Download Important Number Series questions for LIC AAO exam. Top 20 Reasoning questions with answers based on previous year asked questions.<\/p>\n
Download LIC AAO Number Series Questions PDF<\/a><\/p>\n 10 LIC AAO Mock Tests – Just Rs. 117<\/a><\/p>\n Download LIC AAO Previous Papers PDF<\/a><\/p>\n Take a free mock test for LIC AAO<\/a><\/p>\n Question 1:\u00a0<\/b>Find the missing number in series: a)\u00a014<\/p>\n b)\u00a016<\/p>\n c)\u00a018<\/p>\n d)\u00a024<\/p>\n Question 2:\u00a0<\/b>How many 7\u2019s are there in the following series which are preceded by 6, which is not preceded by 8 ? a)\u00a0Two<\/p>\n b)\u00a0Nil<\/p>\n c)\u00a0One<\/p>\n d)\u00a0Three<\/p>\n Instructions<\/b><\/p>\n Study the following arrangement of letters\/numbers\/symbols carefully and answer the questions given below :<\/p>\n J & B L E H $ 3 C % 8 I F M 4 1 A R 6 V $\\uparrow$ 9 D 5 K U 2 T ? Z<\/p>\n Question 3:\u00a0<\/b>How many such symbols are there in the above arrangement each of which is immediately preceded by a digit and also immediately followed by a consonant?<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n Download LIC AAO Previous Papers PDF<\/a><\/p>\n 100 Free Computer Awareness Tests for LIC<\/a><\/p>\n Question 4:\u00a0<\/b>On the basis of the positions in the above arrangement if H is related to I, then F is related to<\/p>\n a)\u00a0$\\uparrow$<\/p>\n b)\u00a09<\/p>\n c)\u00a0D<\/p>\n d)\u00a06<\/p>\n e)\u00a0R<\/p>\n Question 5:\u00a0<\/b>If all the numbers are deleted from the above arrangement, which of the following would be the seventh from the left end?<\/p>\n a)\u00a0C<\/p>\n b)\u00a0$<\/p>\n c)\u00a0H<\/p>\n d)\u00a0%<\/p>\n e)\u00a0I<\/p>\n Download LIC GK Material PDF<\/a><\/p>\n 100 Free GK Tests for Insurance Exams<\/a><\/p>\n Question 6:\u00a0<\/b>If all the alphabets are deleted from the above arrangement, which of the following will be the third from the right end?<\/p>\n a)\u00a04<\/p>\n b)\u00a0$\\uparrow$<\/p>\n c)\u00a01<\/p>\n d)\u00a05<\/p>\n e)\u00a09<\/p>\n Question 7:\u00a0<\/b>The position of how many digits will remain the same if the digits in the number 7029351648 are rearranged in the ascending order from the left to right?<\/p>\n a)\u00a0None<\/p>\n b)\u00a0One<\/p>\n c)\u00a0Two<\/p>\n d)\u00a0Three<\/p>\n e)\u00a0More than three<\/p>\n Question 8:\u00a0<\/b>What should come in the place of question mark (?) in the following number series? a)\u00a04<\/p>\n b)\u00a03<\/p>\n c)\u00a06<\/p>\n d)\u00a05<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n What will come in the place of question mark (?) in the following number series ?<\/p>\n Question 9:\u00a0<\/b>13 14 22 31 95 120 ?<\/p>\n a)\u00a0336<\/p>\n b)\u00a0332<\/p>\n c)\u00a0325<\/p>\n d)\u00a0225<\/p>\n e)\u00a0425<\/p>\n Question 10:\u00a0<\/b>9 ? 161 504 1233 2564<\/p>\n a)\u00a023<\/p>\n b)\u00a034<\/p>\n c)\u00a032<\/p>\n d)\u00a036<\/p>\n e)\u00a0None of these<\/p>\n LIC AAO REASONING QUESTIONS PDF<\/a><\/p>\n LIC AAO Free Mock Test<\/a><\/p>\n Question 11:\u00a0<\/b>11 ? 