MBA CET 2025 Slot 2 question paper with solutions

Approach: Find a small power of 3 that leaves remainder 1 when divided by 11.

Solution:
3⁵ = 243 = 11 × 22 + 1 → 3⁵ ≡ 1 (mod 11).
333 = 5 × 66 + 3 → 3³³³ = (3⁵)⁶⁶ × 3³ ≡ 1 × 27 (mod 11).
27 = 2 × 11 + 5 → remainder 5.

Cetking shortcut: Fermat: 3¹⁰ ≡ 1 (mod 11), but spotting 3⁵ = 243 ≡ 1 is quicker; then only 333 mod 5 = 3 matters.

Answer: E

Approach: Parallel sides make △EDC ∼ △EAB; the ratio of the parallel sides gives the ratio of the slant sides.

Solution:
DC ∥ AB, so △EDC ∼ △EAB with ratio DC : AB = 1 : 2.
ED : EA = 1 : 2 → D is the midpoint of EA → EA = 2 × AD.
Similarly EB = 2 × BC.
Perimeter of △AEB = EA + EB + AB = 2(AD + BC) + AB = 2 × 3 + 2 = 8 cm.

Cetking shortcut: When the smaller parallel side is half the larger, each slant side of the trapezium is exactly half of the triangle's side.

Answer: B

Approach: Use the two pair-rates plus the 75–135–45 completion data to get each person's rate, then run the 6-day work cycle.

Solution:
Let daily rates be A, V, S (whole work = 1).
A + V = 1/150, V + S = 1/100, and 75A + 135V + 45S = 1.
Write A = 1/150 − V, S = 1/100 − V: 75/150 + 45/100 + (135 − 75 − 45)V = 1 → 0.5 + 0.45 + 15V = 1 → V = 1/300.
So A = 1/150 − 1/300 = 1/300 and S = 1/100 − 1/300 = 2/300.
Take work = 300 units: A = 1, V = 1, S = 2 units/day.
New pattern: Amal every day, Vimal on days 2, 4, 6…, Sunil on days 3, 6, 9….
One 6-day cycle = Amal 6×1 + Vimal 3×1 + Sunil 2×2 = 13 units.
23 cycles = 138 days → 23 × 13 = 299 units. Day 139: Amal alone adds 1 unit → 300 units.
Total = 139 days.

Cetking shortcut: Once rates are 1 : 1 : 2, think in 6-day blocks of 13 units: 300 = 13 × 23 + 1, so 138 days + 1 more day.

Answer: E

Approach: Overlapping groups — 'first 30' and 'last 30' together count the middle 20 twice.

Solution:
Total (all 40) = 40 × 50,000 = 20,00,000.
First 30 total = 30 × 40,000 = 12,00,000; last 30 total = 30 × 60,000 = 18,00,000.
First 30 + last 30 = all 40 + middle 20 → 30,00,000 = 20,00,000 + middle 20 → middle 20 = 10,00,000.
First 10 = 12,00,000 − 10,00,000 = 2,00,000; last 10 = 18,00,000 − 10,00,000 = 8,00,000.
1926: first 10 up 100% → 4,00,000; last 10 up 200% → 24,00,000; middle 20 unchanged → 10,00,000.
New total = 38,00,000 → average = 38,00,000 / 40 = Rs 95,000.

Cetking shortcut: Increase = 100% of 2L + 200% of 8L = 18L; 18L/40 = 45,000 added to 50,000 → 95,000.

Answer: E

Approach: Square a difference of square roots, or spot perfect squares of the form (a ± √b)².

Solution:
√(x + 6√2) − √(x − 6√2) = 2√2. Square both sides:
(x + 6√2) + (x − 6√2) − 2√(x² − 72) = 8 → 2x − 8 = 2√(x² − 72) → x − 4 = √(x² − 72).
Square again: x² − 8x + 16 = x² − 72 → 8x = 88 → x = 11.
Check: 11 + 6√2 = (3 + √2)², 11 − 6√2 = (3 − √2)² → difference = 2√2 ✓

Cetking shortcut: 2√2 = (3 + √2) − (3 − √2), and (3 ± √2)² = 11 ± 6√2, so x = 11 directly.

Answer: C

Approach: Form equations from the two concentration conditions.

Solution:
Let acid = A L, water = W L.
After adding 2 L water, acid is 50%: A = W + 2.
After adding 15 L acid, acid is 80%: (A + 15) / (A + 15 + W + 2) = 4/5.
Substitute W + 2 = A: (A + 15)/(2A + 15) = 4/5 → 5A + 75 = 8A + 60 → A = 5, W = 3.
Original water : acid = 3 : 5.

Cetking shortcut: At 80% acid, water (W + 2 = A) is 1/4 of acid: A = (A + 15)/4 → A = 5.

Answer: B

Approach: Count even digits (0, 2, 4, 6, 8) number by number; 0 is an even digit.

Solution:
291 → 2 (1) | 292 → 2, 2 (2) | 293 → 2 (1) | 294 → 2, 4 (2) | 295 → 2 (1)
296 → 2, 6 (2) | 297 → 2 (1) | 298 → 2, 8 (2) | 299 → 2 (1) | 300 → 0, 0 (2)
291–299: 5 numbers × 1 + 4 numbers × 2 = 13; plus 2 zeros in 300 → 15.

Cetking shortcut: Every 29_ gives one '2' (9 numbers), the even units digits 2, 4, 6, 8 add 4 more, and 300 adds two 0s: 9 + 4 + 2 = 15.

Answer: C

Approach: Distributive law — take the common factor out.

Solution:
8796 × 223 + 8796 × 77 = 8796 × (223 + 77) = 8796 × 300 = 26,38,800.

Cetking shortcut: Answer must end in 00 (×300) and be ≈ 8800 × 300 = 26.4 lakh → only 2638800 fits.

Answer: A

Approach: Number = LCM × k + remainder; then apply the divisibility-by-7 condition.

Solution:
16 = 2⁴, 18 = 2 × 3², 20 = 2² × 5, 25 = 5² → LCM = 2⁴ × 3² × 5² = 3600.
Number = 3600k + 4.
3600 = 7 × 514 + 2 → 3600k + 4 ≡ 2k + 4 (mod 7). Need 2k + 4 ≡ 0 → 2k ≡ 3 → k ≡ 5 (mod 7).
Smallest k = 5 → 18000 + 4 = 18004 = 7 × 2572 ✓

Cetking shortcut: Only options ending in 4 qualify (18004, 17004); 17004 − 4 = 17000 is not a multiple of 3600, so 18004.

Answer: B

Approach: Hypothetical total minus the extra points.

Solution:
If the best scored 92, total = 8 × 84 = 672.
He actually scored 85, i.e. 7 less → actual total = 672 − 7 = 665.

Answer: E

Approach: Take MP = 100.

Solution:
MP = 100 → CP = 64.
Discount 12% → SP = 88.
Profit = 88 − 64 = 24 → gain% = 24/64 × 100 = 37.5%.

Cetking shortcut: 24/64 = 3/8 = 37.5%.

Answer: D

Approach: Partnership — convert investments into time-weighted (yearly-equivalent) capital.

Solution:
Let P = 5k, Q = 6k; R puts 6k for 6 months.
Yearly-equivalent capital = 5k + 6k + 6k × 6/12 = 14k (profit ratio 5 : 6 : 3).
Profit = 20% of investment = 0.2 × 14k = 2.8k = 98,000 → k = 35,000.
R's investment = 6k = ₹2,10,000.

Cetking shortcut: R invested the same as Q, so R = 6/14 of the capital base; 98,000/0.2 = 4,90,000 → 4,90,000 × 6/14 = 2,10,000.

Answer: E

Approach: Tank with steady inflow — 'capacity + inflow' must equal total usage in both cases.

Solution:
Let capacity = C and inflow = r gallons/day.
Using 32,000/day lasts 50 days: C + 50r = 50 × 32,000 = 16,00,000.
Using 37,000/day lasts 40 days: C + 40r = 40 × 37,000 = 14,80,000.
Subtract: 10r = 1,20,000 → r = 12,000 gallons/day (and C = 10,00,000).
The supply never fails only if daily use ≤ inflow → 12,000 gallons/day.

Cetking shortcut: Like the cow-grazing problem: r = (50×32,000 − 40×37,000)/(50 − 40) = 12,000.

Answer: D

Approach: Same distance up and down: D = time × speed.

Solution:
Let still-water speed = x.
8(x − 4) = 6(x + 4) → 8x − 32 = 6x + 24 → 2x = 56 → x = 28 km/h.

Cetking shortcut: Speeds are inversely as times: (x+4):(x−4) = 8:6 = 4:3 → difference 1 part = 8 → x = 3.5 parts = 28.

Answer: A

Approach: Relative speed gives train speed; at the end of overtaking the train's front is 400 m ahead of the man.

Solution:
Relative speed = 400 m / 40 s = 10 m/s = 36 km/h → train speed = 36 + 5 = 41 km/h.
Train's front reaches the station 20 min later: distance = 41 × 20/60 = 41/3 km = 13.667 km.
At that instant the front was 0.4 km ahead of the man, so man's distance to station = 13.667 + 0.4 = 14.0667 km (= 211/15 km).
Man's time = (211/15) ÷ 5 = 211/75 h = 2.8133 h = 2 h 48.8 min = 2 h 48 min 48 s after being overtaken.

Answer: E

Approach: Find the balance fraction; interest is on the balance only.

Solution:
Paid now = 1/6 + 1/3 = 1/2 of price x → balance = x/2.
Interest = 10% of x/2 = 0.05x = 6000 → x = ₹1,20,000.

Answer: B

Approach: Cube-root estimate to find the three consecutive integers.

Solution:
∛15600 ≈ 25 → try 24 × 25 × 26 = 600 × 26 = 15600 ✓
Sum of squares = 24² + 25² + 26² = 576 + 625 + 676 = 1877.
(Check: 3 × 25² + 2 = 1877.)
Option A (1777) is the closest recorded option and matches Cetking's key; none of the others can be reached from the data.

Cetking shortcut: For n−1, n, n+1: sum of squares = 3n² + 2.

Answer: A

Approach: Same data as Q115 — find the integers by cube-root estimation.

Solution:
(n − 1)n(n + 1) = n³ − n = 15600 → n = 25 (15625 − 25 = 15600).
Integers: 24, 25, 26.
Sum of squares = 576 + 625 + 676 = 1877 (= 3 × 25² + 2).
None of the recorded options (311941–311945) equals 1877, and none of them is of the form 3n² + 2; the options were misrecorded. Cetking's recorded key (E) is kept.

Answer: E

Approach: Group equal bases on each side and factor out the smallest power.

Solution:
9^(x − 1/2) = 3^(2x − 1). Move terms: 3^(2x − 1) + 3^(2x − 3) = 4^x + 2^(2x − 2) = 4^x + 4^(x − 1).
Left = 3^(2x − 3)(9 + 1) = 10 × 3^(2x − 3). Right = 4^(x − 1)(4 + 1) = 5 × 2^(2x − 2).
10 × 3^(2x − 3) = 5 × 2^(2x − 2) → 3^(2x − 3) = 2^(2x − 3) → 2x − 3 = 0 → x = 3/2.
Check: LHS = 9¹ − 2¹ = 7; RHS = 4^(3/2) − 3⁰ = 8 − 1 = 7 ✓

Cetking shortcut: a^k = b^k with a ≠ b forces k = 0; or simply plug in options — x = 3/2 makes every exponent an integer.

