MBA CET 2025 Slot 6 question paper with solutions

Approach: Trains meeting rule: S_P/S_Q = √(t_Q/t_P), where t = time taken after meeting.

Solution:
Time of P after meeting t_P = 25 h; time of Q after meeting t_Q = 16 h.
S_P/S_Q = √(16/25) = 4/5
100/S_Q = 4/5 ⇒ S_Q = 100 × 5/4 = 125 km/h

Cetking shortcut: The train that takes LESS time after meeting is FASTER — so Q > 100; only 120, 125, 140, 115 qualify, and √(25/16) = 5/4 gives 125.

Answer: B

Approach: Ages with a past ratio — add the years back to each part.

Solution:
5 years ago: Anish = 3x, Ajay = 4x, Anvit = 5x.
Present ages: 3x+5, 4x+5, 5x+5 ⇒ sum = 12x + 15 = 75 ⇒ 12x = 60 ⇒ x = 5.
Ajay now = 4x + 5 = 20 + 5 = 25 years.

Cetking shortcut: Sum 5 years ago = 75 − 15 = 60; Ajay's share = 4/12 × 60 = 20; add 5 ⇒ 25.

Answer: E

Approach: Distance = Speed × Time; same distance both ways.

Solution:
Let one-way distance = D km.
Time going = D/12, time returning = D/6.
D/12 + D/6 = 6 ⇒ (D + 2D)/12 = 6 ⇒ 3D = 72 ⇒ D = 24 km.

Cetking shortcut: Average speed for equal distances = 2×12×6/(12+6) = 8 km/h; total distance = 8 × 6 = 48 km, one way = 24 km.

Answer: C

Approach: Probability = favourable cards / 52.

Solution:
Red suits: hearts and diamonds (13 each).
Numbered cards greater than 6: 7, 8, 9, 10 ⇒ 4 per suit.
Favourable = 4 × 2 = 8 ⇒ P = 8/52 = 2/13.

Why the other options are wrong:
A 4/13: counts all four suits (16 cards).
C, D, E: include face cards or the 6 itself.

Cetking shortcut: Per suit 4 of 13 cards qualify; red = half the deck ⇒ (4/13) × (1/2) = 2/13.

Answer: B

Approach: Addition rule for mutually exclusive events.

Solution:
Total balls = 10 + 5 + 5 = 20.
Red or white = 5 + 5 = 10.
P = 10/20 = 1/2.

Cetking shortcut: P(red or white) = 1 − P(black) = 1 − 10/20 = 1/2.

Answer: B

Approach: Vowels together: tie the vowels into one block, divide by repeats.

Solution:
ESPECIALLY = E, S, P, E, C, I, A, L, L, Y (10 letters).
Vowels: E, E, I, A (4); consonants: S, P, C, L, L, Y (6, L repeated).
Units to arrange = 1 vowel block + 6 consonants = 7 ⇒ 7!/2! = 2520 (2! for the two Ls).
Inside the block: 4!/2! = 12 (2! for the two Es).
Total = 2520 × 12 = 30240.

Why the other options are wrong:
B 25200, C 10080: forget one of the repeat divisions or the block's internal arrangement.
A, E: no standard method gives them.

Cetking shortcut: Answer must be divisible by 7!/2! = 2520 — only 25200, 10080 and 30240 are; 2520 × 12 = 30240.

Answer: D

Approach: Percent-of chain: fix Face Value = 100 and convert each statement.

Solution:
Let Face Value (FV) = ₹100 ⇒ Market Value (MV) = 35% of 100 = ₹35.
MV is 30% less than SP ⇒ MV = 0.70 × SP ⇒ SP = 35/0.70 = ₹50.
Taking FV as the cost (the share was bought at par), loss = 100 − 50 = ₹50.
Loss % = 50/100 × 100 = 50% loss.

Why the other options are wrong:
B 50% profit: sign reversed — SP (50) is below cost (100).
C, D, E: arithmetic slips; no route gives 49%, 48% or 45%.

Cetking shortcut: SP = 35/0.7 = 50 = half the face value ⇒ 50% loss straight away.

Answer: A

Approach: Odd days: each ordinary year adds 1 odd day, a leap year adds 2 — but only if its 29 February falls inside the interval.

Solution:
12 Feb 1882 → 12 Feb 1883: no 29 Feb in between ⇒ 1 odd day.
12 Feb 1883 → 12 Feb 1884: 29 Feb 1884 comes AFTER 12 Feb 1884 ⇒ 1 odd day.
12 Feb 1884 → 12 Feb 1885: includes 29 Feb 1884 ⇒ 2 odd days.
Total = 1 + 1 + 2 = 4 odd days ⇒ Sunday + 4 = Thursday.

Why the other options are wrong:
A Wednesday: only 3 odd days (ignores the leap day).
C Friday: 5 odd days — counts the leap day twice or counts 1884 from 1 Jan.
D, E: wrong count.

Cetking shortcut: 3 years apart = 3 odd days + 1 for every 29 Feb crossed (only 29 Feb 1884) = 4.

Answer: B

Approach: Curved surface area of a cylinder = 2πrh; cost = area × rate (π = 3.14, which the options are built on).

Solution:
CSA of one roller = 2 × 3.14 × 9 × 12 = 678.24 cm².
CSA of 8 rollers = 678.24 × 8 = 5425.92 cm².
Cost = 5425.92 × 15 = ₹81388.8.

Why the other options are wrong:
With π = 22/7 the cost is ≈ ₹81463 — not offered, so the paper clearly uses π = 3.14.
A, B, C, E: do not match 216π × 8 × 15 for any standard π.

Cetking shortcut: Cost = 2πrh × 8 × 15 = 25920π; 25920 × 3.14 = 81388.8.

Answer: D

Approach: Number of tiles = area of floor / area of one tile (same units).

Solution:
Courtyard area = 30 m × 20 m = 600 m².
Brick = 15 cm × 10 cm = 0.15 m × 0.10 m = 0.015 m².
Number of bricks = 600 / 0.015 = 40000.

Cetking shortcut: Work in cm: (3000 × 2000)/(15 × 10) = 200 × 200 = 40000.

Answer: C

Approach: Simple interest: SI = P × R × T / 100, with SI = Amount − Principal.

Solution:
Amount = 3.25X ⇒ SI = 3.25X − X = 2.25X.
2.25X = X × 3.25 × Y / 100 ⇒ 225 = 3.25Y.
Y = 225 / 3.25 = 69.23 years.

Why the other options are wrong:
Using SI = 3.25X (forgetting to subtract principal) gives 100 — not an option; A, B, C, E are calculation slips.

Cetking shortcut: Y = (Amount/P − 1) × 100 / R = 2.25 × 100 / 3.25 = 900/13 ≈ 69.23.

Answer: D

Approach: Weighted average (alligation) of concentrations.

Solution:
Acid % = (43 × 10 + 82 × 9)/(10 + 9)
= (430 + 738)/19 = 1168/19 = 61.47%.

Why the other options are wrong:
A 60.45%, B 61.25%, C 64.25%, D 63.2%: arithmetic slips; 62.5% would be the simple average, which ignores the 10:9 weights.

Cetking shortcut: Answer must be slightly below the simple average 62.5% (more of the weaker mixture): 43 + 39 × 9/19 = 43 + 18.47 = 61.47%.

Answer: E

Approach: Trains in opposite directions: relative speed = sum of speeds; distance = sum of lengths.

Solution:
Relative speed = 70 + 50 = 120 km/h = 120 × 5/18 = 100/3 m/s.
Distance to cover = 1500 m (total length).
Time = 1500 ÷ (100/3) = 45 s.

Why the other options are wrong:
Using the difference of speeds (20 km/h) gives 270 s — not offered; A, B, D, E are slips.

Cetking shortcut: 1500 × 18/(120 × 5) = 1500 × 3/100 = 45 s.

Answer: C

Approach: Downstream speed = boat speed + stream speed (same units).

Solution:
Stream = 25 km/h = 25 × 5/18 = 6.944 m/s.
Downstream = 15 m/s ⇒ boat in still water = 15 − 6.944 = 8.056 ≈ 8.06 m/s.

Why the other options are wrong:
E 8.56, D 7.05, B 6.05, C 5.23: arithmetic or conversion slips.

Cetking shortcut: Convert 15 m/s to km/h: 54 km/h; boat = 54 − 25 = 29 km/h = 29 × 5/18 = 8.06 m/s.

Answer: A

Approach: Gap between two runners in the same direction = (difference of speeds) × time; max gap is between fastest and slowest.

Solution:
Speeds in m/s: A = 15; B = 67 × 5/18 ≈ 18.61; C = 1020/60 = 17.
Fastest = B, slowest = A ⇒ speed difference = 18.61 − 15 = 3.61 m/s.
Time = 8 h 30 min 39 s = 30600 + 39 = 30639 s.
Maximum gap = 3.61 × 30639 ≈ 110606.8 m ≈ 110.606 km (rounding the speed to 3.61 m/s; the unrounded 3.6111 gives 110.64 km, still closest to E).

Why the other options are wrong:
Ignoring the 39 seconds gives ≈ 110.47 km — not offered.
A, B, C, D: other pairs/slips; B−C and C−A gaps are far smaller (≈ 49 km and ≈ 61 km).

Cetking shortcut: Only the fastest-minus-slowest pair matters: 3.61 m/s × 30639 s.

Answer: E

Approach: Convert men and women to a common efficiency unit.

Solution:
5 men ≡ 7 women (both finish in 28 days) ⇒ 1 man = 7 units/day, 1 woman = 5 units/day.
Total work = 5 × 7 × 28 = 980 units.
7 men + 5 women per day = 49 + 25 = 74 units.
Time = 980 / 74 ≈ 13.24 days ⇒ about 13 days.

Why the other options are wrong:
A 9, C 11, D 12, E 10 days: all are further from 13.24 than 13 is; no other reading of the data gives them.

Cetking shortcut: Express everything in women: 7M + 5W = 9.8W + 5W = 14.8W; days = 7 × 28/14.8 ≈ 13.24.

