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

Concept: Find one rule that changes the first number into the second for both given pairs.

1. $(19,200)$: $19\times 10+10=200$, i.e. $(19+1)\times 10$.
2. $(15,160)$: $(15+1)\times 10=160$ ✓. Rule: $(n+1)\times 10$.
3. Check options: $(22+1)\times 10=230$ ✓; $(23+1)\times 10=240\ne 300$; $(25+1)\times 10=260\ne 250$; $(18+1)\times 10=190\ne 200$.

Trick: Second number ends in 0 and is about 10 times the first → try $(n\pm 1)\times 10$.

Answer: (B) (22, 230)

Concept: Use reverse alphabetical positions (Z = 1, Y = 2, …, A = 26), i.e. reverse value $=27-\text{position}$, and add them.

1. FLY: F = 21, L = 15, Y = 2 → $21+15+2=38$ ✓.
2. NUT: N = 13, U = 6, T = 7 → $13+6+7=26$ ✓.
3. GOD: G = 20, O = 12, D = 23 → $20+12+23=55$.

Trick: If forward position sums do not match, try reverse positions ($27-n$) before anything else.

Answer: (A) 55

Concept: Code = $2\times$ (sum of alphabetical positions of the letters).

1. PEN: $16+5+14=35$, $35\times 2=70$ ✓.
2. NEST: $14+5+19+20=58$, $58\times 2=116$ ✓.
3. PEON: $16+5+15+14=50$, $50\times 2=100$.

Trick: If the code is even and about double the letter-sum, test "sum × 2" first.

Answer: (C) 100

Concept: Replace the letters by signs and solve with BODMAS.

1. P = $÷$, Q = $\times$, R = $+$, S = $-$: $174÷2-45+(16÷4)\times 5-26$.
2. Bracket and division/multiplication: $174÷2=87$, $(16÷4)\times 5=4\times 5=20$.
3. $87-45+20-26=36$.

Trick: Solve the bracket first, then all ÷ and ×, and only then + and − from left to right.

Answer: (C) 36

Concept: Interchange the two signs given in each option and test with BODMAS.

1. Option (B), swap ÷ and −: $9+6\times 3÷9-1$.
2. $6\times 3÷9=18÷9=2$, so $9+2-1=10$ ✓.
3. Others fail, e.g. (A) $9+6÷3-9\times 1=2$; (C) $9-6\times 3+9÷1=0$.

Trick: Look for a swap that makes a big product shrink (here $18÷9$) — the RHS 10 is small.

Answer: (B) ÷ and −

Concept: Write the alphabetical positions and check the gaps between letters.

1. OVS: $15\to 22\to 19$ = $+7,-3$. LSP: $12\to 19\to 16$ = $+7,-3$. NUR: $14\to 21\to 18$ = $+7,-3$.
2. JQM: $10\to 17\to 13$ = $+7,-4$.
3. So JQM breaks the $+7,-3$ pattern.

Trick: Equivalently, the 1st and 3rd letters differ by 4 in the group (O–S, L–P, N–R), but by 3 in J–M.

Answer: (C) JQM

Concept: Facing the centre, a person's right is the anticlockwise direction and left is clockwise.

1. Fix A. B is second to the right of A. D and C sit on either side of A.
2. F is fourth to the left of D. If D were on A's left, F would land on C's seat, so D is immediately right of A; then F is opposite A.
3. E is a neighbour of F; B already sits on one side of F, so E takes the seat between F and C.
4. Order (going right from A): A, D, B, F, E, C. E is immediately to the left of C.

Trick: With 6 people, "fourth to the left" equals "second to the right" — saves counting.

Answer: (B) Immediate left of C

Concept: Find the rule linking the three numbers in each given set.

1. $(21,105,35)$: $21\times 5=105$ and $105÷3=35$.
2. $(27,135,45)$: $27\times 5=135$, $135÷3=45$ ✓.
3. $(39,195,65)$: $39\times 5=195$, $195÷3=65$ ✓. In (B) $65\ne 55$, in (C) $39\times 5\ne 185$, in (D) $37\times 5\ne 195$.

Trick: Third number $=\frac{5}{3}\times$ first: $\frac{5}{3}\times 39=65$.

Answer: (A) (39, 195, 65)

Concept: Replace the letters by signs and solve with BODMAS.

1. A = $÷$, B = $\times$, C = $+$, D = $-$: $455÷5-17+36\times 4$.
2. $455÷5=91$, $36\times 4=144$.
3. $91-17+144=218$.

Trick: Do ÷ and × first, then add/subtract from left to right.

Answer: (D) 218

Concept: Replace the letters by signs and solve with BODMAS.

