Download Top-20 IBPS RRB Clerk Number Series Questions PDF. Number Series questions based on asked questions in previous year exam papers very important for the IBPS RRB Assistant exam.<\/p>\n
Download Number Series Questions For IBPS RRB Clerk<\/a><\/p>\n 35 IBPS RRB Clerk Mocks @ Rs. 149<\/a><\/p>\n 70 IBPS RRB (PO + Clerk) Mocks @ Rs. 199<\/a><\/p>\n Question 1:\u00a0<\/b>In the number series below which element doesn\u2019t fit in? a)\u00a0101<\/p>\n b)\u00a0193<\/p>\n c)\u00a0169<\/p>\n d)\u00a0147<\/p>\n e)\u00a0None of these<\/p>\n Question 2:\u00a0<\/b>Find the wrong number in the series. a)\u00a051<\/p>\n b)\u00a031<\/p>\n c)\u00a019<\/p>\n d)\u00a042<\/p>\n e)\u00a0None of these<\/p>\n Question 3:\u00a0<\/b>Find out the wrong number in the series: a)\u00a020<\/p>\n b)\u00a047<\/p>\n c)\u00a0130<\/p>\n d)\u00a0371<\/p>\n e)\u00a0None of these<\/p>\n Question 4:\u00a0<\/b>Find the wrong number in the series. a)\u00a088<\/p>\n b)\u00a091<\/p>\n c)\u00a086<\/p>\n d)\u00a081<\/p>\n e)\u00a0None of these<\/p>\n Question 5:\u00a0<\/b>Find the wrong number in the series. a)\u00a023<\/p>\n b)\u00a031<\/p>\n c)\u00a044<\/p>\n d)\u00a056<\/p>\n e)\u00a0None of these<\/p>\n Question 6:\u00a0<\/b>Find out the wrong number in the series: a)\u00a050<\/p>\n b)\u00a053<\/p>\n c)\u00a047<\/p>\n d)\u00a049<\/p>\n e)\u00a0None of these<\/p>\n Question 7:\u00a0<\/b>Find out the wrong number in the series: a)\u00a08<\/p>\n b)\u00a012<\/p>\n c)\u00a020<\/p>\n d)\u00a034<\/p>\n e)\u00a0None of these<\/p>\n Question 8:\u00a0<\/b>Find the wrong number in the series. a)\u00a091<\/p>\n b)\u00a055<\/p>\n c)\u00a079<\/p>\n d)\u00a041<\/p>\n e)\u00a0None of these<\/p>\n Question 9:\u00a0<\/b>Find the wrong number in the series. a)\u00a048<\/p>\n b)\u00a0110<\/p>\n c)\u00a072<\/p>\n d)\u00a0162<\/p>\n e)\u00a0None of these<\/p>\n Free Mock Test for IBPS RRB PO<\/a><\/p>\n IBPS RRB Clerk Previous Papers<\/a><\/p>\n Question 10:\u00a0<\/b>Find the number which should replace \u2018?\u2019 a)\u00a010<\/p>\n b)\u00a08<\/p>\n c)\u00a012<\/p>\n d)\u00a07<\/p>\n e)\u00a09<\/p>\n Question 11:\u00a0<\/b>Find the number which should replace \u2018?\u2019<\/p>\n 52, 50, 45, 43, 38, 36, ?<\/p>\n a)\u00a033<\/p>\n b)\u00a034<\/p>\n c)\u00a028<\/p>\n d)\u00a031<\/p>\n e)\u00a035<\/p>\n Question 12:\u00a0<\/b>Find the number which should replace \u2018?\u2019 a)\u00a037<\/p>\n b)\u00a038<\/p>\n c)\u00a039<\/p>\n d)\u00a040<\/p>\n e)\u00a041<\/p>\n Question 13:\u00a0<\/b>Find the number which should replace \u2018?\u2019 a)\u00a061<\/p>\n b)\u00a056<\/p>\n c)\u00a059<\/p>\n d)\u00a052<\/p>\n e)\u00a065<\/p>\n Instructions<\/b><\/p>\n What will come in place of question mark \u2018?\u2019 in the following number sequences?<\/p>\n Question 14:\u00a0<\/b>2, 5, 14, 71, ?<\/p>\n a)\u00a0632<\/p>\n b)\u00a0679<\/p>\n c)\u00a0711<\/p>\n d)\u00a0634<\/p>\n e)\u00a0None of the above.<\/p>\n Question 15:\u00a0<\/b>15, 35, 63, ?, 143<\/p>\n a)\u00a099<\/p>\n b)\u00a0117<\/p>\n c)\u00a091<\/p>\n d)\u00a095<\/p>\n e)\u00a0None of the above<\/p>\n Question 16:\u00a0<\/b>5, 12, 31, 86, ?<\/p>\n a)\u00a0168<\/p>\n b)\u00a0192<\/p>\n c)\u00a0249<\/p>\n d)\u00a0207<\/p>\n e)\u00a0256<\/p>\n Question 17:\u00a0<\/b>4, 9 , 25, 49, 121, ?<\/p>\n a)\u00a0144<\/p>\n b)\u00a0155<\/p>\n c)\u00a0164<\/p>\n d)\u00a0196<\/p>\n e)\u00a0169<\/p>\n Question 18:\u00a0<\/b>2 , 6 , 30 , 260 , ?<\/p>\n a)\u00a01029<\/p>\n b)\u00a01030<\/p>\n c)\u00a03129<\/p>\n d)\u00a03130<\/p>\n e)\u00a0None of the above<\/p>\n Instructions<\/b><\/p>\n Find the number which should replace \u2018?\u2019<\/p>\n Question 19:\u00a0<\/b>27, 108, 36, 144, 48, 192, 64, ?