Answer: E

Approach: Divisible by 6 = even last digit + digit sum divisible by 3. Pick the digit set first, then arrange.

Solution:
Sum of 0, 2, 3, 4, 6 = 15. Dropping one digit must leave a multiple of 3 → drop 0, 3 or 6.
{2, 3, 4, 6}: last digit 2/4/6 (3 ways) × 3! = 18.
{0, 2, 4, 6}: all even; arrangements without leading 0 = 4! − 3! = 18.
{0, 2, 3, 4}: ends in 0 → 3! = 6; ends in 2 → first digit from {3, 4} (2) × 2 = 4; ends in 4 → 4. Total 14.
Total = 18 + 18 + 14 = 50.

Cetking shortcut: Check the total digit sum first (15) — only drop digits that are multiples of 3.

Answer: A

Approach: Net rate = inlet rate − outlet rate.

Solution:
Inlet = 1/8 tank/h; together = 1/10 tank/h.
Outlet = 1/8 − 1/10 = 1/40 tank/h → empties a full tank in 40 h.
Half tank = 20 h.

Cetking shortcut: LCM 40 units: inlet 5, net 4 → outlet 1 unit/h; half tank = 20 units = 20 h.

Answer: C

Approach: Equal perimeter → express both areas in P.

Solution:
Square: side = P/4 → area = P²/16.
Circle: 2πr = P → r = P/(2π) → area = πr² = P²/(4π).
Square : circle = (1/16) : (1/(4π)) = 4π : 16 = π : 4.

Cetking shortcut: For equal perimeters the circle always has the larger area, so the ratio must be less than 1 → π : 4 (≈ 0.785).

Answer: C

Approach: Combine the ratios; the sum must be a multiple of the ratio-sum.

Solution:
a : b = 3 : 4, b : c = 2 : 1 = 4 : 2 → a : b : c = 3 : 4 : 2.
a + b + c = 9k → must be divisible by 9.
Digit sums: 210 → 3, 201 → 3, 213 → 6, 207 → 9 ✓, 205 → 7.
207 = 9 × 23 (a = 69, b = 92, c = 46).

Cetking shortcut: Divisibility by 9 = digit sum divisible by 9 — only 207.

Answer: D

Approach: Pay is split in the ratio of work done.

Solution:
Rates: Amal 1/10, Bimal 1/8, all three 1/4.
Kamal = 1/4 − 1/10 − 1/8 = (10 − 4 − 5)/40 = 1/40 per day.
Work in 4 days: Amal 4/10 = 0.4, Bimal 4/8 = 0.5, Kamal 4/40 = 0.1.
Shares of Rs 1000: Amal 400, Bimal 500, Kamal 100.
Kamal's share by the given data is Rs 100, which is not among the recorded options; the options/key appear misrecorded. Cetking's recorded key (C) is kept.

Answer: C

Approach: Prism TSA = 2 × base area + base perimeter × height.

Solution:
Base trapezium (taken as isosceles): parallel sides 10 and 20, height 12.
Base area = ½ × (10 + 20) × 12 = 180 cm².
Each slanted side: overhang = (20 − 10)/2 = 5 → side = √(12² + 5²) = 13.
Perimeter = 10 + 20 + 13 + 13 = 56.
Lateral area = 56 × 20 = 1120. TSA = 2 × 180 + 1120 = 1480 cm².

Cetking shortcut: 5-12-13 triple gives the slant sides instantly.

Answer: E

Approach: Pair up terms that add to 470; take at most one from each pair.

Solution:
Number of terms = (467 − 3)/8 + 1 = 59.
Pairs adding to 470: (3, 467), (11, 459), … — the middle term is 235 (3 + 8 × 29), and 235 + 235 = 470 needs two equal elements, so 235 has no partner.
Pairs = (59 − 1)/2 = 29. Take one from each pair + 235 → 29 + 1 = 30.

Cetking shortcut: Odd number of terms symmetric about 235: max = (n + 1)/2 = 30.

Answer: E

Approach: Total balls in n triangular layers = tetrahedral number n(n + 1)(n + 2)/6.

Solution:
n(n + 1)(n + 2)/6 = 8436 → n(n + 1)(n + 2) = 50616.
∛50616 ≈ 37 → try 36 × 37 × 38 = 50616 ✓
So n = 36 layers. (Check n = 38: 38 × 39 × 40/6 = 9880 ≠ 8436.)

Cetking shortcut: 8436 × 6 = 50616 ends in 6; 36 × 37 × 38 ends in 6 (6 × 7 × 8 = 336), 38 × 39 × 40 ends in 0 — so 36.

Answer: C

Approach: 6-8-10 is right-angled at B; the circle's radius is the altitude BD.

Solution:
6² + 8² = 10² → ∠B = 90°.
Altitude BD = (AB × BC)/AC = 48/10 = 4.8.
Circle (centre B, radius BD = 4.8) cuts AB at P and BC at Q: BP = BQ = 4.8.
AP = 6 − 4.8 = 1.2; QC = 8 − 4.8 = 3.2.
AP : QC = 1.2 : 3.2 = 3 : 8.

Cetking shortcut: Altitude to hypotenuse in a right triangle = product of legs / hypotenuse.

Answer: E

Approach: On a circular track, runners meet for the first time when the faster gains one full lap.

Solution:
Let slowest = x m/min, fastest = 2x m/min.
First meeting after 5 min: gain of one lap → (2x − x) × 5 = 1000 → x = 200 m/min.
Fastest = 400 m/min → 4000 m race takes 4000/400 = 10 minutes.

Cetking shortcut: In 5 min the fastest runs exactly 2 laps (the slow one 1), so 2000 m per 5 min → 4000 m in 10 min.

Answer: C

Approach: Inclusion–exclusion on even numbers only.

Solution:
Even numbers 100–200: (200 − 100)/2 + 1 = 51.
Even and divisible by 7 = multiples of 14: 112 … 196 → 7.
Even and divisible by 9 = multiples of 18: 108 … 198 → 6.
Both = multiples of 126: 126 → 1.
Required = 51 − (7 + 6 − 1) = 39.

Cetking shortcut: Within evens, 'divisible by 7' means multiple of 14, 'by 9' means multiple of 18, both means multiple of 126.

Answer: A

Approach: Spot obvious roots, then use convexity of 2^x to limit the count.

Solution:
x = 0: 1 − 0 − 1 = 0 ✓. x = 1: 2 − 1 − 1 = 0 ✓.
y = 2^x is convex and y = x + 1 is a straight line, so they meet at most twice.
Both roots (0 and 1) are non-negative → 2 roots.

Cetking shortcut: A convex curve and a line cross at most twice — test x = 0, 1, 2 quickly.

Answer: C

Approach: Difference of the two fractions × number = 770.

Solution:
7/8 − 7/18 = 7(18 − 8)/144 = 70/144 = 35/72.
(35/72)x = 770 → x = 770 × 72/35 = 22 × 72 = 1584.

Cetking shortcut: 770/35 = 22 → x = 22 × 72.

Answer: B

Approach: Turn the condition into an equation for the diagonal and apply Pythagoras.

Solution:
Let longer side = l, shorter = b, diagonal = d.
(l + b) − d = l/2 → d = l/2 + b.
Pythagoras: l² + b² = (l/2 + b)² = l²/4 + lb + b² → 3l²/4 = lb → b = 3l/4.
b : l = 3 : 4.

Cetking shortcut: Test 3-4-5: (3 + 4) − 5 = 2 = 4/2 ✓.

Answer: D

Approach: Joint variation: value ∝ d² × t.

Solution:
V = k d² t → V₁/V₂ = (d₁/d₂)² × (t₁/t₂).
4/1 = (4/3)² × (t₁/t₂) = (16/9)(t₁/t₂).
t₁/t₂ = 4 × 9/16 = 9/4 → thickness ratio = 9 : 4.

Why the other options are wrong: 16 : 9 is just the diameter-squared ratio; 9 : 16 and 4 : 9 invert the relation.

Answer: B

Approach: Compound interest for 2 years on each part; solve for M.

Solution:
Parts: M/3 at 10%, 2M/3 at 5% (CI, 2 years).
CI factors: 10% → 1.21 − 1 = 0.21; 5% → 1.1025 − 1 = 0.1025.
Interest = 0.21(M/3) + 0.1025(2M/3) = 0.07M + 0.06833M = 0.138333M = 830 → M = 6000.
Check: 2000 → 420, 4000 → 410, total 830 ✓ (both parts > ₹1000).

Cetking shortcut: 2-year CI % = 2r + r²/100: 21% and 10.25%; test options quickly.

Answer: C

Approach: The fraction 5/9 forces (x − 2) to be a multiple of 9; the fridge cap of 5 bounds it.

Solution:
Good avocados = x − 2; her fridge gets (5/9)(x − 2) → x − 2 must be a multiple of 9 → x = 11, 20, 29, ….
Fridge capacity 5: (5/9)(x − 2) ≤ 5 → x − 2 ≤ 9 → x = 11.
Check x = 11: 9 good, 5 to her fridge, 4 left; she eats 2 and 2 go to the hostel fridge ✓
Options 20 and 29 put 10 and 15 in a 5-capacity fridge; 9 and 14 do not give a whole number.

Answer: B

Approach: Right triangle BEC gives BE; AD ∥ BC makes triangles AED and BEC similar.

Solution:
∠BEC = 90°: BE = √(1.3² − 0.5²) = √(1.69 − 0.25) = √1.44 = 1.2 km.
AD ∥ BC → ΔAED ~ ΔBEC with ratio ED/EC = 2.5/0.5 = 5 → AE = 5 × 1.2 = 6 km.
Route A → B → C → D: AB = AE + EB = 7.2, BC = 1.3, CD = CE + ED = 3.0.
Total = 7.2 + 1.3 + 3.0 = 11.5 km.
The recorded options (15, 108, 88, 112, 125) do not contain 11.5; they appear misrecorded. Cetking's recorded key (A) is kept.

Cetking shortcut: 1.3, 0.5, 1.2 is the 5-12-13 triple scaled by 0.1.

Answer: A

Approach: All meet at the start after LCM of lap times.

Solution:
LCM(60, 90, 40, 80) s: 60 = 2²·3·5, 90 = 2·3²·5, 40 = 2³·5, 80 = 2⁴·5 → 2⁴·3²·5 = 720 s = 12 min (within 30 min).
Laps: A 720/60 = 12, B 720/90 = 8, C 720/40 = 18, D 720/80 = 9.
Total = 12 + 8 + 18 + 9 = 47.

Answer: E

Approach: Numbers = HCF × co-prime ratio terms.

Solution:
Ratio 1 : 4 with HCF 21 → numbers 21 and 84.
Check: LCM(21, 84) = 84 ✓, and 21 × 84 = HCF × LCM = 21 × 84 ✓.
Larger = 84.

Answer: E

Approach: 3 independent customers × 3 colours = 27 equally likely outcomes; count failures (demand > stock).

Solution:
Stock: W 2, B 2, R 1. The store fails if 2 or more customers want red, or all 3 want white, or all 3 want black.
Exactly 2 red: C(3,2) × 2 = 6; 3 red: 1; WWW: 1; BBB: 1 → failures = 9.
P(satisfy all) = (27 − 9)/27 = 18/27 = 2/3.

Why the other options are wrong: 1/3 is the probability of failure, not success.

Answer: B

Approach: Use the 12 km condition to find the stream speed, then apply the round-trip time bound.

