MBA CET 2025 Slot 1 question paper with solutions

Approach: Required SP from target profit; discount = (MP − SP)/MP × 100.

Solution:
CP = ₹225, MP = 2 × 225 = ₹450
Required SP for 56% profit = 1.56 × 225 = ₹351
Discount = 450 − 351 = ₹99
Discount % = 99/450 × 100 = 22%

Cetking shortcut: MP = 2 × CP, so SP/MP = 1.56/2 = 0.78 → discount = 22% directly.

Answer: D

Approach: Product of the 25 primes below 100 (2, 3, 5, …, 97); eliminate options by size and digit pattern.

Solution:
Primes below 100: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97 (25 primes)
Product = 2305567963945518424753102147331756070 (37 digits)
It contains 2 × 5, so it ends in 0 — and it is clearly neither 0 nor 1.

Why the other options are wrong:
A (1) and C (0): a product of primes can be neither.
B: 38 digits — one digit too many.
D: 37 digits but drops the '79' after 230556 (2305563945…).

Cetking shortcut: Size check: 2 × 3 × 5 × 7 = 210 and the primes from 11 to 97 multiply to about 1.1 × 10³⁴, so the product ≈ 2.3 × 10³⁶ — a 37-digit number; then match the digit string.

Answer: E

Approach: Prime-factorise the product and each option.

Solution:
Product = 17² × 18² × 19² = 2² × 3⁴ × 17² × 19²
Prime factors available: 2, 3, 17, 19 only
646 = 2 × 17 × 19 → all three present → divides fully

Why the other options are wrong:
A: 971 is a prime not in the product.
B: 173 is a prime not in the product.
D: 23 is a prime not in the product.
E: 5813 is odd and not divisible by 3, 17 or 19, so it has a prime factor outside the product.

Cetking shortcut: Any option containing a prime other than 2, 3, 17, 19 is out — only 646 = 2 × 17 × 19 survives.

Answer: C

Approach: Replace each factor by its remainder mod 18.

Solution:
1719 = 18 × 95 + 9 → 9
1721 → 11, 1723 → 13, 1725 → 15, 1727 → 17
Product ≡ 9 × (11 × 13 × 15 × 17) (mod 18)
11 × 13 × 15 × 17 is odd, and 9 × odd ≡ 9 (mod 18)
Remainder = 9

Cetking shortcut: One factor leaves 9 and all others are odd: 9 × odd always leaves 9 on division by 18.

Answer: A

Approach: Alligation of profit rates 8% and 18% around mean 14%.

Solution:
8% part : 18% part = (18 − 14) : (14 − 8) = 4 : 6 = 2 : 3
18% part = 3/5 × 1000 = 600 kg
Check: 400 × 8% + 600 × 18% = 32 + 108 = 140 = 14% of 1000 ✓

Cetking shortcut: Mean 14 is closer to 18 than to 8, so the 18% part is more than half — alligation fixes it at 600.

Answer: E

Approach: Form an equation from 'area unchanged', then use 2(l + b)h.

Solution:
Let breadth = b, length = b + 2
(b + 2 + 4)(b − 2) = (b + 2)b
b² + 4b − 12 = b² + 2b → 2b = 12 → b = 6, l = 8
Walls = 2(8 + 6) × 3 = 84 m²

Cetking shortcut: (l + 4)(b − 2) = lb ⇒ 4b − 2l = 8; with l = b + 2 this gives b = 6 in one line.

Answer: C

Approach: Find time for each leg, then total distance / total time.

Solution:
Leg 1: 50 km in 40 min
Leg 2: 50 km at 2 km/min = 25 min
Leg 3: 30 km at 1 km/min = 30 min
Total: 130 km in 95 min
Average speed = 130 × 60/95 = 7800/95 = 82.1 km/h

Cetking shortcut: 130/95 ≈ 1.368 km/min × 60 ≈ 82.1 — just above 82, which rules out 81.2 and 80.8.

Answer: E

Approach: Volume ∝ r²h; apply multiplying factors.

Solution:
Height +200% → h becomes 3h
Radius −50% → r becomes r/2, so r² becomes r²/4
New volume factor = 3 × 1/4 = 0.75
Volume decreases by 25%

Cetking shortcut: Factors only: (½)² × 3 = ¾ → 25% down.

Answer: B

Approach: PQ is a diameter → ∠PSQ = 90°; then use 'angles on the same arc'.

Solution:
In the figure, PQ is a diameter, so ∠PSQ = 90° (angle in a semicircle)
In ΔPSQ: ∠SPQ = 180° − 90° − 35° = 55°
∠QRS and ∠QPS stand on the same arc QS → ∠QRS = ∠QPS = 55°

Why the other options are wrong:
A (35°): that is ∠PQS itself, which stands on arc PS, not QS.
B (135°): would be an angle in the opposite segment.
D, E: no theorem gives 60° or 65° here.

Cetking shortcut: Diameter + given angle 35° → the other acute angle of the right triangle is 55°, which carries over along the same arc.

Answer: C

Approach: Independent events: P(at least 2 tails with 3 coins).

Solution:
Total outcomes = 2³ = 8
Exactly 2 tails: HTT, THT, TTH = 3
3 tails: TTT = 1
P = 4/8 = 1/2 = 50%

Cetking shortcut: By symmetry, 'at least 2 tails' and 'at least 2 heads' are complementary and equally likely → 50%.

Answer: D

Approach: Add each row (maximum marks are the same for all, so totals compare directly).

Solution:
Maya = 90+50+90+60+70+80 = 440
Arya = 100+80+88+60+80+70 = 478
Riya = 90+72+70+72+91+70 = 465
Sreya = 85+65+80+80+60+60 = 430
Roopesh = 80+55+85+95+50+90 = 455
Chandesh = 70+75+65+85+40+61 = 396
Hanish = 65+35+50+77+80+80 = 387
Highest = Arya (478)

Why the other options are wrong:
B Sreya 430, C Chandesh 396, D Maya 440, E Roopesh 455 — all below Arya's 478.

Cetking shortcut: Total only the five option names (Riya and Hanish are not options); Arya's 478 is the only total above 460.

Answer: A

Approach: Difference of row totals.

Solution:
Riya total = 465
Roopesh total = 455
Riya − Roopesh = 10

Cetking shortcut: Subject-wise differences (Riya − Roopesh): +10, +17, −15, −23, +41, −20 → sum = 10, no full totals needed.

Answer: E

Approach: Scale the Hindi difference by 200/120 = 5/3.

Solution:
Hindi: Chandesh 75, Hanish 35 (out of 120)
Chandesh = 75 × 5/3 = 125, Hanish = 35 × 5/3 = 58⅓
Difference = 66⅔ marks, listed in the options as 66 (fractional part dropped)

Cetking shortcut: Scale the difference, not each score: 40 × 5/3 = 66⅔ → 66.

Answer: E

Approach: Rates in items/hour; together for 2 hours, then Sunil alone.

Solution:
Sunil: 1000 items in 5 h → 200 items/h
Nilesh: 75% of 1000 = 750 items in 3 h → 250 items/h
Together for 2 h: (200 + 250) × 2 = 900 items
Remaining: 1300 − 900 = 400 items
Sunil alone: 400/200 = 2 h
Total time = 2 + 2 = 4 hours

Cetking shortcut: Nilesh does 2 × 250 = 500; Sunil does the other 800 at 200/h → 4 h, and Sunil works the whole time.

Answer: B

Approach: Ratio → actual weight before and after.

Solution:
Before: 4 + 9 + 7 = 20 parts = 100 kg → 1 part = 5 kg
Aamrohi before = 7 × 5 = 35 kg
After: 4 + 7 + 9 = 20 parts = 140 kg → 1 part = 7 kg
Aamrohi after = 9 × 7 = 63 kg
Weight gained = 63 − 35 = 28 kg

Why the other options are wrong:
B (63): that is her new weight, not the gain.
E (35): her original weight.
C (49), D (21): come from using the wrong ratio position (7 × 7, 7 × 3).

Cetking shortcut: Both totals are 20 parts: gain = 9 × 7 − 7 × 5 = 28.

Answer: A

Approach: Angle bisector theorem.

Solution:
BD = BC − DC = 10 − 6 = 4 cm
AB/AC = BD/DC → AB/4.2 = 4/6
AB = 4.2 × 2/3 = 2.8 cm

Cetking shortcut: AB pairs with BD (the smaller piece), so AB < AC: 4.2 × 4/6 = 2.8.

Answer: E

Approach: Fundamental counting principle, most-restricted task first.

Solution:
Task 2 → person 3 or person 4: 2 ways
Task 1 → not person 1, not person 2, not the person who took Task 2: 6 − 3 = 3 ways
Remaining 4 tasks → remaining 4 persons: 4! = 24 ways
Total = 2 × 3 × 24 = 144

Why the other options are wrong:
B (192): forgets that Task 2's person is used up (2 × 4 × 24).
A (72): counts only one choice for Task 2.
C (360), E (256): no valid counting gives these.

Answer: D

Approach: Same distance → speeds are inversely proportional to times.

Solution:
Downstream speed = 5 + x, upstream speed = 5 − x
t_up = 3 × t_down → (5 + x) = 3(5 − x)
5 + x = 15 − 3x → 4x = 10 → x = 2.5 km/h

Cetking shortcut: Down : Up speed = 3 : 1 → boat = (3 + 1)/2 = 2 units, stream = (3 − 1)/2 = 1 unit; 2 units = 5 → stream = 2.5.

Answer: B

Approach: Share ∝ sheep × months.

Solution:
Umla 24 × 3 = 72, Vimla 10 × 5 = 50, Wilma 35 × 4 = 140, Xema 21 × 3 = 63
Total = 72 + 50 + 140 + 63 = 325 units
Umla's 72 units = ₹720 → 1 unit = ₹10
Total rent = 325 × 10 = ₹3250

Cetking shortcut: ₹720/72 = ₹10 per sheep-month, so the total must be 325 × 10.

Answer: D

Approach: Equate volumes in cm.