<\/p>\n a)\u00a0216<\/p>\n b)\u00a0123<\/p>\n c)\u00a0212<\/p>\n d)\u00a0256<\/p>\n e)\u00a0272<\/p>\n Question 20:\u00a0<\/b>2, 10, 30, 68, 130, ?<\/p>\n a)\u00a0213<\/p>\n b)\u00a0197<\/p>\n c)\u00a0267<\/p>\n d)\u00a0313<\/p>\n e)\u00a0222<\/p>\n Quantitative Aptitude formulas PDF<\/a><\/p>\n 520 Banking Mocks – Just Rs. 499<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 78 + 23 = 101 2)\u00a0Answer\u00a0(A)<\/strong><\/p>\n The difference between the first term and the second term = 2 3)\u00a0Answer\u00a0(C)<\/strong><\/p>\n The difference between the first term and the second term = $3^1 = 3$ 4)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $T_{2} – T_{1}$ = 81 – 85 = -4 18000 Free Solved Questions – Banking Study Material<\/a><\/p>\n 35 IBPS RRB Clerk Mocks @ Rs. 149<\/a><\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n We can see that Hence we can say that \u201844\u2019 is the incorrect term in the sequence. (Answer :C)<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 1st<\/sup> term is increased by 1, 2nd<\/sup> term is decreased by 2, 3rd<\/sup> term is increased by 3 and so on. 6th<\/sup> term must be decreased by 6. So, 7th<\/sup> term should be 48. Hence, 7th<\/sup> term that is 47 is incorrect.<\/p>\n 7)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Difference between 1st and 2nd term is 2. 8)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 43+ 12 = 55 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 32 * 1.5 = 48 10)\u00a0Answer\u00a0(E)<\/strong><\/p>\n Finding out difference between two consecutive term 11)\u00a0Answer\u00a0(D)<\/strong><\/p>\n Calculating the difference between consecutive terms,<\/p>\n $T_{1} – T_{2} = 52 – 50 = 2$<\/p>\n $T_{2} – T_{3} = 50 – 45 = 5$<\/p>\n $T_{3} – T_{4} = 45 – 43 = 2$<\/p>\n $T_{4} – T_{5} = 43 – 38 = 5$<\/p>\n $T_{5} – T_{6} = 38 – 36 = 2$<\/p>\n By analysing the difference between consecutive terms we can figure out that number 2, 5 are being subtracted alternatively so next term will be = 36 – 5 = 31<\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We can see that difference between two consecutive terms is increasing. 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We can see that every term is sum of square of 2 consecutive natural numbers starting from 1. 14)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $1 + (1^{0}+0) = 2$ 15)\u00a0Answer\u00a0(A)<\/strong><\/p>\n We can see that 16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $3^{1} + 2 = 5$ 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n We can see that above mentioned sequence is square of prime numbers. 18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n We can see that 19)\u00a0Answer\u00a0(D)<\/strong><\/p>\n The series follows the following pattern:<\/p>\n 27 20)\u00a0Answer\u00a0(E)<\/strong><\/p>\n The series follows the following pattern:
\n78, 101, 124, 147, 169, 193<\/p>\n
\n14, 16, 19, 24, 31, 42, 51<\/p>\n
\n8, 11, 20, 47, 130, 371<\/p>\n
\n85, 81, 88, 84, 91, 86, 94<\/p>\n
\n16, 18, 23, 31, 44, 56<\/p>\n
\n51, 52, 50, 53, 49, 54, 47<\/p>\n
\n6, 8, 12, 20, 34, 68<\/p>\n
\n41, 55, 67, 79, 91<\/p>\n
\n32, 48, 72, 110, 162<\/p>\n
\n99, 63, 38, 22, 13, ?<\/p>\n
\n2, 5, 12, 23, ?<\/p>\n
\n5,13,25,41, ?<\/p>\n
\n101 + 23 = 124
\n124 + 23 = 147
\n147 + 23 = 170
\n170 + 23 = 193
\nHence, option C is the right answer.<\/p>\n
\nThe difference between the second term and the third term = 3
\nThe difference between the third term and the fourth term = 5
\nThe difference between the fourth term and the fifth term = 7
\nThe difference between the fifth term and the sixth term = 11
\nSo, the difference between consecutive terms is a series of prime number starting from 2.