Solution:
Still-water speed = 5 km/h, stream = v.
12/(5 − v) − 12/(5 + v) = 1 → 24v = 25 − v² → v² + 24v − 25 = 0 → (v + 25)(v − 1) = 0 → v = 1.
Upstream 4 km/h, downstream 6 km/h.
Round trip: N/4 + N/6 = 5N/12 ≥ 2 → N ≥ 24/5 = 4.8 km.
Minimum N = 4.8 km.

Cetking shortcut: Try v = 1: 12/4 − 12/6 = 3 − 2 = 1 ✓ instantly.

Answer: E

Approach: Equate pairs of line sums to get linear relations among x, y, z.

Solution:
Grid: [3x+4y, 2x, 2x+y+z] / [2x², 4y, y²+z] / [y+z, 3x+2z, z−1].
Row 1 = 7x + 5y + z; Column 2 = 5x + 4y + 2z → z = 2x + y.
Row 3 = 3x + y + 4z − 1; equate with Row 1 → 4x + 4y − 3z + 1 = 0 → with z = 2x + y: y = 2x − 1, z = 4x − 1.
Anti-diagonal = 2x + 6y + 2z; equate with Row 1 → 5x = y + z = 6x − 2 → x = 2, y = 3, z = 7.
N = 7(2) + 5(3) + 7 = 36. Check Row 2: 8 + 12 + 16 = 36 ✓; Column 1: 18 + 8 + 10 = 36 ✓; Main diagonal: 18 + 12 + 6 = 36 ✓

Cetking shortcut: Use only linear lines (Row 1, Column 2, Row 3, anti-diagonal) — avoid the x² and y² cells until the check.

Answer: B

Approach: Work units with LCM; run the 3-day roster cycle.

Solution:
B = 40 days; A twice B → 20 days; A thrice C → C = 60 days.
Work = 120 units: A 6, B 3, C 2 units/day.
Day 1 (A+B) 9, Day 2 (B+C) 5, Day 3 (C+A) 8 → 22 units per 3 days.
5 cycles = 15 days → 110 units. Day 16 (A+B) → 119. Day 17 (B+C) → 124 ≥ 120, job done.
A works 2 days per cycle × 5 = 10, plus day 16 → 11 days.

Cetking shortcut: 120 = 22 × 5 + 10; the leftover 10 needs day 16 (A+B, 9) and day 17 (B+C) — A adds just one day.

Answer: A

Approach: Equal-volume exchange — whatever water entered the glass equals the milk that left into the cup.

Solution:
After step 1: cup = 500 water + 150 milk = 650 ml.
150 ml of cup mixture → water = 150 × 500/650 = 1500/13 ml, milk = 150 × 150/650 = 450/13 ml.
Water in glass = 1500/13 ml. Milk in cup = 150 − 450/13 = 1500/13 ml.
Ratio = 1 : 1.

Cetking shortcut: Each vessel ends with 500 ml, so the water that moved into the glass must equal the milk left in the cup — always 1 : 1.

Answer: A

Approach: By symmetry each position holds each digit equally often; average digit × 1111.

Solution:
Digits 1, 1, 2, 4 → distinct arrangements = 4!/2! = 12.
In each place, the average digit = (1 + 1 + 2 + 4)/4 = 2.
Average number = 2 × 1000 + 2 × 100 + 2 × 10 + 2 = 2 × 1111 = 2222.

Cetking shortcut: Mean of all permutations = (digit sum / number of digits) × 111…1.

Answer: C

Approach: Push the middle two numbers as high as the distinctness constraint allows.

Solution:
Order a < b < c < d < e < f. a + b = 28, e + f = 56.
To raise c, d, make e as large as possible: e < f → e = 27, f = 29.
Then d ≤ 26, c ≤ 25 → c + d = 51 (a, b e.g. 13, 15 stay below c).
Max average = (28 + 51 + 56)/6 = 135/6 = 22.5.

Cetking shortcut: e is at most 27 (just under half of 56), so the middle pair is 26 + 25.

Answer: A

Approach: Add the two equations — the coefficients become equal.

Solution:
2o + 3b + 4a = 15 and 3o + 2b + a = 10.
Add: 5o + 5b + 5a = 25 → o + b + a = 5.
3 oranges + 3 bananas + 3 apples = 3 × 5 = Rs 15.

Cetking shortcut: When individual prices can't be found, look for a combination of the equations that matches the target.

Answer: D

Approach: 'At least one' → 1 − P(none).

Solution:
Total ways = C(10, 3) = 120. No sweater (all caps) = C(6, 3) = 20.
P(at least one sweater) = 1 − 20/120 = 5/6.

Answer: C

Approach: Man-hours are proportional to work done.

Solution:
Phase 1: N × 7 × 10 = 70N man-hours → 35% of work.
Phase 2: (N − 10) × 10 × 14 = 140(N − 10) → 65% of work.
70N/35 = 140(N − 10)/65 → 2N = 28(N − 10)/13 → 26N = 28N − 280 → N = 140.

Cetking shortcut: Test options: N = 140 → 9800 man-hours for 35% (280 per %); 130 × 140 = 18200 for 65% (280 per %) ✓

Answer: E

Approach: Same remainder 4 → number = LCM × k + 4; then impose divisibility by 11.

Solution:
LCM(9, 10, 12, 25) = 2² × 3² × 5² = 900 → x = 900k + 4 < 5000 → k = 0 to 5.
900 ≡ 9 (mod 11) → 9k + 4 ≡ 0 (mod 11) → 9k ≡ 7 → k ≡ 2 (mod 11).
Only k = 2 → x = 1804 (= 11 × 164 ✓).
Teams of 12: 1804 ÷ 12 = 150 remainder 4 → maximum 150 teams.

Cetking shortcut: List 4, 904, 1804, 2704, 3604, 4504 and test divisibility by 11 (alternating digit sums): only 1804.

Answer: A

Approach: Net rate equation with f fill pipes and d drain pipes.

Solution:
f/6 − d/12 = 1/2 → 2f − d = 6 → d = 2f − 6.
n = f + d = 3f − 6 → n must be a multiple of 3 (and d ≥ 1 → f ≥ 4 → n ≥ 6).
From 24, 16, 33, 13, 9, 8: 24 (f = 10, d = 14), 33 (f = 13, d = 20), 9 (f = 5, d = 4) work; 16, 13, 8 do not.
Count = 3.

Cetking shortcut: n = 3f − 6 → just pick the multiples of 3 (≥ 6) from the list.

Answer: D

Approach: Alligation on the fraction of milk.

Solution: Milk fraction: Bottle 1 = 7/9, Bottle 2 = 9/13, target = 3/4.
Bottle 1 − target = 7/9 − 3/4 = (28 − 27)/36 = 1/36.
Target − Bottle 2 = 3/4 − 9/13 = (39 − 36)/52 = 3/52.
Bottle 1 : Bottle 2 = (target − B2) : (B1 − target) = 3/52 : 1/36.
Multiply by 468 (LCM of 52 and 36): 3 × 9 : 1 × 13 = 27 : 13.
Check: 27 parts of 7/9 → 21 milk, 6 water; 13 parts of 9/13 → 9 milk, 4 water; total 30 : 10 = 3 : 1 ✓.

Why the other options are wrong:
C (14:27) / E (17:27): reverse ratios – Bottle 1 is closer to the target, so it must be used more.
A, D: off by one step in the cross-multiplication.

Cetking shortcut: Bottle 1 (0.778) is nearer to 0.75 than Bottle 2 (0.692), so Bottle 1's share is larger – eliminate C and E at once.

Answer: B

Approach: Identical items with minimums: give the minimums first, then distribute the rest freely (stars and bars).

Solution: Minimums: Aai 1 + Bai 2 + Cai 3 = 6 pens.
Remaining = 8 − 6 = 2 pens, distributed freely among 3 people.
Non-negative solutions of x + y + z = 2: C(2 + 3 − 1, 3 − 1) = C(4, 2) = 6.
(List: 200, 020, 002, 110, 101, 011.)

Cetking shortcut: Ways to give r identical items to n people (zero allowed) = C(r + n − 1, n − 1).

Answer: D

Approach: Use the meeting-point distance from A to find the bike's speed, then add the car's distance.

Solution: Let bike speed = x km/h; car speed = 2x.
Bike time (1:00 → 3:40 pm) = 2 h 40 min = 8/3 h. Meeting point is 168 km from A, so (8/3)x = 168 → x = 63 km/h.
Car time (2:00 → 3:40 pm) = 1 h 40 min = 5/3 h. Car distance = 126 × 5/3 = 210 km.
AB = 168 + 210 = 378 km.

Why the other options are wrong:
A, B, C, D: none matches 168 + 210; D (375) is a near-miss trap.

Cetking shortcut: Distances ∝ speed × time = 1 × 8 : 2 × 5 = 8 : 10. Bike's 8 parts = 168 → 1 part = 21 → total 18 parts = 378.

Answer: E

Approach: Check each sentence for count nouns, preposition with time, relative pronoun and collective nouns.

Solution:
A ✗ — 'cars' is countable: 'fewer cars', not 'less cars'.
B ✓ — correct comparative 'nicer than'.
C ✗ — with the present perfect, a starting point takes 'since': 'I have been here since Monday'.
D ✓ — 'know how to swim' is correct.
E — borderline: for a person 'who' is preferred ('the girl who won'), though 'that' is tolerated; the intended correct set excludes it.
F ✓ — 'media' as a plural takes 'are'.
Correct set = B, D & F.

Why the other options are wrong: A (B, C & E) and B (A, D & E) include the 'from Monday' or 'less cars' errors; C (A, D & F) includes 'less cars'; E (D, E & F) leaves out the clearly correct B and includes E.

Cetking shortcut: B is clearly correct → only options A and D contain B; A contains C ('from Monday') → D.

Answer: D

Approach: Fixed prepositions with common verbs/adjectives (discuss, good at, emphasize, superior to).

Solution: Check each sentence:
A. 'asked for additional funding' – correct.
B. 'discussed about' – wrong; discuss takes a direct object: 'discussed the new project'.
C. 'good in' – wrong; idiom is 'good at data interpretation'.
D. 'transitioning into a new phase' – correct.
E. 'emphasized for the need' – wrong; 'emphasized the need' (or 'laid emphasis on').
F. 'superior to the previous one' – correct ('superior' takes 'to', not 'than').
Correct set = A, D, F.

Why the other options are wrong:
A (C, E, F): C and E are wrong.
B (B, C, F): B and C are wrong.
C (A, B, C): B and C are wrong.
D (B, C, D): B and C are wrong.

Cetking shortcut: Spot B ('discussed about') and C ('good in') as wrong – every option except E contains at least one of them.

Answer: E

Approach: Find the opener that introduces the idea, then follow pronoun/reference links.

Solution: B opens the topic: the final ingredient for a habitable world is an energy source.
A applies it to Europa: whatever 'that' (the energy source) is there, it will not be sunlight.
E reacts: 'That is a bit of a problem.'
C explains why it is a problem: on Earth almost everything depends on photosynthesis.
D gives the exception: 'Their inhabitants' (of places without sunlight) do not use sunlight directly.
Order: B, A, E, C, D.

Why the other options are wrong:
A (D…): D begins with 'Their', which needs an antecedent – cannot open.
B (C, E…): 'That is a bit of a problem' after C makes no sense – photosynthesis on Earth is not a problem.
C (A…): A's 'that' refers to the energy source in B, so A cannot open.
E (E…): E is a reaction and cannot open the paragraph.

Cetking shortcut: Pair A–E: 'it will not be sunlight' → 'That is a bit of a problem'. Only option D has B-A-E together.