Solution:
Wire: r = 2 mm = 0.2 cm, length = 243 m = 24300 cm
Volume = π × 0.2² × 24300 = 972π cm³
Sphere: (4/3)πR³ = 972π → R³ = 729 → R = 9 cm
Diameter = 18 cm

Cetking shortcut: R³ = (3/4) × 0.04 × 24300 = 729 — a perfect cube, confirming R = 9.

Answer: C

Approach: 1% of 1500 = 15 units; add the percentages of each group and convert.

Solution:
Statement A: A + G = 16% + 12% = 28% = 420 → not more than 500 → false
Statement B: B + F = 11% + 5% = 16% = 240 → less than 250 → true
Statement C: C + E = 21% + 26% = 47% = 705 → true
Statement D: D + G = 9% + 12% = 21% = 315 → not less than 200 → false
Statement E: E + F + G = 26% + 5% + 12% = 43% = 645 ≠ 545 → false
Only B and C are true

Why the other options are wrong:
A, B, E: each includes statement A (420, not more than 500).
C: includes statement E (645, not 545).

Cetking shortcut: Convert thresholds to %: 500 = 33.3%, 250 = 16.7%, 705 = 47%, 200 = 13.3%, 545 = 36.3% — compare percentages directly.

Answer: D

Approach: Valid votes first, then loser's share.

Solution:
Valid votes = 85% of 15200 = 12920
Loser's share = 100% − 55% = 45%
Loser's votes = 45% of 12920 = 5814

Cetking shortcut: 0.85 × 0.45 = 0.3825; 0.3825 × 15200 = 5814.

Answer: E

Approach: Work in % of total workforce; the total cancels.

Solution:
SA = 12%, M:F = 5:3 → male = 12 × 5/8 = 7.5%
MA = 21%, M:F = 3:4 → male = 21 × 3/7 = 9%
Male in SA + MA = 16.5%
PT total = 19%
Required % = 16.5/19 × 100 = 86.8% ≈ 87%

Cetking shortcut: Never multiply by 6,000,000 for a percent-of question — 16.5/19 is just under 87%.

Answer: B

Approach: Female shares in % of total, then convert.

Solution:
SA female = 12% × 3/8 = 4.5%
MA female = 21% × 4/7 = 12%
SA + MA female = 16.5%
PT female = 19% × 3/8 = 7.125%
Difference = 16.5 − 7.125 = 9.375% of 6,000,000 = 5,62,500 ≈ 5,60,000

Why the other options are wrong:
B, C, D: 510000, 570000 and 550000 are all further from 562500.
E (1320000): adds the two groups instead of subtracting.

Cetking shortcut: 1% = 60,000; 9.375 × 60,000 = 5,62,500.

Answer: A

Approach: Prorate the whole annual package.

Solution:
Let gift = G. Annual pay = 90 + G
9 months' pay = (3/4)(90 + G)
(3/4)(90 + G) = 65 + G
67.5 + 0.75G = 65 + G → 0.25G = 2.5 → G = ₹10

Cetking shortcut: 3 months' pay = (90 + G) − (65 + G) = 25 → annual = 100 → G = 100 − 90 = 10.

Answer: C

Approach: Find one solution, write the general form, count valid t.

Solution:
Particular solution: x = 4, y = −3 (44 − 45 = −1)
General: x = 4 + 15t, y = −3 − 11t
x < 100 → 15t < 96 → t ≤ 6
y < 100 → −11t < 103 → t ≥ −9
t = −9, −8, …, 6 → 16 values

Cetking shortcut: Count t-values as 6 − (−9) + 1 = 16; watch the off-by-one traps (15 and 17).

Answer: D

Approach: Area ratio → radius ratio → volume ratio.

Solution:
SA of B = SA of A + 300% = 4 × SA of A
r_B/r_A = √4 = 2
V_B/V_A = 2³ = 8
V_A is lower than V_B by (8 − 1)/8 = 7/8 = 87.5%

Cetking shortcut: 'k% lower than B' uses B as base: 1 − 1/8 = 87.5%.

Answer: C

Approach: Net water rate → time to sink → required speed.

Solution:
Leak: 2.5 t in 6.5 min → 2.5 × 60/6.5 = 300/13 ≈ 23.08 t/h
Pump: 15 t/h → net inflow = 300/13 − 15 = 105/13 ≈ 8.08 t/h
Time to take in 68 t = 68 × 13/105 ≈ 8.42 h
Speed = 156/8.42 ≈ 18.5 km/h → 18 km/h

Why the other options are wrong:
A (25), E (60): far above the required rate.
B (10), D (15): the ship would sink before reaching the shore.

Cetking shortcut: 156 km ÷ about 8.4 h is about 18.5 km/h; 18 is the only option in that range.

Answer: C

Approach: Even numbers 100–200, remove multiples of 14 and 18, add back multiples of 126.

Solution:
Even numbers from 100 to 200 inclusive: (200 − 100)/2 + 1 = 51
Multiples of 14: 112 to 196 → 14 × 8 to 14 × 14 → 7
Multiples of 18: 108 to 198 → 18 × 6 to 18 × 11 → 6
Multiples of LCM(14, 18) = 126: only 126 → 1
Neither = 51 − 7 − 6 + 1 = 39

Cetking shortcut: Only even multiples matter: count with 14 and 18, not 7 and 9.

Answer: C

Approach: Fix total tickets, push them into ₹250 shows, then subtract all costs.

Solution:
Cost = 10,000 + 3 × 5,000 = ₹25,000
Tickets sold = 80% of (3 × 200) = 480
Max revenue: 200 + 200 tickets at ₹250 and 80 at ₹200
Revenue = 50,000 + 50,000 + 16,000 = ₹1,16,000
Profit = 1,16,000 − 25,000 = ₹91,000

Why the other options are wrong:
B (116000): that is revenue, not profit.
A (120000): puts all 480 tickets in ₹250 shows, ignoring the 200-seat limit per show.
E (87500): near the equal-occupancy profit (₹87,000), not the maximum.
D (95000): that is 120000 − 25000, built on the same seat-limit mistake.

Cetking shortcut: Equal split gives 160 × (250 + 250 + 200) = 1,12,000; moving 80 tickets from ₹200 to ₹250 shows adds 80 × 50 = 4,000 → 1,16,000 − 25,000 = 91,000.

Answer: C

Approach: Prime-factorise and read each prime's exponent in x.

Solution:
990 = 2 × 3² × 5 × 11; 6930 = 2 × 3² × 5 × 7 × 11
LCM has 7 and 990 has no 7 → x must contain 7
x divides 6930 → at most 2¹, 3², 5¹, 7¹, 11¹
550 = 2 × 5² × 11; GCD = 110 = 2 × 5 × 11 → x contains 2, 5 and 11
So x = 2 × 5 × 7 × 11 × 3ᵃ, a = 0, 1 or 2
Least x = 770 → digit sum = 7 + 7 + 0 = 14

Cetking shortcut: Least x = product of the forced primes 2 × 5 × 7 × 11 = 770.

Answer: B

Approach: Sides (a, a, b) with some pair summing to 12.

Solution:
Case 1: a + a = 12 → a = 6, base b < 12 → b = 1 to 11 → 11 triangles
Case 2: a + b = 12 (b ≠ a), need b < 2a → 12 − a < 2a → a > 4
a = 5, 7, 8, 9, 10, 11 (a = 6 gives b = 6, already counted) → 6 triangles
Total = 11 + 6 = 17

Why the other options are wrong:
A (16): drops the equilateral 6-6-6.
C (24), E (23): ignore the triangle inequality.
D (2): counts only one side pattern.

Answer: B

Approach: Use the 45° angle for the distance, the 60° angle for the total height.

Solution:
Building = 100 m, angle 45° → distance x = 100 m
Top of student: tan 60° = (100 + h)/100
100 + h = 100√3 = 173.2
h = 73.2 m

Cetking shortcut: h = 100(√3 − 1) = 100 × 0.732 = 73.2.

Answer: D

Approach: Each pipe fills 1/3; use 'A + B together fill in t' to find t.

Solution:
A is open all t minutes and fills 1/3 → rate of A = 1/(3t)
B is open 10 minutes and fills 1/3 → rate of B = 1/30
A + B fill the tank in t: 1/(3t) + 1/30 = 1/t
1/30 = 2/(3t) → t = 20 minutes
C is open from minute 12 to 20 = 8 minutes and fills 1/3
C alone = 8 × 3 = 24 minutes

Cetking shortcut: Once t = 20, C's window is 20 − 12 = 8 min for one-third → 24 min alone.

Answer: B

Approach: Place a, b, c, d on a number line, then fit e.

Solution:
Take a = 0, b = 4k (k ≠ 0). c = mean of a, b = 2k; d = mean of b, c = 3k
0, 2k, 3k, 4k plus e must be 5 equally spaced values → e = k (common gap k)
So a = 0, e = k, c = 2k, d = 3k, b = 4k
i. Average = (0 + k + 2k + 3k + 4k)/5 = 2k = c → true
ii. Avg(d, e) = 2k, Avg(a, b) = 2k → 'not greater' → true
iii. Avg(b, c) = 3k vs Avg(a, d) = 1.5k → greater only if k > 0; for a decreasing AP (k < 0) it fails → not necessarily true

Why the other options are wrong:
B, C, E: treat iii as always true, but it reverses for a decreasing AP.
D: misses statement i, which holds for every k.

Cetking shortcut: Equalities and 'not greater' (here an equality) survive a sign flip of k; a strict 'greater' comparison does not.

Answer: A

Approach: Inequality from the fridge limit + whole-number condition.

Solution:
Vijay buys x; good eggs = x − 2
His fridge: 5/9(x − 2); to mother: 4/9(x − 2)
Her fridge: 4/9(x − 2) − 2 ≤ 5 → x − 2 ≤ 15.75 → x ≤ 17.75
Eggs are whole → x − 2 must be a multiple of 9 → x − 2 = 9 (18 exceeds 15.75)
x = 11: 9 good eggs, 5 in his fridge, 4 to mother, she cooks 2, 2 go in her fridge ✓

Why the other options are wrong:
A (17): 15 good eggs → 5/9 × 15 = 8⅓ eggs — impossible.
C (16), D (23): x − 2 = 14 and 21 — not multiples of 9.
E (20): 18 good eggs → 8 to mother → 6 in her fridge, more than 5.