Answer: B

Approach: Efficiency method for pipes.

Solution:
Efficiency ratio first : second = 8 : 1 ⇒ together 9 units/min.
Tank = 9 × 10 = 90 units.
Second pipe alone = 90 / 1 = 90 minutes.

Cetking shortcut: Second pipe does 1/9 of the joint rate ⇒ takes 9 × 10 = 90 minutes.

Answer: D

Approach: Partnership: profit share ∝ capital × time.

Solution:
Vijay joins after 3 months ⇒ invests for 12 − 3 = 9 months.
Ajay joins 6 months after Vijay (month 9) ⇒ invests for 3 months.
A × 3 : V × 9 = 2 : 3 ⇒ 3A × 3 = 2 × 9V ⇒ 9A = 18V ⇒ A : V = 2 : 1.

Why the other options are wrong:
D 1:2: inverts the ratio.
B, C, E: wrong durations (e.g. 6 months for Ajay).

Cetking shortcut: Capital ratio = profit ratio ÷ time ratio = (2/3) ÷ (3/9) = 2 ⇒ 2 : 1.

Answer: A

Approach: Milk stays constant when water is added.

Solution:
Initially milk = 7x, water = 11x.
After adding 7 L water: 7x/(11x + 7) = 7/13 ⇒ 91x = 77x + 49 ⇒ 14x = 49 ⇒ x = 3.5.
Milk = 7 × 3.5 = 24.5 L.

Cetking shortcut: Milk part fixed at 7; water goes 11 → 13, so 2 parts = 7 L ⇒ 1 part = 3.5 L ⇒ milk = 24.5 L.

Answer: D

Approach: Assume SP = 100 and translate each statement.

Solution:
Discount = 40% of SP = ₹40.
CP = 2 × discount = ₹80.
Profit = 100 − 80 = ₹20 ⇒ Profit % = 20/80 × 100 = 25% profit.

Why the other options are wrong:
D 20% profit: divides profit by SP instead of CP.
A, C: SP is above CP, so it cannot be a loss.
E: arithmetic slip.

Cetking shortcut: CP = 0.8 × SP ⇒ SP/CP = 5/4 ⇒ 25% profit.

Answer: B

Approach: Percentage of sales to total cars held (opening stock + production) over 7 days; the garbled hourly figure is 8 cars.

Solution:
Production = 8 cars/hour × 8 hours = 64 cars/day ⇒ 7 days = 448 cars.
Total cars held in 7 days = 1450 + 448 = 1898.
Cars sold in 7 days = 80 × 7 = 560.
Sale % = 560/1898 × 100 = 29.5%.

Why the other options are wrong:
With 6 cars/hour (doc's reading) the figures give 31.35% (sales/1786) or 45.68% (sales/closing stock) — neither is an option, so 6 is not the intended value.
A 28.6%, B 27.5%, D 24.54%, E 26.74%: correspond to non-integer hourly production (≈ 9.1, 10.5, 14.9, 11.5 cars) — not sensible.

Cetking shortcut: Back-solve: 560/0.295 ≈ 1898 ⇒ production = 448 = 7 × 64 = 7 × 8 × 8 — a clean 8 cars/hour, so C.

Answer: C

Approach: Ratio split of a total.

Solution:
p : q = 4 : 20 = 1 : 5.
Sum = 10 years 6 months = 10.5 years ⇒ 6 parts = 10.5 ⇒ 1 part = 1.75.
q = 5 × 1.75 = 8.75 years = 8 years 9 months.

Why the other options are wrong:
Only B (8 years) carries q's age in whole years; A 5, C 4, D 7, E 6 years do not fit 5/6 of 10.5.

Cetking shortcut: q = 5/6 × 10.5 = 8.75 ⇒ 8 years (9 months).

Answer: B

Approach: Average = total runs / number of players; recompute the total for the next match.

Solution:
Total runs per match = 3 × 160 = 480; ratio 4 : 6 : 2 (12 parts) ⇒ 160, 240, 80.
Next match: lowest scorer (80) scores 100; the other two score the same 160 and 240.
New total = 160 + 240 + 100 = 500.
Average for the match = 500/3 = 166.66.

Why the other options are wrong:
C/E 162.5, A 160.5, D 168.5: none equals 500/3; combining both matches (980/6 = 163.33) is not what is asked and is not an option either.

Cetking shortcut: Only the lowest scorer changes by +20 ⇒ average rises by 20/3 = 6.66 ⇒ 160 + 6.66 = 166.66.

Answer: B

Approach: List the primes, add, divide by 3, take the square root.

Solution:
Primes greater than 2 and less than 20: 3, 5, 7, 11, 13, 17, 19.
Sum = 75 ⇒ one-third = 25 = 5².

Cetking shortcut: One-third must be a perfect square: 25, 36, 49… ⇒ sum 75, 108, 147…; the sum of odd primes below 20 is 75 ⇒ 5.

Answer: A

Approach: Equate the two linear population expressions.

Solution:
P after x years = 78000 − 1800x.
Q after x years = 54000 + 600x.
78000 − 1800x = 54000 + 600x ⇒ 24000 = 2400x ⇒ x = 10 years.

Cetking shortcut: Gap 24000 closes at 1800 + 600 = 2400 per year ⇒ 10 years.

Answer: C

Approach: BODMAS — innermost brackets first.

Solution:
23 − 20 = 3 ⇒ 42 × 3 = 126.
5 × 80 − 100 = 300 ⇒ 2 × 300 = 600.
Inside: 126 + 2 − 600 + 16 − 4 = −460.
26 + (−460) = −434.

Why the other options are wrong:
B 434: sign error.
C −443, D −453, E 435: arithmetic slips.

Cetking shortcut: Sign check: −600 dominates ⇒ answer must be negative, leaving only A, C, D.

Answer: A

Approach: Remove brackets from the innermost outward, flipping signs after a minus.

Solution:
{2a³ − 7a³} = −5a³ ⇒ −{…} = +5a³.
Inside [ ]: 3a³ + 4a² + 5a³ + 5a³ − 7a² = 13a³ − 3a².
2a³ − [13a³ − 3a²] = 2a³ − 13a³ + 3a² = −11a³ + 3a².

Why the other options are wrong:
C 11a³ − 3a²: whole sign reversed.
D −11a³ − 3a²: forgets to flip the a² sign.
B, E: wrong coefficients.

Cetking shortcut: Put a = 1: 2 − [3 + 4 + 5 + 5 − 7] = 2 − 10 = −8; only A gives −11 + 3 = −8.

Answer: A

Approach: Fraction removed × capacity = quantity drawn.

Solution:
3/4 − 11/16 = 12/16 − 11/16 = 1/16 of the drum.
1/16 × capacity = 50 L ⇒ capacity = 800 L.

Cetking shortcut: Drop in fraction = 1/16 ⇒ 16 × 50 = 800.

Answer: C

Approach: Form two linear equations from the two exchanges.

Solution:
Let Sujit = S, Suraj = R.
Suraj gives ₹10 ⇒ equal: S + 10 = R − 10 ⇒ R = S + 20.
Sujit gives ₹10 ⇒ R + 10 = 2(S − 10) ⇒ S + 30 = 2S − 20 ⇒ S = 50.
Check: R = 70; 70 + 10 = 80 = 2 × 40 ✓.

Why the other options are wrong:
A ₹70: that is Suraj's money.
B, C, E: fail the checks.

Cetking shortcut: Test options: S = 50 ⇒ R = 70 ⇒ 80 vs 40 ✓ twice.

Answer: D

Approach: Identity (a + b)³ = a³ + b³ + 3ab(a + b) with a = 0.081, b = 0.009.

Solution:
3ab = 3 × 0.081 × 0.009 = 0.002187 and a + b = 0.09.
So 0.081³ + 0.009³ + 0.002187 × 0.09 = a³ + b³ + 3ab(a + b) = (a + b)³.
(0.09)³ = 0.000729.

Why the other options are wrong:
B, C, D, E: same digits 729 with the decimal misplaced — (0.09)³ has 6 decimal places.

Cetking shortcut: Spot 0.002187 = 3 × 0.081 × 0.009 and 0.09 = 0.081 + 0.009 ⇒ cube of 0.09.

Answer: A

Approach: Two-variable coin equation (or alligation on coin values).

Solution:
₹8.75 = 875 paise. Let 10-paise coins = x ⇒ 5-paise coins = 105 − x.
10x + 5(105 − x) = 875 ⇒ 5x + 525 = 875 ⇒ x = 70.
5-paise coins = 105 − 70 = 35.

Cetking shortcut: If all 105 were 5-paise: 525 p; extra 350 p comes from 10-paise coins at 5 p each ⇒ 70 of them ⇒ 35 five-paise coins.

Answer: E

Approach: HCF = product of common primes with lowest powers.

Solution:
405 = 3⁴ × 5; 585 = 3² × 5 × 13; 765 = 3² × 5 × 17; 900 = 2² × 3² × 5².
Common: 3² × 5 = 45.

Why the other options are wrong:
B 90: 405 is odd, so 2 cannot be a factor.
C 35: 7 does not divide any.
D 15: a common factor, but not the highest.
E 50: 405 is not divisible by 2 or 25.

Cetking shortcut: Check options from the largest: 90 fails (405 odd), 50 fails, 45 divides all four ⇒ 45.

Answer: A

Approach: Largest equal bottle size = HCF of the volumes; bottles = sum of volumes ÷ HCF.

Solution:
558 = 2 × 3² × 31; 682 = 2 × 11 × 31; 589 = 19 × 31 ⇒ HCF = 31 L.
Bottles = 558/31 + 682/31 + 589/31 = 18 + 22 + 19 = 59.

Cetking shortcut: Differences 682 − 589 = 93 = 3 × 31 and 589 = 19 × 31 point straight to 31; total 1829/31 = 59.

Answer: B

Approach: Meeting again at the start = LCM of individual lap times.

Solution:
12 = 2² × 3; 18 = 2 × 3²; 24 = 2³ × 3.
LCM = 2³ × 3² = 72 minutes.

