SSC GD 2025 PYQs

Concept: A word present in two sentences gets the code present in both code groups.

1. Sentence 1: drink clean water → rd tm kg
2. Sentence 2: clean your house → db fo tm
3. Only common word = clean; only common code = tm.

Trick: Circle the repeated word and the repeated code – match them directly, no need to decode the rest.

Answer: (A) tm

Concept: Rearrange letters alphabetically, locate the asked positions, then count alphabet letters between them.

1. INJURED → D, E, I, J, N, R, U (7 letters)
2. 2nd from left = E; 4th from right = J (U, R, N, J)
3. Letters between E(5) and J(10) in alphabet = 10 − 5 − 1 = 4 (F, G, H, I)

Trick: Letters between two letters = difference of their positions − 1.

Answer: (C) 4

Concept: A face has exactly four neighbours; the one letter that never appears beside it is opposite.

1. T is seen in both positions.
2. Neighbours of T: F, U (view 1) and A, L (view 2) = 4 faces.
3. The only letter left, E, must be opposite T.

Trick: Find the letter common to both views, list its 4 neighbours – the 6th letter is its opposite.

Answer: (A) T

Concept: Differences follow two alternating patterns.

1. Differences: +1, −10, +2, −10, +3, −10
2. Additions grow 1, 2, 3 → next is +4
3. ? = 286 + 4 = 290

Trick: When a constant step (−10) keeps repeating, look at the other steps separately – they usually rise by 1.

Answer: (C) 290

Concept: $\text{Code}=(\text{sum of alphabet positions})\times 3$.

1. JQBT: $10+17+2+20=49$ → $49\times 3=147$ ✔
2. IWOG: $9+23+15+7=54$ → $54\times 3=162$ ✔
3. SDLO: $19+4+12+15=50$ → $50\times 3=150$

Trick: Both codes are multiples of 3 (147, 162) – that hints at a ×3 rule on the letter sum.

Answer: (B) 150

Concept: Each letter of a cluster moves forward by 3 in the next cluster.

1. H→K→N→Q→T (+3)
2. N→Q→T→W→Z (+3)
3. B→E→H→K→N (+3)

Trick: Check only the last letter: K + 3 = N – only option (B) ends in N with TZ.

Answer: (B) TZN

Concept: Decode each symbol and join the links.

1. B # L → B is mother of L
2. L $ C → L is husband of C
3. So B is the mother of C's husband (C # Q only adds C's child).

Trick: Read the chain from the asked person (C) back to B: C → husband L → his mother B.

Answer: (D) Husband's mother

Concept: Track every element separately: position change and rotation.

1. θ shifts clockwise by one place per figure, so in figure 5 it sits at the middle of the right side.
2. Ψ shifts the same way, so in figure 5 it reaches the lower-left corner.
3. The dot-and-square piece rotates 90° clockwise each time and alternates corners → back to top-left with the dot on the left.

Trick: Follow one symbol at a time and eliminate options that fail even one symbol.

Answer: (D)

Concept: Steps alternate: +9, then ×2 + 3.

1. $23+9=32$; $32\times 2+3=67$
2. $67+9=76$; $76\times 2+3=155$
3. $155+9=164$; $164\times 2+3=331$

Trick: The jump roughly doubles every other step, so the answer is about $2\times 164=328$ → only 331 fits +3.

Answer: (B) 331

Concept: Find the shift of each position in the given pairs and repeat it.

1. GTBK → KRFI: +4, −2, +4, −2 (NSFX → RQJV same)
2. EKVO → (E→I), (K→I), (V→Z), (O→M)
3. Result = IIZM

Trick: First letter E + 4 = I – only option (B) starts with I.

Answer: (B) IIZM

Concept: Three clusters share the same gap pattern between letters.

1. MOT: +2, +5; SUZ: +2, +5; LNS: +2, +5
2. DFO: D→F = +2 but F→O = +9
3. DFO breaks the pattern.

Trick: First and last letters of the group members differ by 7 (M–T, S–Z, L–S); D–O differs by 11.

Answer: (A) DFO

Concept: Swap the signs first, then apply BODMAS.

1. After swapping: $20+7\times 4-20÷4$
2. $=20+28-5$
3. $=43$

Trick: Rewrite the whole expression with new signs before calculating – never swap mentally midway.

Answer: (A) 43

Concept: The series is a repeating 5-letter block.

1. Block = AZPLM, repeated 4 times (20 letters)
2. AZPLM | AZPLM | AZPLM | AZPLM
3. Blanks: Z, M, A, P, A, Z, L, M

Trick: Fill the first blank only: A_P → Z (from AZP later in the string). Only (B) and (C) start with Z; then second blank M after L kills nothing, third blank A picks (B).

Answer: (B) ZMAPAZLM

Concept: Each letter moves +8, wrapping after Z.

1. K→S, M→U, O→W, Q→Y: +8 each
2. YACE → (Y→G), (A→I), (C→K), (E→M)
3. Result = GIKM

Trick: YACE has equal gaps of 2 (Y,A,C,E), so the answer must also have gaps of 2 – only GIKM.

Answer: (A) GIKM

Concept: A vertical mirror swaps left and right only; shading and up–down stay the same.

1. The figure is symmetric about its vertical centre line.
2. So its image in MN looks identical to the original.
3. Options that change a solid part to hollow (or vice versa) are wrong.

Trick: A mirror never changes shading – reject any option where fill has changed.

Answer: (A)

Concept: Some A are B + All B are C ⇒ Some A are C; All B are C + No C is D ⇒ No B is D.

1. Some papers are books, all books are shelves ⇒ some papers are shelves → I follows.
2. All books are shelves, no shelf is library ⇒ no book is a library.
3. So 'All books are libraries' is false → II does not follow.

Trick: Some + All = Some; All + No = No. Use these two rules in 5 seconds.

Answer: (C) Only conclusion (I) follows.

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

1. Seats 1–6 clockwise, U = 1. W is 2 seats to U's right → W = 5.
2. One between A and W (seat 1 is U) → A = 3.
3. Two between A and B → B = 6; C beside U → C = 2; V = 4.
4. From V's left (clockwise): W, B, U, C, then A → 4 people.

Trick: In a 6-seat circle, counting the long way between two people = 4 − (short way count).

Answer: (C) Four

Concept: Each fold doubles the holes; unfolding reflects holes across the fold line.

1. Two folds → one punch makes $2\times 2=4$ holes.
2. Unfold the horizontal fold: hole mirrored top–bottom.
3. Unfold the vertical fold: both holes mirrored left–right → one hole in each quarter.

Trick: Count first: $\text{holes}=2^{\text{number of folds}}\times \text{punches}$. Options with 2 holes are out instantly.

Answer: (B)

Concept: Fix the end positions first, then place the rest.

1. Four between R and Q → they sit at the ends.
2. S is immediately left of Q → Q = seat 6, S = 5, R = 1.
3. One between S and Y → Y = 3.
4. U left of Y, T right of Y → U = 2, T = 4. Order: R U Y T S Q

Trick: 'Four people between' in a row of six always means the two are at the ends.

Answer: (C) Y

Concept: Check each pair: can they overlap? must one contain the other?

1. A mother may or may not be a wife, and vice versa → partial overlap.
2. A princess can be a mother and/or a wife, but need not be → overlaps both.
3. No class lies fully inside another → three mutually intersecting circles.

Trick: If every pair 'can be but need not be', choose the three-overlapping-circles figure.

Answer: (B)