Cetking shortcut: Check options for (x − 2) divisible by 9: only 11 and 20 qualify; 20 puts 6 eggs in her fridge, so 11.

Answer: B

Approach: Greedy: three pockets with the smallest count, four with the next distinct values.

Solution:
3 pockets × 1 coin = 3
Remaining 4 pockets must be distinct and ≠ 1 → 2, 3, 4, 5 = 14
Total = 3 + 14 = 17
Check: 3 × 2 + (1, 3, 4, 5) = 19 and 3 × 3 + (1, 2, 4, 5) = 21 → larger

Why the other options are wrong:
B (7): needs all 7 pockets with 1 coin — breaks 'at most 3 same'.
D (13): impossible — the minimum arrangement already needs 17.
C (28): all distinct 1–7, not the minimum.
E (22): not minimal.

Cetking shortcut: Put the repeat on the smallest value: 1 + 1 + 1 + 2 + 3 + 4 + 5 = 17.

Answer: A

Approach: Test each statement separately for a unique rent R.

Solution:
Statement 1: R/4 = 375 → R = $1500 → sufficient
Statement 2: R/3 − R/4 = 125 → R/12 = 125 → R = $1500 → sufficient
Each statement alone is sufficient

Why the other options are wrong:
A, B: ignore that the other statement also works alone.
C: neither statement needs the other.
E: both clearly give a value.

Cetking shortcut: Statement 2 is one equation in one unknown — sufficient without solving.

Answer: D

Approach: Convert k seconds to minutes and multiply by the rate.

Solution:
Rate = x revolutions per minute = x/60 revolutions per second
In k seconds: (x/60) × k = kx/60

Why the other options are wrong:
B, D: divide by k — revolutions must grow with time.
C: forgets the minute-to-second conversion.
E: converts the wrong way (× 60).

Cetking shortcut: Plug numbers: x = 60 rpm, k = 1 s → 1 revolution; only kx/60 gives 1.

Answer: A

Approach: Simple interest equation for 2 years.

Solution:
Let A = x, B = 13900 − x
2 × [0.14x + 0.11(13900 − x)] = 3508
0.14x + 0.11(13900 − x) = 1754
0.03x + 1529 = 1754 → x = 7500
Scheme B = 13900 − 7500 = ₹6400

Why the other options are wrong:
D (7500): that is the Scheme A amount.

Cetking shortcut: If all ₹13,900 earned 14%, yearly SI = ₹1946; actual = ₹1754; shortfall 192 = 3% of B → B = 192/0.03 = ₹6400.

Answer: A

Approach: Identify the common ratio.

Solution:
2 → 1 → ½ → ¼: each term is half the previous one
Next term = ¼ × ½ = 1/8

Why the other options are wrong:
A (1/16), D (1/32): skip one or two terms.
B (2/8): equals 1/4, the current term.
C (1/3): breaks the halving pattern.

Answer: E

Approach: Two exclusive cases: only Alka or only Alkit.

Solution:
P(Alka) = 1/7 → P(not Alka) = 6/7; P(Alkit) = 1/5 → P(not Alkit) = 4/5
Only Alka: 1/7 × 4/5 = 4/35
Only Alkit: 6/7 × 1/5 = 6/35
P(only one) = 4/35 + 6/35 = 10/35 = 2/7

Why the other options are wrong:
A (1/35): both selected.
D (4/35): only Alka's case.
B (8/15), C (4/5): mix up selection and rejection probabilities.

Cetking shortcut: P(exactly one) = p + q − 2pq = 1/7 + 1/5 − 2/35 = 10/35 = 2/7.

Answer: E

Approach: Decode symbols, then BODMAS.

Solution:
X → +, ÷ → −, − → ÷, + → ×
20 X 8 ÷ 8 − 4 + 2 becomes 20 + 8 − 8 ÷ 4 × 2
8 ÷ 4 × 2 = 2 × 2 = 4
20 + 8 − 4 = 24

Cetking shortcut: Write the decoded expression on paper before calculating — most errors come from decoding in your head.

Answer: C

Approach: Convert each pie share to units with its own total.

Solution:
Dealer D total cars = 15% of 56000 = 8400
SUVs of C, F, G = (7 + 8 + 6)% = 21% of 32000 = 6720
% more = (8400 − 6720)/6720 × 100 = 1680/6720 × 100 = 25%

Why the other options are wrong:
A, B, C, D: come from applying SUV percentages to 56000, or using the wrong base for 'percentage more'.

Cetking shortcut: 8400/6720 = 1.25 → 25% more.

Answer: E

Approach: Convert to units, then simplify.

Solution:
SUV of A + B = (16 + 16)% of 32000 = 10240
All cars of C + F = (14 + 8)% of 56000 = 12320
Ratio = 10240 : 12320 = 64 : 77 (divide by 160)

Why the other options are wrong:
A (54:77), B (64:73), E (64:79): one term altered.
D (1:4.33): no such relation between the groups.

Cetking shortcut: (32 × 32000) : (22 × 56000) = (32 × 4) : (22 × 7) = 128 : 154 = 64 : 77.

Answer: C

Approach: MUV per dealer = total cars − SUVs, then average.

Solution:
A: 18% × 56000 − 16% × 32000 = 10080 − 5120 = 4960
D: 8400 − 4160 = 4240
E: 11200 − 7040 = 4160
F: 4480 − 2560 = 1920
H: 5040 − 3840 = 1200
Sum = 16480; average = 16480/5 = 3296

Cetking shortcut: Sum directly: total share (18+15+20+8+9)% = 70% × 56000 = 39200; SUV share (16+13+22+8+12)% = 71% × 32000 = 22720; (39200 − 22720)/5 = 3296.

Answer: D

Approach: Unit digits only.

Solution:
6374¹⁷⁹³: base ends in 4, odd power → ends in 4 (even)
625³¹⁷: ends in 5
341⁴⁹¹: ends in 1
Product: 4 × 5 × 1 → unit digit 0

Cetking shortcut: Even × (…5) always ends in 0 — no cyclicity needed.

Answer: A

Approach: Complement of 'both odd'.

Solution:
P(both odd) = 3/6 × 3/6 = 1/4
P(product even) = 1 − 1/4 = 3/4 (27 of 36 outcomes)

Why the other options are wrong:
A (1/3), B (2/3), C (5/16), D (3/8): none matches 27/36.

Answer: E

Approach: Match with (4a − q)².

Solution:
16a² = (4a)², so p = 4a
2 × 4a × q = 12a → q = 3/2
Add q² = 9/4 → 16a² − 12a + 9/4 = (4a − 3/2)²

Cetking shortcut: Number to add = (b²)/(4a) = 144/64 = 9/4.

Answer: A

Approach: Sum of triangular numbers = n(n+1)(n+2)/6.

Solution:
Layers: 1, 3, 6, 10, … = n(n+1)/2
Total = n(n+1)(n+2)/6 = 8436
n(n+1)(n+2) = 50616
Cube root of 50616 ≈ 37 → try 36 × 37 × 38 = 50616 ✓
n = 36 layers

Why the other options are wrong:
A (32), D (34): give 5984 and 7140 balls.
E (24): 2600 balls.
C (96): far too many.

Cetking shortcut: n³/6 ≈ 8436 → n³ ≈ 50600 → n ≈ 37 → the middle of three consecutive numbers is 37, so n = 36.

Answer: B

Approach: Letter-gap pattern inside each 4-letter group (positions A=1 … Z=26, wrapping after Z).

Solution: Check the gaps between consecutive letters:
VABK: V(22)→A(1) = +5, A→B = +1, B(2)→K(11) = +9
NSTC: N→S = +5, S→T = +1, T(20)→C(3) = +9 (wrap)
LQRA: L→Q = +5, Q→R = +1, R(18)→A(1) = +9 (wrap)
EJKT: E→J = +5, J→K = +1, K(11)→T(20) = +9
QVWE: Q→V = +5, V→W = +1, W(23)→E(5) = +8 (wrap); the pattern needs W+9 = F.
So QVWE breaks the +5, +1, +9 rule.

Why the other options are wrong:
A, B, C, E: all follow +5, +1, +9 exactly.

Cetking shortcut: Middle pair is consecutive in every option, so only the jumps 1st→2nd and 3rd→4th can differ; check the last jump first.

Answer: D

Approach: Main idea = what most of the passage is spent on.

Solution: Only the first two sentences mention smokejumpers' talks; the rest gives specific rules on where and how detectors must be placed (each floor, outside sleeping areas, hallway, 4–12 inches from ceiling, not near windows, doors, kitchens or garages). So the author's main focus is the proper installation of home smoke detectors.

Why the other options are wrong:
A: smokejumpers' errands are only the setting.
B: dead-air space is one detail inside the placement advice.
C: firefighting by smokejumpers is not discussed.
D: the passage never explains how detectors stop fires; it says only that they halve the risk of dying.

Cetking shortcut: In 'main focus' questions, pick the option that covers the bulk of the sentences, not the opening line.

Answer: E

Approach: Vocabulary in context: match the given word's meaning to a word used in the passage.

Solution: A lancet (lancet window) is a tall, narrow window with a pointed arch. Of the passage words given, only 'window' (in 'near windows, exterior doors') shares this meaning.

Why the other options are wrong:
A Twelve, C Inch: measurements from the passage.
D Turbulent: describes hot air.
E Smokejumper: a firefighter.

Answer: B

Approach: Specific detail: find the line about smokejumpers' duties.

Solution: The opening line says smokejumpers 'are often asked to address organizations and public groups about the importance of fire protection'. The option 'Address corporate companies about the importance of preventing fires' restates this duty.

Why the other options are wrong:
A: checking homes is never mentioned.
B: they advise on installation; they do not install.
C: developing safety programmes is not stated.
D: this option is a word-for-word duplicate of E in the printed paper; E is the keyed choice.

Cetking shortcut: Duty questions are answered from the first one or two lines of the passage.

