SSC CHSL 2025 PYQs

Concept: The second figure is the first flipped about a horizontal line (top ↔ bottom).

1. In pair 1, the parts at the top go to the bottom and vice versa; left-right positions stay the same.
2. Flip the third figure the same way: the upright triangle becomes inverted, its top dot goes to the bottom vertex and the stem with the other dot now points upward.
3. This matches option (B).

Trick: In a top-bottom flip, nothing moves sideways — reject options where elements shift left or right.

Answer: (B) Option (B)

Concept: Track how the row of shapes on the diagonal shifts from one figure to the next.

1. Each step, the five diagonal shapes slide one place towards the bottom-right.
2. The shape that falls off the end goes to the top-right corner; a new shape enters at the top-left.
3. Figure 4: triangle, bar, 4-point star, circle, square on the diagonal → in figure 5 the square leaves to the corner and the rest shift down → option (D).

Trick: Only the corner shape needs checking: it is always the last shape of the previous diagonal — here the square (口).

Answer: (D) $\_\_\_\_$口

Concept: Swap the two given signs (− and ÷) everywhere, then solve by BODMAS.

1. $40+20\times 15-5÷5$ → $40+20\times 15÷5-5$.
2. $20\times 15÷5=300÷5=60$.
3. $40+60-5$ = 95.

Trick: Rewrite the whole expression after swapping before calculating — most errors come from swapping only one sign.

Answer: (D) 95

Concept: Insert the signs in order and evaluate with BODMAS.

1. Using ×, +, ÷, −: $24\times 4+30÷6-7$.
2. $24\times 4=96$; $30÷6=5$.
3. $96+5-7$ = 94 ✓.

Trick: 94 is close to $24\times 4=96$, so the first sign is likely ×; then you need $+5-7$, i.e. $30÷6$.

Answer: (A) $\times ,+,÷$,

Concept: Check letters and numbers separately; the pair that breaks either rule is odd.

1. Letters: 1st letter −1, 2nd letter −2 in all pairs (T→S, B→Z; Q→P, L→J …).
2. Numbers: 287 + 294 = 581; 231 + 294 = 525; 410 + 294 = 704.
3. But 320 + 294 = 614 ≠ 604 → QL320 – PJ604 is odd.

Trick: If the letters agree everywhere, go straight to the numbers — just compare the difference in each pair.

Answer: (B) QL320 - PJ604

Concept: Find the common letter shift and number rule from the majority.

1. Letters: 1st letter −3, 2nd letter −6 (R→O, C→W; L→I, A→U; K→H, S→M).
2. Numbers: × 3 − 1: 206 → 617, 194 → 581, 187 → 560.
3. ER313 – BN939: R→N is only −4, and $313\times 3-1=938\ne 939$ → odd.

Trick: Backward shift past A: count C, B, A, Z, Y, X, W — or use $C(3)-6+26=23$ = W.

Answer: (D) ER313 - BN939

Concept: $\text{Second number}=\text{first number}÷4$.

1. $2652÷4=663$ ✓; $1548÷4=387$ ✓; $2376÷4=594$ ✓.
2. 1981 is odd, so not divisible by 4 → 1981 – 492 is the odd pair.

Trick: Divisibility by 4: last two digits must be divisible by 4 — 52, 48, 76 yes; 81 no.

Answer: (D) 1981-492

Concept: In each group: $\text{2nd}=2\times \text{1st}$ and $\text{3rd}=2.5\times \text{1st}$ (ratio 2 : 4 : 5).

1. 48 → 96, 120 ✓; 90 → 180, 225 ✓; 72 → 144, 180 ✓.
2. 66 → 132 ✓ but $66\times 2.5=165$, not 170 → odd group.

Trick: 3rd ÷ 2nd should be $\frac{5}{4}=1.25$: $170÷132$ is not 1.25 (165 would be).

Answer: (B) (66, 132, 170)

Concept: Apply the same letter shift and number operation to the third cluster.

1. Letters: 1st −1, 2nd +1 (B→A, M→N; C→B, S→T).
2. Numbers: 112 + 450 = 562; 280 + 450 = 730.
3. GX96 → G−1 = F, X+1 = Y, 96 + 450 = 546 → FY546.

Trick: Letters alone settle it: G→F, X→Y gives 'FY' — only one option starts with FY.

Answer: (C) FY546

Concept: Find the letter shift and number rule from the first two pairs.

1. Letters: both −6 (P→J, Q→K; G→A, O→I).
2. Numbers: $8^2+2=66$; ${19}^2+2=361+2=363$.
3. BX13 → B−6 = V, X−6 = R; ${13}^2+2=169+2=171$ → VR171.