1. A = $+$, B = $-$, C = $\times$, D = $÷$: $29+37-14\times 3+(84÷4)+46$.
2. $14\times 3=42$, $84÷4=21$.
3. $29+37-42+21+46=91$.

Trick: Group all positives and subtract the negatives once: $(29+37+21+46)-42=133-42=91$.

Answer: (B) 91

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

1. 4 is on top in all three views and its neighbours are 3, 2, 5 and 6, so 1 is opposite 4.
2. Turning the dice about the 4–1 axis, the side faces come round in the order 3, 2, 5, 6.
3. In this ring of four, alternate faces are opposite: 3 ↔ 5 and 2 ↔ 6.

Trick: With a common top face, the four side faces form a ring — the 1st and 3rd of the ring are opposite, as are the 2nd and 4th.

Answer: (C) 2

Concept: Trace the path step by step; a left turn while going West means facing South, and a left turn while going South means facing East.

1. From A, 6 km West.
2. Left turn → South, 8 km.
3. Left turn → East, 6 km to B. The 6 km West and 6 km East cancel.
4. So B is 8 km directly South of A.

Trick: Equal and opposite moves cancel; only the unpaired 8 km South remains.

Answer: (B) South

Concept: Read the direction of the relation carefully: the first number is related to the second.

1. $3\to 2$ and $9\to 6$: the second number is $\frac{2}{3}$ of the first.
2. We need the number $x$ that is related to 12, i.e. $\frac{2}{3}\times x=12$.
3. $x=12\times \frac{3}{2}=18$. Check: $\frac{2}{3}\times 18=12$ ✓.

Trick: Here 12 plays the role of 2 and 6, so reverse the rule: multiply by $\frac{3}{2}$.

Answer: (B) 18

Concept: Note what is added at each step; here an element is added every step, alternating between curved arcs inside the circle and vertical lines beside it.

1. Fig 1: plain circle. Fig 2: two arcs added inside.
2. Fig 3: a vertical line added on the left. Fig 4: a third arc added inside.
3. Next step should add a vertical line again, on the right, keeping the left line and all three arcs.
4. Option (D) has both lines and three arcs; (C) has only two arcs, (A) and (B) have extra lines.

Trick: Alternating additions: arc, line, arc, line — count both elements in the options.

Answer: (D)

Concept: Check each letter position separately.

1. 1st letters: H, I, J → next K ($+1$).
2. 2nd letters: M, N, O → next P ($+1$).
3. 3rd letters: W, X, Y → next Z ($+1$).

Trick: Every letter moves one step forward, so the next cluster is simply each letter $+1$.

Answer: (A) KPZ

Concept: Code = sum of alphabetical positions of the letters.

1. JAM: $10+1+13=24$ ✓.
2. GEL: $7+5+12=24$ ✓.
3. LAW: $12+1+23=36$.

Trick: Remember EJOTY: E = 5, J = 10, O = 15, T = 20, Y = 25 — W = 23 is two after U = 21.

Answer: (C) 36

Concept: Find the rule linking the three numbers of each given set.

1. $(4,12,48)$: $4\times 3=12$, $12\times 4=48$.
2. $(9,27,108)$: $9\times 3=27$, $27\times 4=108$ ✓.
3. $(13,39,156)$: $13\times 3=39$, $39\times 4=156$ ✓. (A) has 146, (B) and (D) have 36.

Trick: Third number $=12\times$ first: $13\times 12=156$.

Answer: (C) (13, 39, 156)

Concept: Decide for each pair of classes whether one lies inside the other, they overlap, or they are separate.

1. Every mother is a woman, so Mother lies completely inside Women.
2. A husband is a man, so Husband has nothing in common with Women or Mother — a separate circle.
3. Option (A) shows one circle inside a bigger one and a third separate circle.

Trick: Gender-based classes (husband vs woman) are always disjoint.

Answer: (A)

Concept: Two "All" statements in a chain give Roads ⊂ Routes ⊂ Bridges.

1. I: "Some routes are roads" — since all roads are routes, the reverse "some" is always true (conversion).
2. II: "All roads are bridges" — all roads are routes and all routes are bridges, so all roads are bridges.
3. Both conclusions are certain.

Trick: All A are B + All B are C ⇒ All A are C; and All A are B ⇒ Some B are A.

Answer: (C) Both conclusions I and II follow

Concept: Each letter is written as its reverse alphabetical position ($27-\text{position}$).

1. CITY: C = 24, I = 18, T = 7, Y = 2 → 241872 ✓.
2. DRUG: D = 23, R = 9, U = 6, G = 20 → 239620 ✓.
3. GOLF: G = 20, O = 12, L = 15, F = 21 → 20121521.

Trick: Reverse position: A ↔ 26, M ↔ 14, N ↔ 13; pairs of letters always add to 27.

Answer: (B) 20121521