SSC CHSL 2025 PYQs

Concept: Replace the coded symbols with real signs, then apply BODMAS.

1. Decoded: $7\times 6+9÷4$ (% = ×, ∧ = ÷, # = +).
2. Multiply/divide first: $7\times 6=42$; $9÷4=\frac{9}{4}$.
3. $42+\frac{9}{4}=\frac{168+9}{4}=\frac{177}{4}$.

Trick: With a ÷4 left over, the answer must be a fraction with denominator 4 → only $\frac{177}{4}$ fits.

Answer: (D) $\frac{177}{4}$

Concept: Convert both positions to the same end, then between = difference − 1.

1. Amit from front = 72 − 19 + 1 = 54.
2. Ravi = 27th from front.
3. Between = 54 − 27 − 1 = 26.

Trick: Between (no overlap) = Total − (front rank + back rank) = 72 − 46 = 26.

Answer: (B) 26

Concept: Left rank + right rank = total + 1.

1. 9 + R = 35 + 1
2. R = 36 − 9 = 27

Trick: Right rank = Total − Left + 1.

Answer: (B) 27th

Concept: Terms decrease by a fixed amount.

1. −701 → −715 → −729 … each step −14.
2. Next = −785 − 14 = −799.

Trick: Last digits go 1, 5, 9, 3, 7, 1, 5 → next ends in 9 → only −799.

Answer: (B) -799

Concept: Each term ≈ double the previous; check the extra added.

1. $35\times 2=70(+4)=74$; $74\times 2=148(+3)=151$
2. 151 → 306 (+4), 306 → 615 (+3), 615 → 1234 (+4), 1234 → 2471 (+3)
3. Next: $2471\times 2+4=4946$.

Trick: Terms alternate even/odd; 2471 is odd so next is even and = 4942 + 4.

Answer: (D) 4946

Concept: Find which signs make the given equations true; apply to the third.

1. Try $ = +, ! = ÷: $452+45÷15=452+3=455$ ✓
2. $578+84÷12=578+7=585$ ✓
3. $317+65÷13=317+5=322$.

Trick: Answer is just above the first number → '+'; middle ÷ last is a clean whole number → '÷'.

Answer: (B) 322

Concept: A vertical mirror reverses left–right: order is reversed and each symbol is flipped.

1. Object: F b 3 e 5 g (mirror on the right).
2. Image order: g 5 e 3 b F, each laterally flipped.
3. So b looks like d, 3 looks like Ɛ, F faces left — only option (D) shows all these.

Trick: Check just one asymmetric letter: 'b' must turn into 'd' — eliminate options keeping 'b'.

Answer: (D) P c $\epsilon$ d 7

Concept: Unfold step by step; every cut is mirrored across each fold line.

1. Two folds → 4 layers → each cut appears 4 times.
2. Undo the last fold first, then the diagonal fold, reflecting each cut.
3. The triangle at the centre joins into one square; house shapes appear on four sides pointing inward; circles land in two opposite corners — matches (B).

Trick: Count holes first: each cut × 4. Discard options with the wrong count before checking positions.

Answer: (B)Option figure (b)

Concept: Each fold doubles the layers; opened pattern is symmetric about every fold line.

1. Vertical fold then horizontal fold → 4 layers.
2. 3 squares + 1 circle punched → 12 squares + 4 circles on opening.
3. Pattern must be symmetric about both centre lines — only (D) satisfies count and symmetry.

Trick: Hole count = cuts × 2^folds = 4 × 4 = 16; then test symmetry about both creases.

Answer: (D)Option figure (d)

Concept: Fix the definite clues first.

1. Years: 2010–2014. Riya oldest → 2010.
2. Only leap year among them = 2012 → Sameer.
3. Tanu not 2011/2014 → 2013; Nikhil younger than Vaibhav → Vaibhav 2011, Nikhil 2014.

Trick: Leap year = divisible by 4 → 2012 is the only one; question answered in one step.