87 258 771 2310<\/p>\n a)\u00a024<\/p>\n b)\u00a030<\/p>\n c)\u00a036<\/p>\n d)\u00a045<\/p>\n e)\u00a054<\/p>\n Question 12:\u00a0<\/b>14 9 11 18.5 ? 99.5 300.5<\/p>\n a)\u00a037.5<\/p>\n b)\u00a039<\/p>\n c)\u00a036<\/p>\n d)\u00a038.5<\/p>\n e)\u00a0None of these<\/p>\n Question 13:\u00a0<\/b>36.2 37.6 39.4 41.6 ? 47.2 50.6<\/p>\n a)\u00a044.2<\/p>\n b)\u00a044.4<\/p>\n c)\u00a044.6<\/p>\n d)\u00a042.2<\/p>\n e)\u00a0None of these<\/p>\n Instructions<\/b><\/p>\n Find the missing number (?) in the given series from the options provided.<\/p>\n Question 14:\u00a0<\/b>2, 4, 7, 11, 16, ?<\/p>\n a)\u00a022<\/p>\n b)\u00a021<\/p>\n c)\u00a020<\/p>\n d)\u00a023<\/p>\n e)\u00a024<\/p>\n Question 15:\u00a0<\/b>9, 27, 45, 63, ?<\/p>\n a)\u00a079<\/p>\n b)\u00a080<\/p>\n c)\u00a081<\/p>\n d)\u00a082<\/p>\n e)\u00a083<\/p>\n Question 16:\u00a0<\/b>3, 8, 15, 24, 35, ?<\/p>\n a)\u00a063<\/p>\n b)\u00a049<\/p>\n c)\u00a080<\/p>\n d)\u00a048<\/p>\n e)\u00a056<\/p>\n Question 17:\u00a0<\/b>-3, -1, 11, 39, 89, ?<\/p>\n a)\u00a0198<\/p>\n b)\u00a0189<\/p>\n c)\u00a0210<\/p>\n d)\u00a0174<\/p>\n e)\u00a0167<\/p>\n Question 18:\u00a0<\/b>9, 12, 19, 34, 65, ?<\/p>\n a)\u00a0127<\/p>\n b)\u00a0128<\/p>\n c)\u00a0129<\/p>\n d)\u00a0130<\/p>\n e)\u00a0134<\/p>\n Instructions<\/b><\/p>\n Find the missing number (?) in the given series from the options provided.<\/p>\n Question 19:\u00a0<\/b>3, 12, 25, 42, 63,?<\/p>\n a)\u00a088<\/p>\n b)\u00a077<\/p>\n c)\u00a078<\/p>\n d)\u00a083<\/p>\n e)\u00a087<\/p>\n Question 20:\u00a0<\/b>6, 24, 60, 120, 210, ?<\/p>\n a)\u00a0345<\/p>\n b)\u00a0336<\/p>\n c)\u00a0326<\/p>\n d)\u00a0316<\/p>\n e)\u00a0320<\/p>\n Download General Science Material PDF<\/a><\/p>\n Answers & Solutions:<\/strong><\/span><\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 2)\u00a0Answer\u00a0(D)<\/strong><\/p>\n There are 3 such 7 preceeded by\u00a0 which is not preceeded by 8. 4)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 5)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 6)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 8)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 10)\u00a0Answer\u00a0(D)<\/strong><\/p>\n 11)\u00a0Answer\u00a0(C)<\/strong><\/p>\n 12)\u00a0Answer\u00a0(B)<\/strong><\/p>\n 13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n 14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\Rightarrow$ $T_{2} – T _{1} = 4 – 2 = 2$ 15)\u00a0Answer\u00a0(C)<\/strong><\/p>\n $\\Rightarrow$ $T_{1} = 9 = 5^2 – 4^2$ 16)\u00a0Answer\u00a0(D)<\/strong><\/p>\n $\\Rightarrow$ $T_{1} = 3 = 2^2 – 1$ 17)\u00a0Answer\u00a0(E)<\/strong><\/p>\n $\\Rightarrow$ $T_{1} = -3 = 1^3 – 2^2$ 18)\u00a0Answer\u00a0(B)<\/strong><\/p>\n $\\Rightarrow$ $T_{1} = 9 = 2^2 + 5$ 19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n $\\Rightarrow$ $T_{1} = 3 = 3*1$ 20)\u00a0Answer\u00a0(B)<\/strong><\/p>\n We can see that
\n2, 5, 9, ?, 20, 27<\/p>\n
\n8 7 6 7 8 6 7 5 6 7 9 7 6 1 6 7 7 6 8 8 6 9 7 6 8 7<\/p>\n
\n3 4 4 3 4 4 5 3 4 4 5 6 ? 4 4 5 6 7 3 4 4 5 6 7 8<\/p>\n
\n8 7 6 7<\/u> <\/strong>8 6 7 5 6 7<\/u><\/strong> 9 7 6 1 6 7<\/strong><\/u> 7 6 8 8 6 9 7 6 8 7
\nTherefore, the correct option is option D.
