Q&A: practice questions with answers

Concept: $(l+b+h{)}^2=l^2+b^2+h^2+2(lb+bh+hl)$.

1. $l+b+h=p$, ${\text{diagonal}}^2=l^2+b^2+h^2=q^2$.
2. $p^2=q^2+\text{TSA}$ (since $\text{TSA}=2(lb+bh+hl)$).
3. $\text{TSA}=p^2-q^2$.

Trick: Test a unit cube: $l=b=h=1\to p=3,q^2=3,\text{TSA}=6=9-3=p^2-q^2$ ✓.

Answer: (B) $p^2-q^2$

Concept: For equal distances, average speed = $\frac{2xy}{x+y}$.

1. $\frac{2x(x+20)}{2x+20}=60\to x(x+20)=60(x+10)$.
2. $x^2-40x-600=0$.
3. $x=\frac{40+\sqrt{1600+2400}}{2}=\frac{40+63.25}{2}\approx 51.6$ km/h.

Trick: Average speed lies between x and x + 20 and closer to the smaller one — check 51.6: $\frac{2\times 51.6\times 71.6}{123.2}\approx 60$ ✓.

Answer: (B) 51.6

Concept: Mean proportional of a and b = $\sqrt{ab}$.

1. Numbers = $26\times \frac{4}{13}=8$ and $26\times \frac{9}{13}=18$.
2. $\sqrt{8\times 18}=\sqrt{144}=12$.

Trick: Ratio 4 : 9 → mean proportional ratio = $\sqrt{4\times 9}=6$ parts; 1 part = $\frac{26}{13}=2$ → 12.

Answer: (B) 12

Concept: SP after successive discounts $=\text{MP}\times (1-d_1)(1-d_2)$.

1. SP of X = $1200+172=$ ₹1372.
2. $\text{MP}\times 0.20\times 0.56=1372\to 0.112\text{MP}=1372\to \text{MP}=$ ₹12,250.
3. SP of Y = 12% of 12250 = ₹1470.
4. Profit = 270 → $\frac{270}{1200}\times 100=22.5\%$.

Trick: $\frac{\text{SP}(Y)}{\text{SP}(X)}=\frac{0.12}{0.112}=\frac{15}{14}\to \text{SP}(Y)=1372\times \frac{15}{14}=1470$.

Answer: (A) 22.5%

Concept: x as % of y = $\frac{x}{y}\times 100$ (base = the quantity after 'of').

1. B's marks as % of A's = $\frac{50}{80}\times 100$.
2. $=62.5\%$.

Trick: Maximum marks (110) is a distractor — ignore it. $\frac{50}{80}=\frac{5}{8}=62.5\%$.

Answer: (D) 62.5%

Concept: Common letters in two words share the common digits; the leftover letter gets the leftover digit.

1. SNOW → 1732, KNOW → 5317.
2. Common letters N, O, W ↔ common digits 1, 3, 7.
3. So S = 2 and K = 5.

Trick: Digit present in KNOW's code but not in SNOW's code = code of K → 5.

Answer: (B) 5

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

1. QK → RL: +1, +1. HB → IC: +1, +1. MG → NH: +1, +1.
2. JD → LE: +2, +1 → different.

Trick: In the three alike pairs the second cluster is just the 'next letters' of the first — JD's next letters would be KE, not LE.

Answer: (C) JD - LE

Concept: Each letter moves back by the same number of places.

1. POK → GFB: P(16) − 9 = G(7), O − 9 = F, K − 9 = B.
2. FEA → WVR: F − 9 = W, E − 9 = V, A − 9 = R (wrapping round).
3. WVR → W − 9 = N, V − 9 = M, R − 9 = I → NMI.

Trick: −9 is the same as +17; or note WVR itself appears in the given pair, so just apply the same −9 to it.

Answer: (A) WVR−NMI

Concept: Track each position separately (mod 26).

1. 1st letters: S → H → W → L → A (each +15, i.e. −11).
2. 2nd letters: J → Y → N → C → R (+15).
3. 3rd letters: T → I → X → M → B (+15).
4. Missing term = ARB.

Trick: +15 = −11 in a circular alphabet; L(12) − 11 = A(1), C(3) + 15 = R(18), M(13) − 11 = B(2).

Answer: (B) ARB

Concept: Vowel → next letter; consonant → previous letter.

1. A → B, M → L, B → A, L → K, E → F, R → Q, S → R.
2. New group: B L A K F Q R.
3. All 7 letters are different → no letter appears twice.

