Q&A: practice questions with answers

Concept: Decode the chain from left to right and build the family tree.

1. T + P: T is the sister of P. P × M: P is the son of M → T is the daughter of M.
2. M − V: M is the husband of V → V is the mother of T.
3. V ÷ S: V is the daughter of S.
4. So T is the daughter of V, and V is the daughter of S → T is S's daughter's daughter (granddaughter).

Trick: T's gender is fixed by "sister" (female), so options with "son" or "brother" as the link are eliminated at once.

Answer: (C) Daughter’s daughter

Concept: Fix the definite days first, then fit the ordered group into the gaps.

1. No one before N → N on Monday. Three between N and O → O on Friday.
2. One person between O and M → M on Wednesday or Sunday; Wednesday is L's → M on Sunday.
3. Tuesday, Thursday, Saturday remain for C, B, A in the order C → B → A.
4. B gets Thursday.

Trick: "No one has exam before N" = N is on the first day; such absolute clues anchor the whole schedule.

Answer: (C) Thursday

Concept: Check the alphabetical gap from the first cluster to the second.

1. AB → DE: A(+3)D, B(+3)E.
2. TU → WX: T(+3)W, U(+3)X.
3. KL → NO: K(+3)N, L(+3)O.
4. XY → YZ: X(+1)Y, Y(+1)Z – gap is +1, not +3.

Trick: Write the place numbers (A=1 … Z=26) and subtract – the pair with a different gap is the odd one.

Answer: (B) XY - YZ

Concept: Remove every digit and then count from the left in the remaining series.

1. Series: 5 # % 8 * $ % @ 1 3 @ $ 4 6 9 7 # $ 2 % 5 4.
2. After dropping numbers: # % * $ % @ @ $ # $ %.
3. Fourth from the left = $.

Trick: Strike the digits with a pencil first; never try to count and skip at the same time.

Answer: (A) $

Concept: Replace each symbol by its given meaning and then apply BODMAS.

1. 79 − 8 ÷ 55 + 423 × 3 becomes $79\times 8+55-423÷3$.
2. $423÷3=141$, $79\times 8=632$.
3. $632+55-141=546$.

Trick: Rewrite the whole expression with new symbols first, then solve – do not substitute and calculate in the same step.

Answer: (A) 546

Concept: Facing the centre, ‘right’ means anticlockwise and ‘left’ means clockwise. Start from the definite clue.

1. Q is immediate right of C; O is 3rd left of Q; D is such that C is 3rd left of D → going anticlockwise: C, Q, _, D, _, O, _.
2. R is not next to C, so R takes the seat after Q. Only one person between P and B (from P’s right) → P takes the seat after D and B the last seat, with O between them.
3. Order anticlockwise: C, Q, R, D, P, O, B. From R’s right: D, P, O, then B → three people.

Trick: Fix the pair with an ‘immediate’ relation first (C–Q), then place the ‘third-left’ clues; the negative clue (R not near C) settles the rest.

Answer: (D) Three

Concept: Decode the expression pair by pair and draw the family tree.

1. P + Q: P is husband of Q. Q – R: Q is mother of R → P is father of R.
2. R × S: R is sister of S → S is also P’s child (son).
3. S ÷ T: S is father of T.
4. So P is the father of T’s father → father’s father (paternal grandfather).

Trick: ‘Husband of the mother’ = father; then one more generation down gives grandfather.

Answer: (C) Father’s father

Concept: Letters common to both words get the common digits; the leftover letter gets the leftover digit.

1. BELL → 1582, SELL → 2568.
2. Common letters E, L, L ↔ common digits 2, 5, 8.
3. Leftover in BELL: B ↔ 1; leftover in SELL: S ↔ 6.

Trick: In such questions, only the uncommon letter–digit pair is asked; compare the two codes and pick the digit present in one but not the other.

Answer: (A) 6

Concept: Match the common letters of the two words with the common digits of the two codes.

1. TOUR → 4791, FORT → 1279.
2. Common letters T, O, R ↔ common digits 7, 9, 1.
3. Leftover: U ↔ 4 (in TOUR), F ↔ 2 (in FORT).

Trick: The digit that appears in FORT’s code but not in TOUR’s code is the code of the letter unique to FORT.

Answer: (B) 2

Concept: Find the rule from the complete side and apply it to both blanks.

1. Try the options for %: TSX → each letter −5 gives O, N, S → ONS (option C).
2. Apply the same rule to #: # −5 = LKP ⇒ # = L+5, K+5, P+5 = Q, P, U → QPU.
3. Option C gives exactly # = QPU and % = ONS.

Trick: When both sides have blanks, test the simplest rule (same shift for all letters) on the option values.

Answer: (C) # = QPU, % = ONS

Concept: Look for the pattern ‘number – number – symbol’ and count the middle number.

