Q&A: practice questions with answers

Concept: Form two equations in notebook price $n$ and pen price $p$ and solve by elimination.

1. $5n+3p=146$ … (i), $7n+2p=178$ … (ii)
2. (i)$\times 2$ − (ii)$\times 3$: $10n+6p-21n-6p=292-534\Rightarrow -11n=-242\Rightarrow n=22$
3. $3p=146-110=36\Rightarrow p=12$
4. Cost $=4\times 22+6\times 12=88+72=160$

Trick: Check: $5(22)+3(12)=110+36=146$ ✓. Options differ a lot, so finding $n$ alone almost settles it.

Answer: (B) ₹160

Concept: Same direction: first meeting time $=\frac{\text{track length}}{\text{relative speed}}$ (the faster one gains a full lap).

1. Relative speed $=36-33=3\text{km/h}=3\times \frac{5}{18}=\frac{5}{6}\text{m/s}$
2. Time $=\frac{2025}{5/6}=2025\times \frac{6}{5}=2430$ s

Trick: km/h → m/s: multiply by $\frac{5}{18}$. Opposite directions would use the sum $69$ km/h instead.

Answer: (B) 2430 seconds

Concept: Total work in man-days stays fixed: $\text{men}\times \text{days}$.

1. Total work $=24\times 15=360$ man-days
2. Done in day 1: $24$; remaining $=336$
3. Remaining workers $=16$: days $=\frac{336}{16}=21$

Trick: 'Additional days' here means days needed from now with 16 workers — $21$ (not $21-14=7$ extra over schedule).

Answer: (B) 21 days

Concept: Empirical relation: $\text{Mode}=3\text{Median}-2\text{Mean}$.

1. $\text{Mode}=3\times 50-2\times 41$
2. $=150-82=68$

Yaad rakhein: Mean − Mode $=3(\text{Mean}-\text{Median})$. Remember '3 median, 2 mean'.

Answer: (B) 68

Concept: B's one-day work $=$ (A+B)'s $-$ A's.

1. $\frac{1}{24}-\frac{1}{60}=\frac{5-2}{120}=\frac{1}{40}$ ⇒ B alone takes 40 days
2. One-fourth of the work: $\frac{40}{4}=10$ days

Trick: LCM method: work $=120$; A+B $=5$/day, A $=2$/day, B $=3$/day; $\frac{1}{4}$ work $=30$ units $\Rightarrow 10$ days.

Answer: (D) 10

Concept: Differences follow cubes with alternating signs.

1. $74\to 75$: $+1^3$; $75\to 67$: $-2^3$; $67\to 94$: $+3^3$; $94\to 30$: $-4^3$
2. Next: $30+5^3=30+125=155$

Trick: When terms jump up and down, check alternating $+/-$ of squares or cubes: $1,8,27,64,125$.

Answer: (A) 155

Concept: Compare the letter shifts from the first cluster to the second in each pair.

1. SX–UA: S+2=U, X+3=A; JP–LS: J+2=L, P+3=S; MI–OL: M+2=O, I+3=L — pattern (+2, +3)
2. TC–VE: T+2=V, C+2=E — pattern (+2, +2), different

Trick: Write place values (A=1 … Z=26) and subtract position-wise; the pair with a different difference set is the odd one.

Answer: (D) TC - VE

Concept: Track the position on a grid; each left turn rotates East→North→West→South→East.

1. Start A(0,0) → 28 km East (28,0) → 25 km North (28,25) → 29 km West (−1,25)
2. → 27 km South (−1,−2) → 1 km East: P(0,−2)
3. P is 2 km directly South of A, so he must drive 2 km North

Trick: Net East–West: $28-29+1=0$; net North–South: $25-27=-2$. Zero sideways offset ⇒ answer is purely North/South.

Answer: (D) 2 km to the North

Concept: Substitute the symbols and apply BODMAS.

1. $12\times 9+(98÷7)\times 3-25$
2. $=108+14\times 3-25=108+42-25=125$

Trick: Rewrite the whole expression in symbols first, then solve — never substitute and compute term by term.

Answer: (D) 125

Concept: Remove all letters, then count from the right in the remaining series.

1. Remaining symbols (left to right): * & ^ % $ # @ ! + ? — 10 symbols
2. 9th from the right $=$ 2nd from the left $=$ &

Trick: $n$th from right $=(N-n+1)$th from left: $10-9+1=2$.

Answer: (A) &

Concept: Number seats 1–7 from left (all face north, so left/right are the usual).

