UPSC Prelims PYQs · CSAT (Paper II)

Coding-decoding, series, patterns, cubes, dice and visual reasoning: previous-year questions

95 real UPSC Prelims questions (2011–2026) with the official answer and a clear explanation. 34 of them came in the last five papers.

How often UPSC asks this

Questions per year, counted from the official papers.

Attempt the questions

Newest papers first. Tap an option; the answer and the reason appear instantly.

UPSC Prelims 2024 · CSAT (Paper II)

What will come in place of * in the sequence 3, 14, 39, 84, *, 258?

Answer (b) & explanation

The missing number is 155 (b). Step 1: take the differences: 14 − 3 = 11, 39 − 14 = 25, 84 − 39 = 45. Step 2: take the differences again: 25 − 11 = 14, 45 − 25 = 20. These rise by 6, so the next ones are 26 and 32. Step 3: next first differences: 45 + 26 = 71, then 71 + 32 = 103. So * = 84 + 71 = 155, and 155 + 103 = 258 checks out. Options (a), (c) and (d) fail the check.

Tip: if the first differences show no pattern, take differences of differences. Always confirm with the last given term.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2024 · CSAT (Paper II)

What is the least possible number of cuts required to cut a cube into 64 identical pieces?

Answer (b) & explanation

The least number of cuts is 9 (b). Step 1: 64 identical small cubes means 64 = 4 × 4 × 4. So each of the three directions (length, breadth, height) must be split into 4 parts. Step 2: to split one direction into 4 parts you need 3 cuts, since cuts are always one less than pieces. Step 3: three directions × 3 cuts = 3 + 3 + 3 = 9 cuts. Option (a), 8, is too few. Options (c), 12, and (d), 16, waste cuts.

Tip: for n × n × n identical pieces, the least cuts are 3(n − 1). For 27 pieces: 3 × 2 = 6 cuts. If the cuts are not equal in each direction, use (a − 1) + (b − 1) + (c − 1).

Trap

8 cuts can make at most 4 × 4 × 3 = 48 pieces, so 8 is not enough.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

125 identical cubes are arranged in the form of a cubical block. How many cubes are surrounded by other cubes from each side?

Answer (a) & explanation

27 cubes are surrounded on every side (a). Step 1: 125 = 5 × 5 × 5, so the block is 5 cubes on each side. Step 2: a cube with neighbours on all six sides cannot be on the outer surface. Remove one layer from each face. Step 3: the inside block is (5 − 2) × (5 − 2) × (5 − 2) = 3 × 3 × 3 = 27. Options (b), (c) and (d) miscount.

Tip: for an n × n × n cube, the hidden inner cubes are (n − 2)³. The cubes with paint on just one face are 6(n − 2)².

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

If 7 ⊕ 9 ⊕ 10 = 8, 9 ⊕ 11 ⊕ 30 = 5, 11 ⊕ 17 ⊕ 21 = 13, what is the value of 23 ⊕ 4 ⊕ 15 ?

Answer (a) & explanation

The value is 6 (a). Step 1: find the rule from the examples. Add the three numbers, then add the digits of the sum. Step 2: 7 + 9 + 10 = 26 → 2 + 6 = 8. 9 + 11 + 30 = 50 → 5. 11 + 17 + 21 = 49 → 4 + 9 = 13. The rule works for all three. Step 3: 23 + 4 + 15 = 42 → 4 + 2 = 6. Options (b), (c) and (d) do not follow this rule.

Tip: for made-up symbols, test simple operations (sum, product, digit sum) on all examples. The rule must fit every one.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

If 'ZERO' is written as 'CHUR', then how is 'PLAYER' written ?

Answer (d) & explanation

PLAYER is written SODBHU (d). Step 1: compare ZERO with CHUR. Z → C, E → H, R → U, O → R. Each letter moves 3 places forward, wrapping from Z back to A. Step 2: apply this to PLAYER: P → S, L → O, A → D, Y → B, E → H, R → U. Step 3: the code is SODBHU. Option (a) gets A and Y wrong. Options (b) and (c) go wrong on the last letters.