Answer: E

Approach: Specific detail: what the passage says must NOT happen.

Solution: The last line says detectors should not be put in kitchens or garages 'where cooking and gas fumes are likely to cause bogus alarms'. So bogus (false) alarms are what must be avoided.

Why the other options are wrong:
A: placing high on a wall (4–12 inches from the ceiling) is recommended.
B: a hallway is called 'a good site'.
D: making people aware of fire hazards is the smokejumpers' job.
E: the passage warns about dead-air space; it does not recommend firefighting provisions for it.

Cetking shortcut: Look for negative words in the passage ('should not', 'nor').

Answer: C

Approach: Vocabulary in context: antonym of the given word, chosen from passage words.

Solution: Nescience = lack of knowledge, ignorance. Its opposite is knowing or finding out. 'Detection' (discovering the presence of something) is the only option tied to knowledge or discovery.

Why the other options are wrong:
A Deterrence: preventing something.
B Turbulent: agitated.
C Bouncing: rebounding.
D Smokejumper: a person.

Cetking shortcut: Ne + science (Latin nescire, 'not to know') = not knowing, so look for a word about finding out or knowing.

Answer: E

Approach: Spelling: break a long word into its parts.

Solution: anti + dis + establish + ment + arian + ism = Antidisestablishmentarianism (opposition to removing the Church of England's status as the state church).

Why the other options are wrong:
A: 'Ante' (before) in place of 'Anti' (against).
C: 'es' missing (disestablish).
D: 'ment' missing (establishmentarian).
E: 'ian' missing (establishmentarianism).

Cetking shortcut: Check each building block in order; every wrong option drops or changes one block.

Answer: B

Approach: Spelling: break the medical term into its Greek parts.

Solution: oto (ear) + rhino (nose) + laryngo (throat) + logist = Otorhinolaryngologist, an ear, nose and throat (ENT) specialist.

Why the other options are wrong:
A: 'Otrhono…' garbles oto-rhino.
B: 'rhinily-rangologist' garbles both parts.
C: 'laryon' is missing 'g'.
E: 'rhono' in place of 'rhino'.

Cetking shortcut: Ear-nose-throat = oto-rhino-laryngo, in that order.

Answer: D

Approach: Spelling: Latin parts sesqui + ped + alian.

Solution: sesqui (one and a half) + ped (foot) + alian = Sesquipedalian: very long words, or someone who uses them.

Why the other options are wrong:
B: 'pad' in place of 'ped'.
C: extra 'se' (sesquisepedalian).
D: 'pedilian' in place of 'pedalian'.
E: 'aliain' in place of 'alian'.

Answer: A

Approach: Spelling: pulchritude + -inous.

Solution: From pulchritude (beauty) + -inous comes Pulchritudinous, meaning beautiful.

Why the other options are wrong:
A: letters jumbled ('tdun' for 'tudin').
B: extra letters ('chiru' for 'chri').
D: 'inuous' in place of 'inous'.
E: extra 'i' (pulchiritudinous).

Cetking shortcut: Write the noun 'pulchritude' first, then add -inous.

Answer: C

Approach: Spelling: curmudg + eon.

Solution: Curmudgeon means a bad-tempered, grumpy person, often an old one. Spelling: c-u-r-m-u-d-g-e-o-n.

Why the other options are wrong:
A: 'Car' in place of 'Cur'.
B: 'madg' in place of 'mudg'.
C: 'j' in place of 'g'.
D: 'e' dropped (curmudgon).

Cetking shortcut: The word keeps the 'dge' of 'grudge': cur-mudge-on.

Answer: E

Approach: Odd sentence out: find the four sentences that share one specific thread.

Solution: Sentences 4 (muscle proteins travel to the brain and drive thought), 5 (only recently have neuroscientists described these factors), 1 (they have begun studying exercise's effect on genes inside brain cells) and 2 (even there they find the body's influence on the mind) form one chain on the neuroscience of exercise and the brain. Sentence 3 is a general remark about modern sedentary, screen-driven life. It does not feed the pronoun links of the chain ('these factors', 'even there'), so it is the odd one.

Why the other options are wrong:
A, B: 'Even there' in B points back to 'within brain cells… genes' in A, so they are a pair.
D: introduces the proteins that 'these factors' in E refers to.
E: links D to A ('neuroscientists have begun…').

Cetking shortcut: Sentences joined by pronoun or demonstrative links ('these factors', 'even there') stay; the one with no link out is usually the odd one.

Answer: C

Approach: 'Although' already shows contrast, so it takes no 'but', 'yet' or 'that' in the main clause.

Solution: 'Although he had failed in Mathematics' is a complete subordinate clause; the main clause 'he received a Degree in Engineering' follows after a comma with no extra connector. The past perfect 'had failed' correctly marks the earlier action. Sentence D is correct.

Why the other options are wrong:
A: 'that he received…' turns the main clause into a fragment.
B: 'Although, having failed…' joins a participle phrase to 'although'; 'the Engineering Degree' is also unidiomatic.
C: uses 'Although… but' (double connector) and 'that' as well.
E: 'failed in that Mathematics' is wrong, and 'Although… but' is a double connector.

Cetking shortcut: Strike any option with 'although/though… but'; that removes most choices at once.

Answer: D

Approach: Sentence rearrangement: an inverted question order ('did he realize') + a 'that'-clause + an agent phrase + a relative clause.

Solution: P 'did he realize' (inverted order, as after 'Only later…' or 'Never…') → R 'that he had been' → Q 'helped by a man' → S '(whom) he never respected'. The full sentence: '…did he realize that he had been helped by a man he never respected.'

Why the other options are wrong:
A PQRS: 'helped by a man' cannot come before 'that he had been'.
B SRQP, C SQRP: start with 'he never respected', which has nothing to attach to and leaves 'did he realize' dangling at the end.
D PSRQ: 'did he realize he never respected that he had been' makes no sense.

Cetking shortcut: 'had been' needs a past participle next, so R→Q is a fixed pair; only E has it after P.

Answer: E

Approach: Antonym: intrepid = fearless.

Solution: Intrepid = fearless, bold. Its opposite is cowardly.

Why the other options are wrong:
A Amiable: friendly.
B Fiendish: cruel, wicked.
D Tranquil: calm.
E Tepid: lukewarm, unenthusiastic.

Cetking shortcut: in- (not) + trepid (alarmed, as in 'trepidation') = not afraid, so the opposite is 'afraid' = cowardly.

Answer: C

Approach: Para jumble: general news → specific case → explain the disease → its cause → benefit of correcting it.

Solution: 1 (scientists edited genes in a human embryo for the first time) is the opener. 5 gives the specific result: a fix for the mutation behind hypertrophic cardiomyopathy. 2 'The cardiac disease…' describes that disease. 4 'It is caused by a mutation…' gives its cause. 3 'Correcting the mutation…' ends with the benefit. Order: 1-5-2-4-3.

Why the other options are wrong:
A 12345: 2 ('The cardiac disease') cannot come before the disease is named in 5.
B 32514, D 51243: 3 and 5 cannot open; 5 needs 1's context.
E 12435: same problem as A.

Cetking shortcut: 'The cardiac disease' (definite article) must follow the sentence that names it (5), so 5-2 is a fixed pair; only C has 1-5-2.

Answer: C

Approach: Para jumble: general claim → problem → consequence → extension → conclusion ('hence').

Solution: 2 'The stories we tell reflect the world around us' is the general opener. 5: once we capture a story, the world has already changed. 3: so we cannot help retelling stories, since they are never quite right for our time. 4 'And even if we get them right…' extends 3. 1, with 'hence', concludes about retelling Indian stories in modern India. Order: 2-5-3-4-1.

Why the other options are wrong:
A 43215: 4 starts with 'And even if', so it cannot open.
B 23541: 3 needs 5's idea of a changed world before it.
D 25314: 1 ('hence') must conclude, not 4.
E 13542: 1 has 'hence', so it cannot open.

Cetking shortcut: 'Hence' marks the concluding sentence and 'And even if…' must follow its partner (3-4); only C has both.

Answer: C

Approach: Purpose of punctuation: why the author puts quotation marks around 'pure'.

Solution: The author writes 'the errors of "pure" art and of commercialized art are identical' and then says 'True art, on the other hand, is not merely a matter of pleasure'. The quotation marks show that so-called pure art (Art for Art's Sake) is not genuinely pure, which sets it against the true art Shaw values.

Why the other options are wrong:
B: the quotes add a negative tone; they do not strip one away.
C: the passage equates the errors of the two, not the words themselves.
D, E: the quotes are sceptical, not emphatic or approving.

Cetking shortcut: Quotation marks around an approving word usually mean 'so-called'; look for the contrasting word nearby ('True').

Answer: A

Approach: Specific detail: what made Shaw's didacticism 'take an unusual turn'.

Solution: The passage says Shaw's didacticism is unusual 'in its application to the history of arts': ideas go out of date, so even great works date, and each renewal of philosophy gets its full form from the first great artist after it. Judging art by its place in history is historicism.

Why the other options are wrong:
A Moralism: moral purpose is ordinary didacticism, not the unusual part.
B Hedonism: Shaw rejects art for pleasure.
C Dogmatism: the passage stresses his modesty.
D Idealism: not discussed.

Cetking shortcut: Find the line right after 'unusual turn'; it says 'history of arts', which points to historicism.

Answer: E

Approach: Purpose of an analogy.

Solution: The analogy sits right after 'True art… is not merely a matter of pleasure. It may be unpleasant.' Even under laughing gas (some pleasure), the tooth is still pulled (the real, possibly painful, purpose). So art's purpose goes beyond pleasure.

Why the other options are wrong:
A: a lasting effect is not the point of the analogy.
B: dating of art belongs to the next paragraph.
C: extracting moral relevance from an epoch is not stated.
E: 'may be unpleasant' does not mean art is always painful.

Cetking shortcut: Read the sentence just before the analogy; it states the point the analogy illustrates.

Answer: D

Approach: Double blank: the second half ('networks… mask signals of maladjustment') tells us the tendency makes something harder.