Cetking shortcut: Check multiples of 24: 24, 48, 72 — 72 is the first divisible by 18.

Answer: D

Approach: Spots touch the ground together after the LCM of the time per revolution; LCM of fractions = LCM(numerators)/HCF(denominators).

Solution:
Time per revolution: 1/25, 1/21, 1/16 minute.
LCM = LCM(1, 1, 1)/HCF(25, 21, 16) = 1/1 = 1 minute = 60 seconds.

Why the other options are wrong:
A 58, C 62, D 59, E 61 s: none is a whole number of revolutions for all three wheels (e.g. in 59 s the 25-rpm wheel turns 24.58 times).

Cetking shortcut: In 1 minute every wheel completes a whole number of turns (25, 21, 16); 25, 21, 16 share no common factor, so nothing shorter works.

Answer: B

Approach: LCM(a, b) = a means b divides a.

Solution:
LCM(X, Y) = X ⇒ Y is a factor of X.
LCM(Y, Z) = Y ⇒ Z is a factor of Y.
So Z | Y | X ⇒ Z divides all three and is the largest such number ⇒ HCF = Z.

Why the other options are wrong:
A X, B Y: X and Y do not divide Z (unless equal).
D XY, E YZ: an HCF cannot exceed the smallest number.

Cetking shortcut: Try numbers: X = 12, Y = 6, Z = 3 ⇒ HCF = 3 = Z.

Answer: C

Approach: Count primes below 100 decade by decade.

Solution:
1–10: 2, 3, 5, 7 (4); 11–20: 11, 13, 17, 19 (4); 21–30: 23, 29 (2); 31–40: 31, 37 (2); 41–50: 41, 43, 47 (3);
51–60: 53, 59 (2); 61–70: 61, 67 (2); 71–80: 71, 73, 79 (3); 81–90: 83, 89 (2); 91–100: 97 (1).
Total = 4 + 4 + 2 + 2 + 3 + 2 + 2 + 3 + 2 + 1 = 25.

Cetking shortcut: Fact to memorise: 25 primes below 100 (and 168 below 1000).

Answer: D

Approach: Sum of first n even natural numbers = n(n + 1).

Solution:
2 + 4 + … + 200 = 2(1 + 2 + … + 100) = 2 × 100 × 101/2 = 100 × 101 = 10100.

Why the other options are wrong:
D 10050: uses n(n+1)/2 × 2 wrongly or 100 × 100.5.
B, C, E: arithmetic slips.

Cetking shortcut: n(n + 1) = 100 × 101 = 10100.

Answer: A

Approach: Number of ways to write N as a product of two co-prime factors = 2^k−1, k = number of distinct primes of N.

Solution:
3234 = 2 × 3 × 7² × 11 ⇒ distinct primes: 2, 3, 7, 11 ⇒ k = 4.
Each prime power must go wholly to one factor ⇒ 2⁴ = 16 ordered splits.
Unordered pairs = 16/2 = 8 (including 1 × 3234).

Why the other options are wrong:
B 9, D 10, E 11: overcount; A 7: excludes the pair 1 × 3234, which is co-prime.

Cetking shortcut: Just count distinct primes: k = 4 ⇒ 2³ = 8.

Answer: C

Approach: Scaling: every term is 3 times an odd number, so each square is 9 times.

Solution:
3² + 9² + 15² + 21² + 27² = 3²(1² + 3² + 5² + 7² + 9²)
= 9 × 165 = 1485.

Why the other options are wrong:
D 495: multiplies by 3 instead of 9.
C 990: multiplies by 6.
B, E: slips.

Cetking shortcut: Factor out 3² = 9 ⇒ 9 × 165 = 1485.

Answer: A

Approach: Match with (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

Solution:
Square terms give 2a, 3b, c (signs to fix).
+12ab = 2(2a)(3b) ⇒ 2a and 3b have the same sign.
−4ac = 2(2a)(−c) and −6bc = 2(3b)(−c) ⇒ c has the opposite sign.
So the expression = (2a + 3b − c)² ⇒ square root = 2a + 3b − c.

Why the other options are wrong:
B / C 2a − 3b − c: would make the ab term −12ab.
A 2a + b − 3c, E 2a + b − 2c: squares give b², not 9b².

Cetking shortcut: Only the sign pattern matters: ab positive, ac and bc negative ⇒ c alone is negative ⇒ D.

Answer: D

Approach: Use x + y + z = 0 to replace each pair sum by the negative of the third.

Solution:
x + y = −z, y + z = −x, x + z = −y.
(x + y)(y + z)(x + z) = (−z)(−x)(−y) = −xyz.

Why the other options are wrong:
D xyz: misses one minus sign (three negatives give a negative).
A, B, E: B and E are 0 here; A is unrelated.

Cetking shortcut: Take x = 1, y = 1, z = −2: (2)(−1)(−1) = 2 and −xyz = 2 ✓.

Answer: C

Approach: Multiples of 5 and 3 = multiples of LCM 15; count terms of the AP.

Solution:
First multiple of 15 above 55 = 60; last below 3815 = 3810 (15 × 254).
Count = (3810 − 60)/15 + 1 = 250 + 1 = 251.

Why the other options are wrong:
B 250: forgets the +1.
D 252: includes an extra end term.
A 249, E 238: slips.

Cetking shortcut: ⌊3815/15⌋ − ⌊55/15⌋ = 254 − 3 = 251.

Answer: C

Approach: Three numbers in AP: take them as a − d, a, a + d.

Solution:
Sum = 3a = 48 ⇒ a = 16.
Sum of squares = 3a² + 2d² = 800 ⇒ 768 + 2d² = 800 ⇒ d² = 16 ⇒ d = 4.
Numbers: 12, 16, 20 ⇒ smallest = 12.

Why the other options are wrong:
A 10 ⇒ 10, 16, 22: squares sum 840.
C 11, D 9, E 13: squares sum 818, 866, 786 — not 800.

Cetking shortcut: Middle term = 48/3 = 16; test options: 12, 16, 20 ⇒ 144 + 256 + 400 = 800 ✓.

Answer: B

Approach: Last-digit check, then a rough size check.

Solution:
Units digit: 7 × 3 = 21 ⇒ product ends in 1 ⇒ only A (…511) qualifies.
Size check: 968427 × 3493 ≈ 9.68 × 10⁵ × 3.49 × 10³ ≈ 3.38 × 10⁹ ✓.
Exact: 968427 × 3493 = 3382715511.

Why the other options are wrong:
B …523, C …879, D …965, E …673: wrong units digit.

Cetking shortcut: Units digit 7 × 3 → 1 eliminates every option except A.

Answer: A

Approach: Dividend = Divisor × Quotient + Remainder.

Solution:
30879 = 67 × d + 327.
67d = 30879 − 327 = 30552.
d = 30552/67 = 456 (remainder 327 < 456 ✓).

Why the other options are wrong:
A 476, C 466, D 444, E 457: 67 × each + 327 ≠ 30879 (e.g. 67 × 457 + 327 = 30946).

Cetking shortcut: Units digit: 67 × d must end in 2 (…879 − …327) ⇒ 7 × d ends in 2 ⇒ d ends in 6: options 476, 456, 466 — then 30552/67 = 456.

Answer: B

Approach: aⁿ − bⁿ is divisible by (a + b) when n is even.

Solution:
n = 176 is even ⇒ 354¹⁷⁶ − 346¹⁷⁶ is divisible by 354 + 346 = 700.
Remainder = 0.

Why the other options are wrong:
A, C, D, E: non-zero remainders are impossible since 700 divides the expression exactly.

Cetking shortcut: Spot 354 + 346 = 700 and the even power ⇒ remainder 0 instantly.

Answer: B

Approach: A factor ends in 5 ⇔ it is odd and a multiple of 5. Count factors with power of 2 = 0 and power of 5 ≥ 1.

Solution:
19! = 2¹⁶ × 3⁸ × 5³ × 7² × 11 × 13 × 17 × 19 (powers: 3 → ⌊19/3⌋ + ⌊19/9⌋ = 8; 5 → 3; 7 → 2).
Choices: 2 → 1 (only 0); 3 → 9; 5 → 3 (1 to 3); 7 → 3; 11, 13, 17, 19 → 2 each.
Count = 1 × 9 × 3 × 3 × 2 × 2 × 2 × 2 = 1296.

Why the other options are wrong:
A, C, D, E: off by small amounts — usually from including 5⁰ (gives 1728) or miscounting the power of 3.

Cetking shortcut: Odd part of 19! has (8+1)(3+1)(2+1)·2⁴ = 1728 factors; drop the 5⁰ ones (1/4) ⇒ 1728 × 3/4 = 1296.

Answer: B

Approach: Divisible by 33 ⇔ divisible by 3 and by 11.

Solution:
Divisibility by 11: alternating sum of 3, 3, 1, 3, A = 3 − 3 + 1 − 3 + A = A − 2 ⇒ A = 2.
Divisibility by 3: 3 + 3 + 1 + 3 + 2 = 12 ✓.
Check: 33132 = 33 × 1004 ✓.

Why the other options are wrong:
A 4, D 6: digit sums 14, 16 — fail divisibility by 3.
B 5, E 8: pass the 3-test but fail the 11-test.

Cetking shortcut: Apply the stricter 11-test first — it fixes A = 2 in one line.

Answer: C

Approach: Take out the common factor 3⁴⁵.

Solution:
3⁴⁵ + 3⁴⁶ + 3⁴⁷ + 3⁴⁸ + 3⁴⁹ = 3⁴⁵(1 + 3 + 9 + 27 + 81) = 3⁴⁵ × 121 = 3⁴⁵ × 11².
So the sum is divisible by 3 (every term is a power of 3).

Why the other options are wrong:
B 5, C 2, E 4: the sum is odd and has no factor 5 (3⁴⁵ × 11² only).
D 11: also divides the sum (121 = 11²), so two options are valid; A is the stored key.

Cetking shortcut: Factor 3⁴⁵ out ⇒ 3⁴⁵ × 121: prime factors are only 3 and 11.

