83 practice questions on Number & Letter Series from the Logical Reasoning section of the UPSC Prelims syllabus.
83 come with a written explanation and 47 are actual previous year questions.
Try the sample set below - the answer stays hidden until you ask for it.
22 Easy50 Medium11 Hard47 from past papers
Sample questions
Q1
Previous year questionmedium
Consider the following matrix. The top row is: 3, 8, 10, 2, ?, 1 and the bottom row is: 6, 56, 90, 2, 20, 0. What is the missing number at '?' in the matrix?
(a) 5
(b) 0
(c) 7
(d) 3
A5
B0
C7
D3
Show answer and explanation
Correct answer: A - 5
In each column the bottom number equals the top number multiplied by one less than itself, i.e. n x (n - 1). Check: 3x2 = 6, 8x7 = 56, 10x9 = 90, 2x1 = 2, 1x0 = 0. For the missing entry the bottom value is 20, so n x (n - 1) = 20, which gives n = 5 (since 5x4 = 20). The missing number is 5.
Q2
Previous year questionmedium
The letters of the word "INCOMPREHENSIBILITIES" are arranged alphabetically in reverse order. How many positions of the letter/letters will remain unchanged?
ANone
BOne
CTwo
DThree
Show answer and explanation
Correct answer: C - Two
INCOMPREHENSIBILITIES has 21 letters with counts: I x 5, E x 3, N x 2, S x 2, and one each of C, O, M, P, R, H, B, L, T. In reverse alphabetical order the word becomes T S S R P O N N M L I I I I I H E E E C B, so the block of five I's occupies positions 11 to 15. In the original word, I's sit at positions 1, 13, 15, 17 and 19; comparing position by position, only positions 13 and 15 hold an I in both arrangements, and no other letter lands on its original position. So exactly two positions are unchanged - option (c). 'None' (a) misses the overlapping I block, 'One' (b) catches only one of the two coincidences, and 'Three' (d) overcounts.
Q3
Previous year questionmedium
40 children are standing in a circle and one of them (say child-1) has a ring. The ring is passed clockwise. Child-1 passes on to child-2, child-2 passes on to child-4, child-4 passes on to child-7 and so on. After how many such changes (including child-1) will the ring be in the hands of child-1 again?
A14
B15
C16
D17
Show answer and explanation
Correct answer: B - 15
The k-th pass moves the ring k places clockwise, so after k passes the ring sits at position 1 + (1 + 2 + ... + k) = 1 + k(k+1)/2, taken modulo 40. The ring returns to child-1 when k(k+1)/2 is a multiple of 40, i.e. k(k+1) is a multiple of 80. Testing: k = 14 gives 210 (no), k = 15 gives 240 = 3 x 80 (yes). So after 15 changes the ring is back with child-1 - option (b), as per the official key. Option (c) 16 is the trap of counting the 16 holders (including child-1 at the start) instead of the 15 passes; options (a) and (d) fail the divisibility test.
Q4
Previous year questioneasy
In the English alphabet, the first 4 letters are written in opposite order; and the next 4 letters are written in opposite order and so on; and at the end Y and Z are interchanged. Which will be the fourth letter to the right of the 13th letter?
AN
BT
CH
DI
Show answer and explanation
Correct answer: B - T
Reversing each block of 4 letters and swapping Y and Z gives the new order: DCBA HGFE LKJI PONM TSRQ XWVU ZY. Counting along this arrangement, the 13th letter is P (the blocks contribute 4 letters each, so positions 13-16 are P, O, N, M). The fourth letter to the right of position 13 is the letter at position 17, which is T - option (b). Option (a) N is at position 15, option (c) H is at position 5, and option (d) I is at position 12, so none of them fits.
Q5
Previous year questionmedium
You are given two identical sequences in two rows:
Sequence-I: 8, 4, 6, 15, 52.5, 236.25
Sequence-II: 5, A, B, C, D, E
What is the entry in the place of C for the Sequence-II?
A2.5
B5
C9.375
D32.8125
Show answer and explanation
Correct answer: C - 9.375
The rule in Sequence-I uses multipliers that grow by 1 at each step: 8 x 0.5 = 4, 4 x 1.5 = 6, 6 x 2.5 = 15, 15 x 3.5 = 52.5, 52.5 x 4.5 = 236.25. Since Sequence-II is built by the identical rule starting from 5: A = 5 x 0.5 = 2.5, B = 2.5 x 1.5 = 3.75, C = 3.75 x 2.5 = 9.375. So option (c) is correct. Option (a) 2.5 is the value of A, not C. Option (b) 5 is just the first term of Sequence-II. Option (d) 32.8125 is the value of D (9.375 x 3.5), one step too far.
Q6
Previous year questionmedium
What is X in the sequence 4, 196, 16, 144, 36, 100, 64, X?
A48
B64
C125
D256
Show answer and explanation
Correct answer: B - 64
Split the sequence into two interleaved sub-sequences. The terms in odd positions are 4, 16, 36, 64 = 2^2, 4^2, 6^2, 8^2 (squares of even numbers). The terms in even positions are 196, 144, 100, X = 14^2, 12^2, 10^2, X. The pattern is decreasing even squares: 14, 12, 10, then 8, so X = 8^2 = 64. Answer (b) 64.
Q7
Previous year questionhard
What comes at X and Y respectively in the following sequence?
January, January, December, October, X, March, October, Y, September
AJuly, May
BJuly, April
CJune, May
DJune, April
Show answer and explanation
Correct answer: B - July, April
Number the months (Jan=1 ... Dec=12) and subtract an increasing offset from a fixed reference. Reading the positions, each term is obtained by going back 0, 1, 2, 3, 4, 5, 6, 7, 8 months from January (treated cyclically): position 1 = Jan (Jan - 0), position 2 = Jan (Jan - 0 again as the base), position 3 = December (back 1 from Jan), position 4 = October (back further), and so on, with the step between successive terms growing. Following this decreasing-month pattern, the 5th term X = July and the 8th term Y = April. Hence X = July, Y = April. Official UPSC 2025 key: B.
Q8
Previous year questioneasy
What is the missing number 'X' of the series 7, X, 21, 31, 43?
(a) 11
(b) 12
(c) 13
(d) 14
A11
B12
C13
D14
Show answer and explanation
Correct answer: C - 13
The gaps between successive terms increase by 2 each time: starting from 7 the differences run +6, +8, +10, +12. So 7 + 6 = 13 (= X), 13 + 8 = 21, 21 + 10 = 31, 31 + 12 = 43, which fits the rest of the series. Hence X = 13.
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