◆ DILR · Logical Reasoning

Logical Reasoning, approach + real CAT sets

CAT LR comes in SETS: a caselet plus 4-6 linked questions. Winning is 90% set-selection and clean table-building. Master the approach cards, then drill real CAT sets with the book's worked answers.

16approach cards
31CAT sets
★★★priority

Approach & Set-Type Sheet

How to read a caselet, pick the right sets, and tackle each family. A-to-Z for CAT LR.

1Two-pass set selection
  • Pass 1 (skim, ~20s/set): read only the caselet + count the variables. Tag each set Easy / Medium / Skip. Do NOT solve yet.
  • Pass 2: solve your Easy/Medium picks first. A doable set gives 3-6 marks; a trap eats 12 minutes for zero.
  • In a 4-set, 40-min paper, aim to fully crack 2-3 sets rather than half-solve all four.
  • Rule: if after ~5 minutes you still have no clean grid, abandon and move on, come back fresh later.
2Reading the caselet
  • Read the common data twice before touching the questions. The trap is usually a constraint you skimmed.
  • List the variables (people, days, items) and the attributes (colour, rank, score) as column/row headers.
  • Create your own shorthand symbols for clues so you can combine them quickly.
  • Order clues by usefulness: start from the most restrictive / direct clue, then link indirect clues onto it.
3Linear arrangements
  • People in a row / seats / queue / floors. Fix a direction (left→right, bottom→top) and stick to it.
  • Anchor the most constrained person first; place "immediately next to / between / ends" clues early.
  • Use a blank slot diagram and pencil tentative placements; branch only when forced.
  • Watch "left of" vs "immediately left of", gaps are allowed in the former.
4Circular arrangements
  • Note whether people face centre or outward, it flips left/right.
  • Fix one person's seat to kill rotational symmetry, then place neighbours.
  • "Opposite" only works cleanly when the count is even.
  • For mixed in/out facing, mark each person's facing direction next to them.
5Distribution & grouping
  • Items/people split into groups or buckets (e.g. idlis & vadas per friend, committees).
  • Build a person × attribute grid; fill forced cells, then test the rest.
  • Use totals as a check: row sums and column sums must match the given total.
  • When numeric, set up equations from the constraints (see Mathematical Reasoning).
6Selection with conditions
  • "Choose a team / menu / mixture such that…", translate each rule into if-then form.
  • "If A then B" and its contrapositive "if not B then not A" are both usable.
  • Eliminate options by testing them against rules, often faster than building from scratch.
  • Watch quantifiers: all / some / none / at least / at most change everything.
7Ordering & ranking
  • Rank by height/score/age/finish. Draw a vertical or horizontal order line.
  • Convert comparisons into a chain: A > B, B > C ⇒ A > B > C
  • "X is taller than exactly two people" pins X's absolute position.
  • Track which gaps are fixed vs flexible; many ranking sets have multiple valid orders.
8Mathematical reasoning
  • Clues hide arithmetic, percentages, ratios, averages, products of ages, costs.
  • Assign variables, write one equation per clue, solve simultaneously.
  • Use number properties & factor lists (e.g. age-product puzzles).
  • Plug answer options back in when direct algebra is messy.
9Games & tournaments
  • Knockout / elimination: n teams ⇒ n − 1 matches total (byes included).
  • Round-robin (all-play-all): ⁿC₂ matches for n teams.
  • Seeding: in round 1, seed k plays seed (N+1−k); favourites meet late. An upset = lower seed beats higher.
  • "M did not beat N" ≠ "N won", a draw/tie may be possible. Read win/loss logic carefully.
10Scheduling & timetables
  • Match activities/people to days, slots, rooms, or years (e.g. vendor contracts 2010-2019).
  • Grid: rows = entities, columns = time periods; mark contracts/sessions as spans.
  • Use "exactly N in period X" and "no overlap / multi-year" constraints to lock cells.
  • Single-year vs multi-year spans behave differently, track durations explicitly.
11Networks & routes
  • Nodes (junctions) + directed edges with costs/times. List every valid path S→T.
  • Total cost of a route = sum of edge costs + tolls/penalties at nodes.
  • For "make all routes equal-cost", set path-cost expressions equal and solve for tolls.
  • Congestion games: travel time rises per extra car, equilibrium = no car can switch and gain.
12Set theory / Venn
  • 2 sets: n(A∪B) = n(A) + n(B) − n(A∩B)
  • 3 sets: n(A∪B∪C) = Σn(A) − Σn(A∩B) + n(A∩B∩C)
  • A+B+C = (exactly 1) + 2(exactly 2) + 3(exactly 3)
  • For "minimum in all sets", maximise the people outside: min(all four) ≥ 100 − Σ(don't-have %).
13Binary & conditional logic
  • Truth-tellers vs liars, true/false statements, two-valued (Hi/Lo, Yes/No) bids.
  • Assume one value, propagate consequences, check for contradiction; flip if it breaks.
  • Tabulate every round/statement against each player and verify the running totals.
  • Coded-arithmetic (cryptarithm): use carries, leading digit can't be 0, X+X gives parity clues.
14Grids & layouts
  • Spatial sets: m×n grid of heights/beads, line-of-sight "can reach" rules.
  • Re-state each rule as a cell-by-cell condition; then scan systematically row by row.
  • Build a parallel "reachability count" or colour grid rather than re-deriving each time.
  • Adjacency & "between" rules (e.g. one Green between two Blues) cap the pattern density.
15Data-sufficiency style LR
  • "What can be concluded?", answer only what the clues force, never assume the rest.
  • "Must be true / must be false / cannot be determined", test each option against ALL valid cases.
  • If two arrangements both satisfy the clues, anything differing between them is "cannot be determined".
  • Beware option-set trap: if "A, B and C" are all possible but options only pair them, pick "None of these".
16Exam-day LR tactics
  • Spend the first 2-3 min doing Pass-1 selection across the whole section before solving.
  • TITA (non-MCQ) sets carry no negative marks, attempt freely once the grid is built.
  • One clean grid usually answers all sub-questions, invest in the grid, not in each question.
  • Don't fall for sunk cost: a half-built table you've struggled with for 7 min is a signal to move on.
31 CAT sets · 131 questions

Real CAT LR Sets

21 actual CAT previous-year sets (87 questions) from the book. Each set shows the caselet, then its linked questions with the book's answers. Difficulty: Easy Moderate Hard. Click any question to reveal the solution.