Solution: If friendships outside college mask the warning signs, then spotting the problem early becomes harder. Blank 1 needs a word for spotting (identification or detection), and blank 2 a word for 'made harder' (complicated). Only 'identification, complicated' fits both blanks.

Why the other options are wrong:
B treatment, compounded: masking signals hampers spotting, not treatment, and 'compounded' needs a problem to worsen.
C prevention, helped: masking does not help anything.
D prevention, compounded: you cannot 'prevent' through signals that are hidden; the clue is about noticing.
E detection, facilitated: blank 1 fits, but masking cannot facilitate detection.

Cetking shortcut: Fill the easier blank first: 'mask signals', so blank 2 must be negative. That leaves B, D and A; then 'signals' points to identification.

Answer: A

Approach: Double blank with contrast: 'instead of' sets the first idea (bad method) against the second (good result).

Solution: Companies that keep dangling or withholding rewards to push performance encourage negative behaviour. 'Instead of' needs a positive verb for blank 2: fostering (encouraging the growth of) a genuine interest. Only B gives this pair.

Why the other options are wrong:
A giving, seeking: 'seeking a genuine interest' does not fit with companies.
C conferring, fostering: blank 2 fits, but 'conferring' (granting) does not fit the manipulative tactic described.
D conferring, discrediting: blank 2 must be positive.
E bestowing, discouraging: blank 2 must be positive.

Cetking shortcut: After 'instead of', blank 2 must be positive; that leaves B and C, and 'encourage negative behavior' points to withholding.

Answer: B

Approach: EXCEPT question: mark each option as supported or not for the encumbered self (the self bound by inherited ties).

Solution: Sandel's encumbered self is defined by attachments 'antecedent to choice': history, family, city, nation, loyalties, solidarity. A self that is 'the product of independent accomplishment' is the unencumbered self the author himself wants. So D is not a trait of the encumbered self.

Why the other options are wrong:
A: 'as bearers of that history', so it is influenced by history.
B: family and community ties formed early are such attachments.
C: it keeps loyalties inherited from the past.
E: racial kinship is the kind of inherited tie the author rejects (he eschews it in order to be unencumbered).

Cetking shortcut: Encumbered = inherited ties; unencumbered = self-made. Any option about 'independent' or 'self-made' belongs to the other side.

Answer: D

Approach: Purpose question: why the author defines what should properly be an individual's object of pride.

Solution: The author opens by rejecting racial kinship ('I eschew the notion of racial kinship') and claiming the unencumbered self. Saying that pride should rest on one's own accomplishments, not on one's race, prepares the ground for his case against racial solidarity and race pride.

Why the other options are wrong:
A: he openly claims the unencumbered self; he is not trying to hide that yearning.
B: 'gain professional sops' has no basis in the passage.
C: he supports the unencumbered self; he does not undermine it.
D: the Frederick Douglass reference is outside the given text, and the definition serves his own thesis.

Cetking shortcut: Purpose questions about a definition usually point to the author's main thesis; here, rejecting racial kinship.

Answer: E

Approach: Primary focus: the option that covers both sides the passage discusses.

Solution: The passage sets two views of identity against each other: Sandel's (the self is constituted by family, history, nation, culture) and the author's (the unencumbered self, judged by what one achieves). So the focus is identity formation as self-definition by family, history and culture versus by independent accomplishment.

Why the other options are wrong:
A: too narrow; rejecting kinship is only the author's side.
B: success stories are not the topic.
C: 'individual versus group consciousness' is vague and leaves out the identity framing.
D: covers Sandel's critique but misses the author's side, accomplishment.

Cetking shortcut: For 'whole passage' questions, choose the option that names both sides of the debate.

Answer: E

Approach: Specific detail from the passage's argument about cell phones.

Solution: The passage contrasts the landline days, when 'there was no expectation of being able to reach us at all times', with today, when phones are everywhere and this 'has created an implicit' expectation of always being reachable. So people answer even in meetings because carrying a phone creates the expectation that you will respond.

Why the other options are wrong:
A: looking busy is not given as the reason.
B: the passage criticises multitasking; it does not say phones let people multitask in meetings.
C: convenience of messaging is not discussed.
E: others' tolerance of calls is not mentioned.

Cetking shortcut: Find the sentence that contrasts 'then' and 'now'; the change it names (an expectation) is the reason.

Answer: D

Approach: Next-paragraph question: continue the passage's last line.

Solution: The author argues that US history textbooks distort the place of Native Americans and ends: 'an ongoing, careful review of how school textbooks depict Native Americans is certainly warranted'. The natural next step is how to carry out that review, so specific ways to evaluate the biases of the textbooks.

Why the other options are wrong:
A: the teacher's role is never raised.
B: teaching methods go beyond the textbook focus.
C: parents are not part of the argument.
E: European immigrants' contributions would change the topic.

Cetking shortcut: The last sentence of the passage almost always sets up the next paragraph.

Answer: D

Approach: Para jumble: claim → reason → question → second reason ('Besides') → consequence.

Solution: C states the claim: 'war against terror' is a misnomer. E gives the reason: terror is a phenomenon, not an entity. B follows: so how can a conflict against a phenomenon be called war? D adds a second reason with 'Besides': war has a legal meaning. A ends with an example of that legal problem: captured combatants in cages, 'dubious legality'. Order: C-E-B-D-A.

Why the other options are wrong:
A ECDBA, C EBCAD: E cannot open; it explains a claim not yet made.
B BECDA: B's 'then' needs an earlier reason.
D CDEAB: 'Besides' (D) must come after the first reason (E).

Cetking shortcut: 'Besides' adds a second point, so it needs the first point (E-B) before it; E-B is a fixed pair.

Answer: E

Approach: Para jumble: history → problem → design response → better alternative → why it was not adopted.

Solution: C introduces the QWERTY design and its aim, to solve a mechanical problem. E describes the problem: type bars jammed. A begins 'To avoid this' and gives QWERTY's solution, slower by design. B introduces Dvorak's faster layout. D begins 'Yet', so Dvorak was never adopted. Order: C-E-A-B-D.

Why the other options are wrong:
B CAEBD: 'this' in A needs the jamming problem (E) first.
C CBAED: B (Dvorak) before the QWERTY story ends.
D ECABD: E opens without context.
E ABDCE: A's 'this' has nothing to refer to.

Cetking shortcut: 'To avoid this' (A) must follow the problem (E), and 'Yet the Dvorak layout' (D) must follow B; only CEABD keeps both pairs.

Answer: A

Approach: Idiom meaning.

Solution: To catch a tartar = to take on someone or something that turns out far tougher or harder to handle than expected. So it means to deal with someone or something difficult to handle.

Why the other options are wrong:
A: 'tricky path' describes a situation, not a tough opponent.
C: reputation is not involved.
D: the literal reading; the idiom is about unexpected difficulty, not danger.
E: being stuck is too general and misses the idea of a formidable opponent.

Answer: B

Approach: Weaken: attack the premise the claim rests on.

Solution: The claim 'it would be a miracle if there weren't any miracle cures' rests on the premise that most diseases improve by themselves, fluctuate or are self-limiting, so chance alone produces some 'cures' after any treatment. Evidence that improvement is impossible in the course of many illnesses removes that natural variation. Then 'miracle cures' could not arise by chance, and the claim is weakened.

Why the other options are wrong:
B, C: patients recovering without intervention, or illnesses curing themselves, support the claim.
D: crackpot treatments sometimes working does not touch the chance argument.
E: fewer fraudulent practitioners does not change how diseases behave.

Cetking shortcut: To weaken, negate the key premise ('diseases improve by themselves'), so look for the option that says improvement does not happen.

Answer: A

Approach: Apply the author's reasoning to a new case.

Solution: The author says most conditions improve by themselves, so any treatment started at the right time gets the credit. Faced with testimonials from people 'cured by a blessing', he would say these people would have recovered anyway, without the healer.

Why the other options are wrong:
A: the author does not doubt that patients improved; he doubts the cause.
B: self-promotion ethics is not his argument.
C: he would deny healers have special power.
E: unreliable eyewitnesses is a different argument from the one in the passage (natural recovery).

Cetking shortcut: Match the reply to the passage's core mechanism (natural recovery plus false credit).

Answer: D

Approach: Relevance of new information to the passage's point.

Solution: The passage argues that even outlandish cures are 'often difficult to refute conclusively'. If experts still debate whether certain alternative medicine is scientific or pseudo-scientific, that shows science and pseudo-science cannot always be told apart, which strengthens the passage's claim.

Why the other options are wrong:
A: medicine versus science is not the passage's distinction.
B: not a relevance statement, and unsupported.
C: the debate does not weaken the usefulness of the scientific method.
D: the debate does not touch 'pet theories'.

Cetking shortcut: An ongoing expert debate means 'hard to tell apart', so look for the 'cannot always be distinguished' option.

Answer: E

Approach: Author's likely view of an extreme position, read from the passage's stance.

Solution: The author holds that some claims can be judged: he shows how fraudulent cures exploit chance and calls certain cures outlandish. Quine is known for holding that any statement can be kept true 'come what may' if enough other beliefs are adjusted. An author who thinks some beliefs can be shown true or false would call such a view extreme.

Why the other options are wrong:
A: 'do not apply to any particular situation' is not the author's objection.
B: 'callous' is an emotional judgement the author does not make.
D: the author's concern is beliefs that are hard to refute, not why people reject beliefs.
E: the author does not treat all beliefs as equally valid.

Cetking shortcut: The author debunks fraud, so he believes some claims are testable; choose the option that says beliefs can be proven true or false.

Answer: C

Approach: Antonym: fiendish = extremely cruel or wicked.

Solution: Fiendish = devilishly cruel. Its opposite is someone who feels for others: sympathetic (kind, compassionate).

Why the other options are wrong:
A Bashful: shy.
B Callous: unfeeling, closer to a synonym.
C Complacent: self-satisfied.
E Incomplete: unrelated.

Cetking shortcut: Fiend = devil, so the opposite is kind or compassionate.

Answer: D

Approach: Synonym: elegiac relates to an elegy (a lament for the dead).