Answer: A

Approach: Spelling check of each bold word.

Solution: Full, updated and commentary are spelt correctly and fit the sentence. 'Contemparary' is a misspelling — the correct form is contemporary (con-tem-po-rary), meaning modern / of the present time.

Why the other options are wrong:
A: 'full' is correct and fits ('full of').
B: 'updated' is correct.
D: 'commentary' is correct.
E: one word is misspelt.

Cetking shortcut: Split long words into syllables: con-tem-po-ra-ry — the 'po' syllable is the usual trap.

Answer: C

Approach: Word form — an adjective needs a noun after it.

Solution: 'Stiff' is an adjective and must qualify a noun: 'face stiff competition from foreign companies'. 'Competing' is a verb/participle form and makes the phrase ungrammatical. Manufacturing and similar are correct.

Why the other options are wrong:
A: 'stiff competition' is the standard collocation.
C: 'companies manufacturing similar products' is a correct participle phrase.
D: 'similar products' is correct.
E: 'competing' is wrong.

Cetking shortcut: Remember the collocation 'stiff competition' — the noun is fixed.

Answer: B

Approach: Spelling check of each bold word.

Solution: Industries, facing and huge are correct. The sentence prints 'presure', which is misspelt — the correct spelling is pressure with double 's'. So the fourth bold word is the answer.

Why the other options are wrong:
A: 'industries' is correct (y → ies).
B: 'facing' is correct.
C: 'huge' is correct.
E: one word is misspelt.

Answer: D

Approach: Spelling and usage check of each bold word.

Solution: 'Benefit', 'technology', 'adapted' (= modified to suit) and 'needs' are all spelt correctly and fit the meaning: technology can be quickly modified to suit clients' needs. Hence all are correct.

Why the other options are wrong:
A: 'benefit' — correct spelling (single 'f', single 't').
B: 'technology' — correct.
C: 'adapted' is the right word (adapt = modify to suit); 'adopted' would be wrong here.
D: 'needs' — correct.

Cetking shortcut: Adapt (modify) vs adopt (take up) is the usual trap in such sentences — here 'adapted to needs' is right.

Answer: E

Approach: Spelling check of each bold word.

Solution: In the paper the first bold word is printed 'Traditionlly' — the 'a' is missing. The correct spelling is traditionally (tradition + al + ly). Boost, festival and season are correct.

Why the other options are wrong:
B: 'boost' — correct.
C: 'festival' — correct.
D: 'season' — correct.
E: one word is misspelt.

Cetking shortcut: Words ending in -ally: adjective in -al + ly (tradition-al-ly), so both 'a' and 'll' must be there.

Answer: A

Approach: Vocabulary in context — read the sentence where the word appears.

Solution: The passage says 'The economy is beset by a consumption slowdown'. Beset = troubled/surrounded by problems from all sides. Besiege (surround, overwhelm) carries the same sense.

Why the other options are wrong:
B: terminate = end — unrelated.
C: filter — unrelated.
D, E: tiny / huge are size adjectives, not verbs of attack/trouble.

Cetking shortcut: Beset and besiege share the 'be-' prefix (all around) — the only verb with that sense.

Answer: A

Approach: Multi-statement true/false — check each statement against the passage.

Solution: I: The RBI rate cut is only 'widely expected this week' — not definite. False.
II: The passage says the rural economy 'remains in distress' (2.9% agriculture growth); it never says rural is better than urban. False.
III: Government spending 'was cut back in the last quarter of 2018-19' to meet the 3.4% fiscal deficit target — not increased. False.
None of the statements is true.

Why the other options are wrong:
A, B, D: each includes II, which is false.
C: includes I and III, both false.

Cetking shortcut: One firm reversal (III says 'increased', passage says 'cut back') kills options C and D at once.

Answer: E

Approach: Multi-statement — which items does the passage give as signs of slowdown?

Solution: I: 'consumption slowdown as reflected in the falling sales of ... consumer durables' — a slowdown indicator. True.
II: Core sector growth 'fell to 2.6%' from 4.7% — slower but still positive, not negative growth. False.
III: Inflation undershooting the target is called 'the good news'. False.
Only I.

Why the other options are wrong:
B: III is presented as good news.
C, D: II says 'negative growth', but growth was +2.6%.
E: includes II and III.

Cetking shortcut: Watch for 'negative growth' vs 'fall in growth' — a classic CET distortion.

Answer: A

Approach: Locate the 'money in the hands of people' line and see what it links to.

Solution: Passage: 'she has to boost consumption, which means putting more money in the hands of people. That, in turn, means cutting taxes'. So a decrease in tax rates puts money in people's hands — I is true. Inflation is not linked to this at all (II false). Private investment is a separate medium-term goal, not a way to put money in people's hands (III false).

Why the other options are wrong:
B, D: III — private investment is mentioned as a separate goal.
C, E: II — inflation is not linked to people's money.

Answer: A

Approach: NOT-mentioned question — tick each option against the reforms list.

Solution: Reforms listed: 'land acquisition and labour, corporate taxes by reducing exemptions and dropping rates, and nursing banks back to health ... further recapitalisation of the ailing banks, and consolidation'. Land acquisition (A), corporate tax adjustment (C) and bank recapitalisation (D) are all mentioned. Privatisation is mentioned only as a way to raise money, and never for banks — privatisation of banks is NOT mentioned.

Why the other options are wrong:
A: land acquisition — mentioned.
C: corporate tax (exemptions and rates) — mentioned.
D: recapitalisation of banks — mentioned.
E: not valid, since B is not mentioned.

Cetking shortcut: In EXCEPT/NOT questions, the answer often merges two passage words (privatisation + banks) that never appear together.

Answer: B

Approach: Find the common theme; the odd sentence breaks theme or tone.

Solution: A–D all argue that digital education harms students: it hurts weaker students (A), leaves them to self-study (B), a study shows they learn less (C), and gives easy grades without real learning (D). E praises digital platforms for high achievers — it goes against the critical tone, so it does not contribute.

Why the other options are wrong:
A: states the main claim (harm to less proficient students).
B: explains a drawback.
C: gives evidence of the drawback.
D: adds another drawback.

Cetking shortcut: Spot the lone positive sentence in a paragraph of negatives.

Answer: E

Approach: Find the common theme; the odd sentence shifts focus.

Solution: A–D are about Amazon Prime's popularity and benefits for viewers: better than TV (A), freedom to choose shows (B), strong content library (C), personalised favourites and pausing live TV (D). E talks about investors being unaware of the company — a shift from viewers to investors.

Why the other options are wrong:
A: introduces the theme (Prime more beneficial than TV).
B: supports the streaming trend.
C: a viewer benefit.
D: another viewer benefit.

Cetking shortcut: Change of stakeholder (viewers → investors) marks the odd one out.

Answer: E

Approach: Find the common theme; the odd sentence contradicts it.

Solution: The paragraph is about climate change threatening biodiversity: threat (A), habitats disrupted (B), rising temperatures affect ecosystems (D), extinction risk (E). C calls climate change beneficial — it contradicts the theme.

Why the other options are wrong:
A: topic sentence.
B: effect on animals.
D: cause–effect on habitats.
E: extinction risk — supports the threat.

Answer: C

Approach: Find the common theme; the odd sentence contradicts it.

Solution: The paragraph gives a balanced view of social media: it changes interaction (A), people share ideas (B), comparison can cause loneliness and anxiety (D), balance is important (E). C says it has 'only positive effects' — contradicting D and the balanced theme.

Why the other options are wrong:
A: introduces the topic.
B: a positive use — fits a balanced view.
D: a negative effect — part of the balance.
E: the conclusion.

Cetking shortcut: Absolute words like 'only' usually signal the odd sentence.

Answer: C

Approach: Find the common theme; the odd sentence contradicts it.

Solution: A, B, D and E all say early childhood education builds skills and drives later success. C says it has little impact — directly against the theme.

Why the other options are wrong:
A: main claim.
B: how preschool helps.
D: foundational skills matter.
E: restates the conclusion.

Answer: C

Approach: Multi-statement — verify each against the passage.

Solution: 1: The ocean plastic story is 'mostly originating from where we stand, the land'. True.
2: 'global production and consumption of plastics have continued to rise' — production has not declined. False.
3: 'A simple walk on any beach, anywhere, and the plastic waste spectacle is present' — true.
So only 1 and 3.

Why the other options are wrong:
A: misses statement 3.
B: 2 is false.
C: 1 and 3 are true.
D: includes 2.

Cetking shortcut: Statement 2 reverses the 'continued to rise' trend — eliminate every option containing 2 first.

Answer: E

Approach: Find the stated cause in the passage.

Solution: Passage: Air India 'has failed to maintain its market dominance as a slew of carriers including InterGlobe Aviation and SpiceJet started to offer ultra-cheap, on-time flights more than a decade ago.' So the main reason is competition from other airlines that it could not match.

Why the other options are wrong:
A: a consequence (it survives on a bailout), not the stated reason.
B: no lost 'deposit' is mentioned.
C: no depositors or investors earning deposits are mentioned.
E: D is stated.

Answer: D

Approach: Antonym in context — 'market dominance' means a superior, leading position.

Solution: 'Failed to maintain its market dominance' = superiority/leading position over rivals. The opposite of a superior position is inferiority (a lower, weaker position compared with others).

Why the other options are wrong:
B: powerlessness = no power at all — opposite of 'power', too extreme for market position.
C: submission = yielding to someone's control — opposite of dominating a person, not of market leadership.
D: ascendency is a synonym of dominance.
E: A fits.

Cetking shortcut: Ask what 'dominance' compares — Air India vs rivals — so the antonym must also be comparative: inferiority.

Answer: A

Approach: Antonym in context — check how 'frigid' is used.

Solution: Passage: infant penguins were thrown 'into the frigid water' — frigid = extremely cold. The opposite is hot.

Why the other options are wrong:
A, B: flexible / soft oppose 'rigid', not 'frigid'.
C: lukewarm is mildly warm — between cold and hot, not the opposite.
D: rigid only sounds like frigid.