\n3)\u00a0Answer\u00a0(A)<\/strong><\/p>\n
\n$\\Rightarrow$ $T_{3} – T _{2} = 7 – 4 = 3$
\n$\\Rightarrow$ $T_{4} – T _{3} = 11 – 7 = 4$
\n$\\Rightarrow$ $T_{5} – T _{4} = 16 – 11 = 5$
\nHere we can see the pattern that difference between consecutive terms are in A.P. and it is increasing by 1. Therefore,
\n$\\Rightarrow$ $T_{6} = 6 + T _{5} = 6+16 = 22$
\nHence, option A is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{2} = 27 = 6^2 – 3^2$
\n$\\Rightarrow$ $T_{3} = 45 = 7^2 – 2^2$
\n$\\Rightarrow$ $T_{4} = 63 = 8^2 – 1^2$
\nHence, we can say that $\\Rightarrow$ $T_{n} = (n+4)^2 – (5-n)^2$
\n$\\Rightarrow$ $T_{5} = (5+4)^2 – (5-4)^2 = 81$
\nHence, option C is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{2} = 8 = 3^2 – 1$
\n$\\Rightarrow$ $T_{3} = 15 = 4^2 – 1$
\n$\\Rightarrow$ $T_{4} = 24 = 5^2 – 1$
\n$\\Rightarrow$ $T_{5} = 35 = 6^2 – 1$
\nHence, we can say that $\\Rightarrow$ $T_{n} = (n+1)^2 – 1$
\n$\\Rightarrow$ $T_{6} = (6+1)^2 – 1 = 48$
\nHence, option D is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{2} = -1 = 2^3 – 3^2$
\n$\\Rightarrow$ $T_{3} = 11 = 3^3 – 4^2$
\n$\\Rightarrow$ $T_{4} = 39 = 4^3 – 5^2$
\n$\\Rightarrow$ $T_{5} = 89 = 5^3 – 6^2$
\nHence, we can say that $\\Rightarrow$ $T_{n} = n^3 – (n+1)^2$
\n$\\Rightarrow$ $T_{6} = 6^3 – 7^2 = 167$
\nHence, option E is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{2} = 12 = 2^3 + 4$
\n$\\Rightarrow$ $T_{3} = 19 = 2^4 + 3$
\n$\\Rightarrow$ $T_{4} = 34 = 2^5 + 2$
\n$\\Rightarrow$ $T_{5} = 65 = 2^6 + 1$
\nHence, we can say that $\\Rightarrow$ $T_{n} = 2^{n+1} + (6 -n)$
\n$\\Rightarrow$ $T_{6} = 2^{6+1} + (6 – 6) = 128$
\nHence, option B is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{2} = 12 = 4*3$
\n$\\Rightarrow$ $T_{3} = 25 = 5*5$
\n$\\Rightarrow$ $T_{4} = 42 = 6*7$
\n$\\Rightarrow$ $T_{5} = 63 = 7*9$
\nHence, we can say that $\\Rightarrow$ $T_{n} = (n+2)*(2n -1)$
\n$\\Rightarrow$ $T_{6} = 8*11 = 88$
\nHence, option A is the correct answer.<\/p>\n
\n$\\Rightarrow$ $T_{1} = 6 = 1*2*3$
\n$\\Rightarrow$ $T_{2} = 24 = 2*3*4$
\n$\\Rightarrow$ $T_{3} = 60 = 3*4*5$
\n$\\Rightarrow$ $T_{4} = 120 = 4*5*6$
\n$\\Rightarrow$ $T_{5} = 120 = 5*6*7$
\nHence we can generalise that $T_{n} = (n)*(n+1)*(n+2)$
\n$\\Rightarrow$ $T_{6} = 6*7*8 = 336$
\nHence, option B is the correct answer.<\/p>\n