Solution: Elegiac = expressing sorrow or lament, so it means mournful.

Why the other options are wrong:
A Peaceful, C Confused, E Restless: unrelated moods.
D Cheerful: opposite.

Cetking shortcut: Elegy = a mourning poem, so elegiac = mournful.

Answer: B

Approach: Antonym: acme = the highest point.

Solution: Acme = the peak or highest point of achievement. Its opposite is the lowest point (nadir).

Why the other options are wrong:
A, B: puns on 'acne'.
C Volcano: a mountain, not a meaning of acme.
D Palatial: grand, like a palace.

Cetking shortcut: Acme and zenith mean peak; nadir means lowest. Ignore look-alike 'acne' traps.

Answer: E

Approach: Word meaning; the options contain only the word's opposite.

Solution: Emancipate = set free from legal, social or political restraint. No option means 'set free', and the only word directly linked to emancipate is its opposite, enslave. The stem asks for the meaning, but the options hold only the opposite, so D is the keyed answer.

Why the other options are wrong:
A Rigorous, B Healthy, C Royal: unrelated.
E Long rope: 'giving long rope' means allowing freedom, an idiom and not a meaning of emancipate.

Cetking shortcut: When no synonym is offered, check whether the option set is really an antonym set.

Answer: D

Approach: Antonym: sanguine = optimistic.

Solution: Sanguine = cheerfully optimistic, even in a bad situation. Its opposite is pessimistic.

Why the other options are wrong:
A Bellicose: aggressive, warlike.
B Red skinned: a trap from 'sanguine' meaning blood-red (complexion).
D Gladiatorial: combative.
E Technical: unrelated.

Cetking shortcut: Sanguine is tied to sang (blood), but in exams it nearly always means optimistic.

Answer: C

Approach: Antonym: wrench = pull or twist out suddenly and violently.

Solution: Wrench means to tear or twist something free. Yank, tug and rip are synonyms. The opposite action, closing or fastening firmly in place, is seal.

Why the other options are wrong:
B Yank, D Tug, E Rip: all near-synonyms of wrench.
C Abstract: 'to remove', also close to wrench.

Cetking shortcut: Remove every synonym (yank, tug, rip, abstract); the one word left is the antonym.

Answer: A

Approach: Idiom meaning (British informal).

Solution: (As) thick as two short planks = very stupid. 'Thick' means dull-witted, and a short plank is thick for its length. So it means having a lack of intelligence. Example: 'He's a skilled footballer, but he's as thick as two short planks.'

Why the other options are wrong:
A: trouble is not implied.
C, D: unrelated.
E: 'thick with someone' (close friends) is a different idiom.

Cetking shortcut: 'Thick' in idioms usually means stupid ('thick-headed').

Answer: B

Approach: Idiom meaning.

Solution: On the breadline = extremely poor, with barely enough money to live (from queues for free bread). So it means being in a condition of poverty.

Why the other options are wrong:
A, D: diet or weight puns on 'bread'.
B: a risky position is 'on the line' or 'on thin ice'.
C: boredom is unrelated.

Answer: E

Approach: One-word substitution.

Solution: Euphemism = a mild or indirect word used in place of a harsh or blunt one (e.g. 'passed away' for 'died'). This matches 'use of mild words in place of words required by truth'.

Why the other options are wrong:
B Cinephile: lover of cinema.
C Misobiblic: hating books.
D Pessimistic: expecting the worst.
E Rapacious: greedy.

Cetking shortcut: eu- = good or well (euphony, eulogy), so euphemism = 'speaking well' of something harsh.

Answer: A

Approach: Synonym: vituperate = abuse in words.

Solution: Vituperate = blame or insult someone in harsh, violent language. Its synonym is abuse.

Why the other options are wrong:
A Teach, B Warn: neutral.
D Appreciate, E Hail: opposites (praise).

Cetking shortcut: 'Vituperative' remarks are bitterly abusive.

Answer: C

Approach: Synonym: contrite = feeling sorry for wrongdoing.

Solution: Contrite = remorseful, repentant. Its synonym is penitent (feeling sorrow for one's sins).

Why the other options are wrong:
B Bashful: shy.
C Joyful: unrelated.
D Remorseless: opposite.
E Conniving: secretly scheming.

Cetking shortcut: Contrition, penitence and remorse form one word family.

Answer: A

Approach: Inference from one specific line.

Solution: The passage says that in legal and health care services, guarantees 'may mislead clients by suggesting that lawsuits or medical procedures will have guaranteed outcomes'. For that to mislead, it must be true that such outcomes cannot be guaranteed in advance. So the result of a lawsuit or medical procedure cannot necessarily be determined beforehand by the professionals.

Why the other options are wrong:
A: compensation is not discussed.
B: cost prediction is not discussed.
C: professional standards being violated is not mentioned.
D: the passage says guarantees may look like begging for business; it does not say all marketing undermines dignity.

Cetking shortcut: 'Misleading by suggesting X' implies X is not true. Look for the option that negates X.

Answer: E

Approach: Specific detail on the 1920s road narrative.

Solution: The passage says the 1920s highway created the road narrative, which highlights male protagonists 'discovering themselves' and emphasises 'the independence of road travel'. Celebrating the freedom that travel gives restates this.

Why the other options are wrong:
A: the narratives emphasised independence; they did not 'develop the desire' for it.
B: women travellers are not discussed.
C: 'discovering themselves' does not mean being different from others.
E: they valued rural folk traditions but did not show the travellers taking part in them.

Cetking shortcut: Match the option to the exact phrase 'emphasizing the independence of road travel' = freedom of travel.

Answer: D

Approach: Word meaning: Greek root of a phobia name.

Solution: Greek homichle = fog or mist, so homichlophobia is the fear of fog.

Why the other options are wrong:
A: fear of humans is anthropophobia.
B: fear of being alone is monophobia or autophobia.
C: fear of heights is acrophobia.
D Fox: unrelated to the root homichle.

Cetking shortcut: Do not let 'homi' pull you to 'homo sapiens'.

Answer: E

Approach: Word meaning: Greek root of a phobia name.

Solution: Greek aichme = point or spear-tip, so aichmophobia is the fear of sharp, pointed objects such as knives and needles. Of the options, it is knives.

Why the other options are wrong:
A Computers: cyberphobia.
B Text books: bibliophobia.
C Back aches, E Tooth aches: not phobia objects here.

Answer: D

Approach: EXCEPT question: check each option against the passage.

Solution: The passage says grammar rules exist so that communication works; when they break down, 'confusion and misunderstanding result'. So the purpose of grammar is clarity, not keeping sentences simple. 'The primary purpose of grammar is to ensure that sentences remain simple' cannot be inferred.

Why the other options are wrong:
A: inferable: 'Must you write complete sentences each time? Perish the thought.'
B: inferable: a sentence has 'a subject (noun) and a predicate (verb)'.
C: inferable: 'Grammar Police' is a figure for strict rule-enforcers.
E: the passage opens with parts of speech (words) and rules that organise them, so it does not support 'words play no role' either; D is the keyed statement, since it contradicts the passage's stress on clarity.

Cetking shortcut: In EXCEPT questions, the answer usually states a purpose or claim that the passage never makes ('simple').

Answer: D

Approach: Word-to-number coding; test letter-count and alphabet-value rules against all three examples.

Solution:
Letter counts: BOOK = 4, TABLE = 5, BOTTLE = 6, COMPUTER = 8.
Alphabet sums: BOOK = 2+15+15+11 = 43, TABLE = 40, BOTTLE = 74, COMPUTER = 111.
No common rule (letter count, alphabet sum, digit sum, vowel/consonant split, reverse values) produces 8, 7 and 1 for all three examples together.
The key used for this question builds the code from the letter count of COMPUTER: 8 letters → 8 ÷ 2 = 4.

Cetking shortcut: In the exam, give such a question at most 30 seconds; if no rule fits all examples, mark and move on.

Note: The examples as printed do not fix a unique rule; 4 is the key's answer.

Answer: D

Approach: Caselet DI: % of total → state total → split by ratio.

Solution:
State 4 total = 35% of 8000 = 2800; ratio 7:5:2 (14 parts) → 1 part = 200 → Diesel = 7 × 200 = 1400.
State 3 total = 30% of 8000 = 2400; ratio 4:5:3 (12 parts) → 1 part = 200 → Electric = 3 × 200 = 600.
Required ratio = 1400 : 600 = 7 : 3.

Why the other options are wrong:
B 7:4, C 13:7 — do not match 1400:600.
D 4:7, E 3:7 — reversed / wrong figures.

Cetking shortcut: Both states happen to have 1 part = 200 cars, so the ratio is simply 7 parts : 3 parts = 7:3.

Answer: A

Approach: Caselet DI: State 4 electric cars, then 55% of them.

Solution:
State 4 total = 35% of 8000 = 2800; ratio 7:5:2 → Electric = 2/14 × 2800 = 400.
AC = 45% of 400 = 180, so Non-AC = 55% of 400 = 220.
The question refers to the electric cars of State 4, so the total number of non-AC cars = 220.

Why the other options are wrong:
E 180 — this is the AC count.
A 380, C 200, D 240 — do not come from 400 split 45:55.

Cetking shortcut: 55% of 400 = 4 × 55 = 220.

Answer: B

Approach: Linear arrangement — option elimination using fixed positions.

Solution:
D is 2nd from left; 5 people between D and B → B at position 8.
Check A (ADHECGFB): positions A1 D2 H3 E4 C5 G6 F7 B8.
C at 5 = 4th from right ✓; E (4) next to C ✓; A (1) and E (4) have two between ✓; H (3) and C (5) have one between ✓; G–F adjacent ✓.
All conditions hold, so A is correct.

Why the other options are wrong:
B ADCEHFGB — C is 3rd from left and 6th from right, not 4th from an end.
C GDAFCEHB — G and F are not adjacent.
D ADGHFCEB — C at 6 is 3rd from right.
E ADCEHFBG — B is at 7, so only 4 people sit between D and B.

Cetking shortcut: First fix D at 2 and B at 8 — this alone kills option E; then check C's position.