CAT 2001

SET, Coloured leaves (mathematical reasoning). On her walk through the park, Hansa collected 50 coloured leaves, all either maple or oak. She sorted them and found: A. The number of red oak leaves with spots is even and positive. B. The number of red oak leaves without any spot equals the number of red maple leaves without spots. C. All non-red oak leaves have spots, and there are five times as many of them as there are red spotted oak leaves. D. There are no spotted maple leaves that are not red. E. There are exactly 6 red spotted maple leaves. F. There are exactly 22 maple leaves that are neither spotted nor red.

ModerateCAT 2001

How many oak leaves did she collect?

  • (1) 22
  • (2) 17
  • (3) 25
  • (4) 18
Show solution
(2) 17. Let red-not-spotted oak = x, red-spotted oak = y. Non-red oak (all spotted) = 5y; red-not-spotted maple = x (clue B); red-spotted maple = 6; maple neither = 22. Total: 6 + x + 22 + y + x + 5y = 50 ⇒ x + 3y = 11. With y even & positive ⇒ y = 2, x = 5. Oak = y + x + 5y = 2 + 5 + 10 = 17.
HardCAT 2001

I have a total of ₹1,000. Item A costs ₹110, B ₹90, C ₹70, D ₹40, E ₹45. For every D I must buy two B; for every A, one C; for every E, two D and one B. Each item earns 1000 points; every rupee not spent costs 1500 points. To maximise points, how many items must I purchase?

  • (1) 13
  • (2) 14
  • (3) 15
  • (4) 16
Show solution
(2) 14. Valid bundles: Set1 (A,C)=₹180 [2 items]; Set2 (D,2B)=₹220 [3 items]; Set3 (E,2D,B)=₹215 [4 items]. To maximise items within ₹1000: 3×Set1 + 1×Set3 = 540 + 430 = ₹970, giving 3×2 + 4 = 14 items, very close to budget. Hence 14.
HardCAT 2001

Ashok, Bashir, Chirag, Deepak go shopping. Ashok has less than three times Bashir's money; Chirag has more than Bashir; Deepak = |Bashir − Chirag|; Ashok = 3 × Deepak. Items cost ₹200 (shirt), ₹400 (shawl), ₹600 (sweater), ₹1000 (jacket). Chirag borrows ₹300 from Ashok and buys a jacket. Bashir buys a sweater after borrowing ₹100 from Ashok and is left with nothing. Ashok buys three shirts. What is the costliest item Deepak could buy with his own money?

  • (1) A shirt
  • (2) A shawl
  • (3) A sweater
  • (4) A jacket
Show solution
(2) A shawl. Bashir + 100 = 600 ⇒ Bashir = 500, so Ashok < 1500. Ashok lent 300 + 100 and bought 3 shirts (₹600) ⇒ Ashok ≥ 1000. So 1000 ≤ Ashok ≤ 1500 ⇒ Deepak = Ashok/3 lies between ~333 and 500. Deepak can afford a shawl (₹400) but not a sweater.

CAT 2003

SET, Paint mixing (Rang Barsey). RBPC buys RED (₹20), YELLOW (₹25), WHITE (₹15), ORANGE (₹22), PINK (₹18) per litre. ORANGE = RED + YELLOW (equal parts); PINK = RED + WHITE (equal parts). It sells CREAM = WHITE:YELLOW = 70:30, AVOCADO = equal ORANGE + PINK, WASHEDORANGE = equal ORANGE + WHITE.

ModerateCAT 2003

The cheapest way to manufacture AVOCADO paint would cost:

  • (1) ₹19.50/L
  • (2) ₹19.75/L
  • (3) ₹20.00/L
  • (4) ₹20.25/L
Show solution
(2) ₹19.75/L. AVOCADO = ORANGE + PINK. Cheapest: ORANGE (₹22) + (RED+WHITE)/2 = 22 + (20+15)/2 = 22 + 17.5, averaged = (22 + 17.5)/2 = ₹19.75. Other combinations give 20, 20.25, all higher.
ModerateCAT 2003

WASHEDORANGE can be manufactured by mixing:

  • (1) CREAM and RED, 14:10
  • (2) CREAM and RED, 3:1
  • (3) YELLOW and PINK, 1:1
  • (4) RED, YELLOW, WHITE, 1:1:2
Show solution
(4) RED, YELLOW, WHITE in 1:1:2. WASHEDORANGE = ORANGE + WHITE (1:1) = (RED+YELLOW) + 2·WHITE = RED:YELLOW:WHITE = 1:1:2.
HardCAT 2003

Assume AVOCADO, CREAM and WASHEDORANGE each sell at the same price. Which is the most profitable to manufacture?

  • (1) AVOCADO
  • (2) CREAM
  • (3) WASHEDORANGE
  • (4) Insufficient data
Show solution
(2) CREAM. Least costs: AVOCADO = ₹19.75; CREAM = (7×15 + 3×25)/10 = ₹18; WASHEDORANGE (O:W=1:1) = (22+15)/2 = ₹18.5. Same selling price ⇒ lowest cost = highest profit ⇒ CREAM.

SET, Basketball baskets. Facts: (a) Ganesh = Ashish − 8; (b) Dhanraj + Ramesh = 37; (c) Jugraj = Dhanraj + 8; (d) Ashish = Dhanraj + 5; (e) Ashish + Ganesh = 40.

EasyCAT 2003

Which of the following statements is true?