Trick: Watch pairs like B→A and A→B or S→R and R→Q — they swap letters but don't create repeats here.

Answer: (B) Zero

Concept: Check gaps within each cluster and between clusters.

1. PS - TW: P→S +3, T→W +3, P→T +4.
2. JM - NQ and QT - UX follow the same +3, +3, +4 pattern.
3. MP - RT: M→P +3 but R→T +2 → odd.

Trick: Just check the second cluster's gap: TW, NQ, UX are all +3; RT is +2.

Answer: (A) MP - RT

Concept: Fix the definite positions first, then place pairs.

1. Two people left of L → L at position 3 (from left).
2. Three between L and H → H at 7.
3. I is 2nd right of J: (2, 4) works; (4, 6) fails as I would be H's neighbour → J = 2, I = 4.
4. M, K adjacent (K right of M) → M = 5, K = 6; N = 1.

Trick: Once L and H are fixed, only one J–I placement survives the 'not neighbour of H' clue.

Answer: (A) N

Concept: Convert 'from top' to 'from bottom' (7 boxes: nth from top = (8 − n)th from bottom).

1. E = 5th from bottom; C = 5th from top = 3rd from bottom.
2. One box between D and C → D = 1 or 5; 5 is E → D = 1.
3. F just below B → free consecutive slots only 6, 7 → F = 6, B = 7.
4. A, G in 2 and 4; A not below G → A = 4, G = 2.
5. Between G (2) and A (4): only C → one box.

Trick: Fill fixed positions first; the B–F pair then has just one possible place.

Answer: (D) One

Concept: Vowel → next letter; consonant → previous letter.

1. T → S, H → G, A → B, L → K, E → F, R → Q, S → R.
2. New group: S G B K F Q R.
3. All letters are different → 0 letters appear twice.

Trick: Repeats can only come from 'chains' like S→R and R→Q; here the original R and S both shift, so no clash.

Answer: (B) 0

Concept: Find the gap pattern between consecutive letters.

1. PSQ: P→S +3, S→Q −2.
2. UXV: +3, −2; SVT: +3, −2.
3. RKL: R→K −7, K→L +1 → odd.

Trick: In the alike clusters the 3rd letter is just after the 1st (P-Q, U-V, S-T); in RKL it isn't.

Answer: (A) RKL

Concept: Swap the signs first, then apply BODMAS.

1. $4÷2+12\times 4-3$ becomes $4\times 2-12÷4+3$.
2. $=8-3+3$.
3. $=8$.

Trick: Rewrite the whole expression with new signs before calculating — avoids sign mix-ups.

Answer: (C) 8

Concept: Check each vowel's left and right neighbours.

1. Vowels: E(1st, no left), A (K A R ✓), U (F U C ✓), I (K I E ✗), E (I E N ✗), O (N O A ✗), A (O A K ✗).
2. Only A and U qualify → 2.

Trick: The cluster 'I E N O A' at the end has vowels next to each other — none of them can count.

Answer: (D) 2

Concept: Join the linked clues into one block, then place it.

1. G just above T, T above R, R just above H, one box (R) between T and H → block G-T-R-H (top to bottom).
2. Only one box below Y → Y = 6th from top.
3. Block cannot cover slot 6 and G is not at top → G = 2, T = 3, R = 4, H = 5.
4. Top is not B → F = 1, B = 7.
5. Above T: F and G → 2.

Trick: A 4-box block in 7 slots with slot 6 taken has only 2 options; 'G not at top' kills one.

Answer: (C) 2

Concept: Look for the pattern Even – Odd – Even.

1. 2 9 4 ✓ (positions 6–8).
2. 4 1 6 ✓ (8–10).
3. 8 5 4 ✓ (11–13).
4. Others (2 9 1, 4 3 7, 8 1 3 …) fail → total 3.

Trick: Scan only the even digits and look two steps ahead: E _ E with an odd digit in between.

Answer: (D) Three

Concept: Start from the end-fixed person.

1. No one right of K → K = 7 (from left).
2. Two between K and L → L = 4; J immediate right of L → J = 5.
3. Three between D and J → D = 1.
4. E, C adjacent (C right of E) in 2, 3 → E = 2, C = 3; F = 6.
5. Between F (6) and C (3): positions 4, 5 → two.

Trick: People between positions a and b = |a − b| − 1 = 6 − 3 − 1 = 2.

Answer: (C) Two