Answer: D

Approach: Use the tone of the passage (irony about elders judging the young) to fill all four blanks.

Solution: Blank 1: 'each ___ generation grows up' → successive (one after another).
Blank 2: we forget what it feels like to be young.
Blank 3: most adults remember their own youthful indiscretions with a wink.
Blank 4: but they condemn the same behaviour in the next generation – that is hypocritical, so 'hypocritically sound the alarm'.
Set: successive, young, adults, hypocritically.

Why the other options are wrong:
A: 'next' (blank 2) makes no sense; 'naturally' misses the irony.
B: 'juvenile' is a noun/adjective misfit here and 'carefully' misses the double standard.
D: 'youngsters' cannot recall 'their own youthful indiscretions'; 'sincerely' contradicts the irony.
E: 'preceding' generation growing up is illogical; 'cautiously' misses the irony.

Cetking shortcut: The last blank decides it: wink at your own faults but alarm at others' = hypocrisy. Only C has it.

Answer: C

Approach: Main argument: ranking people by common criteria produces homogeneity, so hiring top scorers does not give the best (diverse) team. Attack that link.

Solution: The author's case rests on the premise that top scorers on common tests are alike, so a team of them lacks diversity. If top scorers themselves possessed multidisciplinary knowledge and could see a problem from several perspectives, hiring the 'best' would automatically bring diversity – the conflict between merit and diversity disappears and the argument collapses.

Why the other options are wrong:
A: making assessors 'insensitive' is irrelevant to the argument.
B: a better algorithm than random forests removes only one example, not the argument.
D: diverse teams being slow does not show they produce worse solutions; it only adds a cost.
E: more rigorous tests still rank on common criteria – the author says no test can capture a whole domain.

Cetking shortcut: To invalidate, deny the key premise ('ranking → homogeneity'); look for the option where top scorers are themselves diverse.

Answer: C

Approach: EXCEPT question: four options are reasons the author gives; find the one the author argues against.

Solution: Author's reasons: problems are multidimensional and need varied skills (para 1); problems are complex (para 1); no test can measure expertise in a deep domain (para 2); optimal hiring depends on context and optimal teams are diverse, which leads to breakthroughs (paras 2–4).
D – 'an ideal team comprises the best individuals from diverse fields' – is exactly the position of 'believers in a meritocracy' in para 2, which the author says 'suffers from a similar flaw'. It is not his reason; it is what he rejects.

Why the other options are wrong:
A: stated – complex problems need diverse knowledge bases, tools and skills.
B: stated – 'optimal hiring depends on context', diverse ideas rank high-impact.
C: stated – 'the complexity of modern problems'.
E: stated – 'no test or criteria applied to individuals will produce the best team… no test can exist'.

Cetking shortcut: In EXCEPT questions, look for the option that repeats the opponent's view – here the 'best mathematicians, best oncologists' line.

Answer: D

Approach: Apply the passage: an effective team is diverse across fields and not built by picking top scorers.

Solution: Obesity has causes in nutrition, psychology, transport/infrastructure and media (para 1). So the team must span these fields – options A and D both do.
The difference: D picks people who 'scored a distinction' – i.e. the top scorers in each category, exactly the 'best within each category' approach the author rejects. A only says members performed well, i.e. they are competent, not selected by ranking. So A fits the author's view best.

Why the other options are wrong:
B: one field only (psychology) and one shared experience – no diversity.
C: nutritionists only; knowing the random-forest algorithm is a misread of an example.
D: diverse fields but selected as top scorers – the meritocratic flaw the author criticises.
E: 'specialised team of top nutritionists' – single field and ranked by merit.

Cetking shortcut: Eliminate single-field teams (B, C, E), then reject the one that stresses top scores (D).

Answer: A

Approach: Purpose of an example = the claim it supports, stated just before it.

Solution: Just before the example the author says: 'Each of these domains possesses such depth and breadth, that no test can exist. Consider the field of neuroscience.' Neuroscience (50,000 papers a year, molecules to networks) is an illustration of a general point: every field is too vast for a meaningful ranking of merit. So E.

Why the other options are wrong:
A: oncology and biostatistics are not linked to neuroscience in the passage.
B: opposite – the author says such assessment is impossible.
C: anxiety of workers is never discussed.
D: 'unlike other fields' is wrong – neuroscience is a representative example, not an exception.

Cetking shortcut: 'Consider X' examples usually illustrate the general sentence right before them – read that sentence.

Answer: E

Approach: Draw a conclusion that stays within the sentence: engaging with words is a rich, many-sided encounter with the world.

Solution: The sentence calls engagement with words 'captivating, enchanting and sometimes daunting' – an encounter with the world through words. The passage adds that words give warmth or coldness, joy, light, and let us 'see a world in a grain of sand'. So words can show the world in all its varied shades – option C.

Why the other options are wrong:
A: manipulation of others is not suggested.
B: faithfulness of words to the author is never discussed.
D: a caution about using words is advice, not a conclusion from the statement.
E: 'atrocities' is too narrow and negative; 'daunting' is only one shade.

Cetking shortcut: Three adjectives of different shades (captivating, enchanting, daunting) → the answer is the option about variety.

Answer: C

Approach: Read the phrase with its own sentence: 'the world within the word is the abode of faith'.

Solution: The author contrasts scripture ('the word is the world') with his own view that the world inside words is 'the abode of faith'. Calling it a home of faith gives words a sacred, spiritual meaning – a spiritual exposition. So B.

Why the other options are wrong:
A: exploring cultures is not mentioned.
C: inspiration and energy are added – the line speaks of faith only.
D: love for the global community is not mentioned.
E: mystery is in the opening line, but the phrase itself is tied to faith, not mystery.

Cetking shortcut: Key word 'abode of faith' + 'Scriptures say' → spiritual.

Answer: B

Approach: Central theme = what every sentence of the passage is about.

Solution: Every line describes what words do for humans: they convince us of our existence, carry emotion, light up darkness, save us like rafts, let us explore the universe and give form to the formless. The theme is the impact of words on human existence – C.

Why the other options are wrong:
A: storytelling is never discussed.
B: syntax is not mentioned.
D: no history of language is given.
E: cognition is not the topic; the mystery is of words, not the mind.

Answer: C

Approach: Fill each blank by grammar (part of speech) and meaning.

Solution: I: 'he ___ from his current job' needs a past-tense verb taking 'from' → resigned.
II: 'more ___ than a lie' needs an adjective → shocking (truth is often more startling than a lie).
III: willpower that becomes ___ as we use it → depleted (used up).
Set: resigned, shocking, depleted.

Why the other options are wrong:
B: 'he depleted from his job' is meaningless.
C: 'he shocks from his job' is meaningless.
D: 'he shocked from his job' is meaningless; 'truth is more depleted' makes no sense.
E: 'depleted from his job' is meaningless; willpower does not become 'resigned'.

Cetking shortcut: Blank I alone settles it: only 'resigned' goes with 'from his job'.

Answer: A

Approach: Check each sentence against standard rules: fewer/less, since/from, who/that for persons, collective 'media'.

Solution: A. 'less cars' – wrong; countable noun → 'fewer cars'.
B. 'nicer than her sister' – correct comparative.
C. 'been here from Monday' – wrong; with perfect continuous/perfect tense and a point of time use 'since Monday'.
D. 'I know how to swim' – correct.
E. 'the girl that won' – in exam grammar the relative pronoun for a person is 'who': 'the girl who won'.
F. 'The media are divided' – correct; 'media' is the plural of 'medium' and takes a plural verb.
Correct: B, D, F.

Why the other options are wrong:
A (B, D, E): E uses 'that' for a person.
B (B, C, E): C needs 'since'.
C (C, D, F): C needs 'since'.
D (A, D, E): A needs 'fewer'.

Cetking shortcut: Kill A and C first (fewer/since) – that leaves options A and E; decide between E and F: 'media are' is textbook-correct.

Answer: E

Approach: Use time markers (1992, 1997, 1999) and linking words ('But', 'That … also').

Solution: C introduces the topic: an external shock – entry into the euro in 1999.
E gives the background: Germany designed the 1992 Maastricht criteria to keep Italy out.
A states those tests seemed impossible for Italy to pass by 1999.
D – 'But' in 1997 Prodi decided Italy must join anyway.
B – what helped: Belgium's similar debt and a special euro tax.
Order: C, E, A, D, B.

Why the other options are wrong:
A (E, D, B, A, C): A (tests seemed impossible) after B (what helped Italy pass) breaks the logic.
B (C, E, B, A, D): B ('helped') comes before Italy even decided to join (D).
C (E, B, C, A, D): B has nothing to have 'helped' yet.
D (C, B, D, E, A): B again too early; E (1992 background) after D (1997) breaks chronology.

Cetking shortcut: B ('That Belgium also … helped') must follow D (Prodi's decision) – only option E has D→B.

Answer: E

Approach: Replace the made-up word with what the context demands.

Solution: A 'sensitive traveller' sees streets crowded with ragged beggar women and their children begging for alms. A sensitive person feels sadness and pain at such a sight – it is distressing. (The original line, from Swift, reads 'It is a melancholy object…'.)

Why the other options are wrong:
A: 'irritating' suggests annoyance, not a sensitive person's pity.
C: 'refuting' (disproving) does not fit grammatically or logically.
D: 'disgusting' is contempt, opposite to sensitivity.
E: 'amusing' is the opposite of the tone.

Cetking shortcut: The word 'sensitive' tells you the reaction is sympathy/sorrow.

Answer: B

Approach: Locate the line on inequalities.

Solution: The passage says: 'A stark discrepancy emerges between preparation for board examinations and competitive tests, amplifying the inequalities that plague the education system.' So inequalities already exist and the competitive-exam system enlarges them – A.

Why the other options are wrong:
B: board exams hiding or revealing inequality is not discussed.
C: opposite – competitive exams amplify inequality.
D: the system is 'plagued' by inequalities, not 'accommodating' them.
E: opposite – the author says there is a stark discrepancy.

Cetking shortcut: 'Amplifying' = 'enlarged' – direct paraphrase.

Answer: A

Approach: Locate the line on online platforms and do not go beyond it.

Solution: 'Technology emerges as a potential equaliser… Online platforms now offer preparation opportunities from the comfort of one's home.' So, as per the author, online platforms help/encourage students to prepare for exams – C.

Why the other options are wrong:
A: harm to mental health is said of social media, not educational platforms.
B: opposite – the author sees them as increasing accessibility.
D: 'only means' is an extreme claim not made.
E: 'potential equaliser' only; guaranteeing equal opportunity to all irrespective of caste/creed overstates it.

Cetking shortcut: Words like 'potential' and 'could' rule out absolute options such as D and E.

Answer: C

Approach: NOT-a-challenge question: find the item the passage presents as a help, not a problem.

Solution: The challenges named: social-media algorithms harming mental health and attention (B); misunderstood, unexamined idea of merit (C); stress, sorrow and self-doubt of students and parents (E); a high-pressure exam environment. The passage ends by presenting online platforms as a 'potential equaliser' offering preparation from home – a solution, not a challenge. So D.

Why the other options are wrong:
A: the passage never lists a lack of competitive atmosphere as a challenge, but D is the item it explicitly presents as a remedy (a 'potential equaliser'), so D is the clear answer.
B: stated challenge – social media sows loneliness and hurts attention.
C: stated challenge – merit is misunderstood and unexamined.
E: stated challenge – sorrow, inadequacy, self-doubt.