  • (1) Ramesh 18, Dhanraj 19
  • (2) Ganesh 24, Ashish 16
  • (3) Jugraj 19, Dhanraj 27
  • (4) Dhanraj 11, Ashish 16
Show solution
(1) Ramesh 18, Dhanraj 19. From (a)+(e): Ashish = 24, Ganesh = 16. (d): Dhanraj = 19. (b): Ramesh = 18. (c): Jugraj = 27. So Ramesh 18 and Dhanraj 19.
EasyCAT 2003

Which of the following statements is true?

  • (1) Dhanraj + Jugraj = 46
  • (2) Ganesh 18, Ramesh 21
  • (3) Dhanraj 3 more than Ramesh
  • (4) Ramesh + Jugraj = 29
Show solution
(1) Dhanraj + Jugraj = 46. Dhanraj 19 + Jugraj 27 = 46.

SET, Idli-vada breakfast. Five friends each eat a different number of idlis (1,4,5,6,8) and vadas (0,1,2,4,6). Clues: (a) Ignesh's vadas = 3× the vadas of the person eating 4 idlis; (b) three persons (incl. the 4-vada one) eat without chutney; (c) Sandeep takes no chutney; (d) the 1-idli person eats no vada/chutney and is not Mukesh; (e) Daljit eats idli with chutney and also eats vada; (f) Mukesh (no chutney) eats half as many vadas as the person eating twice his idlis; (g) Bimal eats 2 more idlis than Ignesh, but Ignesh eats 2 more vadas than Bimal.

HardCAT 2003

Which one statement is true?

  • (1) Daljit eats 5 idlis
  • (2) Ignesh eats 8 idlis
  • (3) Bimal eats 1 idli
  • (4) Bimal eats 6 idlis
Show solution
(1) Daljit eats 5 idlis. Final grid: Ignesh 6 idli/6 vada (chutney), Mukesh 4/2 (no), Sandeep 1/0 (no), Bimal 8/4 (no), Daljit 5/1 (chutney). So Daljit = 5 idlis.
HardCAT 2003

Which statement is true?

  • (1) Sandeep eats 2 vadas
  • (2) Mukesh eats 4 vadas
  • (3) Ignesh eats 6 vadas
  • (4) Bimal eats 2 vadas
Show solution
(3) Ignesh eats 6 vadas. From the grid above, Ignesh = 6 vadas.
HardCAT 2003

Which statement is true?

  • (1) Mukesh eats 8 idlis & 4 vadas, no chutney
  • (2) The 5-idli, 1-vada person takes no chutney
  • (3) The person eating equal vadas & idlis also takes chutney
  • (4) The 4-idli, 2-vada person also takes chutney
Show solution
(3) The equal-vadas-and-idlis person takes chutney. Ignesh eats 6 idlis and 6 vadas with chutney, the only person with equal counts. So statement (3) holds.

CAT 2005

SET, Disguised firms (data-sufficiency LR). Revenue (₹ million) of four firms in three states; the firms Honest, Aggressive, Truthful and Profitable are disguised as A, B, C, D in some order.

StateFirm AFirm BFirm CFirm D
UP49828055
Bihar69727065
MP72637265

Also: in MP, Truthful has the highest market share; Aggressive's aggregate revenue differs from Honest's by ₹5 million. (Aggregates: A=190, B=217, C=222, D=185.)

HardCAT 2005

Statement 1: Profitable has the lowest share in MP. Statement 2: Honest's total revenue is more than Profitable. What can be said?

  • (1) S1 true ⇒ S2 necessarily true
  • (2) S1 true ⇒ S2 necessarily false
  • (3) Both true
  • (4) Neither true
Show solution
(2) If S1 is true then S2 is necessarily false. Aggressive & Honest differ by 5 ⇒ they are {A,D}(190,185) or {B,C}(217,222). If S1 true, B is Profitable, so Honest is A or D, whose total is less than B(Profitable). Hence S2 false.
HardCAT 2005

Statement 1: Aggressive's lowest revenues are from MP. Statement 2: Honest's lowest revenues are from Bihar. What can be said?

  • (1) S2 true ⇒ S1 false
  • (2) S1 false ⇒ S2 true
  • (3) S1 true ⇒ S2 true
  • (4) None
Show solution
(3) If S1 is true then S2 is true. S1 true ⇒ Firm B = Aggressive (lowest in MP=63) and Firm C = Honest, whose lowest (70) is in Bihar. So S2 follows.
HardCAT 2005

Statement 1: Honest has the highest share in UP. Statement 2: Aggressive has the highest share in Bihar. What can be said?

  • (1) Both could be true
  • (2) At least one must be true
  • (3) At most one is true
  • (4) None
Show solution
(3) At most one is true. Firm B has the highest share in both UP (82) and Bihar (72). Honest and Aggressive are different firms, so they cannot both be B, at most one statement holds.
ModerateCAT 2005

If Profitable's lowest revenue is from UP, which is true?

  • (1) Truthful's lowest from MP
  • (2) Truthful's lowest from Bihar
  • (3) Truthful's lowest from UP
  • (4) No definite conclusion
Show solution
(3) Truthful's lowest revenues are from UP. Profitable lowest in UP ⇒ Firm D = Profitable, so Firm A = Truthful, whose lowest (49) is in UP.

CAT 2006

SET, Traffic network & tolls. One-way streets from S to T via junctions A, B, C, D with fuel costs: S→A=9, A→T=5, S→B=2, B→A=2, B→C=3, S→D=7, D→T=6, D→C=1, C→T=2. Motorists pick the least-cost route; ties split traffic evenly. The government levies tolls a, b, c, d at A, B, C, D.