Tip: when a letter near Z changes to one near A, the code wraps around. Count forward: Y, Z, A, B is 3 steps.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

What is the middle term of the sequence

Z, Z, Y, Y, Y, X, X, X, X, W, W, W, W, W, …, A ?

Answer (b) & explanation

The middle term is I (b). Step 1: Z appears 2 times, Y 3 times, X 4 times, and so on. The kth letter from Z appears k + 1 times, so A appears 27 times. Step 2: total terms = 2 + 3 + ... + 27 = 377. The middle term is the 189th. Step 3: the first m letters fill m(m + 3)/2 places. For m = 18 this is exactly 189. The 18th letter counting back from Z is I. Options (a) H, (c) J and (d) M are off.

Tip: counting back from Z, letter number k is the letter at position 27 − k. Here 27 − 18 = 9, which is I.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

Consider the following :

  • I. A + B means A is neither smaller nor equal to B.
  • II. A − B means A is not greater than B.
  • III. A × B means A is not smaller than B.
  • IV. A ÷ B means A is neither greater nor equal to B.
  • V. A ± B means A is neither smaller nor greater than B.

Statement : P × Q, P − T, T ÷ R, R ± S

Conclusion-1 : Q ± T

Conclusion-2 : S + Q

Which one of the following is correct in respect of the above Statement and the Conclusions ?

Answer (b) & explanation

Only Conclusion-2 follows (b). Step 1: decode the symbols: + means >, − means ≤, × means ≥, ÷ means < and ± means =. Step 2: the statement becomes P ≥ Q, P ≤ T, T < R, R = S. So S = R > T ≥ P ≥ Q. Step 3: Conclusion-1 (Q = T) need not be true; T could be bigger. Conclusion-2 (S > Q) is true, since S > T ≥ Q. So options (a), (c) and (d) fail.

Tip: decode every symbol first, then write one chain. A single strict ">" in the chain makes the end-to-end link strict.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

Consider the sequence ABC_ _ABC_DABBCD_ABCD that follows a certain pattern. Which one of the following completes the sequence?

Answer (d) & explanation

The sequence is completed by DDCA (d). Step 1: put D, D, C, A into the gaps. The sequence becomes ABCDDABCCDABBCDAABCD. Step 2: split it into blocks of four: ABCD, DABC, CDAB, BCDA, ABCD. Step 3: each block moves the last letter to the front of the previous block. The pattern holds throughout, so (d). Options (a), (b) and (c) break it.

Tip: in letter-series gaps, try each option and split the result into equal blocks. A rotation pattern is very common.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

A cuboid of dimensions 7 cm × 5 cm × 3 cm is painted red, green and blue colour on each pair of opposite faces of dimensions 7 cm × 5 cm, 5 cm × 3 cm, 7 cm × 3 cm respectively. Then the cuboid is cut and separated into various cubes each of side length 1 cm. Which of the following statements is/are correct ?

  • 1. There are exactly 15 small cubes with no paint on any face.
  • 2. There are exactly 6 small cubes with exactly two faces, one painted with blue and the other with green.

Select the correct answer using the code given below :

Answer (a) & explanation

Only statement 1 is correct (a). Step 1: unpainted cubes form the inner block: (7 − 2) × (5 − 2) × (3 − 2) = 5 × 3 × 1 = 15. So statement 1 is true. Step 2: blue faces (7 × 3) and green faces (5 × 3) meet along the four edges of length 3 cm. Step 3: each such edge has 3 − 2 = 1 middle cube with exactly those two colours. Four edges give 4 cubes, not 6. So statement 2 is false. Options (b), (c) and (d) fail.

Tip: two coloured faces meet along an edge. Cubes with exactly two colours on an edge = edge length − 2 (the corners have three colours).

Trap

corner cubes have three painted faces, so leave them out when you count 'exactly two'.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2023 · CSAT (Paper II)

The letters of the word "INCOMPREHENSIBILITIES" are arranged alphabetically in reverse order. How many positions of the letter/letters will remain unchanged ?

