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

Concept: Substitute the signs and apply BODMAS.

1. Expression: $2\times 12÷4+16-7$.
2. $2\times 12=24$, $24÷4=6$.
3. $6+16-7=15$.

Trick: × and ÷ have equal priority — do them left to right.

Answer: (A) $15$

Concept: All face the centre, so 'left' means clockwise and 'right' means anticlockwise (seen from above).

1. Put P at a seat. T is 2nd to the left of P; V is 3rd to the right of P (P is 3rd left of V).
2. W is 2 seats from V: either T's side seat (then R would have no valid seat) or the seat just right of P — so W sits immediately right of P.
3. R avoids the neighbours of V and P, so R sits between T and the seat beside V. S is 2nd right of Q ⇒ Q sits between V and W, S between R and V; Y takes the last seat (left of P).
4. Clockwise: P, Y, T, R, S, V, Q, W ⇒ 4th to the right of P (directly opposite) is S.

Trick: In an 8-seat arrangement, 4th to the left or right is simply the person opposite.

Answer: (C) S

Concept: In a net, faces separated by one face in a straight line are opposite; two opposite faces can never be seen together on a cube.

1. Row u–a–i–z: u ↔ i and a ↔ z are opposite.
2. e (above a) and o (below i) end up as top and bottom ⇒ e ↔ o.
3. B shows o and e together and D shows u and i together — impossible. A (z, i, o) and C (i, a, e) contain no opposite pair ⇒ possible.

Trick: First find the three opposite pairs; then reject any cube showing a pair together.

Answer: (A) Only A and C

Concept: Letters in odd positions move $+1$, letters in even positions move $-2$.

1. GRANTED: G+1=H, R−2=P, A+1=B, N−2=L, T+1=U, E−2=C, D+1=E ✓.
2. IGNORE: I+1=J, G−2=E, N+1=O, O−2=M, R+1=S, E−2=C.
3. Code: JEOMSC.

Trick: Check the first two letters of the options: only (D) starts with 'JE'.

Answer: (D) JEOMSC

Concept: Swap the given pair of signs in each option and evaluate with BODMAS.

1. Swap + and −: $551÷19+16\times 12-121$.
2. $551÷19=29$, $16\times 12=192$.
3. $29+192-121=100$ ✓.

Trick: $551÷19=29$ is a clean division — keep ÷ where it is and try swaps among the other signs.

Answer: (D) + and −

Concept: The differences repeat in the cycle $16,17,18$.

1. $69-53=16,86-69=17,104-86=18$.
2. $120-104=16$, next difference $17$: $120+17=137$.
3. Check: $155-137=18$ ✓.

Trick: Always verify the answer with the term after the blank.

Answer: (D) $137$

Concept: Decode the chain S $ T # W & U % X @ V, where X is a person.

1. S $ T: S is T's daughter. T # W: T is W's mother ⇒ S and W are siblings, T is their mother.
2. W & U: W is U's father. U % X: U is X's son ⇒ X is W's wife.
3. X @ V: X is V's daughter-in-law ⇒ V is W's parent; W's mother is T, so V is male, T's husband — the father of S.

Trick: Daughter-in-law of V = wife of V's son; then V is the parent of that son.

Answer: (D) Father

Concept: Each letter moves back by 4 and the number becomes its square root.

1. E25I → A5E: $E-4=A$, $\sqrt{25}=5$, $I-4=E$ ✓. Z49T → V7P ✓.
2. P144K: $P-4=L$, $\sqrt{144}=12$, $K-4=G$.
3. Result: L12G.

Trick: Spot perfect squares (25, 49, 144) — the square root is the usual link.

Answer: (B) L12G

Concept: If two faces are common to both views, the remaining faces in the matching position are opposite to each other.

1. Common faces in both dice: 2 and 6.
2. Read each corner clockwise: view 1 gives 2 → 6 → 3, view 2 gives 6 → 2 → 1 — the same edge 2–6 seen from its two ends.
3. The faces 3 and 1 lie at the two ends of the common edge 2–6, so they are opposite: 1 ↔ 3.

Trick: Two common faces ⇒ the two uncommon faces are opposite (standard dice rule).

Answer: (D) $3$

Concept: Replace each sign by its meaning, then use BODMAS.

1. $15÷5\times 36-6+14$ becomes $15\times 5+36÷6-14$.
2. $15\times 5=75$, $36÷6=6$.
3. $75+6-14=67$.

Trick: Rewrite the whole expression with the new signs before calculating.

Answer: (D) $67$

Concept: Replace each symbol by its meaning, then apply BODMAS.

1. Given: $65÷35+72\times 8-5$
2. After replacing: $65+35-72÷8\times 5$
3. $72÷8=9$, $9\times 5=45$
4. $65+35-45=55$

Trick: Write the new expression fully before solving; do division and multiplication first.

