SSC CPO Paper I 2020, 2022, 2023, 2024 PYQs

Concept: Each figure has four elements (top, second, bottom-left, bottom-right). From one figure to the next they move in a fixed cycle and exactly one new letter appears.

1. From Fig 1 to Fig 4: bottom-right → bottom-left in the next figure (dot, S, T, B).
2. Bottom-left goes up to the top area (F → 2nd, dot → top, S → 2nd, T → top), alternately into the second and the top slot.
3. The top-area element that is not replaced drops to bottom-right (S, T, B, …), and the other top slot gets a new letter.
4. Fig 5: T (from BL) → top, B (from BR) → BL, V (top of Fig 4) → BR, S is replaced by a new letter H.

Trick: Track one element at a time (follow ‘B’ or ‘T’) instead of comparing whole figures.

Answer: (C) Top T, second H, bottom-left B, bottom-right V

Concept: Look at the differences between consecutive terms.

1. Differences: $36-35=1,39-36=3,48-39=9$.
2. They are powers of 3: $3^0,3^1,3^2$, so the next difference is $3^3=27$.
3. Next term $=48+27=75$.

Trick: Differences multiplying by the same number (×3) point to powers.

Answer: (C) $75$

Concept: Replace the letters by the signs, then solve using BODMAS.

1. M = −, N = ÷, O = ×, P = +.
2. Expression: $185÷5+62-32\times 4+32$.
3. $=37+62-128+32=3$.

Trick: Do ÷ and × first, then add all positives and subtract negatives: $131-128=3$.

Answer: (D) $3$

Concept: Find a repeating block that fits the given letters.

1. Given: D_C_FE_CC_ _E_C_EF_D_CE_E (24 places).
2. Try the block DCCEFE repeated 4 times: DCCEFE DCCEFE DCCEFE DCCEFE.
3. The blanks in order are C, E, D, E, F, D, C, E, C, F → CEDEFDCECF.

Trick: ‘FE’ in the first six letters suggests a block of 6 – check with the second block.

Answer: (D) CEDEFDCECF

Concept: Substitute the signs and apply BODMAS (brackets first).

1. A = +, B = ×, C = −, D = ÷.
2. $(51÷17)+(21\times 9)-78+(16÷4)$.
3. $=3+189-78+4=118$.

Trick: Solve each bracket separately first: 3, 189, 4.

Answer: (C) $118$

Concept: The exact given figure (same size and orientation, no rotation) must be traceable inside the option.

1. Key parts: a circle whose upper-right part is cut flat, with two thin rectangles attached at its top-right, extending to the right.
2. In option (D), at the lower part, the same drop-shaped circle with the two rectangles to its right appears in the same orientation.
3. In the other options the rectangles are placed differently (vertical or detached), so the figure cannot be traced.

Trick: Look first for the most distinctive part (the two parallel rectangles) and check their direction.

Answer: (D) Option figure (D)

Concept: Facing the centre, ‘right’ means moving anticlockwise. Corners and middles alternate.

1. Fix V at a corner; Q is 3rd to the right of V (a middle seat).
2. Two children between Q and S on the other side → S is at a corner; W is 2nd to the left of S.
3. T is a neighbour of both V and P → T beside V, P next to T. U is not next to V, so U is beside S and R takes the seat between V and S.
4. From W moving right: U (1), S (2), R (3).

Trick: In a closed arrangement of 8, ‘3rd to the right’ = ‘5th to the left’ – check options both ways.

Answer: (D) Third to the right of W

Concept: Find one rule that works for both given sets.

1. $(7,13,10)$: $\frac{7+13}{2}=10$.
2. $(15,21,18)$: $\frac{15+21}{2}=18$.
3. Third number = average of the first two. Option (C): $\frac{11+15}{2}=13$ ✓; (A) gives 11.5, (D) gives 7.5, (B) fails.

Trick: When the third number lies midway between the first two, test the average.

Answer: (C) $(11,15,13)$

Concept: Faces seen together with a face are adjacent to it; the one face never seen with it is opposite.

1. X is on top in both views; Z, B (first view) and Y, A (second view) touch X.
2. So X is adjacent to Z, B, Y, A – four faces.
3. The only remaining face W must be opposite X. With W on top, X is at the bottom.

Trick: A face has 4 neighbours and 1 opposite – list neighbours, the leftover is opposite.

Answer: (C) X

Concept: A vertical mirror on the right swaps left and right; top and bottom stay unchanged.