HardCAT 2006

If the street B→C is unusable and all motorists must pay the same total (fuel + toll), a feasible set of tolls at A, B, C, D is:

  • (1) 2, 5, 3, 2
  • (2) 0, 5, 3, 1
  • (3) 1, 5, 3, 2
  • (4) Both (2) and (3)
Show solution
(4) Both (2) and (3). Routes: SAT=14+a, SBAT=9+a+b, SDT=13+d, SDCT=10+c+d. Setting all equal: (0,5,3,1) makes each = 14; (1,5,3,2) makes each = 15. Both work.
HardCAT 2006

If no traffic should flow D→T while equal traffic flows through A and C, a feasible toll set at A, B, C, D is:

  • (1) 1, 5, 3, 3
  • (2) 1, 4, 4, 3
  • (3) 1, 5, 4, 2
  • (4) 0, 5, 2, 2
Show solution
(4) 0, 5, 2, 2. Active routes SAT, SBCT, SBAT, SDCT must equalise: 14+a = 7+b+c = 9+a+b = 10+c+d. (0,5,2,2) satisfies this.
HardCAT 2006

If all routes from S to T should get the same traffic, a feasible toll set at A, B, C, D is:

  • (1) 0, 5, 2, 2
  • (2) 0, 5, 4, 1
  • (3) 1, 5, 3, 3
  • (4) 1, 5, 3, 2
Show solution
(4) 1, 5, 3, 2. All five routes (SAT=14+a, SBCT=7+b+c, SDT=13+d, SBAT=9+a+b, SDCT=10+c+d) equal only for (a,b,c,d) = (1,5,3,2).
HardCAT 2006

If traffic at S should split evenly along S→A, S→B and S→D, a feasible toll set at A, B, C, D is:

  • (1) 0, 5, 4, 1
  • (2) 0, 5, 2, 2
  • (3) 1, 5, 3, 3
  • (4) 1, 5, 3, 2
Show solution
(1) 0, 5, 4, 1. Need 14+a = 9+b = 13+d (so a=0, b=5, d=1) plus the C-routes costlier than 14, which holds only for option (1).
HardCAT 2006

The government wants to minimise total commuter cost per trip while ensuring no more than 70% of traffic passes through B. The cost per trip under this policy is:

  • (1) ₹7
  • (2) ₹9
  • (3) ₹10
  • (4) ₹13
Show solution
(3) ₹10. With C and D toll = 0 and B toll = ₹3, routes SDCT and SBCT both cost ₹10, splitting traffic acceptably, the minimum feasible commuter cost.

CAT 2007

SET, Diet formulations (selection with conditions). Five ingredients can be mixed in proportions. Cost/unit: O 150, P 50, Q 200, R 500, S 100.

IngredientCarb %Protein %Fat %Minerals %
O50301010
P802000
Q10305010
R550405
S455005
ModerateCAT 2007

A balanced diet needs ≥30% each of carb and protein, ≤25% fat, ≥5% minerals. Which equally-mixed pair is feasible?

  • (1) O and P
  • (2) R and S
  • (3) P and S
  • (4) O and S
Show solution
(4) O and S. Equal mix O+S: carb 47.5%, protein 40%, fat 5%, minerals 7.5%, meets all four conditions.
ModerateCAT 2007

For a diet with 10% minerals and ≥30% protein, in how many ways (mixing ≥2 ingredients) can it be prepared?

  • (1) One
  • (2) Two
  • (3) Three
  • (4) Four
Show solution
(1) One. Only equal parts of O and Q give exactly 10% minerals with protein 30%, a single feasible way.
HardCAT 2007

Which two-ingredient mix has the lowest cost per unit for a diet with 10% fat and ≥30% protein?

  • (1) P and Q
  • (2) P and S
  • (3) P and R
  • (4) Q and S
Show solution
(4) Q and S. Q&S in 1:4 gives fat 10%, protein 46%, at cost 6/5 of base, cheaper than R&S (1:3, cost 2). Lowest cost = Q and S.
HardCAT 2007

In what proportion should P, Q and S be mixed for ≥60% carbohydrate at the lowest unit cost?

  • (1) 2:1:3
  • (2) 4:1:2
  • (3) 2:1:4
  • (4) 4:1:1
Show solution
(4) 4:1:1. P:Q:S = 4:1:1 gives carb 62.5% (≥60) at cost ₹83.3/unit vs 4:1:2 at ₹85.7. Lowest cost = 4:1:1.

CAT 2008

SET, Department transfers (mathematical reasoning). 100 employees across five departments. Gross pay = basic + allowances. A person moving from a lower-avg-age dept to a higher one gets +10% of basic as extra allowance; moving the other way changes nothing. Questions are independent.

DeptEmpAvg AgeAvg Basic (₹)Allow (% of basic)
HR545500070
Marketing3035600080
Finance2030650060
Business Dev.3542750075
Maintenance1035550050
HardCAT 2008

A mutual transfer happens between Marketing and Finance, plus one Marketing→HR transfer. Finance's avg age rises by 1 year; Marketing's stays the same. What is HR's new average age?

  • (1) 30
  • (2) 35
  • (3) 40
  • (4) 45
Show solution
(3) 40. Let the moves bring ages x, y, z. Finance: (600 + x − y)/20 = 31 ⇒ x − y = 20. Marketing unchanged: x − y + z = 35 ⇒ z = 15. HR: (45×5 + 15)/6 = 240/6 = 40.
HardCAT 2008

Approximate % change in average gross pay of HR due to transfer of a 40-yr person with basic ₹8000 from Marketing?

  • (1) 9%
  • (2) 11%
  • (3) 13%
  • (4) 15%
Show solution
(3) 13%. Marketing→HR is to higher avg age, so allowance = 80% of 8000 + 10% = 7040; gross = 15040. New HR avg gross = (8500×5 + 15040)/6 = 9590. Change = (9590−8500)/8500 ≈ 12.8% ≈ 13%.
ModerateCAT 2008

If two employees (basic ₹6000 each) move from Maintenance to HR and one (basic ₹8000) from Marketing to HR, what is the % change in HR's average basic pay?

  • (1) 10.5%
  • (2) 12.5%
  • (3) 15%
  • (4) 30%
Show solution
(2) 12.5%. New avg basic = (5000×5 + 6000×2 + 8000)/8 = 45000/8 = 5625. Change = (5625−5000)/5000 = 12.5%.

CAT 2017

SET, Platform grid "can reach". A 5×5 layout of square platforms of given heights. A can reach B if: (a) same row/column; (b) A is lower than B; (c) anyone between them is lower than A. Rows top→bottom, columns left→right. Heights:

61243
95328
78465
39512
17639
ModerateCAT 2017

How many individuals can be reached by just one individual?