Cetking shortcut: The option the passage explicitly calls a remedy is the 'NOT a challenge' answer.

Answer: D

Approach: Strengthen = support the author's claim that merit is nuanced and not captured by one exam.

Solution: The author cites Sandel's 'tyranny of merit' and the Supreme Court's call for a deeper evaluation of the 'idea of merit', 'highlighting its nuanced nature'. A statement that true merit is multifaceted and goes beyond one competitive exam directly supports this – A.

Why the other options are wrong:
B: 'abstract' is vague; it neither supports nor attacks the nuance argument.
C: opposite – merit 'goes unexamined within school curriculum'.
D: direct negation of the author's view – weakens.
E: against the passage; a single exam amplifies, not identifies, inequalities.

Cetking shortcut: D is the exact negation of A – when two options are opposites, one of them is usually the answer.

Answer: A

Approach: Invented-word question: eliminate options that cannot both be right, then use the speaker's tone.

Solution: 'Funny' and 'Mirthful' mean the same thing, so neither can be the single correct answer. Among Lengthy, Didactic and Terse, a quip (a short witty remark) mocking a communiqué best fits one that is curt and clipped – terse (brief and abrupt).

Why the other options are wrong:
A: 'funny' = 'mirthful' – two synonyms cannot both be the key.
B: 'lengthy' – no hint of length in the line.
C: same as A.
D: 'didactic' (preachy) has no clue in the sentence.

Cetking shortcut: When two options are synonyms, eliminate both.

Answer: E

Approach: Root: elegy = a poem of lament for the dead.

Solution: Elegiac = expressing sorrow or lament, like an elegy → mournful.

Why the other options are wrong:
A: testing – unrelated.
C: peaceful – calm, not sorrowful.
D: restless – agitation, not grief.
E: confused – unrelated.

Cetking shortcut: Elegiac ↔ elegy ↔ funeral poem → sad.

Answer: B

Approach: 'Can NOT be drawn' – find the option that contradicts the passage.

Solution: The passage says the professional classes 'have figured out how to pass their advantages on to their children, converting the meritocracy into a hereditary aristocracy', and the wealthy 'have rigged the system'. So wealth and privilege clearly can undermine meritocracy. Option C says they cannot – it contradicts the passage and cannot be inferred.

Why the other options are wrong:
A: drawn – 'Our disagreements are less about the principle itself than about what it requires.'
B: drawn – 'meritocracy is a myth, a distant promise yet to be redeemed'; colleges favour the well-connected.
D: drawn – 'no one defends hereditary privilege outright', yet abolishing inheritance taxes is said to condone inherited privilege.
E: drawn – 'At the level of… political rhetoric, meritocracy has won the day'; centre-left and centre-right both claim it.

Cetking shortcut: Look for the 'but… cannot' twist – C starts like the passage and then reverses it.

Answer: C

Approach: Use the line stating the defenders' position.

Solution: 'Defenders of affirmative action reply that such policies are necessary to make equality of opportunity a reality for members of groups that have suffered discrimination or disadvantage.' So their problem with a purely merit-based system is that it ignores the disadvantage some groups start with – B.

Why the other options are wrong:
A: hereditary privilege is a separate complaint, not the defenders' argument.
C: rewarding outcomes is the nature of meritocracy; the defenders do not object to it.
D: 'intentionally' is too strong; the defenders speak of disadvantage, not deliberate favouritism.
E: 'structural bias' is broader than what the passage attributes to them.

Cetking shortcut: Paraphrase match: 'suffered discrimination or disadvantage' = 'initial disadvantages in opportunities'.

Answer: B

Approach: A conclusion must follow from the passage as a whole, not just repeat one complaint.

Solution: The passage shows that everyone accepts equal opportunity in principle, yet in practice the wealthy rig the system, the professional classes pass advantages to their children and colleges favour the well-connected – meritocracy turns into 'a hereditary aristocracy'. So when meritocracy is executed in an unequal society, existing advantages carry over and inequality is perpetuated – B.

Why the other options are wrong:
A: 'not understood by anyone' – not supported; the debate is about what the principle requires.
C: 'despised by those rejected' – not discussed.
D: the passage says complaints are 'not about the ideal'; calling meritocracy utopian and blaming only the wealthy's rigging is a partial, critic's view, not the passage's conclusion.
E: hypocrisy of individuals is not discussed.

Cetking shortcut: Combine the two halves of the passage (principle accepted + outcome hereditary) – the option that links them is the conclusion.

Answer: B

Approach: Suffix '-aceous' = resembling / of the nature of; 'viola' = violet.

Solution: Violaceous = violet or purplish in colour → associated with colour.

Why the other options are wrong:
B, C: 'violent' is a false link – violaceous has nothing to do with violence or arson.
D, E: unrelated to the root.

Cetking shortcut: Words ending in '-aceous' after a colour/plant name (violaceous, olivaceous, rosaceous) describe colour or appearance.

Answer: A

Approach: Root 'olive' + '-aceous' (resembling).

Solution: Olivaceous = olive-green / of an olive colour → associated with colour.

Why the other options are wrong:
A, B, C, E: theatre, foundry, delinquency and reproduction have no link to 'olive'.

Cetking shortcut: Same pattern as violaceous – colour word + '-aceous'.

Answer: D

Approach: Greek 'emetos' = vomiting (as in 'emetic').

Solution: Emetophobia = fear of vomiting → vomit.

Why the other options are wrong:
A, B, C: escalator, elevator, electricity – only share the letter 'e'.
D: vanity – unrelated.

Cetking shortcut: Emetic = a medicine that causes vomiting – same root.

Answer: E

Approach: Greek 'alektor' = rooster/cock.

Solution: Alektorophobia = fear of chickens → chicken.

Why the other options are wrong:
A, B: electricity and shock – a sound-alike trap ('alek' ~ 'elec').
D, E: giraffe and coal – unrelated.

Answer: C

Approach: Greek 'hypsos' = height.

Solution: Hypsophobia = fear of heights (same as acrophobia) → heights.

Why the other options are wrong:
A, B, C, E: rivers, magicians, prayers and hats have no link to 'hypsos'.

Cetking shortcut: Hypsometer = instrument to measure height – same root.

Answer: D

Approach: Attenuate = to weaken or reduce in force or intensity; find the opposite.

Solution: Opposite of weakening/reducing = increasing in strength → amplify.

Why the other options are wrong:
A: reduce – a synonym, not an antonym.
B: petiteness – a noun meaning smallness; wrong part of speech.
C: pugnacious – eager to fight; unrelated.
D: miserable – unrelated.

Cetking shortcut: Antonym questions often place a synonym (reduce) as the trap.

Answer: E

Approach: Entente = a friendly understanding or informal agreement between parties/countries.

Solution: Opposite of an agreement/understanding → disagreement.

Why the other options are wrong:
A: sentence – sound-alike, unrelated.
B: decadent – morally declining.
C: identity – unrelated.
D: bequeath – to leave by will.

Answer: E

Approach: Martinet = a very strict disciplinarian who insists on exact obedience; find the option most opposite in spirit.

Solution: A martinet is stern, rigid and humourless. Among the options, only mirthful (cheerful, light-hearted, easy-going) stands opposite to that sternness.

Why the other options are wrong:
A: scullion – a kitchen servant; not an opposite.
B: Martian – sound-alike trap.
C: scything – cutting with a scythe; unrelated.
D: enviable – desirable; unrelated.

Cetking shortcut: Eliminate sound-alikes (Martian) and unrelated words; pick the one opposite in temperament.

Answer: E

Approach: Identify the relationship: peel and peal are homophones (same sound, different spelling and meaning).

Solution: Peel (skin of fruit) and peal (ringing of bells) sound alike. Rain (water from clouds) and reign (rule) sound alike too – homophones.

Why the other options are wrong:
A: coat : rind – both coverings; synonyms, not homophones.
B: brain : cranium – thing and its protective case.
D: bright : brighter – degrees of comparison.
E: laugh : bell – 'peal of laughter/bells' association, not the pair's relation.

Answer: C

Approach: Relationship: doggerel is inferior, low-quality work produced in a poet's field.

Solution: Doggerel = badly written, trivial verse; a poet writes verse. Similarly, pulp fiction = cheap, low-quality fiction; a novelist writes fiction. Both pairs: low-quality product : its maker.

Why the other options are wrong:
A: symphony : composer – a high-quality, serious work, not inferior.
C: kitten : veterinary doctor – patient : doctor.
D: wine : vintner – product : maker, but with no sense of poor quality.
E: prediction : astrologer – output : maker, no quality sense.

Cetking shortcut: Spot the extra layer (poor quality) – it separates B from D/E.

Answer: B

Approach: Relationship: a barge is a type of vessel (specific : general).

Solution: Barge → a kind of vessel. Shovel → a kind of implement (tool). Same 'type of' relation.

Why the other options are wrong:
A: mouse : pad – used together, not a type.
B: rim : edge – synonyms.
C: training : preparation – near-synonyms / purpose.
E: book : anthology – part : collection (reversed whole).

Cetking shortcut: Make a sentence: 'A barge is a type of vessel' – only 'A shovel is a type of implement' fits.

Answer: D

Approach: Choose the idiomatic phrase for steady growth over time.

Solution: 'Increases' describes a continuing, gradual rise. The idiom 'year by year' means gradually, with each passing year, and fits naturally: 'the talent is prodigious, and it increases year by year.'

Why the other options are wrong:
A: 'each year' – states frequency only, not the steady, gradual build-up that 'year by year' conveys.
B: 'progressively' – grammatical but vague; does not convey the yearly growth.
C: 'annually' – means 'once a year' (an annual event), not continuous growth.
E: 'yearwise' – not standard usage.

Cetking shortcut: 'X by X' idioms (day by day, year by year) express gradual change – they pair naturally with 'increases'.

Answer: D

Approach: Test each specific reason against the passage; if none is stated, the 'no specific answer' option wins.

Solution: The author's displeasure is spread across many points – patchy coverage, schemes that 'lack depth', compartmentalising the poor, stigma, and the burden shifted from the state to the individual. None of the specific reasons in A–D appears in the passage, so the best answer is that the author gives no single specific answer of the kind offered – E.

Why the other options are wrong:
A: 'fictive use of job cards' is never said; job cards are mentioned only as a delivery system to reform.
B: schemes 'penetrating the cause of debt' is not stated.
C: the passage says schemes are wrongly viewed as favours, but not that they create a bias against below-average earners.
D: the author never blames the schemes for hunger and poverty; he calls them the most consistent support.

Answer: E

Approach: Check each statement against what the author blames for the debt culture.

Solution: Statement 1 – changing lifestyle: never mentioned.
Statement 2 – the author ends by saying the 'welfare model' is no longer about enabling a moral citizen and that the decline in morals will move in one direction: the erosion of moral institutions that once supported prudence is what he blames.
Statement 3 – the author does not say loans are given to every section regardless of ability to repay; he criticises selective incentives and policy, not universal easy credit.
Only statement 2 is supported.

Why the other options are wrong:
A (only 1): lifestyle is not discussed.
C (only 3): universal easy loans are not the author's claim.
D (1 & 2): statement 1 is unsupported.
E (1 & 3): neither 1 nor 3 is the author's point.

Answer: B

Approach: Subject–verb agreement: the subject is 'industry' (singular).

Solution: 'Poorly regulated pharmaceutical industry mean that…' – the head noun 'industry' is singular, so the verb must be 'means'. The error is in part (B).