Answer: A

Approach: Matrix puzzle (person × profession × colour × day): lock direct links first.

Solution:
Direct links: Hotelier = Saturday; Teacher = Friday & Green; Retired = Orange; Electrician = Red; B = Doctor.
Blue goes to the Lawyer (the Saturday cleaner is the Hotelier). Doctor, Plumber and Hotelier are left with Black, White, Silver.
The Plumber cannot clean on Saturday, Thursday (clue) or Friday (Teacher), so the Plumber cleans on Sunday in the setter's grid.
Name clues: Hotelier is not B, C, D; Retired is not A, D, F; Electrician is not E, G.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
D, A, F, G — in the setter's grid they hold the Electrician, Hotelier, Retired, Teacher and Lawyer roles, not the Plumber.

Answer: E

Approach: Read the colour–profession link from the grid.

Solution:
Blue is liked by 'the Lawyer or the Saturday cleaner'. The Saturday cleaner is the Hotelier, who takes Silver in the setter's grid.
So Blue belongs to the Lawyer.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
B Red — Electrician.
C White — Plumber.
D Green — Teacher.
E Black — Doctor (B).

Answer: A

Approach: Direct clue.

Solution:
'The Hotelier cleans his home on a Saturday' — stated directly in the clues.

Why the other options are wrong:
Doctor (B) — excluded from Saturday.
Lawyer — excluded from Saturday.
Teacher — Friday.
Electrician — Saturday is already the Hotelier's.

Answer: A

Approach: Use the day exclusions for the Plumber.

Solution:
Saturday = Hotelier, Friday = Teacher.
Plumber cannot take Saturday or Thursday, and Friday is gone → in the grid the Plumber takes Sunday.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
A Person who likes Green = Teacher → Friday.
C Teacher → Friday.
B (friend A) and E (Red person = Electrician) — not the Sunday cleaner in the grid.

Answer: D

Approach: List colours without the letter E, then find who owns one.

Solution:
Colours without E: Black, Orange, Silver. Colours with E: Blue, Green, White, Red.
B is the Doctor; the Doctor cannot have Blue (not Lawyer, not Saturday), Green (Teacher), Orange (Retired) or Red (Electrician).
In the grid the Doctor likes Black → no E.
Setter's final grid: Hotelier – Silver – Saturday; Doctor (B) – Black; Teacher – Green – Friday; Electrician – Red; Plumber (C) – White – Sunday; Lawyer – Blue; Retired – Orange.

Why the other options are wrong:
A Electrician → Red (has E).
B (friend A) and D (friend E) — their colours in the grid (Red/Green/Blue) contain E.
E Tuesday cleaner → colour contains E in the grid.

Answer: C

Approach: Test each statement against direct clues first.

Solution:
A — Green belongs to the Teacher, not the Lawyer → false.
B — B is the Doctor → false.
C — the Saturday cleaner is the Hotelier, not the Retired man → false.
D — in the setter's grid C is the Plumber and the Plumber likes White → true.
E — the Plumber (C) cleans on Sunday, not Monday → false.

Cetking shortcut: A, B, C contradict a single clue each — eliminate them in 10 seconds, then compare D and E.

Answer: D

Approach: Blood relations — decode the chain from Neena outwards.

Solution:
Neena's father's granddaughter = daughter of Neena's brother (Neena is his daughter, not granddaughter).
That girl's father = Neena's brother.
Brother's grandfather = Neena's grandfather; his daughter-in-law = Neena's mother.
So Deepali's mother-in-law = Neena's mother → Deepali is married to Neena's brother.
Deepali is Neena's brother's wife, i.e. her sister-in-law — the 'sister' option.

Why the other options are wrong:
B mother — would need Deepali's husband to be Neena's father; the chain lands one generation lower.
C mother-in-law, D daughter-in-law, E daughter — all require a generation gap; Deepali is in Neena's own generation.

Cetking shortcut: Count generations: father's granddaughter (−1) → her father (0) → his grandfather (+2) → daughter-in-law (+1). Generation +1 = Neena's mother, so Deepali is in Neena's generation.

Answer: A

Approach: Standard die: opposite faces sum to 7.

Solution:
Face shown = 3 → opposite face = 7 − 3 = 4.
Four 4s multiplied together: 4 × 4 × 4 × 4 = 256.

Why the other options are wrong:
B 81 = 3⁴ (used the wrong face).
D 1296 = 6⁴, E 625 = 5⁴, A 12 = 3 × 4.

Answer: C

Approach: Scan every 3 and test: next number prime AND previous number not even.

Solution:
Series: 3 5 4 8 4 3 7 0 9 8 3 9 4 3 7 7 5 3 1 2 0 7 3 2
3 (1st): no number before, next 5 (prime) ✓
3 (6th): before 4 (even) ✗
3 (11th): before 8 (even) ✗
3 (14th): before 4 (even) ✗
3 (18th): next 1 (not prime) ✗
3 (23rd): before 7, next 2 (prime) ✓
Count = 2.

Cetking shortcut: First strike out every 3 preceded by an even digit — only three 3s remain to check.

Answer: B

Approach: Alternate (interleaved) series.

Solution:
Odd positions: 8, 10, 12 (+2 each).
Even positions: 13, 15, ? (+2 each) → 17.
The missing term is the 6th (even) position = 17.

Why the other options are wrong:
B 14 — continues the odd-position series, but ? is at an even position.
A 13, C 16, D 20 — break both patterns.

Cetking shortcut: Also seen as +5, −3 alternately: 12 + 5 = 17.

Answer: E

Approach: Linear arrangement, all facing the same side; seats 1–8 from left.

Solution:
Roshesh and Urus at seats 1 and 8; their neighbours (seats 2 and 7) are Quila and Tanaav.
Pravesh is 4th to the right of Tanaav → Tanaav cannot be at 7, so Tanaav = 2, Pravesh = 6, Quila = 7 (Pravesh is next to Quila ✓).
Wahaab is immediately left of Pravesh → Wahaab = 5.
Wahaab is 2nd to the right of Smitha → Smitha = 3; Vihaaz takes the last seat, 4.
Order: R/U, Tanaav, Smitha, Vihaaz, Wahaab, Pravesh, Quila, U/R → Wahaab's neighbours are Vihaaz and Pravesh.

Why the other options are wrong:
A, C, D, E — Urus is at an end and Tanaav/Quila sit at seats 2 and 7, far from Wahaab at seat 5.

Answer: B

Approach: Coded blood relations.

Solution:
P $ Q → P is the father of Q.
Q # R → R is the mother of Q.
R % S → R is the sister of S (so R is female, as already known).
P is Q's father and R is Q's mother → R is P's wife.

Why the other options are wrong:
A sister, C niece — no such link between P and R.
B son — R is female (mother).
E mother — R is the mother of Q, not of P.

Answer: D

Approach: Column-wise alphabet series.

Solution:
1st letters: I, L, O, R → +3 each → next U (21).
2nd letters: C, E, G, I → +2 each → next K (11).
3rd letters: F, I, L, O → +3 each → next R (18).
Next group = U K R.
The opening group UNR does not follow this progression (the series runs from ICF onward).

Why the other options are wrong:
E UKS, C ULS — third letter should be R.
D UJR — second letter should be K (I + 2).
A WTR — first letter should be U.

Cetking shortcut: The answer repeats the 1st and 3rd letters of the first term (U _ R), so only the middle letter needs work.

Answer: B

Approach: Direction sense with a circular detour.

Solution:
Stream flows west → east; turns left → now flowing north.
Flowing north, it meets the island at the island's southern point.
Going anticlockwise around the island from the southern point, it first moves east and, after a quarter circle, reaches the eastern point moving north.
Then it takes a right from north → flowing east.

Why the other options are wrong:
E north — the direction before the final right turn.
D west — treats the quarter circle as a left turn and ignores the right turn.
B south, C north-west — no step in the path gives these.

Cetking shortcut: Draw it: anticlockwise round an island you meet from the south = curve to the right, then straighten north.

Answer: A

Approach: Substitution coding.

Solution:
The animal kept as a pet is a dog.
'Dog' is called 'lion' in this language → answer = lion.

Why the other options are wrong:
D dog — the real name, not the code.
B bison, C snake, E mongoose — codes of lion, bison and snake.

Answer: A

Approach: Five in a row, all facing North.

Solution:
Dony = seat 3; Tony is next to Dony → seat 2 or 4.
Tony's other neighbour (seat 1 or 5) cannot be Bony or Mony → it is Sony.
So the row is Sony, Tony, Dony, _, _ (or its mirror); Bony and Mony take the other side.
Ends = Sony and one of Bony/Mony. Among the options only 'Sony, Mony' fits.

Why the other options are wrong:
A, B, C, D — Dony is in the middle and Tony at seat 2/4, so neither can be at an end.

Cetking shortcut: Dony (middle) and Tony (next to middle) can never be at an end — this removes A–D at once.

Answer: E

Approach: Direction sense — resolve into N–S and E–W components.

Solution:
20 m North (facing N).
Right → East 30 m.
Right → South 35 m.
Left (from South) → East 15 m.
Left (from East) → North 15 m.
N–S: +20 − 35 + 15 = 0. E–W: 30 + 15 = 45 m East.
Krishna is 45 m East of the start.

Why the other options are wrong:
A 15 m, B 30 m — add only one eastward leg.
C, D — West is the wrong side; no leg goes west.

Cetking shortcut: North and south legs cancel (20 + 15 = 35), so only the east legs remain: 30 + 15 = 45.

Answer: E

Approach: Coordinate method + Pythagoras.

Solution:
Take K = (0, 0).
J is 1 km north of K → J = (0, 1).
Q is 2 km south of J → Q = (0, −1).
P is 1 km west of Q → P = (−1, −1).
KP = √(1² + 1²) = √2 ≈ 1.41 km.

Why the other options are wrong:
A 1 km — only one leg of the right triangle.
B 1.5, D 1.67, E 1.23 — not equal to √2.

Cetking shortcut: P is 1 km west and 1 km south of K → the diagonal of a unit square = √2.

Answer: C

Approach: Number analogy: n → n³ + n.

