SSC CHSL 2025 PYQs

Concept: The code writes the word backwards.

1. GERMANY read from the end = YNAMREG.
2. BRAZIL read from the end = L-I-Z-A-R-B = LIZARB.

Trick: First letter of the code = last letter of the word: only A starts with L and ends with B correctly.

Answer: (A) LIZARB

Concept: Differences increase by a fixed amount.

1. Differences: 100, 120, 140 (rising by 20).
2. Next difference = 160 → 380 + 160 = 540.

Trick: If first differences aren't equal, take differences of differences.

Answer: (A) 540

Concept: Differences form an arithmetic pattern (+4).

1. Known gaps: 63 − 47 = 16 and 135 − 107 = 28.
2. Pattern 16, 20, 24, 28 → ? = 63 + 20 = 83.
3. Check: 83 + 24 = 107 ✔.

Trick: Two known differences 16 and 28 with two gaps between → step = (28 − 16) ÷ 3 = 4.

Answer: (D) 83

Concept: Split letters and numbers into separate patterns.

1. Alternate letters: C, D, E (forward) and X, W (backward). 5th term is in the forward set → E.
2. Numbers 2, 3, 5, 7 = consecutive primes → next 11.
3. Term = E11.

Trick: Only two options have E; prime after 7 is 11, not 9 → E11.

Answer: (A) E11

Concept: Compare character by character: letters, digits, punctuation.

1. (A) has an extra dot '4.th' ✗.
2. (C) 'Washingtan' ✗; (D) year 1767 ✗.
3. (B) matches 'Washington: July 4th, 1776.' exactly ✔.

Trick: Check the risky spots first – spelling of names, digits of years, and dots.

Answer: (B) Washington: July 4th, 1776.

Concept: In each row, the last number comes from the first three by one fixed rule.

1. First row: product of 5, 3, 4 is 60; adding 5 gives 65 ✔.
2. Second row: product of 3, 7, 8 is 168; adding 5 gives 173 ✔.
3. Third row: product of 7, 5, 3 is 105; so ? = 110.

Trick: When the result is much larger than the inputs, try multiplication first, then adjust by a constant.

Answer: (B) 110

Concept: Last column = 3 × (sum of first three numbers).

1. First row: sum of 8, 7, 2 is 17, and thrice 17 is 51 ✔.
2. Last row: sum of 6, 3, 1 is 10, and thrice 10 is 30 ✔.
3. Middle row: one-third of 60 is 20, so 14 + ? = 20 → ? = 6.

Trick: All results (51, 60, 30) are multiples of 3 – divide by 3 and look for the sum.

Answer: (C) 6

Concept: Count systematically so no triangle is missed or repeated; curves (ovals, diamonds) form no triangles.

1. Ignore the two ovals and the diamonds inside them – only straight lines matter.
2. Count triangles having the top vertex first (big triangle split by the vertical median and the horizontal cross-lines).
3. Then count those formed by the two slanting lines inside and around the square, and the bottom strip.
4. Adding all groups gives 30 triangles.

Trick: Fix one vertex at a time (top apex, bottom-left corner, then inner crossing points) and tick off each triangle as you count.

Answer: (C) 30

Concept: Split the figure into regions and count each group once.

1. Inside the big square (both diagonals + half-lines from centre): 12 triangles.
2. Using the top apex: 6 triangles.
3. Using the right-hand vertex (no apex): 5 triangles.
4. Small cut in the bottom-left corner: 3 triangles. Total = 12 + 6 + 5 + 3 = 26.

Trick: A square with both diagonals alone gives 8 – extra half-lines inside it add more; count them as a separate group.

Answer: (C) 26

Concept: Every 4 right turns = 360° = no change in direction.

1. 6 right turns = 4 + 2 → net effect = 2 right turns = 180° (about-turn).
2. Final direction East is opposite to the starting direction.
3. Opposite of East = West → he started towards West.

Trick: Divide the number of turns by 4 and use only the remainder.

Answer: (D) West

Concept: A right turn and a left turn cancel each other.

