Q&A: practice questions with answers

Concept: A number is divisible by 9 if its digit sum is a multiple of 9.

1. Digit sum $=5+1+3+5+2+n=16+n$
2. Next multiple of 9 is 18 → $n=2$

Yaad rakhein: 9 test: digit sum ÷ 9; 3 test: digit sum ÷ 3.

Answer: (D) $2$

Concept: The age difference stays constant over time.

1. Let brother $=b$, Rahul $=b+10$
2. 5 years ago: $b+5=2(b-5)\Rightarrow b=15$
3. Rahul $=25$ years

Trick: 5 years ago Rahul = 2 × brother and difference 10 → brother 10, Rahul 20 then; now 25.

Answer: (B) $25$ years

Concept: $\text{Savings}=\text{Income}-\text{Expenditure}$; take income $=100$.

1. Before: $I=100$, $E=75$, $S=25$
2. After: $I=120$, $E=75\times 1.1=82.5$, $S=37.5$
3. Change $=\frac{12.5}{25}\times 100=50\%$ increase

Trick: Always assume income 100 in such questions.

Answer: (B) $50\%$ increase

Concept: Take the common factor $2^{x-1}$.

1. $2^x-2^{x-1}=2^{x-1}(2-1)=2^{x-1}$
2. $2^{x-1}=32=2^5\Rightarrow x=6$
3. $\frac{2x-1}{2x+3}=\frac{11}{15}$

Yaad rakhein: $a^x-a^{x-1}=a^{x-1}(a-1)$.

Answer: (B) $\frac{11}{15}$

Concept: $24=3\times 8$: last three digits divisible by 8 and digit sum divisible by 3.

1. Last three digits $44B$: $440$ and $448$ are multiples of 8 → $B=0$ or $8$
2. Digit sum $=26+A+B$ must be a multiple of 3
3. Take $B=8$: $34+A$ → $A=2,5,8$; max $A=8$
4. Max $A+B=8+8=16$

Trick: Choose the largest B first, then the largest A that fits the 3-test.

Answer: (D) $16$

Concept: Check the gap between consecutive letters of each cluster.

1. KHI: K $\stackrel{-3}{\to }$ H $\stackrel{+1}{\to }$ I
2. ZWX: Z $\stackrel{-3}{\to }$ W $\stackrel{+1}{\to }$ X; FCD: F $\stackrel{-3}{\to }$ C $\stackrel{+1}{\to }$ D
3. PNN: P $\stackrel{-2}{\to }$ N $\stackrel{0}{\to }$ N – different pattern.

Trick: A repeated letter (NN) usually signals the odd cluster – verify quickly.

Answer: (A) PNN

Concept: Facing north, “right” is the east end. Fix the person with a unique position first.

1. Only H is to the right of G → H at the right end (7), G at 6.
2. Two between H and Q → Q = 4. Three between G and F → F = 2.
3. Positions 1, 3, 5 left for M, L, O with M < L < O (left to right) → M = 1, L = 3, O = 5.
4. Between M (1) and O (5): F, L, Q = 3 people.

Trick: “Only X is to the right of Y” → Y is second from the right end.

Answer: (B) Three

Concept: Compare each letter of the first cluster with the corresponding letter of the second.

1. VXW → UWV: each letter $-1$. WYX → VXW: each letter $-1$.
2. BDC → ACB: B→A, D→C, C→B, each $-1$ ✓.
3. BCA → ACB, BDC → ZCA and BCA → ZBA break the $-1$ rule.

Trick: Check only the first letter of each option first (B → A); only (A) and (B) survive, then check the rest.

Answer: (A) BDC−ACB

Concept: A word common to two sentences gets the code common to both coded sentences.

1. salt elbow care → zh pv nf
2. night care jar → ab po zh
3. Only “care” and only “zh” are common → care = zh.

Trick: Common word ↔ common code – no need to decode other words.

Answer: (A) zh

Concept: Write the digits in ascending order and compare place by place with the original number.

1. Original: 7 3 9 1 2 6 8 5
2. Ascending: 1 2 3 5 6 7 8 9
3. Only the 7th place has the same digit (8) in both.

Trick: A digit stays fixed only if exactly (its place − 1) digits are smaller than it: 8 has 6 smaller digits and sits 7th ✓.

