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

Concept: In a Fibonacci series each term is the sum of the two terms before it.

1. $1+1=2$, $1+2=3$, $2+3=5$, $3+5=8$, $5+8=13$.
2. Next term $=8+13=21$.

Trick: 1, 1, 2, 3, 5, 8, 13, 21, 34… – learn the first ten Fibonacci numbers by heart.

Answer: (B) $21$

Concept: Track the five positions (four corners + centre). At each step three symbols rotate among three positions, and the two kinds of rotation alternate.

1. Fig 1→2: centre → bottom-left, bottom-left → top-right, top-right → centre (circle and star-6 stay).
2. Fig 2→3: centre → bottom-right, bottom-right → top-left, top-left → centre.
3. Fig 3→4: again centre → bottom-left, bottom-left → top-right, top-right → centre.
4. Fig 4→5 must repeat the second move: centre (4-point star) → bottom-right, bottom-right (arrow) → top-left, top-left (6-point star) → centre; diamond and circle stay.
5. Result: arrow top-left, diamond top-right, 6-point star centre, circle bottom-left, 4-point star bottom-right = option (D).

Trick: Find what moves between Fig 1→2 and Fig 2→3; in SSC figure series such moves usually repeat alternately.

Answer: (D)

Concept: For a mirror placed vertically on the right (MN), left and right are interchanged: the order of the characters reverses and every character is flipped sideways, while top and bottom stay the same.

1. Given: R k g E y w. In the mirror the order becomes w y E g k R (last character first).
2. Each character is laterally inverted: w stays similar, y becomes its mirror form, E faces left (Ǝ), g and k face left, R becomes Я.
3. Nothing is turned upside down. Option (B) shows exactly this; (A) keeps the original order of w and y, (C) and (D) turn some letters upside down.

Trick: Vertical mirror → reverse the order + flip each letter left-right; never flip upside down.

Answer: (B)

Concept: Swap the two signs of each option and check the equation using BODMAS.

1. Swap ÷ and −: $2+10\times 12÷4-24$.
2. $12÷4=3$, $10\times 3=30$.
3. $2+30-24=8$ ✓ (RHS $=8$).
4. Check (A) ÷ and +: $2÷10\times 12-4+24=22.4\ne 8$.

Trick: The number 24 at the end must be subtracted to get a small RHS like 8 – so − must move next to 24.

Answer: (B) ÷ and −

Concept: Draw the route with North up. When moving South, a left turn points East; when moving East, a left turn points North.

1. A → $7\text{km}$ South.
2. Left turn → East $9\text{km}$.
3. Left turn → North $7\text{km}$ to B.
4. The 7 km South and 7 km North cancel, so B is exactly $9\text{km}$ East of A. Distance from B to A $=9\text{km}$.

Trick: Equal opposite moves cancel – only the unmatched 9 km East remains.

Answer: (A) $9\text{km}$

Concept: Find the shift for each letter position; here it is $+3,-4,+3,-4$.

1. CUTE → FQWA: $C(3)+3=F$, $U(21)-4=Q$, $T(20)+3=W$, $E(5)-4=A$.
2. BLUE → EHXA: $B+3=E$, $L-4=H$, $U+3=X$, $E-4=A$ ✓.
3. VOTE: $V(22)+3=Y$, $O(15)-4=K$, $T(20)+3=W$, $E(5)-4=A$.
4. So VOTE → YKWA.

Trick: Both words end in E → A; so the code of VOTE must end in A – only (D) does.

Answer: (D) YKWA

Concept: Study each of the four letter positions as its own series.

1. 1st letters: H, K, ?, Q, T → $+3$ each → N.
2. 2nd letters: E, I, ?, Q, U → $+4$ each → M.
3. 3rd letters: W, R, ?, H, C → $-5$ each → M.
4. 4th letters: C, I, ?, U, A → $+6$ each (U + 6 wraps to A) → O.
5. Missing term: NMMO.

Trick: The four positions move by $+3,+4,-5,+6$ – write the alphabet positions to see it instantly.

Answer: (A) NMMO

Concept: Compare the letters of each pair position by position.

1. ABUSE → ASUBE: the 2nd letter (B) and 4th letter (S) have swapped places; 1st, 3rd and 5th stay.
2. BEGIN → BIGEN: 2nd (E) and 4th (I) swap ✓.
3. OTHER: 2nd letter T and 4th letter E swap → O E H T R.
4. So OTHER → OEHTR.

Trick: When the same letters appear in a new order, look for a swap of positions, not a shift of letters.

Answer: (D) OEHTR

Concept: Write the alphabet positions and look at the gaps between the letters.

1. PMS: $16\to 13\to 19$, i.e. $-3$, then $+6$.
2. URX: $21\to 18\to 24$: $-3,+6$ ✓.
3. JGM: $10\to 7\to 13$: $-3,+6$ ✓.
4. SPU: $19\to 16\to 21$: $-3,+5$ ✗ – this one is different.

Trick: In each group the 3rd letter should be 3 ahead of the 1st (P→S, U→X, J→M); S→U is only 2 ahead.

Answer: (D) SPU

Concept: Check the differences of consecutive terms.

1. $40-25=15$, $70-40=30$, $115-70=45$, $175-115=60$.
2. The differences are multiples of 15: $15,30,45,60$ → next $75$.
3. $175+75=250$.

Trick: Differences $15\times 1,15\times 2,15\times 3\dots$ – next difference is $15\times 5$.

Answer: (C) $250$

Concept: Find how the 2nd and 3rd numbers of each set are made from the 1st number.