  • (1) 3
  • (2) 5
  • (3) 7
  • (4) 8
Show solution
(3) 7. Building the reachability count for every cell, exactly 7 platforms are reachable by precisely one individual.
EasyCAT 2017

Which is true for any individual at a platform of height 1 m?

  • (1) Reached by all in their row & column
  • (2) Reached by ≥4 individuals
  • (3) Reached by ≥1 individual
  • (4) Cannot be reached by anyone
Show solution
(4) Cannot be reached by anyone. To reach B you must be lower than B; nobody is lower than height 1, so a height-1 platform can never be reached.
ModerateCAT 2017

We can find two individuals who cannot be reached by anyone in:

  • (1) the last row
  • (2) the fourth row
  • (3) the fourth column
  • (4) the middle column
Show solution
(3) the fourth column. Checking each line, the fourth column contains two platforms reachable by nobody.
HardCAT 2017

Which statement is true about this layout?

  • (1) Each row has someone reachable by ≥5
  • (2) Each row has someone reachable by no one
  • (3) Each row has ≥2 individuals reachable by an equal number
  • (4) All height-9 individuals are reachable by ≥5
Show solution
(3) Each row has at least two individuals reachable by an equal number of individuals. Testing all options against the reachability grid, only (3) holds throughout.

SET, Congestion routing (games/networks). Four cars go A→B via M or N. A→M and N→B: 1 car takes 6 min, each extra car +3 min/car. A→N: 6 min... actually 1 car 20 min, +1 min/car. M→B: 1 car 20 min, +0.9 min/car. Police assign routes so no car can cut its time by deviating alone.

HardCAT 2017 · TITA

How many cars would be asked to take the A-N-B route? TITA

Show solution
2. With both routes' times nearly equal, splitting 2 cars on A-M-B and 2 on A-N-B is the equilibrium. So 2 cars take A-N-B.
ModerateCAT 2017

If all cars follow the order, what is the difference in travel time between an A-N-B car and an A-M-B car?

  • (1) 1
  • (2) 0.1
  • (3) 0.2
  • (4) 0.9
Show solution
(2) 0.1. AMB (2 cars) = 6+3 + 20+0.9 = 29.9; ANB (2 cars) = 20+1 + 6+3 = 30. Difference = 0.1.
HardCAT 2017 · TITA

A new one-way road M→N is built (1 car 7 min, +1 min/car), giving routes A-M-B, A-N-B and A-M-N-B. How many cars should take A-M-N-B (police never order all on one route)? TITA

Show solution
2. At equilibrium each route time equals 32 min: AMB = (6+6)+20 = 32, AMNB = (6+6)+(7+1)+(6+6) = 32, ANB = 20+(6+6) = 32. The stable allocation sends 2 cars on A-M-N-B.
HardCAT 2017

With the same M→N road, what is the minimum travel time from A to B if all follow the police order (never all on one route)?

  • (1) 26
  • (2) 32
  • (3) 29.9
  • (4) 30
Show solution
(2) 32. All three routes equalise to 32 minutes at equilibrium, so the minimum (and common) A→B time is 32.

SET, Fingerprint pass-key (counting / binary logic). A pass-key is the order of scanning five fingers T, I, M, R, L. The lab relaxes the order requirement in various ways.

HardCAT 2017 · TITA

If at most two of the five scans may be out of place, how many sequences are allowed? TITA

Show solution
11. Choosing which 2 of 5 positions swap: ⁵C₂ = 10, plus the original sequence = 11.
HardCAT 2017

If the input may vary from the original by one place for any finger (e.g. TIMRL → ITRML allowed, LIMRT not), how many sequences are allowed?

  • (1) 7
  • (2) 5
  • (3) 8
  • (4) 13
Show solution
(3) 8. 1 (original) + 4 (any one adjacent pair swapped) + 2 (first pair swapped together with one of the last-three pairs) + 1 (positions 2-3 and 4-5 both swapped) = 8.
HardCAT 2017 · TITA

Now six scans, one finger scanned twice (e.g. TIMTRL). If at most two of the six scans may be out of place (the twice-scanned finger still twice), how many sequences are allowed? TITA

Show solution
15. Counting valid two-position swaps that preserve the duplicate gives 14 variations, plus the original = 15.
HardCAT 2017

Six scans, one finger twice; input may vary by one place for any finger. For original LRLTIM, how many sequences are allowed?

  • (1) 8
  • (2) 11
  • (3) 13
  • (4) 14
Show solution
(3) 13. 1 (original) + 5 (single adjacent swap of LR/RL/LT/TI/IM) + 7 valid double-swaps = 13.

CAT 2018

SET, ATM notes (distribution / counting). An ATM dispenses exactly ₹5000 per withdrawal using ₹100, ₹200, ₹500 notes. The customer's preferred denomination's note-count must exceed the total count of the other denominations.

HardCAT 2018 · TITA

In how many ways can the ATM serve a customer preferring ₹500 notes? TITA

Show solution
7. Valid cases: ₹4000 in 500s (3 ways), ₹4500 in 500s (3 ways), ₹5000 in 500s (1 way) = 7. (₹3500 fails the majority rule.)
HardCAT 2018 · TITA

If only 50 ₹500 notes are in stock, what is the maximum number of the 10 customers who could have preferred ₹500 notes? TITA

Show solution
6. Least ₹500 notes to serve a ₹500-preferring customer is 8, so 50 notes ⇒ ⌊50/8⌋ = 6 customers.
HardCAT 2018

With 50 ₹500 notes and ≤20 notes per withdrawal, what is the max number of customers the ATM can serve?

  • (1) 12
  • (2) 10
  • (3) 13
  • (4) 16
Show solution
(1) 12. To stay ≤20 notes, the minimal valid case uses 4 ₹500 notes + 15 ₹200 notes = 19 notes. So 50/4 ⇒ 12 customers.
HardCAT 2018

Number of ₹500 notes needed to serve 50 customers preferring ₹500 and 50 preferring ₹100, with the smallest total notes?

