Q&A: practice questions with answers

Concept: Take CP $=100$; then MP and the required SP are easy to compare.

1. CP $=100$, MP $=100+136=236$.
2. Required SP for 18% gain $=118$.
3. Discount $=236-118=118$; discount% $=\frac{118}{236}\times 100=50\%$.

Trick: $236=2\times 118$, so the SP is exactly half of MP → 50% discount.

Answer: (D) 50%

Concept: Look at the differences between consecutive terms.

1. $73-71=2$, $71-68=3$, $68-63=5$, $63-56=7$: the differences are consecutive primes.
2. Next prime is 11, so $?=56-11=45$.

Trick: Differences 2, 3, 5, 7, 11, 13 … (primes) is a very common NTPC pattern.

Answer: (A) 45

Concept: Find the pattern in the differences.

1. $122-120=2$, $120-117=3$, $117-112=5$, $112-105=7$: consecutive primes.
2. Next difference $=11$, so $?=105-11=94$.

Trick: Same prime-difference pattern as the series 73, 71, 68, 63, 56 → 45.

Answer: (B) 94

Concept: The angles of a triangle add to ${180}^{\circ }$; a ratio share as a percentage of the total is $\frac{\text{part}}{\text{sum of parts}}\times 100$.

1. Ratio $2:3:4$, sum of parts $=9$; largest angle $=\frac{4}{9}\times 180={80}^{\circ }$.
2. Percentage $=\frac{80}{180}\times 100=\frac{4}{9}\times 100=44.44\%\approx 44\%$.

Trick: You don't even need ${180}^{\circ }$: the largest share is $\frac{4}{9}$ of the total $=44.4\%$.

Answer: (A) 44%

Concept: Add the per-hour work of filling taps and subtract that of the emptying tap.

1. Net part filled per hour $=\frac{1}{4}+\frac{1}{30}-\frac{1}{30}=\frac{1}{4}$.
2. Full cistern takes 4 hours, so one-fourth takes $4\times \frac{1}{4}=1$ hour.

Trick: The 30-hour filler and 30-hour emptier cancel exactly; only the 4-hour tap matters.

Answer: (B) 1

Concept: Average $=\frac{\text{sum of observations}}{\text{number of observations}}$.

1. Sum $=72+68+74+70+76=360$.
2. Average $=\frac{360}{5}=72$.

Trick: Take 72 as base: deviations $0,-4,+2,-2,+4$ add to 0, so the average is exactly 72.

Answer: (D) 72

Concept: If two runners run in the same direction with speeds in the simplest ratio $a:b$, the number of distinct meeting points on the track is $|a-b|$ (for opposite directions it is $a+b$).

1. Speeds: Shahrukh $=\frac{1530}{85}=18\text{m/s}$, Amir $=\frac{1530}{90}=17\text{m/s}$.
2. Ratio $18:17$ is already in lowest terms.
3. Same direction → meeting points $=18-17=1$ (only at the starting point).

Trick: Ratio of speeds = inverse ratio of lap times $=90:85=18:17$; difference 1 → one meeting point.

Answer: (A) 1

Concept: For ungrouped frequency data, the mode is the observation with the highest frequency.

1. Number of plants: 14 cm → 1, 27 cm → 3, 23 cm → 6, 21 cm → 12, 19 cm → 7, 17 cm → 5.
2. Highest frequency is 12, for height 21 cm.

Trick: Mode = "most repeated"; do not average or pick the largest height (27 cm is a distractor).

Answer: (A) 21 cm

Concept: For equal distances at speeds $a$ and $b$, average speed $=\frac{2ab}{a+b}$.

1. Let morning speed $=v$, afternoon $=v-2$. Then $\frac{2v(v-2)}{2v-2}=4.8\Rightarrow \frac{v(v-2)}{v-1}=4.8$.
2. $v^2-2v=4.8v-4.8\Rightarrow v^2-6.8v+4.8=0\Rightarrow (v-6)(v-0.8)=0$.
3. $v=6$ (since $v>2$); speeds 6 and 4 km/h, sum $=10$ km/h.

Trick: Test options: sum 10 → speeds 6 and 4 → $\frac{2\times 6\times 4}{10}=4.8$ ✓ in one line.

Answer: (B) 10

Concept: Compare two coded words: common letters get the common digits; the leftover letter gets the leftover digit.

1. COOK → 8245 and HOOK → 5429 share the letters O, O, K.
2. Common digits of 8245 and 5429 are 2, 4, 5 → these belong to O, O, K.
3. Leftover: C → 8 (from COOK) and H → 9 (from HOOK).

Trick: In such codes the digits are usually jumbled, so match by elimination, not by position.

Answer: (B) 9

Concept: Fix the given floor first, then place people using the 'between' and 'immediately' clues.

1. U = floor 4. Two people between U and Z → Z = 1 (floor 7 does not exist).
2. X lives immediately below U → X = 3.
3. Only T between U and Y → T = 5, Y = 6 (T = 3 is taken by X). Remaining S = 2.
4. Floor 5 is occupied by T.