Why the other options are wrong:
A: 'poorly regulated pharmaceutical' – correct modifiers.
C: 'are freely available to' – agrees with 'antibiotics' (plural).
D: 'those who can afford them' – correct.
E: there is an error.

Cetking shortcut: Find the verb, ask 'who means?' – 'industry' → singular.

Answer: B

Approach: Open with the general claim, then describe the illness from first symptom to worst stage.

Solution: B introduces the topic: India could be under attack from germs causing atypical pneumonias.
D: 'It appears like ordinary flu' but drugs don't work – the deceptive nature.
A: 'It begins with an ordinary fever and a moderate cough' – the start of the disease.
C: 'Slowly' it progresses to bronchitis, pneumonia and respiratory complications – the end stage.
Order: B, D, A, C.

Why the other options are wrong:
A (BCDA): C (worst stage) cannot come before A (how it begins).
B (ABCD): 'It' in A has no antecedent as an opener.
C (ADCB): A cannot open; B is the topic sentence.
D (BDCA): progression (C) before beginning (A) breaks the sequence.

Cetking shortcut: 'Begins' (A) must come before 'slowly progresses' (C) – only E has A before C after B.

Answer: E

Approach: Narrative order: setting and mood → cause of the mood → effect → final picture.

Solution: B sets the scene: that wintry afternoon, waiting in a French restaurant, I felt humiliated and depressed.
D gives the cause: a project had fallen through because of my miscalculations.
C shows the effect: even the prospect of meeting a dear friend did not cheer me.
A closes the image: I sat frowning at the tablecloth, 'chewing the bitter cud of insight'.
Order: B, D, C, A.

Why the other options are wrong:
B (DBCA): D ('Because of…') as an opener leaves the setting for later.
C (CDAB): C's 'Even the prospect…' needs a mood established first.
D (BCAD): the cause (D) placed at the end is illogical.
E (BACD): C ('Even…') should follow the cause, not A.

Cetking shortcut: Mood (B) → reason (D) is the strongest pair; only option A has B immediately followed by D.

Answer: A

Approach: Read the phrase in its full form: 'a female-shaped absent presence'.

Solution: Women's lives are missing from culture and data (absence), but the gap they leave has a definite female shape – it is visible and has consequences everywhere (presence). The phrase is explained by the shape of the absence: because it is female-shaped, the absence is felt as a presence – A.

Why the other options are wrong:
B: 'specificity' is vague and drops the female shape the author stresses.
C: a generic definition of any absent presence; it ignores what makes this one present (its female shape).
D: making a case for presence is the author's purpose, not the meaning of the phrase.
E: 'recognise its presence' does not explain how the absence becomes present.

Cetking shortcut: Pick the option that keeps the passage's own qualifier ('female-shaped') – generic dictionary meanings are the traps.

Answer: A

Approach: A conclusion must combine the passage's two parts: the old gender data gap and the new reliance on data.

Solution: The author says the gap is old and not deliberate; what is new is that the world is 'increasingly reliant on and in thrall to data'. Big data 'corrupted by big silences' yields half-truths or untruths for women, and the consequences can be deadly (car safety, misdiagnosed heart attacks). So emphasis on data-based decisions can be devastating to women given the gap – D.

Why the other options are wrong:
A: the author says the gap is 'not generally malicious, or even deliberate' – 'men ignored women' overstates.
B: a prescription; the passage does not call for revisiting the past.
C: data-based decisions amplify the consequences, not the gap itself.
E: 'never treated as distinct identities' is too absolute and is a cause, not the conclusion.

Cetking shortcut: The 'what is new' sentence usually carries the author's conclusion.

Answer: D

Approach: Greek 'kalos' = beautiful + '-logy' = study.

Solution: Kalology = the science or study of human beauty.

Why the other options are wrong:
B: etiology (aetiology) – study of causes, especially of disease.
C: physiology – study of body functions.
D: cosmology – study of the universe.
E: osteology – study of bones.

Cetking shortcut: 'Kal-' as in calligraphy (beautiful writing) and kaleidoscope (beautiful forms).

Answer: A

Approach: Fixed prepositions with verbs/adjectives.

Solution: 1. 'sitting at the desk' – correct.
2. 'superior to yours' – correct.
3. 'prefer coffee than tea' – wrong; 'prefer X to Y'.
4. 'accused for stealing' – wrong; 'accused of stealing'.
5. 'an exception to the rule' – correct.
6. 'leaving to England' – wrong; 'leaving for England'.
Incorrect: 3, 4, 6.

Why the other options are wrong:
A (1, 4, 6): sentence 1 is correct.
B (1, 2, 6): 1 and 2 are correct.
D (3, 4, 5): 5 is correct.
E (4, 5, 6): 5 is correct.

Cetking shortcut: prefer to, accused of, leave for, superior to, exception to – memorise these pairs.

Answer: C

Approach: 'Into' shows movement from outside to inside; 'in' shows position.

Solution: The boatman threw the oar – an action of motion ending inside the water – so it should be 'threw the cracked oar into the river'. Part (C) has the error.

Why the other options are wrong:
A: 'The angry boatman threw' – correct.
B: 'the cracked oar' – correct.
D: 'and returned home' – correct ('home' needs no preposition).
E: there is an error.

Cetking shortcut: Verbs of motion (throw, jump, fall, walk) + entering → 'into'.

Answer: C

Approach: Build the word from its root: Latin 'pulcher/pulchritudo' = beauty.

Solution: Pulchritude (beauty) + '-inous' → pulchritudinous (beautiful). Spell it: pul-chri-tu-di-nous.

Why the other options are wrong:
A: 'Pulchritudinuous' – extra 'u' before -ous.
B: 'Pulchirutudionous' – 'chiru' and 'dio' wrong.
C: 'Pulchirutudinous' – 'chiru' instead of 'chri'.
E: 'Pulchritudonous' – 'do' instead of 'di'.

Cetking shortcut: Start from the noun 'pulchritude' and add '-inous' (like platitude → platitudinous).

Answer: D

Approach: Pair the collocation 'manners and morals', then the literary term 'recurrent theme'.

Solution: Blank 1: 'manners and morals' is a standard pair describing social behaviour and values of the newly rich.
Blank 2: a topic that keeps appearing in literature is a 'recurrent theme'.
Set: morals … theme.

Why the other options are wrong:
B: 'style … motif' – 'manners and style' is weak; a motif is a recurring image, not a subject.
C: 'recurrent story' is unidiomatic.
D: 'manners and wealth' does not pair; wealth is already in 'nouveau riche'.
E: 'tableau … milieu' – a milieu is a setting, not something recurrent in literature.

Cetking shortcut: 'Recurrent theme' is a fixed phrase – it leaves only A and D; 'manners and morals' wins.

Answer: A

Approach: Check successive differences; here they are squares of even numbers in descending order.

Solution:
789 − 12² (144) = 645
645 − 10² (100) = 545
545 − 8² (64) = 481
481 − 6² (36) = 445 (the series shows 440)
445 − 4² (16) = 429
429 − 2² (4) = 425
Every term fits except 440; the correct term is 445.

Why the other options are wrong: 789, 545, 429 and 425 all sit exactly on the −144, −100, −64, −16, −4 chain. 429 looks odd only if you take 440 as correct; check it from 445 instead.

Cetking shortcut: Differences 144, 100, 64, 41, 11, 4 – two broken differences next to each other point to the single term between them (440).

Answer: C

Approach: The hands coincide 11 times in every 12 hours (once every 12/11 h = 65 5/11 min).

Solution:
10:20 AM, 8 Mar → 10:20 AM, 11 Mar = 3 days = 72 hours = 6 blocks of 12 hours.
Coincidences = 6 × 11 = 66.
Counting full 12-hour blocks gives the intended answer, 66 (the last 55 minutes add one more overlap at about 10:54 AM on 11 March, but 67 is not among the options).

Why the other options are wrong: 45 and 50 undercount; 89 counts more than 11 per 12 hours; 25 is roughly one day's count.

Cetking shortcut: 22 coincidences per day × 3 days = 66.

Answer: D

Approach: Multiplication principle for routes; set up equations in the unknown path counts.

Solution:
Let Q–R = m direct paths and R–S = n direct paths; P–Q = 3, Q–S = 4, P–R = 4, P–S = 0.
P to S: P→Q→S = 3×4 = 12; P→R→S = 4n; P→Q→R→S = 3mn.
12 + 4n + 3mn = 62 → 4n + 3mn = 50 … (1)
Q to R: direct = m; Q→P→R = 3×4 = 12; Q→S→R = 4n.
m + 12 + 4n = 27 → m + 4n = 15 … (2)
(1) − (2): 3mn − m = 35 → m(3n − 1) = 35 = 7 × 5.
3n − 1 = 5 gives n = 2, m = 7; check (2): 7 + 8 = 15 ✓.

Cetking shortcut: From (2), m = 15 − 4n; try n = 1, 2, 3 → only n = 2 (m = 7) satisfies (1): 8 + 42 = 50.

Answer: E

Approach: Express apples and bananas in terms of total T, then apply integer constraints.

Solution:
Mangoes = 0.4T, so bananas + apples = 0.6T. Let apples = A.
Sold = 0.2T (half mangoes) + 96 + 0.4A = 0.5T → 0.4A = 0.3T − 96 → A = 0.75T − 240.
A ≥ 1 → T ≥ 321.4.
Integers: half of 0.4T must be whole → T multiple of 5; 0.75T whole → T multiple of 4; so T is a multiple of 20 (then 0.4A is also whole).
Smallest multiple of 20 above 321.4 = 340.
Check: mangoes 136 (sell 68), apples 0.75×340 − 240 = 15 (sell 6), bananas 340 − 136 − 15 = 189 (sell 96). Sold = 68 + 96 + 6 = 170 = 50% ✓.

Why the other options are wrong: 345 is not a multiple of 4 (0.75T not whole); 225 and 196 give negative apples; 448 is not a multiple of 5 (mangoes 0.4T = 179.2 is not whole), so it is invalid.

Cetking shortcut: Need T > 320 and T a multiple of 20 → among the options only 340 qualifies.

Answer: B

Approach: The age gap d between two people never changes.

Solution:
Let Garima now = x, Rajesh now = x + d.
When Rajesh was x (d years ago), Garima was x − d. Ratio x : (x − d) = 3 : 2 → 2x = 3x − 3d → x = 3d.
Now: Garima = 3d, Rajesh = 4d.
When Garima becomes 4d (d years later), Rajesh = 5d.
Ratio = 5d : 4d = 5 : 4.

Why the other options are wrong: 4 : 5 reverses the order (Rajesh is older); 4 : 3 is the present ratio; 2 : 3 and 7 : 6 do not keep a constant gap.

Cetking shortcut: Past, present, future ages form 2, 3, 4, 5 (in units of d): then Rajesh 3 : Garima 2, now 4 : 3, later 5 : 4.

Answer: A

Approach: Split the series into blocks of four consecutive letters.

Solution:
Block 1: m n o p
Block 2: n o p q (starts one letter later)
Block 3: o p q r
Block 4 must start at p: p q r s.

Why the other options are wrong: q r s t is block 5, not block 4; m n o p repeats block 1; o q r s and m o p q break the consecutive-letter rule.

Cetking shortcut: The first letters of the blocks (m, n, o, …) give the start of the missing block directly.

Answer: E

Approach: Find one rule that turns 6 into 43, then apply it to 4.

Solution:
6 → 6 × 7 + 1 = 43, i.e. n × (n + 1) + 1 (same as n² + n + 1).
4 → 4 × 5 + 1 = 21.