Answer: (C) One

Concept: Swap the signs as directed, then solve using BODMAS.

1. After interchange: $39+15\times 3-124÷4$
2. $=39+45-31$
3. $=53$

Trick: Swap carefully in one pass (write the new expression fully) before calculating.

Answer: (C) $53$

Concept: Number positions 1 (bottom) to 7 (top) and place the fixed box first.

1. D = 3. C above A but below D → A = 1, C = 2.
2. Positions 4–7 left for B, E, F, G. E above F and below B → F is the lowest of them.
3. Two between G and F → F = 4, G = 7; then E = 5, B = 6.
4. Between E (5) and D (3): only F → one box.

Trick: When only one pair of slots fits “two between” inside a 4-slot block, it must be the two ends (4 and 7).

Answer: (B) One

Concept: Find a rule linking the three numbers of each set.

1. $(15,12,37)$: $15+12+10=37$
2. $(25,15,50)$: $25+15+10=50$
3. $(2,8,20)$: $2+8+10=20$ ✓; others give $38\ne 40$, $44\ne 40$, $45\ne 42$.

Trick: Third number − (first + second) is constant (10) – test that difference on each option.

Answer: (A) $(2,8,20)$

Concept: Number positions 1 (bottom) to 7 (top); “only X above Y” fixes both at the top.

1. Only G above E → G = 7, E = 6.
2. Only C between E and A → C = 5, A = 4.
3. Only F below B → F = 1, B = 2; remaining D = 3.
4. Between C (5) and F (1): A, D, B → 3 boxes.

Trick: Boxes between positions $p$ and $q$ = $|p-q|-1=5-1-1=3$.

Answer: (B) 3

Concept: Track each letter position of the clusters separately.

1. 1st letters: B, E, H, K → $+3$ → N
2. 2nd letters: M, P, S, V → $+3$ → Y
3. 3rd letters: S, V, Y, B → $+3$ (Y → Z → A → B) → E

Trick: All three positions move $+3$; after Z, wrap around to A.

Answer: (B) NYE

Concept: Look at the differences between consecutive terms.

1. Differences: $27,54,81,108$ = $27\times 1,27\times 2,27\times 3,27\times 4$
2. Next difference $=27\times 5=135$
3. $321+135=456$

Trick: Differences in a table of 27 – recognise multiples quickly.

Answer: (A) $456$

Concept: Treat each letter position and the number separately.

1. Every letter moves $-3$: G→D→A→X→U, C→Z→W→T→Q, H→E→B→Y→V.
2. Numbers: 10, 21, 32, 43 → $+11$ → 54.
3. Next term: UQV 54.

Trick: Numbers alone (+11 → 54) cut options to (B); confirm letters.

Answer: (B) UQV 54

Concept: Number positions 1 (bottom) to 7 (top).

1. No box below B → B = 1. Two between G and B → G = 4.
2. Only D above C → D = 7, C = 6.
3. Positions 2, 3, 5 for A, F, E with A above F above E → E = 2, F = 3, A = 5.
4. Between D (7) and E (2): C, A, G, F → 4 boxes.

Trick: “No box below B” = B at the bottom; “only D above C” = D top, C second.

Answer: (C) 4

Concept: Compare the change from the first cluster to the second, letter by letter.

1. PS → OR: $-1,-1$; FI → EH: $-1,-1$; LO → KN: $-1,-1$.
2. KN → JL: K→J is $-1$ but N→L is $-2$.
3. So KN − JL is the odd pair.

Trick: In the three alike pairs both clusters have the same internal gap (+3): PS/OR, FI/EH, LO/KN; JL has gap +2.

Answer: (C) KN - JL

Concept: Take A as origin $(0,0)$, East = $+x$, North = $+y$, and track each move.

1. E 29 → $(29,0)$; right → S 16 → $(29,-16)$; right → W 39 → $(-10,-16)$.
2. Right → N 9 → $(-10,-7)$; right → E 6 → $(-4,-7)$; left → N 3 → $(-4,-4)$.
3. Right → E 4 → P $=(0,-4)$.
4. P is 4 km directly south of A, so he must drive 4 km towards North.

Trick: Net E–W: $29-39+6+4=0$; net N–S: $-16+9+3=-4$ → P is 4 km south of A.

Answer: (C) 4 km to the North