1. Two people left of C ⇒ C = 3; D immediately right ⇒ D = 4; B second right of C ⇒ B = 5
2. Four between F and E ⇒ (F,E) at 1&6 or 2&7; F is immediately right of A ⇒ A = 1, F = 2, E = 7; G = 6
3. Row: A F C D B G E. Between A(1) and D(4): F and C ⇒ two

Trick: Start with the fixed-position clue ('only two to the left of C'); such definite clues anchor the whole row.

Answer: (A) Two

Concept: Substitute the letters by the given operators and apply BODMAS.

1. $67+3\times 19+(84÷2)-6$
2. $=67+57+42-6=160$

Trick: Do the bracket and the multiplication first ($84÷2=42$, $3\times 19=57$), then add/subtract left to right.

Answer: (C) 160

Concept: Find the chain of operations in the given sets and test each option.

1. $3\to 9$ ($n^2$), $9\to 27$ ($\times 3$), $27\to 32$ ($+5$); $4\to 16$ ($n^2$), $16\to 48$ ($\times 3$), $48\to 53$ ($+5$)
2. Option B: $5^2=25$, $25\times 3=75$, $75+5=80$ ✓
3. Others fail: $7\to 49\to 147$ (not 98), $8\to 64\to 192$ (not 128), $6\to 36\to 108$ (not 72)

Trick: Pattern is square → ×3 → +5. Check the second step first — it eliminates A, C, D instantly.

Answer: (B) 5 -25-75-80

Concept: Each letter of the first cluster moves back by 4 to give the second cluster.

1. LIF → HEB: L−4=H, I−4=E, F−4=B; WUH → SQD: W−4=S, U−4=Q, H−4=D
2. TQV → PMR: T−4=P, Q−4=M, V−4=R ✓
3. HRP → DOL fails (R−4=N, not O); KOT → GKQ fails (T−4=P, not Q)

Trick: Check the letter that differs: in option A the middle letter and in C the last letter break the −4 rule.

Answer: (D) TQV : PMR

Concept: Track each position separately: letters by shift, numbers by difference.

1. 1st letters W, B, G, L, Q: +5 each ⇒ Q+5 = V
2. 2nd letters O, K, G, C, Y: −4 each ⇒ Y−4 = U
3. 3rd letters G, B, W, R, M: −5 each ⇒ M−5 = H; numbers +16 ⇒ 211+16 = 227

Trick: Only VUH 227 matches all four parts; WVG/VWG with 225 fail on the number alone.

Answer: (D) VUH 227

Concept: Decode each symbol step by step and build the family tree.

1. A ÷ B: A is the father of B; B − C: B is the brother of C ⇒ A is also the father of C
2. C × D: C is the wife of D
3. So D's wife is C and C's father is A ⇒ A is D's wife's father (father-in-law)

Trick: A's gender is fixed (father) — options 'Mother's brother' and 'Wife's mother' are out at once; 'Brother' would need A in C's generation.

Answer: (B) Wife's father

Concept: Plot the path on a grid; each left turn: North→West→South→East→North.

1. A(0,0) → 75 km North (0,75) → 80 km West (−80,75) → 85 km South (−80,−10)
2. → 89 km East (9,−10) → 10 km North: P(9,0)
3. P is 9 km East of A ⇒ drive 9 km West to return

Trick: Net N–S: $75-85+10=0$; net E–W: $-80+89=+9$ (East). So return is 9 km West.

Answer: (C) 9 km to the west

Concept: Common letters share common digits; the leftover letter takes the leftover digit.

1. CROW → 9863, WORK → 3289; common letters R, O, W ⇒ common digits 9, 8, 3
2. C (only in CROW) = 6; K (only in WORK) = 2

Trick: Match the sets, not the positions: digits are jumbled, so look for the digit missing from the other code.

Answer: (A) 2

Concept: Check every symbol: is its left neighbour a number AND its right neighbour a number?

1. Series: * £ 8 € @ 1 8 8 © & 1 1 1 7 * 4 7 2 8 $ 8 &
2. 7 * 4 ✓ and 8 $ 8 ✓; € (8 € @) and © (8 © &) fail on the right side
3. Count = 2

Trick: Scan only 'number–symbol–number' triplets; the first and last elements can never qualify.

Answer: (D) 2

Concept: Each position shifts by +8 from one term to the next (wrapping after Z).

1. 1st letters: X→F→N→V→D (+8 each)
2. 2nd letters: G→O→W→E→M (+8); 3rd letters: O→W→E→M→U (+8)
3. Next term: DMU

Trick: V+8 = D (V=22, 22+8=30−26=4). Notice the 2nd letter of a term equals the 3rd letter of the previous term.

Answer: (B) DMU