Answer: (D) Sameer

Concept: Check the letter order and number pattern together.

1. Letters P, Q, R, S are consecutive; the next should be T, not U.
2. Numbers alternate: 40 → 35 (−5), 50 → 45 (−5); after R35 the next in that chain should be 30.
3. U25 breaks both patterns.

Trick: Look at letters first — the jump S → U immediately exposes the odd term.

Answer: (D) U25

Concept: Differences form an arithmetic sequence.

1. Differences: 4, 6, 8, 10, 12.
2. Next difference = 14.
3. $43+14=57$.

Trick: Terms are $n^2+n+1$: $6^2+6+1=43$, $7^2+7+1=57$.

Answer: (D) 57

Concept: Find the rule linking the first three numbers of a row to the fourth.

1. Row 1: $3\times 4\times 5=60$, $\times 2=120$ ✓
2. Row 2: $6\times 2\times 3=36$, $\times 2=72$ ✓
3. Row 3: $?\times 3\times 4\times 2=96$ → $?\times 24=96$ → $?=4$.

Trick: Once the rule is product × 2, simply divide: $96÷2÷12=4$.

Answer: (C) 4

Concept: Label every intersection and count small triangles first, then combined ones.

1. Mark corners and the two inner crossing points on the diagonal.
2. Single-region triangles: 8.
3. Triangles formed by joining 2 or more regions: 13.
4. Total = 8 + 13 = 21.

Trick: Count systematically by fixing one vertex at a time so no triangle is missed or repeated.

Answer: (A) 21

Concept: Split the figure into parts and count each part separately.

1. Large left triangle with its vertical line and inner segment: 5 triangles.
2. Small square region with its diagonals/lines: 4 more.
3. Two small triangles on the right: 2.
4. Total = 5 + 4 + 2 = 11.

Trick: Go left to right through the figure and tick each triangle as you count.

Answer: (A) 11

Concept: 'Books that are newspapers' = region common to circle and triangle (with or without rectangle).

1. Circle ∩ triangle only (outside rectangle) = 7.
2. Circle ∩ triangle ∩ rectangle = 4.
3. Total = 7 + 4 = 11.

Trick: 'A that are B' = full overlap of A and B — include the all-three region too.

Answer: (A) 11

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

1. Place J. I is 2 seats to J's right (one person between).
2. H is 2 seats to J's left; G is 2 seats right of I, i.e. between J and H.
3. K fills the seat between J and I. Seats clockwise: J, G, H, I, K.
4. Second to the right of G (anticlockwise): J, then K.

Trick: In a 5-seat circle, 2nd to the right = 3rd to the left — count whichever side is easier.

Answer: (C) K

Concept: Build the fixed block first, then test positions against remaining clues.

1. T immediately left of Q and R between P and T → block P R T Q.
2. Block at seats 1–4 leaves U at an end with nobody to its right (or S at an end) ✗; seats 2–5 forces S to an end ✗.
3. So block = seats 3–6; S = 2, U = 1 → U S P R T Q.
4. Ends: U and Q.

Trick: Biggest block first — it has the fewest possible places, so trial is quick.

Answer: (D) U and Q

Concept: Number seats from the persons' own left to right (facing south flips the viewer's view).

1. R immediately left of M, M at an end → M = 8, R = 7, T = 1.
2. N has neighbours Q and R → N = 6, Q = 5.
3. O is 4th left of N → O = 2; P not next to O → P = 4, S = 3.
4. 3rd right of Q (seat 5) → seat 8 = M.

Trick: Work in the persons' own left/right throughout — don't switch to your view mid-way.

Answer: (C) M

Concept: Replace letters with signs and solve by BODMAS.

1. $24\times 10÷5+6-8$
2. $24\times 10=240$; $240÷5=48$
3. $48+6-8=46$

Trick: $÷5$ cancels into 10 first: $24\times 2=48$ — faster.

Answer: (A) 46