Why the other options are wrong: 44 and 45 are near-miss values; 29 = 4 × 7 + 1 reuses the multiplier 7 instead of n + 1; 88 has no link to 4 under this rule.

Cetking shortcut: 43 is one more than 42 = 6 × 7, a product of consecutive numbers; for 4 that is 4 × 5 = 20, plus 1.

Answer: E

Approach: Compare alphabet positions of each letter and its code.

Solution:
D(4)→V(22), I(9)→Q(17), A(1)→Y(25), M(13)→M(13), O(15)→K(11), N(14)→L(12), D(4)→V(22).
Each pair adds to 26, so code = 26 − position.
FEMALE: F(6)→20 T, E(5)→21 U, M(13)→13 M, A(1)→25 Y, L(12)→14 N, E(5)→21 U → TUMYNU.

Why the other options are wrong: TUMZOU uses the plain reverse (27 − n) for A and L; TVNYNV, UVNZOV, UVNYNU shift the letters of F/E/M wrongly.

Cetking shortcut: M stays M (13 + 13 = 26) – so check the option with M in third place: only TUMZOU and TUMYNU; A → Y (not Z) picks TUMYNU.

Answer: E

Approach: Draw the family tree and locate the couple that has both parents and children in the family.

Solution:
A is the son of C; E is the daughter of A; F is the mother of E → A (male) and F (female) are husband and wife.
D is the daughter of F → D and E are the children of A and F.
So the couple with parents and children is A–F. Their parents: C is A's parent and the only remaining member B must be C's spouse.
Parents of the couple = B and C.

Why the other options are wrong: B A and A F include A, who is part of the couple; C F – F is a spouse, not a parent; D E are the children.

Answer: E

Approach: Use the tree from the common information and mark genders.

Solution:
Females certain: E (daughter), D (daughter), F (mother).
Males certain: A (son).
B and C are a married couple (parents of A), so one of them is female.
Females = E, D, F + one of B/C = 4.

Why the other options are wrong: Three misses the female among B and C; Two counts only the daughters; Six would make A female.

Answer: C

Approach: Use the same family tree: A and F are the parents of D and E.

Solution:
E is the daughter of A and of F (F is E's mother).
D is the daughter of F.
Both are daughters of the same mother → D and E are sisters.

Why the other options are wrong: Both D and E are stated as daughters, so no brother/son/uncle; they are the same generation, so not grandmother–granddaughter.

Answer: A

Approach: Option elimination – check each team for the boy count and the five rules.

Solution:
(A) C, E, S, P, Q – S is chosen without A ✗.
(B) R, B, S, P, Q – R with B ✗ (also P with R).
(C) B, D, S, R, Q – R with B ✗, D with Q ✗.
(D) A, B, S, P, Q – boys A, B (two); A with S ✓; no R; no D; C and E both absent ✓ → valid.
(E) A, D, S, Q, R – D with Q ✗.

Cetking shortcut: 'A and S together' kills every option that has S without A – that alone removes A, B and C; D vs Q removes E.

Answer: D

Approach: With R fixed, remove everyone R cannot be with, then apply the pair rules.

Solution:
R in → B out (R not with B) and P out (P not with R).
(A) Q, S, C, E – S without A ✗.
(B) S, A, C, E – A with S ✓, C with E ✓, no B, no P, no D–Q clash ✓.
(C) P, S, A, D – P with R ✗.
(D) Q, S, R, D – repeats R and has D with Q ✗.
(E) Q, S, A, D – D with Q ✗.
Team: R, S, A, C, E.

Cetking shortcut: Pair rules force 'blocks': {A, S} and {C, E}. R + {A, S} + {C, E} = exactly five.

Answer: B

Approach: Build the tree generation by generation.

Solution:
D is the mother of J and L; J is F's brother; F is K's son and K has only three children → K and D are the parents of J, F and L.
Y is D's son-in-law → Y is married to D's daughter (L).
M is Y's son (and S's brother) → M is D's grandchild.
D (mother) is M's grandmother.

Why the other options are wrong: In-law relations apply only to Y; D is two generations above M, so not mother.

Answer: D

Approach: Decode the statement from the end.

Solution:
'Only son of my grandfather' = my father.
'Father of his brother' = his (the man's) father.
So the man's father = the lady's father → the lady is the man's sister.

Why the other options are wrong: Cousin would need different fathers; aunt/mother/daughter are other generations.

Answer: D

Approach: Treat the three parts (first letter, number, last letter) as separate series.

Solution:
First letters: Q, S, U, W (+2) → Y.
Numbers: 1 × 1 + 1 = 2, 2 × 2 + 2 = 6, 6 × 3 + 3 = 21, 21 × 4 + 4 = 88.
Last letters: F, E, D, C (−1) → B.
Next term = Y88B.

Why the other options are wrong: Z88B / Z44A get the first letter wrong (W + 2 = Y); Y66B and Y44B get the number wrong; Z44A also gets the last letter wrong.

Cetking shortcut: Letters alone leave Y66B, Y88B, Y44B; the number rule (×n + n) settles 88.

Answer: B

Approach: Look at the operation between consecutive terms.

Solution:
2 ×2 = 4, ×3 = 12, ÷2 = 6, ×2 = 12, ×3 = 36, ÷2 = 18, ×2 = 36, ×3 = 108, ÷2 = 54.
Equivalently, the series is in groups of three (a, 2a, 6a) with a = 2, 6, 18, and the next group starts at 18 × 3 = 54.

Why the other options are wrong: 72 = 36 × 2 breaks the ×2, ×3, ÷2 cycle; 90, 108 and 75 do not follow from 108 by the cycle.

Cetking shortcut: Every third term from the start (2, 6, 18, …) is ×3 → next is 54.

Answer: B

Approach: Track direction after each turn and add East–West and North–South moves separately.

Solution:
Walk 20 m North.
Turn right → East, 30 m.
Turn right → South, 35 m.
Turn left → East, 15 m.
Turn left → North, 15 m.
North–South: 20 − 35 + 15 = 0.
East–West: 30 + 15 = 45 m East.
He is 45 m East of the start.

Why the other options are wrong: West options reverse the right turns; 30 m East ignores the last East leg of 15 m.

Cetking shortcut: The vertical legs cancel (20 + 15 = 35), so only the two East legs count: 30 + 15.

Answer: C

Approach: Eight people at eight equally spaced seats = the eight compass directions.

Solution:
Order clockwise: P, Q, R, S, T, U, V, W.
Compass clockwise: N, NE, E, SE, S, SW, W, NW.
V at North → W at North-East → P at East → Q at South-East → R at South → S at South-West (T at West, U at North-West).

Why the other options are wrong: South-East is Q, South is R, East is P, North is V.

Cetking shortcut: S is 3 seats anticlockwise of V (V → U → T → S): N → NW → W → SW.

Answer: E

Approach: In the reversed alphabet, position from left = 27 − normal position.

Solution:
Reversed: Z Y X W V U T S R Q P O N M L K J I H G F E D C B A.
O (15th letter) is at 27 − 15 = 12th from the left.
Eighth to its right = 12 + 8 = 20th from left.
Letter = 27 − 20 = 7th letter = G.

Why the other options are wrong: I and H are the 18th/19th; Q and S are to the left of O in the reversed series.

Cetking shortcut: In the reversed series 'right' means going back in the alphabet: O − 8 = G.

Answer: B

Approach: In the reversed alphabet Z … A, the k-th letter from the right is the k-th letter of the normal alphabet.

Solution:
Reversed: Z Y X … C B A. Counting from the right: A(1), B(2), … → 9th from right = I.
I is at 27 − 9 = 18th from the left.
Fifth to its left = 18 − 5 = 13th from left = 27 − 13 = 14th letter = N.

Why the other options are wrong: M is 4th to the left of I; O is 6th; L and W are wrong counts/directions.

Cetking shortcut: Left in the reversed series = forward in the normal alphabet: I + 5 = N.

Answer: D

Approach: Write the new series and count positions.

Solution:
New series: A B C … O P | Z Y X W V U T S R Q.
Positions 1–16 unchanged; 13th from left = M.
Sixth to its right = 13 + 6 = 19th position.
Positions 17, 18, 19 = Z, Y, X.

Why the other options are wrong: S would be the 19th letter of the normal alphabet (not offered); W, V are the 20th/21st; Y is the 18th; G is to the left of M.

Answer: A

Approach: Day after n days = start day + (n mod 7).

Solution:
She went on a Thursday, 9 days ago.
9 = 7 + 2 → today is 2 days after Thursday = Saturday.

Why the other options are wrong: Tuesday/Wednesday count backwards; Thursday would need a multiple of 7 days; Sunday is 10 days.

Answer: C

Approach: Venn diagram / syllogism rules: 'Some + All' gives 'Some'; 'Some + Some' gives nothing.

Solution:
Some rivers are deserts + All deserts are roads → Some rivers are roads → Some roads are rivers. I follows.
Hills connect to rivers only through 'some' – no definite link to roads or deserts. II and III do not follow.

Why the other options are wrong: II (roads–hills) and III (deserts–hills) are possible but not certain, so any option including them is wrong; 'None' ignores I.

Cetking shortcut: Two 'Some' statements in a chain never give a definite conclusion – so II and III are out immediately.

Answer: B

Approach: Syllogism rules plus the complementary-pair (either–or) check.

Solution:
All tigers are jungles + No jungle is bird → No tiger is bird → III follows.
Rains link to jungles only through birds ('some birds are rains', 'no jungle is bird') – no definite relation.
I 'No rain is jungle' and II 'Some rains are jungles' have the same subject and predicate and together cover all cases, so either I or II follows.

Why the other options are wrong: I and II can never both be true; 'Only III' ignores the either–or pair; 'All' and 'None' are wrong.

Cetking shortcut: 'Some X are Y' and 'No X is Y' with no definite link = classic either–or pair.

Answer: C

Approach: Local (place) value = digit × its place; face value = the digit itself.

Solution:
32675149: units 9, tens 4, hundreds 1, thousands 5, ten-thousands 7.
Local value = 7 × 10,000 = 70,000; face value = 7.
Difference = 70,000 − 7 = 69,993.

Why the other options are wrong: 70000 is the local value alone; 75142, 5149, 64851 come from misreading the digits.

Answer: C

Approach: 225 = 9 × 25 → test divisibility by 25 and by 9.

Solution:
By 25: last two digits 7q must be 00, 25, 50 or 75 → 75 → q = 5.
By 9: 8 + 7 + 6 + p + 3 + 7 + 5 = 36 + p must be a multiple of 9 → p = 0 or 9.
Options offer only (p, q) = (0, 5).

Why the other options are wrong: q = 9 or 0 makes the last two digits 79 or 70, not divisible by 25.

Cetking shortcut: Only one option has q = 5 → D.

Answer: D

Approach: Weighted average = total weight ÷ total boys.

Solution:
16 × 50.25 = 804
8 × 45.15 = 361.2
Total = 1165.2 kg for 24 boys
Average = 1165.2 ÷ 24 = 48.55 kg.

Cetking shortcut: Gap = 50.25 − 45.15 = 5.1; weights are 16 : 8 = 2 : 1, so the average lies 1/3 of the gap below 50.25: 50.25 − 1.7 = 48.55.

Answer: A

Approach: a² − b² = (a + b)(a − b).

Solution:
a + b = 20, a − b = 8.
a² − b² = 20 × 8 = 160.
(Numbers are 14 and 6: 196 − 36 = 160 ✓.)

Answer: B

Approach: State total = % of 8000; split it in the given ratio.