1. In the figure, the curved sides of both half-circles face left and the flat side faces right.
2. In the image the curves must face right, flat side on the left.
3. The black cap remains at the top and the small line at the bottom; the small circles stay inside. Only (D) shows this.

Trick: Vertical mirror: only left↔right changes – up/down details (black cap on top) must stay on top.

Answer: (D) Option figure (D)

Concept: Follow each of the three symbols separately: position and orientation.

1. Small circle: bottom-left → top-left → top-right → bottom-right, i.e. one corner clockwise each step; next = bottom-left.
2. Hourglass: top-right → bottom-right → bottom-left → top-left (clockwise) and turns 90° each time; next = top-right, standing upright.
3. Pac-man shape alternates top-left / top-right and turns 90° each time; after four turns it is back to its Fig 1 form at top-left.
4. So Fig 5 is the same as Fig 1 – option (B).

Trick: If every element moves one corner and turns 90°, the 5th figure repeats the 1st.

Answer: (B) Option figure (B)

Concept: Check each letter position separately.

1. 1st letters: P, Q, R, S → T (+1).
2. 2nd letters: T, U, V, W → X (+1).
3. 3rd letters: A, B, C, D → E (+1). Next cluster: TXE.

Trick: Write the clusters one under another and read each column.

Answer: (D) TXE

Concept: Decode each symbol step by step and draw the family tree.

1. D × O: D is the sister of O (D is female).
2. O ÷ E: O is the son of E.
3. E + Y: E is the husband of Y, so Y is O's mother.
4. D, a sister of O, is also Y's child → D is Y's daughter.

Trick: Sister of someone's son = that person's daughter.

Answer: (C) Daughter

Concept: Substitute the signs and use BODMAS.

1. A = +, B = ×, C = −, D = ÷.
2. $32-16÷8+4\times 2$.
3. $=32-2+8=38$.

Trick: ÷ and × first: $16÷8=2$, $4\times 2=8$.

Answer: (A) $38$

Concept: Use alphabet positions for each letter of the pair.

1. First letters: P, R, T, V, X → +2 each → Z.
2. Second letters: L(12), O(15), S(19), X(24), D(30 = 4): differences $+3,+4,+5,+6$.
3. Next difference $+7$: $4+7=11$ → K. Answer ZK.

Trick: When the count passes Z, subtract 26 (X + 6 = 24 + 6 = 30 → D).

Answer: (B) ZK

Concept: The word is split into two halves, the halves are swapped, and every letter is moved 2 places back.

1. CLOUDS → halves CLO | UDS → swap: UDSCLO → each −2: SBQAJM ✓.
2. BRIGHT → halves BRI | GHT → swap: GHTBRI → each −2: EFRZPG ✓.
3. DREAMY → halves DRE | AMY → swap: AMYDRE → −2: YKWBPC.

Trick: Spot that the code's first letter comes from the 4th letter of the word – a sign of swapped halves.

Answer: (B) YKWBPC

Concept: Reverse the word, then shift each letter one place forward.

1. GARDEN → reverse NEDRAG → +1: OFESBH ✓.
2. LADDER → REDDAL → +1: SFEEBM ✓.
3. HANDLE → ELDNAH → +1: FMEOBI.

Trick: The code's first letter (O) is next to the word's last letter (N) – reverse + 1.

Answer: (B) FMEOBI

Concept: Check the differences and their pattern.

1. Differences: $2,6,12$ = $1\times 2,2\times 3,3\times 4$.
2. Next difference $=4\times 5=20$.
3. Next term $=28+20=48$.

Trick: Differences 2, 6, 12, 20 are $n(n+1)$ – a very common SSC pattern.

Answer: (C) $48$

Concept: Relation: living being : the place where it lives/shelters.

1. A bird lives in a nest.
2. Cattle are kept in a barn – same relation.
3. Goat : Grass and Ant : Sugar are food relations; a kennel is a dog's shelter, not a tiger's.

Trick: Make a sentence: ‘A ___ lives in a ___’ and test each option.

Answer: (D) Cattle : Barn

Concept: Draw the path with A at the origin, north up.

1. 3 km east, left → 4 km north, left → 3 km west: he is 4 km straight north of A.
2. Left again → south 2 km: B is $4-2=2$ km north of A.
3. To return to A he must go 2 km towards south.

Trick: Equal east and west moves cancel; only north–south is left: $4-2=2$.

Answer: (A) 2 km towards south