Answer (c) & explanation

Two positions stay the same (c). Step 1: the 21 letters in reverse alphabetical order are: T S S R P O N N M L I I I I I H E E E C B. Step 2: the original word is I N C O M P R E H E N S I B I L I T I E S. Step 3: compare place by place. Only places 13 and 15 match, both with the letter I. So two positions stay unchanged. Options (a), (b) and (d) miscount.

Tip: write both orders in two rows, one under the other. Count the letters carefully first; repeated letters like I (5 times) decide the answer.

Trap

write both rows one above the other and compare letter by letter.

Fell for this one? Most aspirants do. The PYQ Book puts every UPSC trap side by side, so the pattern finally clicks. Try the free sample

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2022 · CSAT (Paper II)

In the series AABABCABCDABCDE..., which letter appears at the 100th place?

Answer (c) & explanation

The 100th letter is I (c). Step 1: the series is in groups: A, AB, ABC, ABCD and so on. Group n has n letters. Step 2: the first 13 groups fill 1 + 2 + ... + 13 = 13 × 14 ÷ 2 = 91 places. Step 3: group 14 starts at place 92. Place 100 is the 9th letter of that group, which is I. Options (a) G, (b) H and (d) J are one or two places off.

Tip: the sum of the first n numbers is n(n + 1)/2. It quickly tells you which group a position falls in.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2022 · CSAT (Paper II)

What is the value of X in the sequence 2, 12, 36, 80, 150, X?

Answer (b) & explanation

X is 252 (b). Step 1: look for a pattern: 2 = 1 × 2, 12 = 4 × 3, 36 = 9 × 4, 80 = 16 × 5, 150 = 25 × 6. Step 2: each term is n² × (n + 1). Step 3: the next term is 6² × 7 = 36 × 7 = 252. Options (a) 248, (c) 258 and (d) 262 do not fit this pattern.

Tip: when terms grow fast but not by a fixed ratio, try squares or cubes times a simple number. Here n² × (n + 1) = n³ + n².

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2022 · CSAT (Paper II)

What is the value of X in the sequence 20, 10, 10, 15, 30, 75, X?

Answer (d) & explanation

X is 225 (d). Step 1: divide each term by the one before it. 10 ÷ 20 = 0.5; 10 ÷ 10 = 1; 15 ÷ 10 = 1.5; 30 ÷ 15 = 2; 75 ÷ 30 = 2.5. Step 2: the multiplier rises by 0.5 each step. The next one is 3. Step 3: X = 75 × 3 = 225. Option (c), 150, uses × 2 again. Options (a) and (b) come from adding instead of multiplying.

Tip: when a sequence falls and then rises sharply, look at the ratios between terms. A steadily growing multiplier is a common UPSC pattern.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2022 · CSAT (Paper II)

If the order of the letters in the English alphabet is reversed and each letter represents the letter whose position it occupies, then which one of the following represents 'LUCKNOW'?

Answer (d) & explanation

LUCKNOW becomes OFXPMLD (d). Step 1: in the reversed alphabet, A ↔ Z, B ↔ Y, and so on. A letter at place n becomes the letter at place 27 − n. Step 2: L (12) → O (15); U (21) → F (6); C (3) → X (24); K (11) → P (16). Step 3: N (14) → M (13); O (15) → L (12); W (23) → D (4). So the code is OFXPMLD. Option (c) ends in E and options (a), (b) start O-G; each has one wrong letter.

Tip: remember pairs that add to 27 (like EJOTY at 5, 10, 15, 20, 25). Check only the letters where options differ.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)

Images of consonants of the English alphabet (Capitals) are observed in a mirror. What is the number of images of these which do not look like their original shapes ?

Answer (b) & explanation

14 consonants look different in a mirror (b). Step 1: a capital letter looks the same in a mirror if it has a vertical line of symmetry. Step 2: among the 21 consonants, these have one: H, M, T, V, W, X, Y. That is 7 letters. Step 3: the rest look different: 21 − 7 = 14. Option (a), 13, wrongly includes another letter. Options (c) and (d) miss some symmetric letters.