Cetking shortcut: Don't be fooled by the rhyme frigid–rigid; context (Antarctic water) fixes the meaning as cold.

Answer: E

Approach: Find the FALSE statements — check each against the passage.

Solution: I: The Dawson-Lambton colony did not die; it 'significantly swelled in numbers'. It was the Halley Bay colony that collapsed. False.
II: 'The birds arrive at the site ... each April to breed.' True.
III: 5,315 pairs in 2016 was the Dawson-Lambton colony, not Halley Bay. False.
False statements: I and III.

Why the other options are wrong:
A: misses III.
B: II is true.
D: includes II.
E: includes II.

Cetking shortcut: In 'which is false' questions, swap-of-names (Halley Bay ↔ Dawson-Lambton) is the standard trap — mark which number belongs to which colony.

Answer: C

Approach: Multi-statement — check each claim about Tuan.

Solution: I: Tuan only stood 'beside the village doctor to help him prepare the medicines and tend to the sick' — he was a helper, not a doctor. False.
II: 'he was a very kind and good-natured person'. True.
III: He 'had a little land' and earned from his crops — not wealthy, and no giving away of wealth. False.
Only II.

Why the other options are wrong:
A, C, E: I is false — he assisted the doctor.
D: III is false — he had a little land.

Answer: B

Approach: Check what the passage says about Juan's change.

Solution: 'From that day, Juan changed himself. Within a year, he could produce much crops in his land through his hard work.' So I (worked hard on his crops) is stated. Nothing is said about him becoming more helpful (II) or starting a charity (III).

Why the other options are wrong:
B, D, E: III — no charity is mentioned.
C: II is not stated; it is an assumption.

Cetking shortcut: Stick to what is written — 'transformed' is explained only through hard work on his land.

Answer: A

Approach: Locate the sequence of events after the boy fell ill.

Solution: The village doctor could not cure the boy. Tuan 'untied his own horse and rode it ... rushed to the town that was ten miles away and fetched a more experienced doctor'. So he rode to the nearest town to get a better doctor.

Why the other options are wrong:
A: the village doctor was already there and could not cure him.
B: the opposite of what Tuan did.
C: the boy was not taken anywhere; the doctor was brought to him.
D: Tuan did not fetch medicine from his house.

Answer: E

Approach: Antonym in context — check how 'doused' is used.

Solution: 'When the flames were doused' — doused = extinguished, put out (a fire). The opposite of putting out a fire is lighted (set alight) — the passage itself uses 'the fire that Juan had lighted'.

Why the other options are wrong:
A, B, C: saturated, deluged, wet relate to the other meaning of douse (to drench) — they are near-synonyms, not antonyms.
E: not valid — only D is opposite.

Cetking shortcut: Douse has two senses (drench / extinguish); the context 'flames were doused' decides it.

Answer: D

Approach: Find the clue the passage actually gives.

Solution: 'When the flames were doused, Tuan saw which direction the fire had started from.' Juan lived on the land next to his — the direction of the fire was the clue. No lighter, kerosene or eyewitness neighbour is mentioned (the neighbour only alerted others and saved the crops).

Why the other options are wrong:
A: no lighter is mentioned.
B: no kerosene is mentioned.
C: the neighbour saved the crops but did not see Juan set the fire.
E: includes C, which is false.

Answer: D

Approach: Fixed phrase: 'make (something) of' = form an opinion about.

Solution: 'What did you make of the lecture?' means 'What was your opinion of the lecture?'. The verb 'make' in this sense always takes 'of'.

Why the other options are wrong:
A: 'make in' — not used in this sense.
B: 'make about' — not idiomatic (we say 'think about').
C: 'make on' — wrong.
D: 'make at' — wrong.

Cetking shortcut: 'What do you make of it?' — learn it as a fixed opinion question.

Answer: E

Approach: First conditional: if + simple present, will + base verb.

Solution: Main clause 'I'll help him' is future (will). In a first conditional the if-clause takes the simple present, so 'if he asks' (third person singular → asks).

Why the other options are wrong:
A: 'ask' — subject 'he' needs 'asks'.
B: 'asked' — past tense goes with 'would help' (second conditional).
C: 'will ask' — 'will' is not used in the if-clause.
E: D is correct.

Cetking shortcut: Never put 'will' in both clauses of a conditional.

Answer: D

Approach: Idiom + tense: the idiom is 'lose one's temper'; habitual action needs simple present.

Solution: 'Whenever he knows things do not take place...' describes a habit in the present, so the main verb must be simple present. Also the idiom is 'lose one's temper', not 'loosen'. Correct: 'He loses his temper whenever he knows things do not take place as per the planning.' The error is in part 1.

Why the other options are wrong:
B: 'whenever he knows' — correct present tense for a habit.
C: 'things do not take place' — correct.
D: 'as per the planning' — acceptable.
E: part 1 has an error.

Cetking shortcut: lose (misplace/fail to keep) vs loose/loosen (not tight) — a frequent CET confusable.

Answer: A

Approach: Fixed conjunction 'so long as' (= provided that).

Solution: The correct expression is 'So long as' / 'as long as'; the comparative 'longer' cannot be used in this conjunction. Correct: 'So long as you are honest and forthright, they will support you in this task.'

Why the other options are wrong:
B: 'you are honest' — correct.
C: 'and forthright, they will' — correct.
D: 'support you in this task' — correct.
E: part 1 has an error.

Answer: A

Approach: Subject–verb agreement with 'One of the + plural noun'.

Solution: The subject is 'One of the party members' — the head word 'one' is singular, so the verb must be singular: 'One of the party members was dismissed'. 'Were' is wrong.

Why the other options are wrong:
A: 'One of the party' — correct.
C: 'for speaking against' — correct gerund after preposition.
D: 'the leader' — correct.
E: part 2 has an error.

Cetking shortcut: 'One of the + plural noun' always takes a singular verb (unless a relative clause 'who/that' follows).

Answer: B

Approach: Relative pronoun after 'anything': use 'that' (or 'whatever'), never 'what'.

Solution: 'What' already contains its own antecedent (= the thing that), so 'anything what' doubles it. Correct: 'give him anything that he wants' or 'give him whatever he wants'. The error is in part 2.

Why the other options are wrong:
A: 'His father promised to' — correct.
C: 'wants if he' — correct.
D: 'passes in the examination' — correct (simple present after 'if'); the error is in part 2.
E: part 2 has an error.

Cetking shortcut: After all / anything / everything / nothing, the relative pronoun is 'that'.

Answer: B

Approach: Article usage — a plural noun made specific by an 'of' phrase takes 'the'.

Solution: The sentence is logical: Gandhi did not solve all future problems, but he did solve those of his own time. Grammatically, the contrast pairs 'all the problems of the future' with 'problems of his own age'. 'Problems of his own age' is a specific, defined set, so it needs the definite article: '...but he did solve the problems of his own age.' The error is in part 4.

Why the other options are wrong:
A: 'did not solve' — correct negative past.
B: 'all the problems of the future' — grammatical and logical (it says what he did NOT do).
C: 'but he did solve' — correct emphatic 'did'.
E: part 4 lacks the article.

Cetking shortcut: Check parallel halves of a contrast ('not ... but') — the missing word in one half is usually the error.

Answer: D

Approach: Spelling — break the word into its roots.

Solution: Quint (fifth) + essence → quintessence: the purest or most typical example of something. Only E keeps 'quint' and the full 'essence' (e-ss-e-nce).

Why the other options are wrong:
A: 'Quentiscence' — wrong root (quen-) and wrong ending.
B: 'Quintecence' — drops 'ss' and the 's'.
C: 'Quintissense' — 'issense' is wrong.
D: 'Quintecsence' — 'cs' is wrong.

Cetking shortcut: Quint + essence: write 'essence' in full inside the word.

Answer: E

Approach: General → specific chain using pronoun and definite-noun links.

Solution: 1 introduces the breakthrough (gene editing in a human embryo). 5 gives specifics: the mutation causing 'hypertrophic cardiomyopathy'. 2 'The cardiac disease' refers back to that condition. 4 'It is caused by a mutation...' explains its cause. 3 'Correcting the mutation...' gives the benefit — the natural close. Order: 1-5-2-4-3.

Why the other options are wrong:
A (1,4,3,2,5): 'It' in 4 has no antecedent after 1.
B (1,5,3,4,2): 3 'correcting the mutation' comes before the disease is described.
C (1,4,5,3,2): same problem as A.
D (1,3,2,5,4): 3 before the mutation is named.

Cetking shortcut: Anchor pair: 5 names the disease → 2 'The cardiac disease' must follow.

Answer: E

Approach: Two-group structure: 'Some ...' → description, 'Others ...' → description, then the conclusion about 'both'.

Solution: 2 introduces 'Some business executives' with a wait-and-see attitude. 4 describes them with an image (deer caught in headlights — frozen, they risk being run over). 3 'Others...' introduces the second group, too busy bailing themselves out. 1 describes them (driver changing a tyre on the highway). 5 'executives in both situations' sums up both groups — the conclusion. Order: 2-4-3-1-5.

Why the other options are wrong:
A, B, D: do not start with 2, the only sentence that introduces the subject ('Some...').
C (2,5,1,3,4): 5 mentions 'both situations' before the second group (3) is introduced.

Cetking shortcut: 'Some ... Others ... both' is a fixed skeleton: Some-group first, Others-group next, 'both' last.

Answer: E

Approach: Find the strongest link pair, then build around it.

Solution: 3 'We have educated ourselves into a world from which wonder has been banished' → 1 'We have paid a terrible price for our education, such as it is' (the word 'education' links them — mandatory pair 3-1). 4 'Of course wonder is costly ... antithesis of the anxiously worshipped security' explains why we banished it. 5 'The magician world view, in so far as it exists, has taken flight into science' tells where wonder survives. 2 'Wonder is marvellous but is also cruel, cruel, cruel' closes with the emotional summary. Order: 3-1-4-5-2.