Answer: (D) $55$

Concept: The letters are taken in pairs: 1st and 2nd letters swap with shifts, middle letter stays, 4th and 5th swap with shifts.

1. CHART: (C, H) → (H−1, C+1) = (G, D); A stays; (R, T) → (T−1, R+1) = (S, S) → GDASS
2. EMPTY: (E, M) → (L, F); P stays; (T, Y) → (X, U) → LFPXU
3. INDEX: (I, N) → (N−1, I+1) = (M, J); D stays; (E, X) → (X−1, E+1) = (W, F)
4. Code = MJDWF

Trick: Whenever the middle letter is unchanged, check pair-wise reversal with ±1.

Answer: (B) MJDWF

Concept: Draw the smallest set inside the bigger sets (Venn method).

1. All phones are gadgets, all gadgets are machines → Phones inside Gadgets inside Machines.
2. I: All phones are machines — true, since the Phones circle lies wholly inside Machines.
3. II: All machines are gadgets — not certain; the Machines circle is bigger than Gadgets.

Trick: All A are B + All B are C ⇒ All A are C; but the reverse (All C are B) never follows.

Answer: (D) Only conclusion I follows

Concept: Track each element separately: every existing bracket rotates ${90}^{\circ }$ clockwise in the next figure and a new element is added.

1. The bracket ⊓ (open downward) becomes ⊐ (open left), then ⊔ (open up), then ⊏ (open right), then ⊓ again — each step ${90}^{\circ }$ clockwise.
2. The chevron (added in figure 3) also turns ${90}^{\circ }$ clockwise in each step.
3. In figure 5 there are two ⊓ brackets and a chevron; in figure 6 both ⊓ become ⊐ and the chevron turns once more.
4. Only option (C) shows both brackets as ⊐ with the chevron in the correct new orientation.

Trick: In figure series, follow one element at a time — rotation direction first, then position.

Answer: (C) Option (C)

Concept: Find the gaps between consecutive letters of each cluster.

1. GKOQ: G(7) → K(11) → O(15) → Q(17): +4, +4, +2
2. AEIK: +4, +4, +2; KOSU: +4, +4, +2
3. QUXZ: Q(17) → U(21) → X(24) → Z(26): +4, +3, +2 — different

Trick: Write the alphabet positions under the letters; the pattern of gaps reveals the odd one quickly.

Answer: (A) QUXZ

Concept: Compare common words and common codes between sentences.

1. 'you are tall' and 'daily workout makes you tall' share 'you, tall' → codes 'la, ma'.
2. 'you workout daily' and 'daily workout makes you tall' share 'you, workout, daily' → 'la, pa, ka'.
3. In the third sentence, the codes left after removing la, ma, pa, ka are 'ja'.
4. So 'makes' = ja.

Trick: The only word that appears in just one sentence gets the only code that also appears just once there.

Answer: (B) ja

Concept: Check each position of the clusters separately.

1. 1st letters: A, C, E, G, I (+2)
2. 2nd letters: Z, B, D, F, H (+2)
3. 3rd letters: X, Z, B, D, F (+2)
4. Next cluster = IHF

Trick: Position-wise series: each column moves +2 with wrap-around after Z.

Answer: (A) IHF

Concept: With 'some' statements, a definite conclusion must be true in every diagram; a possibility needs only one valid diagram.

1. Classes–Theories overlap and Theories–Lessons overlap; Classes and Lessons have no direct relation.
2. I: 'All classes can never be lessons' means no class is a lesson — not certain (they may overlap). I does not follow.
3. II: Theories may lie completely inside Classes without breaking any statement, so the possibility is true. II follows.

Trick: No direct link between two terms → both 'some' and 'no' are only possibilities, never certainties.

Answer: (C) Only conclusion II follows

Concept: Analyse each letter position separately.

1. 1st: P, M, J, G, D (−3)
2. 2nd: Y, V, S, P, M (−3)
3. 3rd: W, Y, A, C, E (+2)
4. 4th: B, D, F, H, J (+2)
5. Missing term = J S A F

Trick: Two columns go −3 and two go +2 — check the 3rd column across Z → A.

Answer: (D) JSAF

Concept: Facing the centre: right = anticlockwise, left = clockwise.

1. Place Salman; Bala is 2nd to his right.
2. Shwetha is 3rd to the left of Bala → she sits next to Salman on the other side.
3. Salman's neighbours are Kiran and Shwetha → Kiran sits between Salman and Bala.
4. Akshay sits next to Bala and Nishitha → order: Bala, Akshay, Nishitha, Shwetha.
5. Shwetha is the neighbour of both Nishitha and Salman.

Trick: Fix the person mentioned most often first, then place others relative to him.

Answer: (C) Shwetha