Q&A: practice questions with answers

Concept: Average of the first $n$ even numbers $=n+1$ (sum $=n(n+1)$).

1. Sum $=164\times 165$.
2. Average $=\frac{164\times 165}{164}=165$.

Trick: First $n$ even → $n+1$; first $n$ odd → $n$.

Answer: (A) 165

Concept: Find each pipe's full-tank time, take tank = LCM units, and add rates (emptying pipe negative).

1. P: $\frac{5}{6}$ in 30 h → full in $36$ h. Q: $\frac{3}{4}$ in 27 h → full in $36$ h.
2. Tank $=36$ units; P = Q = 1 unit/h. In 2 h they fill $4$ units.
3. R empties 4 units in 4 h → R $=-1$ unit/h.
4. P + Q + R $=1+1-1=1$ unit/h → $36$ hours.

Trick: P and R cancel each other, so the time equals Q's time alone: 36 h.

Answer: (C) 36 hours

Concept: If $p\%$ of $a$ = $\frac{m}{n}$ of $b$, then $\frac{a}{b}=\frac{m}{n}÷\frac{p}{100}$.

1. $\frac{51}{100}a=\frac{6}{7}b$.
2. $\frac{a}{b}=\frac{6\times 100}{7\times 51}=\frac{600}{357}$.
3. Divide by 3: $\frac{200}{119}$.

Trick: Ratio $=$ (fraction of second) : (fraction of first), cross-style.

Answer: (D) 200 : 119

Concept: One-day works add: $\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{\text{together}}$.

1. Total work $=418$ units (LCM of 19, 11, 2). A $=22$, B $=38$, all three $=209$ units/day.
2. C $=209-60=149$ units/day.
3. C alone $=\frac{418}{149}=2\frac{120}{149}$ days.

Trick: Together they take 2 days, so C alone must take just over 2 days – only option (B) is near.

Answer: (B) $2rac120149$ days

Concept: Work through successive percentages of the salary to reach savings.

1. After debt: $75\%$ of $S$ remains.
2. Saves $48\%$ of that: $0.75\times 0.48S=0.36S$.
3. $0.36S=288$ → $S=800$.

Trick: Check: $800\to 600\to$ spends $312$, saves $288$ ✓.

Answer: (C) 800

Concept: Divisibility by 11: (sum of digits at odd places) − (sum at even places) must be 0 or a multiple of 11.

1. Number 747579723: odd places from right $3+7+7+7+7=31$, even places $2+9+5+4=20$.
2. $31-20=11$ → divisible by 11.
3. Digit sum $=51$ → not by 9; last three digits 723 odd → not by 8; division by 13 leaves remainder 7.

Trick: Test 11 first whenever the digits look scattered – alternate-sum rule is quick.

Answer: (D) 11

Concept: For divisibility by 11, (sum of odd-place digits) − (sum of even-place digits) $=0$ or $\pm 11$.

1. N42M47: odd places $N+2+4=N+6$, even places $4+M+7=M+11$.
2. $(N+6)-(M+11)=N-M-5\in \{0,11,-11\}$; $N-M=16$ is impossible and $N-M=-6$ matches no option, so $N-M=5$.
3. $N-M=5$ → $M-N=-5$.

Trick: Example: N = 5, M = 0 gives 542047 $=11\times 49277$ ✓.

Answer: (D) M − N = −5

Concept: BODMAS: innermost bracket first, then multiplication, then addition, then division outside.

1. $(7-3)=4$; $5\times 4=20$.
2. $28+20=48$.
3. $960÷48=20$.

Trick: Solve brackets from inside out: ( ) → { } → [ ].

Answer: (B) 20

Concept: Lateral surface area of a cuboid $=2(l+b)h$.

1. Side of each cube $=\sqrt[3]{216000}=60\text{cm}$.
2. Cuboid: $l=5\times 60=300$, $b=h=60$.
3. LSA $=2(300+60)\times 60=43200{\text{cm}}^2$.

Trick: LSA = base perimeter $\times$ height $=720\times 60$.

Answer: (D) 43200

Concept: Check each option for both conditions: sum and LCM.

