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

Concept: When a folded paper is cut, every cut appears once in each layer. On unfolding, the cuts are reflected across every fold line, so the final design is symmetric about each fold.

1. The paper is folded twice (once vertically and once horizontally), so the final piece has 4 layers and every cut appears 4 times.
2. In the folded piece there is a diamond, a circle and a pair of slanting bars. Opening the last fold mirrors them top-to-bottom; opening the first fold mirrors them left-to-right.
3. So the upper half must be the upside-down mirror of the lower half: bars at the outer top corners, then circles, then diamonds, and the lower half the reverse.
4. Only option (D) shows this double mirror symmetry; in (A), (B) and (C) the top half is not the exact reflection of the bottom half.

Trick: Check one fold at a time: the shapes nearest a fold line in the folded piece stay nearest that line in both halves after unfolding.

Answer: (D) the figure with bars at all four outer corners, mirrored about both centre lines

Concept: In this series one new triangle is added around the circle at every step, moving clockwise, and the colours of the added triangles alternate white, black, white, black…

1. Fig 1: one white triangle at the top.
2. Fig 2: a black triangle is added at the upper right. Fig 3: a white triangle is added at the lower right. Fig 4: a black triangle is added at the lower left.
3. Next step: a white triangle must be added at the upper left, keeping all the earlier triangles unchanged.
4. Option (A) has exactly this: top white, upper-right black, lower-right white, lower-left black, upper-left white. (B) has a black upper-left triangle, (C) changes the lower-left one to white, (D) makes the top one black.

Trick: Count the elements (1, 2, 3, 4 → 5) and continue the colour alternation; then reject any option that changes an earlier element.

Answer: (A) five triangles with the new upper-left triangle white

Concept: Draw one circle inside another when one class is a part of the other; draw separate circles when the classes have nothing in common.

1. Every potato is a vegetable, so the Potatoes circle lies completely inside the Vegetables circle.
2. Grapes are fruits, not vegetables, and no grape is a potato, so Grapes is a separate circle.
3. This "one circle inside another + one separate circle" picture is option (B). Option (A) wrongly puts grapes inside vegetables.

Trick: Ask "is every X a Y?" for each pair: yes → circle inside; no common member → separate circle.

Answer: (B) Potatoes inside Vegetables, Grapes separate

Concept: Find the rule that links each pair: here the second number $=\text{first}\times \frac{\text{first}}{5}$, i.e. $\frac{n^2}{5}$.

1. $15\to \frac{15\times 15}{5}=15\times 3=45$ ✓
2. $25\to \frac{25\times 25}{5}=25\times 5=125$ ✓
3. We need $n$ with $\frac{n^2}{5}=80\Rightarrow n^2=400\Rightarrow n=20$.
4. Check: $20\times \frac{20}{5}=20\times 4=80$ ✓. Other options give $\frac{{21}^2}{5}=88.2$, $\frac{{18}^2}{5}=64.8$, $\frac{{22}^2}{5}=96.8$.

Trick: The multiplier is the number divided by 5: $15\to \times 3$, $25\to \times 5$, so $20\to \times 4$.

Answer: (A) $20$

Concept: Each letter is written as $3\times$ its position in the English alphabet.

1. DINESH: $D=4\to 12$, $I=9\to 27$, $N=14\to 42$, $E=5\to 15$, $S=19\to 57$, $H=8\to 24$ ✓
2. VIRAT: $V=22\to 66$, $I=9\to 27$, $R=18\to 54$, $A=1\to 3$, $T=20\to 60$ ✓
3. GAUTAM: $G=7\to 21$, $A=1\to 3$, $U=21\to 63$, $T=20\to 60$, $A=1\to 3$, $M=13\to 39$.
4. So GAUTAM = 21-3-63-60-3-39.

Trick: All code numbers are multiples of 3 – only option (C) has every number divisible by 3.

Answer: (C) 21-3-63-60-3-39

Concept: In an embedded-figure question the given shape must appear inside an option figure with the same size, shape and orientation (no rotation).

1. The given figure is a tall narrow rectangle on top, touching an L-shaped step below it on its right side.
2. In option (B), lower middle: the vertical rectangle, the short link to the square on its right, the square's left side and the lines down to the bottom edge trace exactly this outline in the same orientation.
3. In (A), (C) and (D) the rectangle and the box sit in different relative positions or the shape would have to be rotated or flipped, which is not allowed.

Trick: Fix one distinctive corner of the given shape (here the step of the L) and search for that exact corner in each option.

Answer: (B)

Concept: Find one operation that turns the first number of each set into the second.

1. $349-247=102$ and $363-261=102$, so the rule is: second $=$ first $+102$.
2. (A) $282+102=384\ne 382$; (B) $294+102=396$ ✓; (C) $253+102=355\ne 358$; (D) $265+102=367\ne 369$.

Trick: Subtract the pair once and test that same difference on every option.

Answer: (B) $(294,396)$

Concept: Look at each letter position of the clusters separately.

1. 1st letters: C, D, E → next F ($+1$).
2. 2nd letters: H, I, J → next K ($+1$).
3. 3rd letters: O, P, Q → next R ($+1$).
4. So the next term is FKR.

Trick: Every letter moves one step forward – just shift each letter of EJQ by $+1$.

Answer: (B) FKR

Concept: Put each set of signs in order and evaluate using BODMAS (÷ and × before + and −).

1. Option (C): $13+7+12÷4\times 15=65$.
2. $12÷4=3$, then $3\times 15=45$.
3. $13+7+45=65$ ✓
4. Check (A): $13-7+12÷4\times 15=6+45=51\ne 65$.

Trick: Since the right side is 65, look for an option whose ÷ and × part makes a big number like 45; then add the rest.