Why the other options are wrong:
A (2,1,5,3,4) and C (2,1,3,4,5): 2 → 1 has no link; 1 'our education' needs 3 before it.
B (1,5,4,3,2): 1 cannot open — 'our education, such as it is' refers back to 3.
E (4,2,1,3,5): 'Of course' in 4 cannot open a paragraph.

Cetking shortcut: 'Such as it is' and 'Of course' never open a paragraph — that alone leaves A, C or D, and only D keeps 3-1 together.

Answer: D

Approach: Cause → effect → reaction → resolution timeline.

Solution: 2 the monsoon failed (cause). 4 'So, no grain could be sown' (effect — 'So' needs a cause before it). 1 people feared a famine. 5 farmers looked anxiously for the next monsoon. 3 the monsoon turned out abundant and the danger was averted (resolution). Order: 2-4-1-5-3.

Why the other options are wrong:
A, B, C: do not start with the cause (2).
D (2,3,4,5,1): the danger is averted (3) before it even arises.

Cetking shortcut: 'So' (4) must follow its cause (2); 'the danger was averted' (3) must be last.

Answer: E

Approach: Present perfect for a recent action with a present result.

Solution: 'Rama is not at home' (present result) because he has just gone for shopping (completed action, still relevant now). 'Just' is a present-perfect marker.

Why the other options are wrong:
A: 'is gone' — not standard in formal grammar.
B: 'was just going' — past continuous does not explain the present absence.
C: 'had gone' — past perfect needs another past reference point.
D: 'will have be going' — ungrammatical.

Cetking shortcut: 'Just' / 'already' / 'yet' + present-result context → has/have + V3.

Answer: E

Approach: Direct → indirect for a request: 'said to + Please' → 'requested + object + to-infinitive'; tenses shift back; pronouns change.

Solution: 'Please' signals a request → reporting verb 'requested'. 'Help me' → 'to help her' (the old woman). 'Till he returns' → 'till he returned' (past reporting verb). E is the only option with the right verb, pronoun and tense shift.

Why the other options are wrong:
A: still in direct speech.
B: 'said that she may please help' — wrong structure and meaning.
C: 'women', 'a boy', no tense shift.
D: 'ordered' is wrong for 'please'; 'him' should be 'her'; no 'please' meaning.

Cetking shortcut: 'Please' → requested; imperative → to + verb.

Answer: E

Approach: Verb pattern: force + object + to-infinitive.

Solution: 'Force' takes the pattern force someone to do something. So 'I cannot force her to play.' 'To playing' (gerund) is wrong after force.

Why the other options are wrong:
A: 'in playing' — wrong preposition and form.
B: 'for play' — wrong.
C: 'over the play' — changes meaning, ungrammatical.
D: 'with play' — wrong.

Cetking shortcut: Force / compel / allow / ask / tell + object + to + base verb.

Answer: E

Approach: Phrasal verb meaning: 'boil down to' = can be reduced/summed up to its essence.

Solution: 'Sri Lanka's problem boils down to just one thing — lack of infrastructure for agriculture' = when summarised, the problem comes to one thing. So 'sums up to'.

Why the other options are wrong:
A: 'heats up to' — literal meaning of boil, not the idiom.
B: 'adapt to something' — unrelated.
C: 'as a result' — a connector, not a summary.
D: 'fulfill something' — unrelated.

Cetking shortcut: Literal-sense options (heats up) are planted traps in idiom questions.

Answer: E

Approach: Subject–verb agreement with collective head noun 'bouquet of + plural'.

Solution: The head noun is 'bouquet' (singular); 'of flowers' is only a modifier. So the verb is singular: 'The bouquet of flowers was beautiful.' 'Beautiful' takes 'more beautiful' as comparative, never 'beautifuller'.

Why the other options are wrong:
A: 'were' — agrees wrongly with 'flowers'.
C: 'are' — same error.
D: 'beautifuller' — not a word.
E: 'am' and 'beautifuller' — both wrong.

Cetking shortcut: Cover the 'of + noun' phrase and match the verb to the first noun.

Answer: B

Approach: Idiom meaning.

Solution: 'Work against the clock' = work very fast to finish something before a deadline. So 'work in great hurry'.

Why the other options are wrong:
A: working late at night is not implied.
C: 'against all problems' — takes 'against' literally.
D, E: concentration / dedication may accompany it but are not the meaning — the core is time pressure.

Cetking shortcut: 'Clock' = time pressure → hurry.

Answer: B

Approach: Narrative order: event → background → trigger → action → result.

Solution: B states the incident (tiger killed two fishermen). A gives background (they had gone fishing in a creek). F: they didn't return, so a search party was formed. E: the party combed the regular areas for two days. C: 'But there was no trace' (contrast with E). D: 'At last' they searched the creeks and found the bodies. Order: B-A-F-E-C-D → 5th sentence is C.

Why the other options are wrong:
A (sentence A) is 2nd.
B (sentence B) is 1st.
D (sentence F) is 3rd.
E (sentence E) is 4th.

Cetking shortcut: 'But there was no trace' (C) must follow the search (E), and 'At last' (D) must be last — so C is 5th without solving the rest.

Answer: C

Approach: Main idea — what does the whole passage build towards?

Solution: The passage traces fairness products sold first to women, then to men; notes that fairness is a cultural notion; then says there is 'growing awareness among consumers that companies are exploiting their insecurities', campaigns are 'slowly making a difference', and companies 'cannot continue to bullshit consumers anymore'. Central idea: people who bought into fairness products are now realising they are being taken for a ride.

Why the other options are wrong:
B: 'only men' — women were the original target; too narrow and false.
C: brand evolution is a predicted consequence, not the central idea; 'appease' misreads it.
D: 'frivolous campaigning' is not the passage's point.
E: A fits.

Cetking shortcut: The passage's turn ('However, there is a growing awareness...') usually carries the main idea.

Answer: A

Approach: Direct detail — locate 'Nielsen'.

Solution: 'According to a 2015 research report by Nielsen, urban Indian men believe that fair skin can improve professional prospects.' That is option A.

Why the other options are wrong:
B: 'marriage' prospects are what companies promised women, not the Nielsen finding.
C: questioning advertisements is the Bombay Shaving Company CEO's view.
D: the opposite of the finding.
E: A matches.

Cetking shortcut: Scan for the proper noun (Nielsen) and read only that sentence.

Answer: A

Approach: Direct detail — locate 'Dark is Beautiful'.

Solution: Nandita Das launched the campaign 'to encourage Indians to embrace a wider definition of beauty', and such efforts are 'encouraging consumers to take pride in their natural skin tones'. That is: give up the fixation on fair skin and accept their inherent beauty — option A.

Why the other options are wrong:
B: exposing brands' intentions is not the stated aim.
C: misleading advertisements are not mentioned as the purpose.
D: sympathy towards others is not the purpose — it is about self-acceptance.
E: A matches.

Answer: A

Approach: NOT-true question — verify each statement; the false one is the answer.

Solution: A says men have been the most targeted audience 'since the inception' — false: companies sold fair skin to women 'for generations'; men became a favoured target only 'over the last few years'. B is true ('In India, as in other parts of the world, light skin is the culturally accepted...'). C is true (China and Japan had skin-whitening products for centuries). D is true ('Until the mid-2000s, deodorants and shaving creams were the only grooming products advertised for men').

Why the other options are wrong:
B: stated in paragraph 2.
C: stated (Kiran Khalap's quote).
D: stated in paragraph 1.
E: A is not true.

Cetking shortcut: Time-words like 'since inception' or 'always' are red flags in NOT-true questions.

Answer: A

Approach: Vocabulary in context.

Solution: 'Abhay Deol took to Facebook to trounce his fellow actors who earn millions from endorsing fairness creams' — here trounce = attack/criticise severely. Lambaste = criticise harshly. Same meaning.

Why the other options are wrong:
A: vacillation = indecision.
B: temerity = boldness/audacity.
D: nefarious = wicked (an adjective).
E: C matches.

Cetking shortcut: Check part of speech: trounce is a verb — only 'lambaste' is a verb among the choices.

Answer: C

Approach: Direct detail — find the sentence on the Supreme Court order's purpose.

Solution: 'The point of the Supreme Court order was to make the States share their distress and not to magically fulfil the needs of farmers on both sides.' So the purpose was that the two states share their distress due to water shortage — option B.

Why the other options are wrong:
A, D: the order was not to solve one state's irrigation needs; water is insufficient for both.
C: explicitly rejected — 'not to magically fulfil the needs of farmers on both sides'.
E: B matches.

Cetking shortcut: The passage states the purpose and rejects option C in the same sentence.

Answer: B

Approach: Decode each link, then chain generations.

Solution:
Sunita is Rohit's wife; Puja is Sunita's mother-in-law → Puja is Rohit's mother.
Rohit is Rupali's father → Puja is the mother of Rupali's father.
So Puja is Rupali's paternal grandmother. (Gajanan is the grandfather; Dhanashree, Puja's daughter, is Rupali's aunt — extra information.)

Why the other options are wrong:
A, C: grandson/granddaughter reverse the generation and Puja is female-elder.
D: aunt is Dhanashree, not Puja.
E: Puja is mother-in-law of Sunita, not of Rupali.

Cetking shortcut: Mother-in-law of X's wife = X's mother. One step down from there = grandmother.

Answer: B

Approach: Convert every 'ahead/behind' into a rank from the top, then Total = rank from top + rank from bottom − 1.

Solution:
Samiksha = 12th from top.
Divya is 5 behind Samiksha → 12 + 5 = 17th.
Divya is 12 ahead of Medha → Medha = 17 + 12 = 29th.
Reshma is 29 behind Medha → 29 + 29 = 58th from top.
Reshma is 7th from bottom → Total = 58 + 7 − 1 = 64.

Cetking shortcut: 'Behind' always adds to the rank from top, 'ahead' subtracts. Remember the −1 in the total formula.

Answer: E

Approach: Two-set Venn: Total = only V + only H + both (everyone plays at least one).

