SSC CGL 2025 PYQs

Concept: Each letter of the word is replaced by its next letter.

1. N→O, E→F, H→I, R→S, U→V: every letter +1.
2. G→H, A→B, N→O, D→E, H→I, I→J.
3. GANDHI = HBOEIJ.

Trick: Check only the 3rd letter: N+1 = O – only option (B) has O there.

Answer: (B) HBOEIJ

Concept: Treat the first and second letters as two separate series.

1. First letters: C, E, G → +2 → I.
2. Second letters: H, J, L → +2 → N.
3. Next pair = IN.

Trick: In each pair the second letter is 5 ahead of the first (C→H); I + 5 = N.

Answer: (D) IN

Concept: Each term = previous term × 2 + 1.

1. Double and add one: $15=7\times 2+1$; $31=15\times 2+1$; $63=31\times 2+1$ ✓
2. So the next term = $63\times 2+1=127$.

Trick: Terms are $2^n-1$ ($8-1$, $16-1$, $32-1$, $64-1$) → next $128-1=127$.

Answer: (C) 127

Concept: First term is $n^2$, second is $n^3$ of the same n.

1. $225={15}^2$, $3375={15}^3$.
2. $256={16}^2$ → answer ${16}^3$.
3. ${16}^3=256\times 16=4096$.

Trick: Cube of n = square × n: $256\times 16=4096$. Option 3375 is just a repeat trap.

Answer: (D) 4096

Concept: Each term = previous × 2 + 1 (all terms are one less than a power of 2).

1. $63=64-1$, $127=128-1$, $255=256-1$ … $1023=1024-1$.
2. Next = $2048-1=2047$ (or $1023\times 2+1$).

Trick: Know powers of 2 up to $2^{12}$ (4096) – such series become one-glance questions.

Answer: (A) 2047

Concept: A word common to two sentences takes the code common to them.

1. Lines 2 & 3 share only 'lion' and code 'rso' → lion = rso.
2. Lines 1 & 3 share 'rat' and 'lmn' → rat = lmn; then tiger = tuv.
3. Cat/dog share xya, pyq; codes follow word order (cat dog → xya pyq; dog … cat → pyq … xya) → dog = pyq.
4. dog lion tiger = pyq, rso, tuv.

Trick: Fix the sure codes (rso, tuv) first – only options (A) and (B) remain; the word order then gives dog = pyq.

Answer: (A) pyq, rso, tuv

Concept: SP = $\text{CP}\times \frac{100+\text{profit}\%}{100}$.

1. $\text{SP}=90\times \frac{110}{100}$.
2. = 99, i.e. ₹99.

Trick: 10% profit → just add one-tenth: $90+9=$ ₹99.

Answer: (A) ₹99

Concept: Total stays fixed, so each share = total ÷ number of people.

1. Total = $12\times 50=$ ₹600.
2. Remaining people = $12-4=8$.
3. New share = $600÷8=$ ₹75.

Trick: Share ∝ 1/people: $50\times \frac{12}{8}=50\times \frac{3}{2}=$ ₹75.

Answer: (C) ₹75

Concept: Differences rise by 1 each time (triangular numbers).

1. $6-3=3$, $10-6=4$, $15-10=5$, $21-15=6$.
2. Next difference = 7 → $21+7=28$.

Trick: Triangular numbers $\frac{n(n+1)}{2}$: 3, 6, 10, 15, 21, 28 – memorise the first ten.

Answer: (B) 28

Concept: Each letter is written as the letter just after it.

1. H→I, E→F, A→B, T→U: rule +1.
2. C→D, O→P, L→M, D→E.
3. COLD = DPME.

Trick: Last letter D+1 = E; only (A) and (D) end in E, and L+1 = M (not N) picks (A).

Answer: (A) DPME

Concept: In reverse order, a letter's value = 27 − its usual position.

1. INTELLIGENCE begins with I and ends with E.
2. I is 9th → reverse value 27 − 9 = 18.
3. E is 5th → reverse value 27 − 5 = 22.
4. Sum = 18 + 22 = 40.

Trick: Sum of reverse values of two letters = 54 − (sum of their normal positions) = 54 − 14 = 40.

Answer: (B) 40

Concept: Compare the given address part by part with each option.

1. Flat no., tower, locality, city and PIN are the same in all options.
2. Only the state changes: Gujarat, Rajasthan, Madhya Pradesh, Uttar Pradesh.
3. The original says Rajasthan → only option (B) matches.

Trick: Scan for the one field that differs among options and check only that field.

Answer: (B) Flat 505, Tower-B, Shastri Nagar, Jaipur, Rajasthan - 302001

Concept: Place events on a calendar line using 'after' (+) and 'before' (−).

1. Aurangabad: 2nd March (day 2).
2. Nagpur: day 2 + 3 → 5th March.
3. Nashik comes a day ahead of Nagpur: day 5 − 1 → 4th March.

Trick: Net shift from Aurangabad = +3 − 1 = +2 days → 2 + 2 = 4 March.

Answer: (B) 4th March

Concept: Here a * a means a raised to the power a.

1. 3 * 3 = $3^3=27$ ✓
2. 4 * 4 = $4^4=256$ ✓
3. 5 * 5 = $5^5=25\times 25\times 5=625\times 5=3125$

Trick: $5^n$ always ends in 5 — only 3125 ends in 5 among the options.

Answer: (A) 3125

Concept: A conclusion is definite only if the chain forces it.

1. Path B → F: B ≥ D > E = C ≥ F. One '>' sign appears, so B > F.
2. B > F means B ≥ F is also true → I is true.
3. E = C ≥ F gives only E ≥ F; E may be bigger than F → II is not definite.

Trick: If the chain has even one '>' the result is '>' ; '≥' alone never proves '='.

Answer: (C) Only Conclusion I is true

Concept: Children having the same father are siblings.

1. X is father of both Y and Z.
2. So Y and Z are children of one parent.
3. Gender is unknown, so the safe word is 'siblings'.

Trick: When gender is not given, choose the gender-neutral relation.

Answer: (C) Siblings

Concept: Second term = $(\text{first term}{)}^3+1$.

1. ${16}^3=4096$ → $4096+1=4097$ ✓
2. ${18}^3=18\times 18\times 18=324\times 18=5832$
3. $5832+1=5833$

Trick: Unit digit check: $8^3$ ends in 2, +1 → ends in 3. Only 5833 fits.

Answer: (A) 5833

Concept: From 'All A are B' and 'Some B are C', nothing definite links A and C.

1. Chairs are fully inside Tables.
2. The 'desk' part of Tables can lie away from Chairs → I not certain.
3. 'Some tables are desks' says nothing about ALL desks → II not certain.

Trick: All + Some (middle term in predicate of 'All') gives no conclusion.

Answer: (D) Neither I nor II follows

Concept: Each letter is moved 3 places forward.

1. J→M, A→D, P→S, A→D, N→Q (all +3).
2. CHINA: C→F, H→K, I→L, N→Q, A→D.
3. Code = FKLQD.

Trick: Both A's became D; in CHINA last letter A → D, so the code must end in D. Only FKLQD starts with F and ends with D.

Answer: (B) FKLQD

Concept: Letters go one step, numbers rise by 2 each time.

1. Letters: R, S, T, U → next should be V.
2. Numbers: 4, 6, 8, 10 → next should be 12.
3. W14 breaks both rules → odd pair.

Trick: Number = 2 × (letter position − 16): R(18) → 4, so V(22) → 12, W(23) does not fit.

Answer: (D) W14