1. $(12,36,6)$: $12\times 3=36$ and $\frac{12}{2}=6$.
2. $(18,54,9)$: $18\times 3=54$ and $\frac{18}{2}=9$ ✓.
3. (C) $(28,84,14)$: $28\times 3=84$, $\frac{28}{2}=14$ ✓.
4. (A) has 16, (B) has 86 and (D) starts with 26 ($26\times 3=78$), so they fail.

Trick: Middle $=3\times$ first and last $=\frac{1}{2}\times$ first; equivalently middle $=6\times$ last.

Answer: (C) $(28,84,14)$

Concept: Insert the signs of each option in order and evaluate using BODMAS.

1. Option (D): $10÷2-1\times 8+11=8$.
2. $10÷2=5$ and $1\times 8=8$.
3. $5-8+11=8$ ✓.
4. Check (A): $10÷2-1\times 8-11=5-8-11=-14\ne 8$.

Trick: Options (A) and (D) differ only in the last sign; $5-8=-3$ needs $+11$ to reach $8$.

Answer: (D) $÷,-,\times ,+,=$

Concept: The given figure must be found inside an option in the same size and orientation, without rotation.

1. The given figure is an arrow-like shape pointing left: its left tip is a meeting point of two slanting lines, a small square sits on the upper slanting line, and a stepped line joins the square to the lower-right end.
2. In option (C), look at the right-hand triangle formed by the two diagonals: the diagonals meet at the centre (the tip), the small square sits on the upper diagonal and the stepped line runs down to the right edge – exactly the given figure, same orientation.
3. In (A), (B) and (D) the small square is attached in a different position or on the other side, so the figure would have to be rotated or flipped.

Trick: Match the most distinctive part first (the small square on the slanting line) and check its side and direction.

Answer: (C)

Concept: When terms grow roughly three times, test the rule "$\times 3$ then add/subtract a constant".

1. $7\times 3-5=16$
2. $16\times 3-5=43$, $43\times 3-5=124$, $124\times 3-5=367$ ✓
3. Next: $367\times 3-5=1101-5=1096$.

Trick: Differences $9,27,81,243$ are powers of 3, so the next difference is $729$: $367+729=1096$.

Answer: (B) $1096$

Concept: Decode the expression one symbol at a time and place each person in a family tree.

1. $M\times Q$: M is the wife of Q.
2. $L÷M$: L is the daughter of M, so L is also the daughter of Q.
3. $K-L$: K is the sister of L, so K is also a daughter of M and Q.
4. $G+K$: G is the wife of K's brother. K's brother is a son of Q, so G is the wife of Q's son.
5. Hence G is Q's son's wife (daughter-in-law).

Trick: Start decoding from the end (M × Q) to fix the oldest generation first.

Answer: (D) Son’s wife

Concept: Here the word is first written in reverse order, then the letters are shifted back by $1,2,3,4$.

1. FIND reversed = DNIF; $D-1=C$, $N-2=L$, $I-3=F$, $F-4=B$ → CLFB ✓.
2. HOW reversed = WOH; $W-1=V$, $O-2=M$, $H-3=E$ → VME ✓.
3. WILL reversed = LLIW; $L-1=K$, $L-2=J$, $I-3=F$, $W-4=S$.
4. So WILL → KJFS.

Trick: If the code's first letter is near the word's last letter (C near D, V near W), try reversing the word.

Answer: (D) KJFS

Concept: A face shown next to another face cannot be opposite to it. A face that touches four others is opposite the sixth.

1. 60 appears in both positions.
2. Its neighbours are 40 and 30 (first position) and 20 and 10 (second position).
3. So 60 is adjacent to 10, 20, 30, 40; the only face left is 50.
4. Hence 50 is opposite 60.

Trick: Common face with four different neighbours → the missing number is its opposite.

Answer: (C) $60$

Concept: Compare words letter by letter; at the first letter that differs, the one earlier in the alphabet comes first.

1. All start with "Con". 4th letter: Conflagration has f, the others have s → Conflagration (3) first.
2. Among "Cons-": 5th letter c (Conscience), e (Consequence), i (Considerable, Consignment, Consistence) → 2, 4.
3. Among "Consi-": 6th letter d (Considerable), g (Consignment), s (Consistence) → 5, 6, 1.
4. Order: 3, 2, 4, 5, 6, 1.

Trick: Group the words by their common beginning, then sort each group by the next letter.

Answer: (D) 3, 2, 4, 5, 6, 1

Concept: First interchange the signs as told ( + ↔ × and − ↔ ÷ ), then solve with BODMAS.

1. $14-7\times 10÷1+7$ becomes $14÷7+10-1\times 7$.
2. $14÷7=2$, $1\times 7=7$.
3. $2+10-7=5$.

Trick: Change all signs in one go on paper before calculating.

Answer: (A) $5$

Concept: Each letter is moved forward by its own position number: $+1,+2,+3,+4$.

1. DICE → EKFI: $D+1=E$, $I+2=K$, $C+3=F$, $E+4=I$.
2. PLAN → QNDR: $P+1=Q$, $L+2=N$, $A+3=D$, $N+4=R$ ✓.
3. LUDO: $L+1=M$, $U+2=W$, $D+3=G$, $O+4=S$.
4. So LUDO → MWGS.

Trick: Increasing shifts $+1,+2,+3,+4$ – the first letter of the answer must be M (L + 1), leaving only (C) and (D); the third letter G decides.

Answer: (C) MWGS