Solution:
Violin total = 30, both = 20 → only violin = 30 − 20 = 10.
Only harmonium = 80 − 10 − 20 = 50.
Harmonium total = 50 + 20 = 70.
So: 70, 50 and 10.

Cetking shortcut: Only option A ends with 10 for only-violin (30 − 20) — the rest can be eliminated instantly.

Answer: A

Approach: Assertion–Reason: check each statement's truth, then ask whether R actually causes A.

Solution:
A: Sound is slowest in gases, faster in liquids, fastest in solids → A is true.
R: Gases are much less dense than solids → R is true.
Does R explain A? Speed of sound v = √(elasticity ÷ density). Density sits in the denominator, so lower density by itself would make sound faster. Sound is slower in gases because gases have far lower elasticity (they are easily compressed, molecules weakly bound) — not because of low density.
So both are true but R is not the correct explanation.

Why the other options are wrong:
A: R is not the cause — density works the other way in v = √(E/ρ).
C, D, E: both statements are factually true.

Cetking shortcut: In A–R science questions, first test whether the reason would push the result in the stated direction; here it pushes the opposite way.

Answer: B

Approach: An assumption is implicit only if the statement cannot make sense without it.

Solution:
The statement only reports that the hill trip was abandoned; it gives no reason.
I: Lack of profit is one possible reason, but the statement does not depend on it → not implicit.
II: Bad climate is another possible reason, equally not required → not implicit.
The trip could have been dropped for low bookings, safety, permissions, etc.

Why the other options are wrong:
A, B: each picks one speculative cause.
D: 'either-or' needs the two to be the only exhaustive possibilities — they are not.
E: neither is necessary.

Cetking shortcut: If the statement states a fact with no reason, any assumption giving a specific cause usually fails.

Answer: C

Approach: Write the word and its alphabetical arrangement position by position and count matches.

Solution:

Cetking shortcut: Sort by counting letters (A2, E1, I2, L2, N1, O1, R1, T2) and fill positions; compare only where the same letter could land.

Answer: D

Approach: Compare sentences pairwise; a word common to two sentences has the code common to both.

Solution:
S1: rbm std bro pus = the puppy is beautiful
S2: tnh pus dim std = the sparrow is red
S3: pus dim bro pus cus = the sparrow has the puppy
'the' appears twice in S3 and 'pus' appears twice → the = pus.
S1 & S2 share {the, is} and {std, pus} → is = std.
S2 & S3 share {the, sparrow} and {pus, dim} → sparrow = dim.
S1 & S3 share {the, puppy} and {pus, bro} → puppy = bro.
Left in S3: has = cus.

Why the other options are wrong:
A Dim = sparrow; B Pus = the; C Bro = puppy; D Std = is.

Answer: E

Approach: Each letter → its alphabet position; two-digit positions are reduced to their digit sum.

Solution:
Check: I9, N14→5, S19→10, T20→2, I9, T→2, U21→3, T→2, E5 → 9 5 10 2 9 2 3 2 5 = 9510292325 ✓
GOVERNMENT: G7, O15→6, V22→4, E5, R18→9, N14→5, M13→4, E5, N→5, T→2
→ 7645954552.

Why the other options are wrong:
A, B, C: start with 5, but G = 7.
E: has 11 digits and wrong values.

Cetking shortcut: G = 7 → only D and E start with 7; O→6 kills E.

Answer: D

Approach: Form one linear equation in the youngest age.

Solution:
Let Kailash = x → Riyansh = 2x → Rajesh = 2x + 2.
x + 2x + 2x + 2 = 37 → 5x = 35 → x = 7.
Riyansh = 2 × 7 = 14.

Cetking shortcut: Riyansh must be even and (37 − 2) must split as 5x → test 14: 7 + 14 + 16 = 37 ✓.

Answer: C

Approach: Find the constant ratio between terms.

Solution:
21 × 5 = 105, 105 × 5 = 525, 525 × 5 = 2625, 2625 × 5 = 13125, 13125 × 5 = 65625.
The term given as 2615 should be 2625 → 2615 is wrong.

Cetking shortcut: All correct terms of a ×5 series from 525 onward end in 25; 2615 does not.

Answer: D

Approach: Compare the word and its code letter by letter.

Solution:
INSTITUTE → EOTUJUVUI: first and last letters swap (I…E → E…I); each middle letter moves +1 (N→O, S→T, T→U, I→J, T→U, U→V, T→U).
MEDICOVER: swap ends → R … M; middle E D I C O V E → F E J D P W F.
Code = RFEJDPWFM.

Why the other options are wrong:
A, B, D: middle letters do not follow +1 (e.g. E→F, D→E fails).
C: ends not swapped.

Cetking shortcut: Check just the 2nd and 3rd letters: E→F, D→E → only option E has 'FE' after R.

Answer: E

Approach: Find the category shared by four words.

Solution:
Phi, Gamma, Delta, Beta are letters of the Greek alphabet.
Peso is a currency (Mexico, Philippines, etc.).

Answer: D

Approach: Place points on a grid with E at the origin.

Solution:
E = (0, 0). N is east of E → N = (1, 0).
G is south of E → G = (0, −1). L is west of G → L = (−1, −1).
From L to N: +2 east and +1 north → N is North-East of L.

Why the other options are wrong:
A, C, E: N is above L (north), not south.
B: N is to the east of L, not west.

Answer: D

Approach: Two-row seating with rows facing each other — fix left/right for each row first.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, C: Aman is at seat 2, not beside Suresh (the puzzle says so).
D: Raj is at seat 1.
E: Sunil is in the other row.

Cetking shortcut: Start with the clue that ties two rows (front of Aman / left of Sunil) — it pins Sunil to a corner at once.

Answer: B

Approach: Use the final two-row arrangement.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, B, C: each is correct but incomplete.
E: Sandeep is at R1-6, next to Rohit only.

Answer: D

Approach: Use the final arrangement; 'in front of' = same seat number, other row.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A: Rajesh faces Suresh.
B: Suresh faces Rajesh, not a vacant seat.
D: Raj faces Sunil.
E: Rohit faces Omkar.

Answer: C

Approach: Use the final arrangement.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
A, E: Yash and Suresh are in Row 2.
B: Ramesh is two seats away (vacant seat between).
D: Sandeep is at seat 6.

Answer: C

Approach: Use the final arrangement; opposite = same seat number in the other row.

Solution:
Facing each other: Row 1 faces South (its right = West), Row 2 faces North (its right = East). Number seats 1–6 from West to East.
Person in front of Aman is immediately left of Sunil → that person is at Sunil+1, so corner Sunil = seat 1, Aman = R2 seat 2.
Sunil in front of Raj → Raj = R2 seat 1. Sandeep diagonally opposite Raj → R1 seat 6; Rohit immediately right of Sandeep → R1 seat 5.
Ramesh 2nd right of Rajesh → Rajesh 4, Ramesh 2, R1 seat 3 vacant. Other vacant is a corner → R2 seat 6.
Aman (R2-2) → Rohit (R1-5) is 3 seats across; same for Ramesh (R1-2) → Omkar = R2-5. Suresh not next to Aman → Suresh 4, Yash 3.

Why the other options are wrong:
B, C: Sunil faces Raj, not Yash.
D: Sandeep and Raj are diagonally opposite, not directly opposite.
E: Rohit faces Omkar and Sunil faces Raj.

Answer: A

Approach: Substitute the real operators, then apply BODMAS.

Solution:
+ → ÷, ÷ → −, − → ×, × → +.
10 + 5 ÷ 3 − 2 × 9 becomes 10 ÷ 5 − 3 × 2 + 9.
= 2 − 6 + 9 = 5.

Cetking shortcut: Rewrite the full expression once before calculating — mid-way substitution causes most errors.

Answer: E

Approach: Check the differences between consecutive terms.

Solution:
110 − 20 = 90, 210 − 110 = 100, 320 − 210 = 110.
Differences increase by 10 → next difference 120.
320 + 120 = 440.

Answer: C

Approach: Decode the machine rule from the example, then run the new input.

Solution:
Rule: Step I – digits of each number in ascending order; Step II – numbers in ascending order; Step III – keep the middle two digits; Step IV – add those two digits.
Step I: 2579 4578 2358 2568 1568 2467 2356 2456
Step II: 1568 2356 2358 2456 2467 2568 2579 4578
Step III: 56 35 35 45 46 56 57 57
Step IV: 11 8 8 9 10 11 12 12
2nd + 4th + 6th + 8th = 8 + 9 + 11 + 12 = 40.

Cetking shortcut: After Step I you only need the middle two digits of each number, and Step II order can be read off the sorted 4-digit values directly.

Answer: D

Approach: Track facing direction after each turn.

Solution:
Start facing South, walk 5 km → (0, −5).
Right turn while facing South → now facing West; walk 6 km → (−6, −5).
Left turn while facing West → now facing South; walk 4 km → (−6, −9).
Final point is west and south of start → South-West.

Why the other options are wrong:
A: he is also 9 km south.
B: he is also 6 km west.
C, E: he is never north of the start.

Cetking shortcut: Right of South = West; left of West = South.

Answer: D

Approach: Split into interleaved series of period 3.

Solution:
Positions 1, 4, 7: 12, 22, 32 → +10 each.
Positions 2, 5, 8: 16, 26, 36 → +10 each.
Positions 3, 6, 9: 78, 78, 78 → constant.
Next pair (8th, 9th) = 36, 78.

Cetking shortcut: A repeating number (78) every third term signals a period-3 interleave.

Answer: A

Approach: Build one chain from all comparisons.

Solution:
Vishal > Rahil > Chaitanya (Rahil taller than Chaitanya, shorter than Vishal).
Chaitanya > Sushil > Aniket.
Chain: Vishal > Rahil > Chaitanya > Sushil > Aniket → Vishal is tallest.

Answer: C

Approach: Decode the speaker's statement, then place Ruchi.

Solution:
Neetu's father-in-law is Chandrakant; his only son is Omkar → Omkar is Neetu's husband.
Omkar is Priya's father; Ruchi is Priya's sister → Ruchi is Omkar's daughter.
Chandrakant is Omkar's father → Ruchi is Chandrakant's granddaughter.