  • (1) 900
  • (2) 800
  • (3) 750
  • (4) 1400
Show solution
(1) 900. ₹500-pref minimum uses 10 ₹500 notes ⇒ 500 notes; ₹100-pref minimum (fewest total notes) uses 8 ₹500 notes each ⇒ 400 notes. Total = 900.

SET, Three committees. 24 people on research, teaching, administration committees; no shared members; all three sizes different; each has bureaucrats, educationalists, politicians (≥1 each). (1) Bureaucrats in research = teaching, and research = 75% of admin. (2) Educationalists: teaching < research, and research = average of the other two. (3) 60% of politicians in admin, 20% in teaching.

HardCAT 2018

Which statement MUST be FALSE?

  • (1) Teaching: educationalists = politicians
  • (2) Admin: bureaucrats = educationalists
  • (3) Research size < teaching size
  • (4) Research size < admin size
Show solution
(3) Research size < teaching size, must be FALSE. Solving: bureaucrats research:teaching:admin = 3:3:4, politicians = 1:1:3, educationalists research=3. Research turns out larger than teaching, so (3) is impossible.
HardCAT 2018 · TITA

What is the number of bureaucrats in the administration committee? TITA

Show solution
4. From 10x + 3y + 5z arithmetic, x = 1 ⇒ admin bureaucrats = 4x = 4.
HardCAT 2018 · TITA

What is the number of educationalists in the research committee? TITA

Show solution
3. With y = 3, research educationalists = y = 3.
HardCAT 2018

Which CANNOT be determined uniquely?

  • (1) Size of the teaching committee
  • (2) Size of the research committee
  • (3) Total bureaucrats
  • (4) Total educationalists
Show solution
(1) The size of the teaching committee. Educationalists in teaching can be 1 or 2 (d = 1 or 2), so the teaching committee's size is not unique.

SET, Cryptarithm addition. Two six-digit numbers add to a six-digit sum; digits 0-9 coded by distinct letters A-K. The columns (left to right) are: Row1: A B H A A G F · Row2: A H J F K F · Sum row: A F G C A F (each column shown as top+top=bottom). Find the digit each letter represents.

HardCAT 2018 · TITA

Which digit does A represent? TITA

Show solution
1. The leftmost carry into a new high digit forces A = 1 (units column F+F=F ⇒ F=0, and the carry gives A = 1).
HardCAT 2018 · TITA

Which digit does B represent? TITA

Show solution
9. With F = 0, A = 1, H = 5 (since H+H ends in 0), the column B + 1 + 1 = 11 ⇒ B = 9.
HardCAT 2018

Which among 3, 4, 6, 7 CANNOT be represented by D?

  • (1) 3
  • (2) 4
  • (3) 6
  • (4) 7
Show solution
(4) 7. Given the consistent digit assignment (F=0, A=1, H=5, B=9 and the G/K/J/C constraints), 7 is already used elsewhere, so D cannot be 7.
HardCAT 2018

Which among 4, 6, 7, 8 CANNOT be represented by G?

  • (1) 4
  • (2) 6
  • (3) 7
  • (4) 8
Show solution
(2) 6. From G + K = 11 with J = G − 1 and C = 2, the feasible (G,K) pairs are (8,3),(4,7),(7,4), G = 6 is impossible.

CAT 2020

SET, Vendor contracts (scheduling). Institutes A,B,C,D and vendors W,X,Y,Z, 2010-2019. Each institute had 2 contracts (with 2 vendors); each vendor had 2 contracts (with 2 institutes); no institute had two contracts with the same vendor. Facts: I. Z had ≥1 contract every year. II. X had contracts up to 2015, none after. III. Y had contracts in 2010 & 2019; W only in 2013. IV. Five contracts in 2012. V. Exactly four multi-year contracts: B 7-yr, D 4-yr, A & C 3-yr each; the other four single-year. VI. C had ≥1 in 2012 but none in 2011. VII. B and D each had exactly one contract in 2012; D had none in 2010.

HardCAT 2020

In which year were there two or more contracts?

  • (1) 2015
  • (2) 2018
  • (3) 2017
  • (4) 2016
Show solution
(1) 2015. Reconstructing the timeline (B-Z 2010-2016, X with D 2012-2015, A-X 2010-2012, etc.), 2015 has multiple active contracts.
HardCAT 2020

Which of the following is true?

  • (1) B-Y in 2019
  • (2) D-X in 2011
  • (3) B-X in 2017
  • (4) D-Y in 2019
Show solution
(4) D had a contract with Y in 2019. Since B holds Y in 2010 and Y reappears in 2019 with D, the 2019 Y-contract is D-Y.
HardCAT 2020

In how many years was there only one contract?

  • (1) 3
  • (2) 5
  • (3) 2
  • (4) 4
Show solution
(1) 3. From the completed grid, exactly three years have a single contract.
HardCAT 2020

What BEST can be concluded about the number of contracts in 2010?

  • (1) At least 3
  • (2) At least 4
  • (3) Exactly 4
  • (4) Exactly 3
Show solution
(4) Exactly 3. B-Z, A-X and Y (with B) give exactly three contracts in 2010 (D has none in 2010).
HardCAT 2020

Which institutes had multiple contracts during the same year?

  • (1) B and C only
  • (2) B only
  • (3) A only
  • (4) A and B only
Show solution
(4) A and B only. Only A and B run two overlapping contracts in some common year.
HardCAT 2020

Which institutes and vendors had more than one contract in any year?

  • (1) A, D, W, Z
  • (2) B, W, X, Z
  • (3) A, B, Z, X
  • (4) B, D, W, X
Show solution
(3) A, B, Z and X. These are the institutes/vendors that overlap two contracts within a single calendar year.

SET, Coloured beads grid (layout). 25 beads in a 5×5 grid, each Red/Blue/Green. Along any row or column: (1) adjacent beads differ in colour; (2) ≥1 Green between any two Blues; (3) ≥1 Blue and ≥1 Green between any two Reds.