1. Series: 2 < 8 5 8 7 4 9 $ % % % & @ ? & @ 5 9 6 ?
2. 4 9 $ → 9 qualifies (preceded by 4, followed by $).
3. 9 6 ? → 6 qualifies (preceded by 9, followed by ?).
4. No other number has a number before it and a symbol after it → two.

Trick: Such numbers can only sit at the end of a run of numbers just before a symbol – check only those spots.

Answer: (B) Two

Concept: A chain of ‘All A are B’ statements can be drawn as nested circles.

1. Crabs ⊂ Bats ⊂ Wigs ⊂ Dogs.
2. (I) All crabs are dogs – the innermost circle lies inside the outermost → follows.
3. (II) Some dogs are bats – the bats circle lies inside dogs, so some dogs are certainly bats → follows.

Trick: ‘All A are B’ always gives ‘Some B are A’ (conversion) – so conclusion II is a direct conversion of ‘All bats are dogs’.

Answer: (D) Both conclusions (I) and (II) follow

Concept: Find the three operations in the given sets and test them on the options.

1. 17 → 7 (−10), 7 → 21 (×3), 21 → 33 (+12). Check: 21 → 11 (−10), 11 → 33 (×3), 33 → 45 (+12). ✓
2. Option C: 14 → 4 (−10), 4 → 12 (×3), 12 → 24 (+12). ✓
3. A: 16−8 is −8; B: 13×3 ≠ 26; D: 33+12 ≠ 72.

Trick: Check the first operation (−10) on all options first – it eliminates A at once; then verify the rest.

Answer: (C) 14-4-12-24

Concept: Number positions 1 (top) to 6 (bottom) and fill from the definite clue.

1. F at position 3. Two boxes between F and D → D at 6 (position 0 does not exist).
2. A above F but below E → E = 1, A = 2.
3. C is not at position 5 → C = 4, so B = 5.
4. Order: E, A, F, C, B, D. Between A (2) and B (5): F and C → two boxes.

Trick: ‘Only one position possible’ clues (like F third from top + D two away) should be placed before the relative ones.

Answer: (A) Two

Concept: Check each statement alone, then both together.

1. I alone: D > C > A, B – but E is unknown; E could be anywhere → not sufficient.
2. II alone: D > E > C – A and B unknown; they could be above E → not sufficient.
3. Together: D > E > C > A, B → second heaviest is E.

Trick: For ‘second heaviest’, every box must be placed below someone except the top two – a statement missing any box is insufficient.

Answer: (C) Both statements I and II together are sufficient

Concept: Look at the successive differences.

1. $30\to 28\to 25\to 20\to 13$: differences $2,3,5,7$ – consecutive primes.
2. Next prime is $11$: $13-11=2$.

Trick: Differences 2, 3, 5, 7, 11, 13 … (primes) is a favourite pattern – check it whenever gaps grow irregularly.

Answer: (C) 2

Concept: Find the position-wise shift from the first cluster to the second.

1. LFGK → RKKN: L+6=R, F+5=K, G+4=K, K+3=N (shifts +6, +5, +4, +3).
2. Check GMOP → MRSS: G+6=M, M+5=R, O+4=S, P+3=S ✓.
3. COLU: C+6=I, O+5=T, L+4=P, U+3=X → ITPX.

Trick: Decreasing shifts (+6, +5, +4, +3) are common; verify with the second pair before applying.

Answer: (A) ITPX

Concept: Elastic potential energy of a spring $U=\frac{1}{2}k{x}^2$, where $x$ is the compression or extension from equilibrium.

1. At equilibrium $x=0$, so $U=0$ (minimum).
2. Compressing increases $x$, and since $x^2$ is always positive, $U$ increases – work done against the spring force is stored in it.

Trick: Stretch or compress – both store energy; $U$ depends on $x^2$, not on direction.

Answer: (C) increase

Concept: Software is broadly system software (runs/manages the hardware) and application software (does user tasks).

1. Operating system (Windows, Linux) and device drivers control hardware and provide a platform for other programs – that is system software.
2. Utility software (antivirus, disk cleanup) is a sub-type of system software, so "utility software only" is too narrow.

Trick: OS + drivers + utilities + language translators = System software.

Answer: (A) System software

Concept: Part IXA (Articles 243P–243ZG, added by the 74th Amendment, 1992) deals with Municipalities.

1. Article 243X – power to impose taxes by, and funds of, the Municipalities: the State Legislature may authorise a municipality to levy taxes, duties, tolls and fees and assign grants.
2. Article 243P – definitions; 243U – duration of municipalities; 243ZB – application to Union Territories. 243Y is the Finance Commission for municipalities.

Trick: Panchayat finance = 243H; Municipality finance = 243X (both "money" articles end in H/X).

Answer: (D) Article 243 X