Answer: (C) $+,+,÷,\times ,=$

Concept: Break the statement from the end: "my grandfather's son" usually means "my father".

1. The man says: the groom is the brother of the son of his grandfather's son.
2. Grandfather's son = the man's father.
3. Father's son = the man himself.
4. So the groom is the brother of the man – the groom is his brother.

Trick: Grandfather's son's son = "me" (or my brother); that person's brother is my brother.

Answer: (C) Brother

Concept: Swap the two signs given in an option and evaluate the equation with BODMAS; the correct swap makes both sides equal.

1. Swap × and ÷: $10\times 63÷7+90-80$.
2. $63÷7=9$, $10\times 9=90$.
3. $90+90-80=100$ ✓ (RHS $=100$).
4. Check (C) + and −: $10÷63\times 7-90+80$ gives a fraction, not 100.

Trick: $63$ and $7$ sit around ×; making it $63÷7=9$ gives a clean whole number – a strong hint.

Answer: (D) × and ÷

Concept: Replace every coded sign with its real meaning first, then solve using BODMAS.

1. Given: $+\to -$, $-\to \times$, $\times \to ÷$, $÷\to +$.
2. $22\times 11-450÷50+500$ becomes $22÷11\times 450+50-500$.
3. $22÷11=2$, $2\times 450=900$.
4. $900+50-500=450$.

Trick: Rewrite the whole expression once with new signs before calculating – never decode while calculating.

Answer: (D) $450$

Concept: Compare the alphabet positions of the letters of the first cluster with the second.

1. CARS → HFWX: $C(3)+5=H(8)$, $A(1)+5=F(6)$, $R(18)+5=W(23)$, $S(19)+5=X(24)$.
2. BOAT → GTFY: every letter $+5$ as well ✓.
3. FUSE: $F(6)+5=K$, $U(21)+5=Z$, $S(19)+5=X$, $E(5)+5=J$.
4. So FUSE → KZXJ.

Trick: One fixed shift ($+5$) for every letter – check the first and last letters to eliminate options fast: only (D) starts with K and ends with J.

Answer: (D) KZXJ

Concept: A face that is seen next to a given face in any position cannot be opposite to it. If one face is adjacent to four others, the sixth face is opposite to it.

1. Face 6 is on top in both positions.
2. Around 6 we see 2 and 3 (first position) and 4 and 5 (second position).
3. So 6 is adjacent to 2, 3, 4 and 5; the only face left is 1.
4. Therefore 1 is opposite 6.

Trick: Common face + four different neighbours → the remaining number is opposite the common face.

Answer: (D) $6$

Concept: When people face the centre, "to the right" means moving anticlockwise and "to the left" means clockwise.

1. Place Gita. Ram is 2nd to her right and Aman is 3rd to her right, i.e. directly opposite Gita.
2. Pranay sits next to both Gita and Bobby. Gita's right-side neighbour seat would put Bobby on Ram's seat, so Pranay is on Gita's left and Bobby next to him.
3. The last seat (immediate right of Gita) goes to Chintu.
4. Order to the right of Gita: Gita, Chintu, Ram, Aman, Bobby, Pranay. Aman sits between Ram and Bobby.

Trick: Seat the fixed clues (Ram, Aman) first, then test the two possible seats for the "neighbour" clue.

Answer: (C) Aman

Concept: Draw the path step by step, taking North as up; a right turn while going North points East, and so on.

1. From the school: $7\text{km}$ North.
2. Right turn → East $10\text{km}$; right turn → South $5\text{km}$. She is now $7-5=2\text{km}$ north and $10\text{km}$ east of the school.
3. Final right turn → West. She must drive $10\text{km}$ west to come directly above the gate and then $2\text{km}$ south.
4. Distance still to drive $=10+2=12\text{km}$.

Trick: North–South net $=7-5=2$, East–West $=10$; the road back along the turns is $10+2$.

Answer: (D) $12\text{km}$

Concept: A class that is a part of another is drawn inside it; an unrelated class is drawn as a separate circle.

1. Every eagle is a bird, so Eagle lies inside Birds.
2. A cheetah is a mammal, not a bird, so Cheetah is a separate circle.
3. "One circle inside another plus one separate circle" is option (C).

Trick: Animal of a different group (mammal vs bird) → always a separate circle.

Answer: (C) Eagle inside Birds, Cheetah separate

Concept: Check the differences between consecutive terms.

1. Differences: $6-1=5$, $15-6=9$, $28-15=13$ – they increase by 4.
2. Next differences: $17,21,25$.
3. $28+17=45$, $45+21=66$, $66+25=91$ ✓ (matches the last term).

Trick: These are hexagonal numbers $n(2n-1)$: $1,6,15,28,45,66,91$.

Answer: (C) $45,66$

Concept: Classes with no common member are shown as separate, non-touching circles.

1. Bananas, apples and grapes are three different fruits.
2. No banana is an apple or a grape, and no apple is a grape.
3. So all three circles are separate – option (B).

Trick: Three different items of the same kind (three fruits, three colours, three cities) → three separate circles.

Answer: (B) three separate circles

Concept: Find the shift of each letter position; here the pattern is $-3,-4,-3,-4$.

1. LAND → IWKZ: $L(12)-3=I$, $A(1)-4=W$ (wrap back), $N(14)-3=K$, $D(4)-4=Z$.
2. GOLD → DKIZ: $G-3=D$, $O-4=K$, $L-3=I$, $D-4=Z$ ✓.
3. TEAM: $T(20)-3=Q$, $E(5)-4=A$, $A(1)-3=X$, $M(13)-4=I$.
4. So TEAM → QAXI.

Trick: Going back from A wraps to Z: $A-3=X$, $A-4=W$.

Answer: (A) QAXI