Tip: the vowels A, I, O and U are also symmetric, but the question asks only about consonants. There are 26 letters, 5 vowels, so 21 consonants. Memorise "AHIMOTUVWXY" as the mirror-same capitals.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)

Replace the incorrect term by the correct term in the given sequence

3, 2, 7, 4, 13, 10, 21, 18, 31, 28, 43, 40

where odd terms and even terms follow the same pattern.

Answer (a) & explanation

The incorrect term 2 should be replaced by 0 (a). Step 1: odd places: 3, 7, 13, 21, 31, 43. The gaps are 4, 6, 8, 10, 12. Step 2: even places: 2, 4, 10, 18, 28, 40. The gaps from 4 on are 6, 8, 10, 12, matching the odd places. So the first gap must also be 4. Step 3: the first even term should be 4 − 4 = 0, not 2. Options (b), (c) and (d) give a wrong first gap.

Tip: when a question says two sub-series follow the same pattern, find the pattern in the clean one, then fix the other. Each even term here is 3 less than its odd partner.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)
7B 10A 3C
3C 9B 6A
10A 13C ?

Following is a matrix of certain entries. The entries follow a certain trend row-wise. Choose the missing entry (?) accordingly.

Answer (c) & explanation

The missing entry is 3B (c). Step 1: look at the numbers in each row. The third number is the second minus the first: 10 − 7 = 3 and 9 − 3 = 6. So in the last row, 13 − 10 = 3. Step 2: look at the letters. Each row uses A, B and C once each. The last row already has A and C, so the missing letter is B. Step 3: put them together: 3B. Option (a), 9B, gets the number wrong. Options (b) 3A and (d) 3C repeat a letter already in the row.

Tip: in matrix puzzles, split the entry into parts (number and letter) and solve each part on its own. Check your rule on every row before trusting it.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)

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 ?

Answer (c) & explanation

The entry at C is 9.375 (c). Step 1: find the rule in Sequence-I. Each term is the one before multiplied by 0.5, 1.5, 2.5, 3.5, 4.5 in turn. The multiplier grows by 1 each step. Step 2: the two sequences are "identical", so Sequence-II follows the same rule, only starting from 5. Step 3: 5 × 0.5 = 2.5, then 2.5 × 1.5 = 3.75, then 3.75 × 2.5 = 9.375. Option (a), 2.5, is the first step and option (b), 5, is the start. Option (d), 32.8125, is a later term.

Tip: when a sequence grows unevenly, divide each term by the one before it. The pattern often hides in these ratios.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)

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 ?

Answer (b) & explanation

The letter is T (b). Step 1: reverse the alphabet in blocks of four: DCBA HGFE LKJI PONM TSRQ XWVU. Then Y and Z swap, giving ZY at the end. Step 2: count to the 13th letter. The first three blocks have 12 letters, so the 13th is the first letter of the 4th block: P. Step 3: the fourth letter to the right of the 13th is the 17th letter. That is the first letter of the 5th block: T. Option (a), N, comes from the normal alphabet. Options (c) and (d) are other miscounts.

Tip: write out the new order fully before counting. "Fourth to the right of the 13th" simply means position 13 + 4 = 17.

Answer as per the official UPSC answer key.

Open this question's page →

UPSC Prelims 2021 · CSAT (Paper II)

What is the value of 'X' in the sequence 2, 7, 22, 67, 202, X, 1822 ?

Answer (c) & explanation

X is 607 (c). Step 1: look at how each term is made from the one before. 2 × 3 + 1 = 7; 7 × 3 + 1 = 22; 22 × 3 + 1 = 67; 67 × 3 + 1 = 202. The rule is × 3, then + 1. Step 2: so X = 202 × 3 + 1 = 607. Step 3: check with the next term: 607 × 3 + 1 = 1822. It matches. Options (a) 603, (b) 605 and (d) 608 fail this check.

Tip: when terms grow about three times each step, try "× 3 ± a small number". Always confirm using the term after the blank.

Answer as per the official UPSC answer key.

Open this question's page →