^n + 1)$ are added where n is even number<\/p>\n 9 $+ (2^0 + 1)$ = 11<\/p>\n 11\u00a0$+ (2^2 + 1)$ =\u00a016<\/p>\n 16\u00a0$+ (2^4 + 1)$ =\u00a033<\/p>\n 33\u00a0$+ (2^6 + 1)$ =\u00a098<\/p>\n 98\u00a0$+ (2^8 + 1)$ =\u00a0355<\/strong><\/p>\n 16)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We can note that consecutive prime numbers are either added or subtracted starting from 5.<\/p>\n 65+5 = 70<\/p>\n 70-7 = 63<\/p>\n 63+11 = 74<\/p>\n 74-13 = 61<\/p>\n 61+17 = 78<\/p>\n Option A is the right answer.<\/p>\n 17)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 13*1 + 1 = 14<\/p>\n 14*2 +2 = 28+2 = 30<\/p>\n 30*3 + 3 = 90+3 =93<\/p>\n 93*4 + 4 = 372+4 = 376<\/p>\n Option D is the right answer.<\/p>\n 18)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 14 – 8 = 6<\/p>\n 25 – 14 =11<\/p>\n 46 – 25 = 21<\/p>\n 82 – 46 = 36<\/p>\n Let us take a look at the second order difference.<\/p>\n 11 – 6 = 5<\/p>\n 21 – 11 = 10<\/p>\n 36 – 21 = 15<\/p>\n The next term will be 20. The difference will be 36 + 20 = 56.<\/p>\n The required term is 82 + 56 = 138.<\/p>\n Option C is the right answer.<\/p>\n 19)\u00a0Answer\u00a0(E)<\/strong><\/p>\n 14*0.5 + 1 = 7+1= 8<\/p>\n 8*1 – 1 = 8-1 = 7<\/p>\n 7*1.5 + 1 = 10.5 + 1 = 11.5<\/p>\n 11.5*2 – 1 = 23 – 1 = 22<\/p>\n 22*2.5 +1 = 55+1 =56<\/p>\n Option E is the right answer.<\/p>\n 20)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The given equation can be written as $\\frac{120}{8}+\\frac{75}{5} – 25 =x$<\/p>\n $15+15-25=5$<\/p>\n Option A is the right answer.<\/span><\/p>\n 21)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The given equation can be written as$ \\frac{\\sqrt{625}*18}{15}$<\/p>\n =$\\frac{25*6}{5}$<\/p>\n = 30.<\/span><\/p>\n Option A is the right answer.<\/span><\/p>\n 22)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 80.04 ~ 80<\/p>\n 150.16 ~ 150<\/p>\n 60.02 ~ 60<\/p>\n 50.07 ~ 50<\/p>\n Now, (80.04% of 150.16 + 60.02% of 50.07) is equivalent to (80% of 150 + 60% of 50) = 120 +30 = 150<\/span><\/p>\n 23)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The given equation can be written as $18^{2} + \\frac{4*90}{5}$<\/p>\n = $324 + 4*18$<\/p>\n = $396$<\/p>\n Option A is the right answer.<\/p>\n 24)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The given statement can be written as $90 + 60 – 151 = \\sqrt[3]{?}$<\/p>\n $9 =\\sqrt[3]{?}$<\/p>\n => ? = 729<\/span><\/p>\n Option D is the answer.<\/span><\/p>\n 25)\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 26)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\frac{26}{2\/1}-1$ = 12<\/p>\n $\\frac{12}{2\/2}-1$ = 11<\/p>\n $\\frac{11}{2\/3}-1$ = 15.5<\/p>\n $\\frac{15.5}{2\/4}-1$ = 30<\/p>\n $\\frac{30}{2\/5}-1$ = 74<\/p>\n 27)\u00a0Answer\u00a0(B)<\/strong><\/p>\n The difference between consecutive terms is doubling.<\/p>\n For example, 9.4 – 8 = 1.4 Hence, the term to come next is 29 + 11.2*2 = 51.4<\/p>\n 28)\u00a0Answer\u00a0(E)<\/strong><\/p>\n 16 = 17*1 – 1<\/p>\n 30 = 16*2 – 2<\/p>\n 87 = 30*3 – 3<\/p>\n 344 = 87*4 – 4<\/p>\n So, the next number is 344*5 – 5 = 1715<\/p>\n 29)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $13 \\times 0.5 + 6.5 = 13$<\/p>\n $13 \\times 1 + 7 = 20$<\/p>\n $20 \\times 1.5 + 7.5 = 37.5$<\/p>\n $37.5 \\times 2 + 8 = 83$<\/p>\n $83 \\times 2.5 + 8.5 = 216$<\/p>\n 30)\u00a0Answer\u00a0(B)<\/strong><\/p>\n +The difference between the consecutive terms is:
\n2 \u00a09 \u00a028 \u00a065 ?<\/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 9 1 8 3 6 5 4 7 2 1 8 3 6 5 4 7 1 8 3 6 5 4<\/p>\n
\n$\\frac{1}{2}$ = 0.5<\/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
\n12.2 – 9.4 = 2.8
\n17.8 – 12.2 = 5.6
\n29 – 17.8 = 11.2<\/p>\n
\n31-29 = 2
\n37 -31 = 6
\n49- 37 = 12
\n69-49 = 20
\nHere, the difference between the consecutive difference terms is
\n6-2 = 4
\n12-6 = 6
\n20-12 = 8
\nHence, the next term in difference series = 10
\nTherefore the next term in series = 69+30 = 99<\/p>\n