Why the other options are wrong:
A: niece is a sibling's daughter.
B: daughter would be Omkar's sister.
C, D: wife / daughter-in-law relate to Chandrakant and Neetu, not Ruchi.

Answer: E

Approach: Two parallel rows facing North: fix positions by the 'not at end' and 'nth to the left/right' clues first.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Cetking shortcut: H8's clue already says it is not at an end — every other option is a corner car.

Answer: D

Approach: Opposite = same position number in the other row.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A: H3 is in H4's own row.
C: H5 faces H6.
D: H1 faces H7.
E: H2 faces H3.

Answer: B

Approach: Read row membership from the final arrangement.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
B, C, D, E: H2, H1, H5, H8 are all Yellow-line cars.

Answer: A

Approach: Same-row relative position; all cars face North.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A, D: there is a car (H7) between them.
B: that is H3 with respect to H4.
E: both are in the Blue line.

Answer: C

Approach: Opposite = same position in the other row.

Solution:
Blue line cars face the back of Yellow line cars → Yellow is the front (north) row. All face North, so left = West. Number positions 1–4 from West.
H3 is 2nd left of H4 and H4 is not at an end → H4 = 3, H3 = 1 (same row).
H5 3rd right of H2 → H2 = 1, H5 = 4 in the other row.
H1 is in a different row from H4 → H1 in the H2/H5 row (position 2 or 3).
H7 is opposite H1, so H7 is in H4's row and that row is Blue; H7 can only be at 2 → H1 = 2.
H8 not at an end → Yellow 3; H6 → Blue 4.

Why the other options are wrong:
A: H1 faces H7.
C, D: H7 and H4 are in H6's own row.
E: H2 faces H3.

Answer: B

Approach: Find the rule linking the side numbers to the centre number in each figure.

Solution:
Figure 1: left 7, right 6 → 7 × 6 + 5 = 47 (centre) ✓
Figure 2: left 12, right 8 → 12 × 8 + 5 = 101 ✓
Figure 3: ? × 8 + 5 = 77 → ? × 8 = 72 → ? = 9.
(Top = 5 × bottom in each figure: 45 = 5×9, 65 = 5×13, 80 = 5×16 — a separate check, not needed for ?.)

Cetking shortcut: Back-solve: (77 − 5) ÷ 8 must be a whole number — only 9 works.

Answer: D

Approach: Find the rotation that maps North to South-East and apply it to East.

Solution:
North → South-East is a clockwise turn of 135° (N 0°, E 90°, SE 135°).
East (90°) + 135° = 225° = South-West.

Why the other options are wrong:
A: NE is a 45° turn.
B: SE is what North became.
D: NW is the opposite (anticlockwise) result.
E: South is a 90° turn.

Cetking shortcut: Each direction moves 3 steps clockwise on the 8-point compass (N→NE→E→SE); East moves E→SE→S→SW.

Answer: C

Approach: Look at a common property of the letter groups.

Solution:
APS, EBT, IRP, OTQ begin with a vowel (A, E, I, O).
PST begins with a consonant → odd one.

Cetking shortcut: Check first letters before computing gaps — vowel/consonant is the quickest test.

Answer: E

Approach: Find the shared category.

Solution:
Kohima (Nagaland), Imphal (Manipur), Ranchi (Jharkhand), Jaipur (Rajasthan) are state capitals.
Indore is not a capital (Madhya Pradesh's capital is Bhopal).

Answer: A

Approach: Identify the relation in the first pair: antonyms.

Solution:
Ally (friend) is the opposite of foe (enemy).
Admit (accept as true) — its opposite is repudiate (deny, reject).

Why the other options are wrong:
B Plaint = complaint.
C Humorous, D Indiscreet, E Tedious have no opposite relation to 'admit'.

Answer: A

Approach: Read the exact Venn region.

Solution:
Chess ∩ Ludo regions: 4 (outside Carrom) and 3 (inside Carrom).
'Chess and Ludo but not Carrom' = the region inside Chess and Ludo but outside Carrom (and outside Table Tennis) = 4.
The total of 600 is not needed.

Why the other options are wrong:
A: 3 is Chess ∩ Ludo ∩ Carrom.
C: 10 is Table Tennis ∩ Chess (or Chess ∩ Carrom).
D: 8 is Carrom ∩ Ludo only.
E: 5 is TT ∩ Chess ∩ Carrom.

Answer: B

Approach: Scan each odd digit and test both neighbours.

Solution:
7 2 8 9 1 3 4 5 7 5 8 9 6 9 1 8 2 4 9 2 6
Odd digits preceded by an even digit: 9 (after 8), 5 (after 4), 9 (after 8), 9 (after 6), 9 (after 4).
Now 'not followed by odd': 8-9-1 ✗, 4-5-7 ✗, 8-9-6 ✓, 6-9-1 ✗, 4-9-2 ✓.
Count = 2.

Cetking shortcut: Look only for the pattern even–odd–even.

Answer: B

Approach: Decode each symbol and note the gender each one fixes.

Solution:
E + F → E is brother of F (E male).
F @ G → F is father of G (F male).
G * H → G is sister of H (G female).
E and F are siblings and F is male → F is the brother of E.

Why the other options are wrong:
A: G is H's sister, not mother.
D: E is male (brother), so cannot be a sister; E is H's uncle.
E: F is male (father), so not a sister.
B: G is female and F is G's father, so G is also F's daughter — this statement is also true from the data.

Answer: C

Approach: Compare letter positions of word and code.

Solution:
S→Q, H→F, R→P, I→G, J→H, E→C, E→C, T→R: every letter moves 2 back.
INTEREST: I→G, N→L, T→R, E→C, R→P, E→C, S→Q, T→R → GLRCPCQR.

Why the other options are wrong:
B, C, E: only 7 letters and no −2 pattern.
D: letters do not follow −2.

Cetking shortcut: I − 2 = G and N − 2 = L → only option A starts with GL.

Answer: A

Approach: Find the rule moving around the star's outer numbers.

Solution:
9 → 17 → 33 → 65: each term = previous × 2 − 1 (9×2−1 = 17, 17×2−1 = 33, 33×2−1 = 65).
Next = 65 × 2 − 1 = 129.
Check with centre: 9 + 17 + 33 + 65 + 129 = 253 ✓

Cetking shortcut: Centre 253 − (9 + 17 + 33 + 65 = 124) = 129 directly.

Answer: B

Approach: Stacked boxes: anchor on the clue chain F → chocolate → G, then split into two cases.

Solution:
Positions 1 (bottom) to 7 (top). Chocolate is just below F, and G is 3 below chocolate (two boxes between, G not above) → G = F − 4. Burger and L are below G → F = 7 or 6.
F = 6: G = 2, L = burger = 1, M = 5, P = 3; H must be 7 (H = 4 would put samosa with P), then H can only hold biscuits — at the top ✗.
F = 7: chocolate 6, G 3; L 2, P 4 (one box between). H needs a box below, so I = 1. M = 5 → burger 1 (I), laddus 4 (P); H = 6 → samosa 5 (M). (M = 6 fails: H = 5 puts samosa with P.)
Left items biscuits, momos, dry fruits for F, G, L: F not biscuits/momos → dry fruits; L not momos → biscuits; G → momos.

Why the other options are wrong:
A P is 4th, C G is 3rd, D L is 2nd, E I is bottom — none is just below F.

Cetking shortcut: G = F − 4 with something below G forces F into the top two slots — only two cases to test.

Answer: B

Approach: Use the final stack.

Solution:
Positions 1 (bottom) to 7 (top). Chocolate is just below F, and G is 3 below chocolate (two boxes between, G not above) → G = F − 4. Burger and L are below G → F = 7 or 6.
F = 6: G = 2, L = burger = 1, M = 5, P = 3; H must be 7 (H = 4 would put samosa with P), then H can only hold biscuits — at the top ✗.
F = 7: chocolate 6, G 3; L 2, P 4 (one box between). H needs a box below, so I = 1. M = 5 → burger 1 (I), laddus 4 (P); H = 6 → samosa 5 (M). (M = 6 fails: H = 5 puts samosa with P.)
Left items biscuits, momos, dry fruits for F, G, L: F not biscuits/momos → dry fruits; L not momos → biscuits; G → momos.

Answer: C

Approach: Use the final stack.

Solution:
Positions 1 (bottom) to 7 (top). Chocolate is just below F, and G is 3 below chocolate (two boxes between, G not above) → G = F − 4. Burger and L are below G → F = 7 or 6.
F = 6: G = 2, L = burger = 1, M = 5, P = 3; H must be 7 (H = 4 would put samosa with P), then H can only hold biscuits — at the top ✗.
F = 7: chocolate 6, G 3; L 2, P 4 (one box between). H needs a box below, so I = 1. M = 5 → burger 1 (I), laddus 4 (P); H = 6 → samosa 5 (M). (M = 6 fails: H = 5 puts samosa with P.)
Left items biscuits, momos, dry fruits for F, G, L: F not biscuits/momos → dry fruits; L not momos → biscuits; G → momos.

Why the other options are wrong:
A H = chocolate, B G = momos, C L = biscuits, E I = burger.

Answer: D

Approach: Use the final stack.

Solution:
Positions 1 (bottom) to 7 (top). Chocolate is just below F, and G is 3 below chocolate (two boxes between, G not above) → G = F − 4. Burger and L are below G → F = 7 or 6.
F = 6: G = 2, L = burger = 1, M = 5, P = 3; H must be 7 (H = 4 would put samosa with P), then H can only hold biscuits — at the top ✗.
F = 7: chocolate 6, G 3; L 2, P 4 (one box between). H needs a box below, so I = 1. M = 5 → burger 1 (I), laddus 4 (P); H = 6 → samosa 5 (M). (M = 6 fails: H = 5 puts samosa with P.)
Left items biscuits, momos, dry fruits for F, G, L: F not biscuits/momos → dry fruits; L not momos → biscuits; G → momos.