Trick: Always begin with the definite clue (U = 4); 'immediately below' kills the other branch at once.

Answer: (D) T

Concept: Write the alphabet positions and look at the pattern inside each pair.

1. RQ−PO: R(18) Q(17) P(16) O(15) – four consecutive letters in reverse order.
2. ZY−XW: 26, 25, 24, 23 and YX−WV: 25, 24, 23, 22 – also four consecutive letters backwards.
3. LK−IK: L(12) K(11) I(9) K(11) – the chain breaks (gap of 2, then K repeats).

Trick: Three options following 'consecutive descending' leave the odd one out immediately – look for the repeated letter.

Answer: (D) LK−IK

Concept: Write the original and the rearranged number one below the other and compare position by position.

1. Original: 7 8 1 4 3 2 9.
2. Ascending order: 1 2 3 4 7 8 9.
3. Position 4 has 4 in both and position 7 has 9 in both; all other positions change.
4. So the position of two digits remains unchanged.

Trick: Only the digit that is already the k-th smallest and sits at the k-th place stays – check each quickly.

Answer: (D) Two

Concept: Facing the centre, 'right' means anticlockwise. In 7 seats, 'four between D and C from D's right' = C is 5th to the right of D.

1. Place D. Going right: R is 3rd to the right of D; C is 5th to the right of D.
2. P is a neighbour of both C and R → P sits in the 4th seat (between R and C).
3. Seats left: 1st, 2nd and 6th to the right of D. B is 3rd to the left of Q → Q at 2nd right, B at 6th right; O takes the 1st seat right of D.
4. From O's left (clockwise): D, B, C, then P → three people between O and P.

Trick: Number the seats 0–6 anticlockwise from D; 'n-th to the right' = +n, 'n-th to the left' = −n (mod 7).

Answer: (D) Three

Concept: Convert letters to positions and find the gap pattern between 1st→2nd and 2nd→3rd letters.

1. PLO: 16, 12, 15 → $-4,+3$.
2. NJM: 14, 10, 13 → $-4,+3$; LHK: 12, 8, 11 → $-4,+3$.
3. JFH: 10, 6, 8 → $-4,+2$ – the second gap differs.

Trick: When three options share a pattern like $(-4,+3)$, the one with a single different step is the answer.

Answer: (C) JFH

Concept: Decode symbol by symbol and build the family tree from the top (oldest) generation.

1. C + D: C is the mother of D. L # C: L is the daughter of C → L and D are siblings.
2. P − L: P is the brother of L → P is also a child of C, i.e. P is D's brother.
3. G # P: G is the daughter of P.
4. So G is the daughter of D's brother → brother's daughter (niece).

Trick: Start from the end of the chain (the mother C) – children of the same mother are siblings; then attach the first person.

Answer: (A) Brother's daughter

Concept: Interchange the signs on both sides of the equation, then apply BODMAS.

1. After interchange: $25\times 2-12÷4+7=?-13$.
2. LHS: $12÷4=3$; $25\times 2=50$; $50-3+7=54$.
3. $?-13=54\Rightarrow ?=67$.

Trick: The '+' on the right side also becomes '−' – forgetting this gives 41, which is not an option.

Answer: (D) 67

Concept: Track the facing direction after each turn: facing West, left = South; facing South, left = East; facing East, left = North; facing North, right = East.

1. Take A as origin. West 46 → $(-46,0)$; left (South) 25 → $(-46,-25)$; left (East) 35 → $(-11,-25)$.
2. Left (North) 67 → $(-11,42)$; right (East) 11 → $(0,42)=P$.
3. P is directly 42 km north of A, so he must drive 42 km towards the South.

Trick: Net East–West: $-46+35+11=0$; net North–South: $-25+67=42$ north – when one component is zero no Pythagoras is needed.

Answer: (C) 42 km to the South

Concept: Scan only the symbols and check the character just before (must be a letter) and just after (must be a number).

1. Symbols in order: $ (W before, D after – fails), % (D before, 9 after – ✓), @ (8 before – fails).
2. # (F before, 6 after – ✓), & (V before, L after – fails), ^ (K before, U after – fails).
3. Required pattern Letter–Symbol–Number occurs twice.

Trick: Mark each symbol and look one step left and one step right only – do not scan the whole series repeatedly.

Answer: (D) Two

Concept: Find the shift applied to each letter position in the given pairs and apply the same to the options.

1. RUB → WSG: R+5=W, U−2=S, B+5=G. BQL → GOQ: B+5=G, Q−2=O, L+5=Q. Pattern $(+5,-2,+5)$.
2. LMV → QKA: L+5=Q, M−2=K, V+5=A (V=22, 22+5=27→1=A) ✓.
3. BCG→GYL (C−4), KDH→NZK (+3,−4,+3), MHV→OCX (+2,−5,+2) do not match.

Trick: Write the three shifts once; check options letter by letter and stop at the first mismatch.

Answer: (A) LMV : QKA