ModerateCAT 2020 · TITA

Total number of possible configurations using beads of only two colours? TITA

Show solution
2. Red can't be used with only two colours (rule 3 needs both Blue and Green between Reds). So Blue/Green alternate: starting Green or starting Blue ⇒ 2 configurations.
HardCAT 2020 · TITA

Maximum number of Red beads in any configuration? TITA

Show solution
9. Placing Reds on a staggered pattern (R G B / B R G / ...) caps the count at 9 Reds while satisfying all rules.
HardCAT 2020 · TITA

Minimum number of Blue beads in any configuration? TITA

Show solution
6. Maximising Reds (9) and Greens forces a minimum of 6 Blue beads in the pattern.
HardCAT 2020 · TITA

Two Reds are at (row2,col3) and (row3,col2). How many more Reds can be placed to maximise the Red count? TITA

Show solution
6. Given those two fixed Reds, six further Red placements are possible (total 8) while keeping every row/column valid.

SET, Hi-Lo game (binary logic). Four players, six rounds; each bids Hi or Lo simultaneously. Scoring: 4 Hi ⇒ all −1; 3 Hi/1 Lo ⇒ Hi +1, Lo −3; 2 Hi/2 Lo ⇒ Hi +2, Lo −2; 1 Hi/3 Lo ⇒ Hi +3, Lo −1; 4 Lo ⇒ all +1. Facts: (1) after 3 rounds Arun 6, Dipak 2, Bankim & Charu −2; (2) after 6 rounds Arun 7, Bankim & Dipak −1, Charu −5; (3) Dipak's round-3 score < round-1 but > round-2; (4) in exactly two of the six rounds Arun was the only one to bid Hi.

HardCAT 2020

What were the bids by Arun, Bankim, Charu and Dipak in the first round?

  • (1) Hi, Lo, Lo, Hi
  • (2) Hi, Hi, Lo, Lo
  • (3) Lo, Lo, Lo, Hi
  • (4) Hi, Lo, Lo, Lo
Show solution
(1) Hi, Lo, Lo, Hi. Reconstructing rounds 1-3 as (2,2,−2,−2), (3,−1,−1,−1), (1,1,1,1), round 1 has Arun & Dipak Hi, Bankim & Charu Lo.
HardCAT 2020 · TITA

In how many rounds did Arun bid Hi? TITA

Show solution
4. Across the six reconstructed rounds, Arun bid Hi in 4.
HardCAT 2020 · TITA

In how many rounds did Bankim bid Lo? TITA

Show solution
4. Bankim bid Lo in 4 of the six rounds.
HardCAT 2020 · TITA

In how many rounds did all four players make identical bids? TITA

Show solution
2. Two rounds had all-Hi or all-Lo (the (1,1,1,1) and a (−1,−1,−1,−1) round).
HardCAT 2020 · TITA

In how many rounds did Dipak gain exactly 1 point? TITA

Show solution
1. Dipak gained exactly +1 (an all-Lo round) in just one round.
HardCAT 2020

In which round was Arun DEFINITELY the only player to bid Hi?

  • (1) Third
  • (2) Second
  • (3) First
  • (4) Fourth
Show solution
(2) Second. Round 2 scored (3,−1,−1,−1), Arun +3 alone Hi, making it the definite "Arun-only-Hi" round.

CAT 2022

SET, Get-together attendance (set theory). 15 girls and some boys. Events: 1-day, 2-day or 3-day (3-day attendees also do 2-day; 2-day also 1-day). 6 singers (4 boys), 10 dancers (4 girls); no dancer is a singer. Facts: (1) all girls & 80% of boys want 1-day; 60% of boys want 2-day. (2) some girls want 1-day but not 2-day; others want both. (3) 70% of boys wanting 2-day are neither singers nor dancers; 60% of girls wanting 2-day are neither. (4) no girl wants 3-day; all male singers & 2 dancers want 3-day. (5) singers wanting 2-day = dancers wanting 2-day + 1.

HardCAT 2022 · Slot 1 · TITA

How many boys are there in the class? TITA

Show solution
50. From 6 + b = (9/50)(10 + x) with b an integer ≤ 4, (10 + x) must be a multiple of 50 ⇒ x = 40, so boys = 10 + 40 = 50.
ModerateCAT 2022 · Slot 1

Which can be determined? I. Boys wanting 1-day who are neither dancers nor singers. II. Female dancers wanting 1-day.

  • (1) Neither
  • (2) Only II
  • (3) Only I
  • (4) Both
Show solution
(2) Only II. Female dancers wanting 1-day = 4 (all girls want 1-day). I cannot be pinned down.
ModerateCAT 2022 · Slot 1

What fraction of the class is interested in a 2-day event?

  • (1) 9/13
  • (2) 2/3
  • (3) 7/10
  • (4) 7/13
Show solution
(4) 7/13. Class = 65. 2-day interested = girls 5 + boys 30 = 35 ⇒ 35/65 = 7/13.
HardCAT 2022 · Slot 1

What BEST can be concluded about male dancers interested in a 1-day event?

  • (1) 6
  • (2) 4 or 6
  • (3) 5
  • (4) 5 or 6
Show solution
(4) 5 or 6. Male dancers wanting 1-day ≥ those wanting 2-day, giving 5 or 6.
ModerateCAT 2022 · Slot 1

How many female dancers are interested in a 2-day event?

  • (1) Cannot be determined
  • (2) 2
  • (3) 1
  • (4) 0
Show solution
(4) 0. Fact 5 (singers = dancers + 1 for 2-day) forces two female singers and zero female dancers in the 2-day group.

SET, Salesmen success rates. Three salesmen Tohri, Hokli, Lahur over two days. Success rate = items sold / households met. (1) Same total households over two days; each sold 100 items total. (2) Lahur met the same households & sold the same items on both days. (3) Hokli sold nothing on day 2 (first household complained). (4) Tohri met 30 more households on day 2 than day 1. (5) Tohri's success rate = 2× Lahur's on day 1, and = 75% of Lahur's on day 2.