\nThe difference between the sixth and the seventh term should be 13.
\nThe seventh term = 42 + 13 = 55 but it is 51. So, the seventh term is incorrect.
\nHence, option A is the correct answer.<\/p>\n
\nThe difference between the second term and the third term = $3^2 = 9$
\nThe difference between the third term and the fourth term = $3^3 = 27$
\nSo, the difference is increasing in power of 3.
\nThe difference between the fourth and the fifth term should be $3^4 = 81$
\nThe fifth term = 47 + 81 = 128.
\nSo, the term \u2018130\u2019 is incorrect.
\nHence, option C is the correct answer.<\/p>\n
\n$T_{3} – T_{2}$ = 88 – 81 = 7
\n$T_{4} – T_{3}$ = 84 – 88 = -4
\n$T_{5} – T_{4}$ = 91 – 84 = 7
\nWe can see the pattern hence for next term
\n$T_{6} – T_{5}$ = -4
\n$\\Rightarrow$ $T_{6}$ = 91 +(- 4) = 87
\nSimilarly $ T_{7} – T_{6}$ = 7
\n$\\Rightarrow$ $ T_{7}$ = 7 + 87 = 94
\nHence we can say that \u201886\u2019 mentioned in the original sequence is incorrect. (Answer :C)<\/p>\n
\n$T_{2} – T_{1}$ = 18 – 16 = 2
\n$T_{3} – T_{2}$ = 23 – 18 = 5
\n$T_{4} – T_{3}$ = 31 – 23 = 8
\nWe can see that the difference is increasing by 3.
\n$T_{5} – T_{4}$ = $T_{5}$ – 31 = 8 + 3
\n$\\Rightarrow$ $T_{5}$ = 42 and $T_{6}$ = 42+(11+3) = 56<\/p>\n
\nDifference between 2nd and 3rd term is 4.
\nDifference between 3rd and 4th term is 8.
\nHence, the difference is getting doubled.
\nSo, difference between 4th and 5th term must be 16 and 5th term should be 36. Hence, 5th term that is 34 is incorrect.<\/p>\n
\n55 + 12 = 67
\n67 + 12 = 79
\n79 + 12 = 91
\nHence, option D is the right answer.<\/p>\n
\n48 * 1.5 = 72
\n72*1.5 = 108
\n108 * 1.5 = 162
\nHence, option B is the right answer.<\/p>\n
\n$T_{1} – T_{2}$ = 99 – 63 = 36
\n$T_{2} – T_{3} $= 63 – 38 = 25
\n$T_{3} – T_{4}$ = 38 – 22 = 16
\n$T_{4} – T_{5}$ = 22 – 13 = 9
\nClearly the difference is square of natural number in decreasing order starting from 6.
\n$T_{5} – T_{6}$ = $2^{2}$
\n$T_{6}$ = 13 – 4 = 9<\/p>\n
\n$T_{2} – T_{1}$ = 5 – 2 =3
\n$T_{3} – T_{2}$ = 12 – 5 = 7
\n$T_{4} – T_{3} $= 23 – 12 = 11
\nClearly the difference is in A.P with common difference of 4.
\nSo $T_{5} – T_{4}$ = 11+4
\n$T_{5}$ = 15 + 23 = 38<\/p>\n
\n$1^{2}+2^{2}$ = 1+4 = 5
\n$2^{2}+3^{2}$ = 4+9 = 13
\n$3^{2}+4^{2}$ = 9+16 = 25
\n$4^{2}+5^{2}$ = 16+25 = 41
\n$5^{2}+6^{2}$ = 25+36 = 61<\/p>\n
\n$2 + (2^{1}+1) = 5$
\n$3 + (3^{2}+2) = 14$
\n$4 + (4^{3}+3) = 71$
\n$5 + (5^{4}+4) = 634$<\/p>\n
\n3 * 5 = 15
\n5 * 7 = 35
\n7 * 9 = 63
\n9 * 11 = 99
\n11 * 13 = 143. Hence answer (a)<\/p>\n
\n$3^{2} + 3 = 12$
\n$3^{3} + 4 = 31$
\n$3^{4} + 5 = 86$
\nSo next term will be $3^{5} + 6 = 249$ .<\/p>\n
\n$2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, \u2026$
\nSo next term will be $13^{2} =169$<\/p>\n
\n$1^{1}+1 = 2$
\n$2^{2}+2 = 6$
\n$3^{3}+3 = 30$
\n$4^{4}+4 = 260$
\nSo $5^{5}+5 = 3125 + 5 = 3130$<\/p>\n
\n27 * 4 = 108
\n108 \/ 3 = 36
\n36 * 4 = 144
\n144 \/ 3 = 48
\n48 * 4 = 192
\n192 \/ 3 = 64
\n64 * 4 = 256<\/p>\n
\n$1 + 1^3 = 2$
\n$2 + 2^3 = 10$
\n$3 + 3^3 = 30$
\n$4 + 4^3 = 68$
\n$5 + 5^3 = 130$
\n$6 + 6^3 = 222$<\/p>\n