1. 8 right turns are cancelled by 8 left turns → net turn = 0°.
2. So he faces his starting direction: South.

Trick: Equal number of right and left turns → direction unchanged, whatever the order.

Answer: (C) South

Concept: The completed figure must be symmetric; copy the lines that run into the missing part.

1. The square holds two rectangles centred on the middle – the missing bottom-right quarter must show both of their corners.
2. The lower X is half-drawn; its other half (a '<'-shaped wedge meeting at the centre) must come from the missing piece.
3. Only option (a) has both nested corners plus the correctly facing wedge.

Trick: Trace each line that touches the cut edge – it must continue in the answer piece.

Answer: (A)

Concept: Use the symmetry of the remaining three quarters.

1. Two rectangles (large and small) are centred in the square – the missing top-left quarter needs both their top-left corners.
2. The small rectangle has diagonals; one half-diagonal (from its top-left corner to the centre) is missing.
3. Option (c) shows exactly two corners plus that one short diagonal.

Trick: Count elements: corners needed = number of rectangles; extra or long diagonals eliminate (a), (b), (d).

Answer: (C)

Concept: Find what changes from figure 1 to figure 2 and apply the same change.

1. Main shape stays; each small symbol keeps its corner but switches side: inside → outside, outside → inside.
2. Figure 3: star and hexagon are inside, diamond and square outside.
3. So answer: star and hexagon outside, diamond and square inside – option (c).

Trick: Track just one symbol (e.g. the star): it must jump outside – this alone removes (d).

Answer: (C)

Concept: Each figure is the previous one rotated 90° clockwise.

1. Follow the symbol R: it moves one quarter clockwise in every step; J, 7 and T do the same.
2. Figures 2, 3, 4 = figure 1 turned 90°, 180°, 270°.
3. Next turn makes 360° → the fifth figure equals figure 1 – option (b).

Trick: Four quarter-turns = full circle, so the 5th figure repeats the 1st.

Answer: (B)

Concept: The whole box (positions and orientation of each object) turns 90° anticlockwise each step.

1. Hexagon: top-right → top-left → bottom-left, i.e. a quarter turn anticlockwise each time.
2. Each object also rotates, so shading and arrow directions turn with it.
3. Missing box = box 1 turned 270° anticlockwise (= 90° clockwise) – option (b).

Trick: 270° anticlockwise = 90° clockwise – easier to visualise.

Answer: (B)

Concept: Check the letter rule and the number rule separately.

1. Letters: every pair moves +5 (B→G, K→P etc.) – same in all.
2. Numbers: 650→980, 720→1050, 810→1140 are all +330.
3. FL540 → KQ920 is +380 ✗ → odd one.

Trick: When the letter rule is common to all, the odd one hides in the numbers.

Answer: (D) FL540 - KQ920

Concept: Code = sum of reverse-alphabet values (27 − position) + number of letters.

1. RING: R 9, I 18, N 13, G 20 → 60; 60 + 4 = 64 ✔.
2. BOOK: 25 + 12 + 12 + 16 = 65; 65 + 4 = 69 ✔.
3. FAIL: F 21, A 26, I 18, L 15 → 80; 80 + 4 = 84.

Trick: Reverse value of any letter = 27 − its position (A = 26, Z = 1).

Answer: (B) 84

Concept: Consonants and vowels follow different rules.

1. TIGER → ROEIP: T→R, G→E, R→P (consonants −2); I→O, E→I (next vowel).
2. LION: L→J, I→O, O→U, N→L.
3. LION = JOUL.

Trick: Vowel cycle A→E→I→O→U: the code of O must be U – only JOUL and JOUR have it; N−2 = L → JOUL.

Answer: (A) JOUL

Concept: Every letter of the target word must be available in the option (with repeats).

1. LANTERN needs A, E, L, N, N, R, T.
2. RENTALN, ENTRALN, NERTALN have exactly these letters.
3. ANTLELN has two L's and no R → LANTERN cannot be made.

Trick: Look for the rare letter (here R) first – the option missing it is the answer.

Answer: (B) ANTLELN