1. All options sum to 66.
2. $\text{LCM}(16,50)=2^4\times 5^2=400$ ✓.
3. Others: $\text{LCM}(21,45)=315$, $\text{LCM}(14,52)=364$, $\text{LCM}(9,57)=171$.

Trick: LCM 400 has only factors 2 and 5 – so both numbers must be made only of 2s and 5s: 16 and 50.

Answer: (C) 16, 50

Concept: Look at the differences; they form an arithmetic pattern.

1. Differences: $2,10,18,26$ (each +8).
2. Next difference $=34$.
3. $77+34=111$.

Trick: Second-level difference constant (8) → AP of differences.

Answer: (B) 111

Concept: Draw a family tree starting from S and move step by step.

1. E is S's father and K is E's wife → K is S's mother.
2. A is K's brother → A is S's mother's brother (maternal uncle).
3. W is A's daughter → W is S's mother's brother's daughter.

Trick: Trace the path from S: mother → brother → daughter.

Answer: (C) Mother's brother's daughter

Concept: Convert letters to positions and compare the gaps.

1. MOJ: 13, 15, 10 → +2, −5. SUP: 19, 21, 16 → +2, −5. PRM: 16, 18, 13 → +2, −5.
2. UWS: 21, 23, 19 → +2, −4.
3. UWS breaks the pattern.

Yaad rakhein: Keep EJOTY (5, 10, 15, 20, 25) in mind for quick letter positions.

Answer: (A) UWS

Concept: Plot each move on a grid (left turn when facing south = east, etc.).

1. A(0, 0) → 7 km south → left (east) 6 km → left (north) 12 km → left (west) 10 km → left (south) 5 km.
2. Net east-west: $6-10=-4$ → 4 km west of A. Net north-south: $-7+12-5=0$.
3. P is 4 km west of A, so he must drive 4 km to the east.

Trick: Add the vertical moves and horizontal moves separately.

Answer: (A) 4 km to the east

Concept: Find the shift applied to each letter.

1. DACB → FCED: each letter +2.
2. YVXW → AXZY: each +2 (Y+2 = A).
3. VSUT → X U W V.

Trick: Same shift for all letters → shift the whole word.

Answer: (A) XUWV

Concept: Check the backward gaps between letters (cyclic alphabet).

1. RMH: 18 → 13 → 8 (−5, −5). DYT: 4 → 25 → 20 (−5, −5). IDY: 9 → 4 → 25 (−5, −5).
2. AVO: 1 → 22 → 15 is −5, then −7.
3. AVO is the odd one.

Trick: Going back from A wraps to Z: A − 5 = V.

Answer: (C) AVO

Concept: Find the gaps between consecutive letters (cyclic).

1. VZFM: +4, +6, +7. PTZG: +4, +6, +7. IMSZ: +4, +6, +7.
2. SWCI: S → W +4, W → C +6, C → I +6.
3. SWCI differs.

Trick: Write positions once: V22 Z26 F6 M13 – the gaps make the rule clear.

Answer: (C) SWCI

Concept: Fix definite positions first (top = 1, bottom = 7).

1. Only T above F → T = 1, F = 2.
2. Two boxes above M → M = 3, A = 4.
3. N directly above O, N not at 5 (3rd from bottom) → N = 6, O = 7, so R = 5.
4. Between R (5) and T (1): F, M, A → three boxes.

Trick: Start with 'only' and number-based clues; they lock positions.

Answer: (A) Three

Concept: Apply the rule digit by digit, then read the required positions.

1. 7856413 → 7+1, 8−2, 5+1, 6−2, 4−2, 1+1, 3+1 = 8664224.
2. Third from left $=6$; third from right $=2$.
3. Sum $=6+2=8$.

Trick: Write the new number fully before counting positions.

Answer: (D) 8

Concept: Differences grow in a pattern.

1. Differences: $3,5,9,17$.
2. Each difference $=$ previous $\times 2-1$ (or differences of differences $2,4,8$ doubling).
3. Next difference $=33$ → $107+33=140$.

Trick: $3,5,9,17,33$ are $2^n+1$.

Answer: (D) 140