Solution:
6³ + 6 = 216 + 6 = 222 ✓.
7³ + 7 = 343 + 7 = 350.

Why the other options are wrong:
E 343 = 7³ only.
C 349 = 7³ + 6 (added the wrong n).
B 225, D 210 — no consistent rule.

Answer: A

Approach: Odd one out — compare the step multipliers ×1.5, ×2, ×2/3.

Solution:
A: 6.4 → 9.6 → 19.2 → 12.8 = ×1.5, ×2, ×2/3.
B: 1 → 1.5 → 3 → 2 = ×1.5, ×2, ×2/3.
C: 0.5 → 0.75 → 1.5 → 1 = ×1.5, ×2, ×2/3.
D: 0.4 → 0.6 → 1.2 → 0.8 = ×1.5, ×2, ×2/3.
E: 0.5 → 1.5 → 3 → 0.25 = ×3, ×2, ×1/12 → breaks the pattern.

Cetking shortcut: Each regular group is 2 : 3 : 6 : 4 of a base value; E's first step is ×3 — spot it at once.

Answer: E

Approach: Ranking with interchange: Total = (position from left) + (position from right) − 1 for the same seat.

Solution:
Before swap: Mimmi's seat is 10th from left.
After swap, Rimmi sits in that seat and is 17th from right.
So that seat is 10th from left and 17th from right → Total = 10 + 17 − 1 = 26.
Check with the other seat: Rimmi's old seat is 9th from right; Mimmi there is 18th from left → 18 + 9 − 1 = 26 ✓.

Why the other options are wrong:
B 27 — forgot to subtract 1.
D 25 — subtracted 2.
A 18, E 35 — do not use the same seat from both ends.

Cetking shortcut: Pair the old rank of one girl with the new rank of the other — they describe the same seat.

Answer: C

Approach: Apply the letter table, then the vowel rule.

Solution:
A (first, vowel) → 9; F → 7; D → 2; Q → 8; E (middle vowel) → 6; N → 4; I (last, vowel) → 9.
A and I are different vowels, so $ is not used.
Code = 9728649.

Why the other options are wrong:
A $72894$ — $ needs the same vowel at both ends.
B, D — start with 6, but A is at the beginning (9).
E 9728679 — N is 4, not 7.

Answer: C

Approach: Apply the letter table, then the vowel rule.

Solution:
O (first, vowel) → 9; L → 3; S → 3; W → 2; Y → 8; C → 1; Z → 2; I (middle vowel) → 6; T → 1.
Code = 933281261.

Why the other options are wrong:
A 932381261 — L, S coded as 3, 2.
B 933281263 — T is 1, not 3.
D 923281261 — L is 3, not 2.
E 932281261 — S is 3, not 2.

Answer: C

Approach: Apply the letter table, then the vowel rule.

Solution:
T → 1; J → 5; I (middle) → 6; T → 1; A (middle) → 6; V → 4; R → 5; Q → 8; X → 7; U (last, vowel) → 9.
Code = 1561645879.

Why the other options are wrong:
A 1561644879 — R is 5, not 4.
B 1521645879 — I in the middle is 6, not 2.
C 1561545879 — A in the middle is 6, not 5.
D 1561645878 — final U is 9, not 8.

Answer: E

Approach: No vowels — straight table lookup.

Solution:
X → 7; L → 3; P → 7; V → 4; R → 5; Y → 8; T → 1; Q → 8; D → 2; C → 1.
Code = 7374581821.

Why the other options are wrong:
A 7374591821 — Y is 8, not 9.
B 7375581821 — V is 4, not 5.
D 7374581921 — Q is 8, not 9.
E 7474581821 — L is 3, not 4.

Cetking shortcut: Compare options digit by digit; each wrong option differs in exactly one place.

Answer: C

Approach: Faulty clock: the clock runs 24 h 10 min while true time runs 24 h.

Solution:
Clock time passed from 8:00 to 11:00 = 3 h = 180 min.
True time = 180 × 1440 ÷ 1450 ≈ 178.76 min = 2 h 58 min 45 s.
True time = 8:00 + 2:58:45 = 10:58:45 AM.

Why the other options are wrong:
A 11:59:35, C 11:58:45 — after 11:00; a fast clock is ahead, so true time is earlier.
B 10:50:45 — far too large a correction.
E 10:58:40 — wrong seconds.

Cetking shortcut: Gain ≈ 10 min/24 h ≈ 25 s per hour → 3 h ≈ 75 s = 1 min 15 s → 11:00 − 1:15 = 10:58:45.

Answer: D

Approach: Classical probability: favourable ÷ 36.

Solution:
Sum 9: (3,6), (4,5), (5,4), (6,3) → 4.
Sum 10: (4,6), (5,5), (6,4) → 3.
Favourable = 7 → P = 7/36.

Why the other options are wrong:
D 5/36 — counts only one of the sums correctly.
E 7/18 — doubled.
A 4/12, B 1/30 — wrong total outcomes.

Cetking shortcut: Ways to get a sum s with two dice = 13 − s for s ≥ 7: 4 for 9 and 3 for 10.

Answer: C

Approach: Average → total → ratio split.

Solution:
Total age = 3 × 50 = 150.
Ratio 3 : 5 : 7 = 15 parts → 1 part = 10.
Youngest = 3 × 10 = 30 years.

Why the other options are wrong:
C 20 — wrong part value.
B 25, E 35, A 40 — not 3 parts of 150/15.

Cetking shortcut: Middle term of 3:5:7 equals the average (5 parts = 50) → 1 part = 10.

Answer: D

Approach: Blood relations — go up then down the family tree.

Solution:
Elder brother of Raj's father = Raj's uncle.
Uncle's granddaughter = daughter of Raj's cousin.
Raj's cousin's daughter is his niece (first cousin once removed) — the 'niece' option.

Why the other options are wrong:
B granddaughter — she is the uncle's granddaughter, not Raj's.
C daughter, D sister, E sister-in-law — different branches/generations.

Answer: A

Approach: Slot counting with minimums (7 experts, one slot each; W at lunch).

Solution:
Before ≥ 2, After ≥ 3, W is at lunch → Lunch ≤ 7 − 2 − 3 = 2.
So After (≥ 3) can never equal Lunch (≤ 2) → statement D is impossible.
Check the rest: A — Before 2, After 4, Lunch 1 (W) is possible.
B — J is only after school, so J is never at lunch (true).
C — Before 2, Lunch 2, After 3 is possible (e.g. N, G before; W, M lunch; J, K, L after).
E — Before 3, After 3, Lunch 1 is possible (N, G, M before; J, K, L after).

Cetking shortcut: Lunch has at most 2 people and After has at least 3 — any option equating them is out.

Answer: D

Approach: Elimination grid for husbands and wives.

Solution:
Fixed: E–W, B–V.
Wives left: R, S, T, U for A, C, D, F.
D is not married to R, S or T → D–U.
C is not married to R or T, and U is gone → C–S.
A cannot marry his sisters R and S → A–T (and F–R).

Why the other options are wrong:
A S, D R — A's sisters.
B V — B's wife.
E U — D's wife.

Answer: C

Approach: Direction renaming = 45° rotation of all labels.

Solution:
SE (135°) is now called E (90°); NW (315°) is now called W (270°) → every label moves back by 45°.
So the old direction 45° (NE) now carries the label 0° = North.
Full renaming: NE → North, SE → East, SW → South, NW → West.
The direction for North is North-East.

Why the other options are wrong:
A North-West — the result of shifting North itself by 45° the other way.
B West, C South-West, E South-East — use the wrong angle (90°/135°/180°).

Cetking shortcut: Both given pairs say 'old diagonal → new main direction one step anticlockwise'; the diagonal one step clockwise of North (NE) becomes North.

Answer: D

Approach: Allocation grid: 4 people × 2 cities each.

Solution:
A visits Delhi and Lucknow (given).
Bangalore: not A, B, C → D.
Mumbai: not C, and A is full → B or D.
Kolkata and Hyderabad: not D, A is full → B or C.
A consistent allocation: D – Bangalore, Chennai; C – Kolkata, Hyderabad; B – Mumbai.
So B can visit Mumbai.

Why the other options are wrong:
C Bangalore — barred for B.
E Delhi — already visited by A.

Note: As printed the clues leave Patna with no eligible visitor, so the grid cannot be completed exactly; Mumbai is the key's answer; Kolkata and Hyderabad are also open to B on these clues.

Answer: D

Approach: Option testing against two conditions.

Solution:
Condition 1: C cannot come immediately after B.
Condition 2: once A is inspected, E cannot come later.
A ABCDE — B then C ✗.
B DCABE — A before E ✗.
C DCBEA — B is followed by E (not C) ✓; A is last, so nothing comes after it ✓.
D DCBAE — A before E ✗.
So DCBEA is a valid order.

Note: CDBEA (option E) also satisfies both conditions as printed; the key marks DCBEA.

Answer: C

Approach: Syllogism chain: All–All gives All.

Solution:
D: All honest persons are obese.
E: All obese persons are good natured.
D + E → All honest persons are good natured = statement A.
So DEA is the valid set.

Why the other options are wrong:
A BEA — 'Some good natured are not honest' + E cannot give 'All honest are good natured'.
B ECF — F follows from C alone (conversion), not from E and C together.
C CEA — 'some' + 'all' cannot give an 'all' conclusion.
D FBD — two 'some' statements give no conclusion.

Cetking shortcut: Look for two 'All' statements with a shared middle term (obese) — DEA.

Answer: E

Approach: Pentagon arrangement: corners in cyclic order, chairs on sides.

Solution:
S is immediately right of P, and R is between P and T → corner order S, P, R, T, Q (cyclic).
Sides: S–P, P–R, R–T, T–Q, Q–S.
Chair A is on P–R; chair B is on T–Q.
D and E are next to B → on R–T and Q–S.
The remaining side S–P gets chair C.

Why the other options are wrong:
B T,S and E P,Q — T–S and P–Q are not sides (not adjacent corners).
C Q,R — not a side.
D S,Q — taken by D or E.

Answer: A