Trick: Moving back past A, add 26: $B(2)-6+26=22$ = V.

Answer: (D) VR171

Concept: On exchange, each person takes the other's exact seat; Total = left rank + right rank − 1.

Solution:

1. Ravi's new seat (24th from left) = Sumit's old seat, which is 15th from right.
2. Total students = 24 + 15 − 1 = 38.
3. Sumit now sits at Ravi's old seat = 10th from left → from right = 38 − 10 + 1 = 29th.

Trick: The seat that has both a left rank and a right rank gives the total in one line.

Answer: (B) 29th

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

Solution:

1. Neha from left = 50 − 24 + 1 = 27th.
2. Arjun = 18th from left.
3. Between them = 27 − 18 − 1 = 8.

Trick: Quick check: 18 + 24 = 42 < 50, so they do not overlap → between = 50 − 42 = 8.

Answer: (C) 8

Concept: Find a rule that fits both given examples, then apply it.

Solution:

1. 56 $ 20: $(56-20)÷2=36÷2=18$ ✔
2. 74 $ 30: $(74-30)÷2=44÷2=22$ ✔
3. 63 $ 18: $(63-18)÷2=45÷2=$ 22.5

Trick: Results are much smaller than the numbers → try subtraction first; decimal options hint at ÷ 2.

Answer: (C) 22.5

Concept: Test the pattern on the given pairs.

Solution:

1. $12\times 20=240$ and $720=3\times 240$ ✔
2. $15\times 20=300$ and $900=3\times 300$ ✔
3. $6\times 25=150$ → $3\times 150=$ 450

Trick: $720÷240=3$ instantly shows the multiplier; answer must be a multiple of 3 and 150 → 450.

Answer: (B) 450

Concept: Check the gaps between terms.

Solution:

1. Gaps: 9, 11, 13 → increase by 2.
2. Next gap = 15 → $52+15$ = 67.

Trick: Terms = $n^2+3$ for n = 4, 5, 6, 7, 8 → $8^2+3=67$.

Answer: (B) 67

Concept: Check the gaps between terms.

Solution:

1. Gaps: 7, 9, 11 → increase by 2.
2. Next gap = 13 → $38+13$ = 51.

Trick: Terms = $n^2+2$ for n = 3, 4, 5, 6, 7 → $7^2+2=51$.

Answer: (D) 51

Concept: Unfold in reverse order; every unfolding mirrors the holes across that fold line.

Solution:

1. Fold 1 (left onto right) + Fold 2 (top onto bottom) leave only the bottom-right quarter.
2. That quarter has 3 square holes and 1 circle.
3. Two unfoldings → each hole appears 4 times: 12 squares and 4 circles, symmetric about both centre lines.

Trick: $\text{Holes after opening}=\text{holes punched}\times 2^{\text{number of folds}}=4\times 2^2=16$; pick the option with full left-right and top-bottom symmetry.

Answer: (D) Option D

Concept: A cut on a fold line becomes a full hole; a cut on a free edge becomes a notch copied across each fold.

Solution:

1. Fold 1 is along the main diagonal; fold 2 is a vertical fold of the triangle.
2. Cut on the folded edge → a diamond hole after opening fold 2; cut on the bottom edge → two symmetric notches.
3. Opening the diagonal fold copies this bottom pattern onto the right edge.

Trick: Diagonal fold acts like a mirror that swaps 'bottom edge' with 'right edge' – look for the option with identical patterns on these two edges.

Answer: (C) Option C

Concept: $\text{Bottom number of each column}=(\text{sum of the three numbers above}{)}^2$.

Solution:

1. Column 1: $11+5+3=19,{19}^2=361$ ✔; Column 2: $17+3+7=27,{27}^2=729$ ✔
2. Column 3: $1225={35}^2$, so $8+15+?=35$
3. $?=35-23$ = 12

Trick: When the bottom row has big numbers like 361, 729, 1225, think of perfect squares first.

Answer: (C) 12

Concept: In each figure: $(\text{top-left}\times \text{top-right})+\text{bottom-right}=\text{bottom-left}$.

Solution:

1. Figure 1: $4\times 7+2=30$ ✔; Figure 3: $3\times 5+4=19$ ✔
2. Figure 2: $4\times 5+?=22$ → $?=22-20$ = 2

Trick: Bigger number in one corner usually comes from multiplying the two smaller ones; the leftover is the missing value.

Answer: (A) 2