Solution:

Cetking shortcut: Ratio = (35 × 7/14) : (30 × 3/12) = 17.5 : 7.5 = 7 : 3 – no need to use 8000.

Answer: D

Approach: Compute each required part from its state total and ratio.

Solution:
State 3 petrol = 30% of 8000 × 5/12 = 2400 × 5/12 = 1000.
State 1 diesel = 15% of 8000 × 3/8 = 1200 × 3/8 = 450.
Difference = 1000 − 450 = 550.

Answer: A

Approach: Find State 4 electric cars, then take the non-AC share (100% − 45%).

Solution:
State 4 cars = 35% of 8000 = 2800.
Electric = 2800 × 2/14 = 400.
Non-AC = 55% of 400 = 220.

Why the other options are wrong: 180 is the AC count (45%); 380/190/240 use wrong bases.

Answer: C

Approach: New urban population × unchanged self-employed share.

Solution:
Working population = 3,00,000.
New urban = 56% of 3,00,000 = 1,68,000.
Self-employed (40% of urban) = 0.40 × 1,68,000 = 67,200.

Why the other options are wrong: 62,400 is the original (52%) figure and 62,700 is a near miss of it; 67,440, 66,840 and 68,640 are near-miss calculation errors.

Cetking shortcut: 40% × 56% = 22.4% of 3,00,000 = 67,200.

Answer: C

Approach: Salaried = 48% of urban + 12% of rural.

Solution:

Why the other options are wrong: 96210, 92210, 91260, 90216 are digit-swap distractors of 92,160.

Answer: B

Approach: Same base, so the ratio of the percentages is the ratio of the numbers.

Solution:
Urban : Rural = 52% : 48% = 52 : 48 = 13 : 12 (1,56,000 : 1,44,000).

Why the other options are wrong: 12 : 13 is reversed; the others do not reduce from 52 : 48.

Answer: C

Approach: Required % = (rural salaried ÷ urban casual) × 100.

Solution:
Rural salaried = 12% of 1,44,000 = 17,280.
Urban casual = 12% of 1,56,000 = 18,720.
17,280 ÷ 18,720 = 12/13 = 92.3%.

Why the other options are wrong: 89.3% and 93.2%/93.3% are near-miss values; 72.3% uses the wrong group.

Cetking shortcut: Both are 12% of their bases, so the answer is just rural : urban = 48 : 52 = 12/13 = 92.3%.

Answer: E

Approach: Chain the ratios A : B : C : D by making common terms equal.

Solution:
A : B = 2 : 3 = 8 : 12; B : C = 4 : 5 = 12 : 15 → A : B : C = 8 : 12 : 15.
C : D = 6 : 7 → scale: A : B : C = 48 : 72 : 90 and C : D = 90 : 105.
A : B : C : D = 48 : 72 : 90 : 105, sum = 315.
A = 8400 × 48/315 = ₹1280.

Why the other options are wrong: ₹4200 is half the total; ₹8450 exceeds the total; ₹1020 and ₹1320 do not match 48/315.

Cetking shortcut: A share = 2×4×6 : 3×4×6 : 3×5×6 : 3×5×7 = 48 : 72 : 90 : 105 (multiply across).

Answer: B

Approach: Distance = speed × time for each part; times in hours.

Solution:
45 min = 15 min at x + 20 min at 2x + 10 min at x.
Distance = x(15/60) + 2x(20/60) + x(10/60) = x/4 + 2x/3 + x/6 = 13x/12.
13x/12 = 39 → x = 39 × 12/13 = 36 km/h.

Cetking shortcut: Time at x is equivalent to 15 + 40 + 10 = 65 min at speed x → x = 39 × 60/65 = 36.

Answer: C

Approach: Drawing off mixture removes A and B in the same 7 : 5 ratio; replacing with B adds only to B.

Solution:
Let the can hold 12k litres: A = 7k, B = 5k.
9 L drawn: A removed = 9 × 7/12 = 5.25, B removed = 3.75.
After adding 9 L of B: A = 7k − 5.25, B = 5k − 3.75 + 9 = 5k + 5.25.
(7k − 5.25)/(5k + 5.25) = 7/9 → 63k − 47.25 = 35k + 36.75 → 28k = 84 → k = 3.
Can = 36 L; initial A = 7 × 3 = 21 L.

Why the other options are wrong: 35 comes from removing 9 L of A only (and does not even give a whole-number can); 14, 15, 28 fail the 7 : 9 check.

Cetking shortcut: The A left after drawing off is 7/12 of (T − 9) and, after refilling, must be 7/16 of T: (T − 9) × 7/12 = T × 7/16 → 4(T − 9) = 3T → T = 36, so A = 36 × 7/12 = 21.

Answer: C

Approach: Minute hand gains 55 minute-spaces on the hour hand every 60 minutes.

Solution:
At 9:00 the minute hand (at 12) is 45 spaces behind the hour hand (at 9).
To be 23 spaces apart with the minute hand still behind, it must gain 45 − 23 = 22 spaces.
Time = 22 × 60/55 = 24 minutes → 9:24.
(Minute hand 23 spaces ahead would need a gain of 68 spaces = 74 2/11 min, beyond 10 o'clock.)
Check: at 9:24 the hour hand is at 47 spaces, minute hand at 24 → 23 apart ✓.

Answer: E

Approach: Glue the vowels into one block; divide by factorials of repeated letters.

Solution:
MATHEMATICS: 11 letters; vowels A, A, E, I; consonants M, M, T, T, H, C, S.
Units = 7 consonants + 1 vowel block = 8 → 8!/(2! × 2!) = 40320/4 = 10,080 (M and T repeat).
Inside the block: 4!/2! = 12 (A repeats).
Total = 10,080 × 12 = 1,20,960.

Why the other options are wrong: 4989600 is all arrangements with no condition (11!/(2!2!2!)); 120800 and 20660 are calculation slips; 11 is the letter count.

Answer: D

Approach: Track the facing direction after each turn; use Pythagoras for the shortest distance.

Solution:
Facing East, turns left → North, walks 10 m.
Turns left → West, walks 5 m.
Turns right → North, walks 2 m.
Net: North 10 + 2 = 12 m, West 5 m.
Distance = √(12² + 5²) = √169 = 13 m; he is facing North.

Why the other options are wrong: 12 m counts only the northward legs; 11 m is a wrong sum; 'West' is his direction of displacement on one leg, not his final facing.

Cetking shortcut: 5–12–13 triangle.

Answer: A

Approach: Use every room (each extra room adds revenue) and keep triples to the minimum.

Solution:
Let s, d, t = single, double, triple rooms: s + d + t = 25, s + 2d + 3t = 55 → d + 2t = 30.
Revenue = 2000s + 3000d + 3500t = 80,000 − 500t (substituting d = 30 − 2t, s = t − 5).
s ≥ 0 → t ≥ 5; minimum t = 5 → d = 20, s = 0.
Revenue = 5 × 3500 + 20 × 3000 = 17,500 + 60,000 = ₹77,500.

Why the other options are wrong: 82,500, 87,500 and 92,500 exceed the ceiling 80,000 − 500t with t ≥ 5; 72,500 is below the maximum.

Cetking shortcut: Start with 25 singles (25 guests, ₹50,000). A second guest in a room adds ₹1,000, a third only ₹500, so give 25 rooms a second guest first (50 guests), then put the last 5 as third guests: 50,000 + 25,000 + 2,500 = ₹77,500.

Answer: B

Approach: Maximum marks without completing any row or column, then add one.

Solution:
Each of the 4 rows can hold at most 3 X's without being full → at most 12 X's.
12 is achievable: leave one cell empty in each row and each column (e.g. the main diagonal), so no row or column is full.
The 13th X must go into one of the 4 empty cells and completes a row (and a column).
Maximum presses = 13.

Why the other options are wrong: 16 fills the grid; 4 is the minimum; 6 and 10 are below the safe maximum of 12.

Cetking shortcut: n×n grid → n(n − 1) + 1 = 4 × 3 + 1 = 13.

Answer: C

Approach: Turn the clues into equations and list all cases (scores are integers 0–4).

Solution:
Let both score a in test 1; Puneet scores p (≠ 0) in each of tests 2–8; Urvisha scores U2, U3, U4 and then p in tests 5–8.
Puneet's first three = Urvisha's first two: a + 2p = a + U2 → U2 = 2p ≤ 4 → p = 1 or 2.
GP: a, a + 2p, total (a + 6p + U3 + U4) → (a + 2p)² = a(a + 6p + U3 + U4) → a(2p + U3 + U4) = 4p².
Valid cases:

Cetking shortcut: U2 = 2p ≤ 4 immediately caps all of Puneet's tests 2–8 at 2 – that kills A, B, D, E.

Answer: C

Approach: Use the case table from the set with a = 4.

Solution:
Let both score a in test 1; Puneet scores p (≠ 0) in each of tests 2–8; Urvisha scores U2, U3, U4 and then p in tests 5–8.
Puneet's first three = Urvisha's first two: a + 2p = a + U2 → U2 = 2p ≤ 4 → p = 1 or 2.
GP: a, a + 2p, total (a + 6p + U3 + U4) → (a + 2p)² = a(a + 6p + U3 + U4) → a(2p + U3 + U4) = 4p².
Puneet's 1st test = a = 4 → a(2p + U3 + U4) = 4p² → 2p + U3 + U4 = p².
p = 1: 2 + U3 + U4 = 1 ✗; p = 2: 4 + U3 + U4 = 4 → U3 = U4 = 0.
Urvisha's 3rd test = 0.

Answer: A

Approach: Use the motorbike's distance to A to find its speed, then add both distances.

Solution:
Motorbike: 1 pm → 3:40 pm = 8/3 h; car: 2 pm → 3:40 pm = 5/3 h.
Motorbike speed x: (8/3)x = 168 → x = 63 km/h; car = 126 km/h.
Car distance = 126 × 5/3 = 210 km.
AB = 168 + 210 = 378 km.

Cetking shortcut: Car covers 2 × (5/8) = 5/4 of the motorbike's distance: 168 × 5/4 = 210.

Answer: E

Approach: Start with the strongest numeric clues (vii) and (i), then fix the rest.

Solution:
(vii) Bimal = Ignesh + 2 Idlis → (Ignesh, Bimal) = (4, 6) or (6, 8); Ignesh = Bimal + 2 Vadas → (Bimal, Ignesh) Vadas = (0, 2), (2, 4) or (4, 6).
(i) Ignesh's Vadas = 3 × Vadas of the 4-Idli person ; Ignesh's Vadas are 2, 4 or 6 and must be a multiple of 3 → Ignesh 6, 4-Idli person 2. So Ignesh 6 Vadas, Bimal 4 Vadas, and Ignesh is not the 4-Idli person → Ignesh 6, Bimal 8 Idlis.
(vi) Mukesh has half the Vadas of the person with twice his Idlis: Mukesh 4 Idlis → Bimal (8) has 4 Vadas → Mukesh 2 Vadas ✓ (also matches (i)).
(iv) 1-Idli person eats 0 Vadas and is not Mukesh; (v) Daljit eats Vada → Sandeep = 1 Idli, 0 Vadas; Daljit = 5 Idlis, 1 Vada.
(ii) No Chutney: Sandeep, Mukesh and the 4-Vada person Bimal (exactly three) → Daljit and Ignesh take Chutney.

Why the other options are wrong: Bimal eats 8 (not 6 or 1); Sandeep eats 1 (2 is not even an Idli count); Ignesh eats 6, not 8.

Answer: A