HardCAT 2022 · Slot 2 · TITA

Total households met by Tohri, Hokli and Lahur on the first day? TITA

Show solution
84. Solving the success-rate equations gives x = 10 (Tohri day 1). Day-1 households = x + (2x+29) + (x+15) = 10 + 49 + 25 = 84.
HardCAT 2022 · Slot 2 · TITA

How many TRICCEK items were sold by Tohri on the first day? TITA

Show solution
40. n = 1500/(x+15) = 60 (day-2 sales), so day-1 sales = 100 − 60 = 40.
ModerateCAT 2022 · Slot 2

How many households did Lahur meet on the second day?

  • (1) more than 35
  • (2) between 30 and 35
  • (3) 20 or less
  • (4) between 21 and 29
Show solution
(4) between 21 and 29. Lahur's daily households = x + 15 = 25, which lies in 21-29.
ModerateCAT 2022 · Slot 2

How many households did Tohri meet on the first day?

  • (1) more than 40
  • (2) 10 or less
  • (3) between 21 and 40
  • (4) between 11 and 20
Show solution
(2) 10 or less. Tohri's day-1 households x = 10.
ModerateCAT 2022 · Slot 2

Which statement is FALSE?

  • (1) Tohri's day-1 rate > day-2 rate
  • (2) Tohri had the highest day-1 rate
  • (3) Tohri had the highest day-2 rate
  • (4) Lahur had the lowest day-1 rate
Show solution
(4) is FALSE, "Lahur had the lowest success rate on day 1". The data table shows Lahur did NOT have the lowest day-1 rate, so statement (4) is the false one.

SET, Supermarket products (set theory). 320 products, each cosmetic or nutrition, each foreign or domestic, each with at least one of FDA / EU approval. (1) Equal domestic and foreign. (2) Half the domestic products were FDA-approved cosmetics. (3) No foreign product had both approvals; 60 domestic had both. (4) 140 nutrition products; half foreign. (5) 200 FDA-approved products: 70 foreign, 120 cosmetic.

HardCAT 2022 · Slot 2 · TITA

How many foreign products were FDA-approved cosmetic products? TITA

Show solution
40. 120 FDA cosmetics − 80 domestic FDA cosmetics = 40 foreign FDA cosmetic products.
ModerateCAT 2022 · Slot 2

How many cosmetic products did not have FDA approval?

  • (1) 50
  • (2) Cannot be determined
  • (3) 60
  • (4) 10
Show solution
(3) 60. Cosmetics without FDA = only-EU foreign cosmetics (50) + only-EU domestic cosmetics (10) = 60.
HardCAT 2022 · Slot 2

Best represents the number of domestic cosmetic products with both approvals?

  • (1) At least 10, at most 80
  • (2) At least 10, at most 60
  • (3) At least 20, at most 70
  • (4) At least 20, at most 50
Show solution
(2) At least 10 and at most 60. Domestic with both approvals = 60 total, so domestic cosmetic-both ≤ 60; only-FDA cosmetic domestic ≤ 70 ⇒ both ≥ 10.
HardCAT 2022 · Slot 2

If 70 cosmetic products had no EU approval, how many nutrition products had both approvals?

  • (1) 30
  • (2) 10
  • (3) 50
  • (4) 20
Show solution
(2) 10. 70 = 40 (only-FDA foreign cosmetic) + only-FDA domestic cosmetic ⇒ 30 only-FDA domestic cosmetic; nutrition with both = 0 + 10 = 10.
HardCAT 2022 · Slot 2 · TITA

If 50 nutrition products had no EU approval, how many domestic cosmetic products had no EU approval? TITA

Show solution
50. 50 = 30 (foreign FDA nutrition) + only-FDA domestic nutrition ⇒ 20; domestic cosmetics without EU = 50.

CAT 2023

SET, Dean election (analytical reasoning). Faculty belong to F&A (9), M&S (7), O&Q (5), B&H (3). Four candidates, Pakrasi, Qureshi, Ramaswamy, Samuel, stood for Dean; only one was from O&Q. Everyone voted; in each department all non-candidates voted for the same candidate. Rules: ≤2 candidates per department; no self-vote; no voting for own department's candidate. Results: Pakrasi 3, Qureshi 14, Ramaswamy 6, Samuel 1. Pakrasi voted Ramaswamy, Qureshi voted Samuel, Ramaswamy voted Qureshi, Samuel voted Pakrasi.

HardCAT 2023 · Slot 1

Which two candidates can belong to the same department?

  • (1) Pakrasi & Samuel
  • (2) Pakrasi & Qureshi
  • (3) Qureshi & Ramaswamy
  • (4) Ramaswamy & Samuel
Show solution
(2) Pakrasi and Qureshi. The only consistent case puts both candidates from M&S as Pakrasi and Qureshi.
HardCAT 2023 · Slot 1

Which can be the number of votes Qureshi received from a single department?

  • (1) 7
  • (2) 8
  • (3) 6
  • (4) 9
Show solution
(4) 9. Qureshi's 13 non-candidate votes came as 9 from F&A + 4 from O&Q, so 9 is possible.
HardCAT 2023 · Slot 1

If Samuel belongs to B&H, which is/are true? A: Pakrasi belongs to M&S. B: Ramaswamy belongs to O&Q.

  • (1) Neither
  • (2) Both A and B
  • (3) Only B
  • (4) Only A
Show solution
(2) Both A and B. Samuel in B&H forces Pakrasi → M&S and Ramaswamy → O&Q; both statements hold.
HardCAT 2023 · Slot 1

What best can be concluded about the candidate from O&Q?

  • (1) Pakrasi or Qureshi
  • (2) Ramaswamy
  • (3) Ramaswamy or Samuel
  • (4) Samuel
Show solution
(3) Ramaswamy or Samuel. O&Q's lone candidate is undetermined between Ramaswamy and Samuel.
HardCAT 2023 · Slot 1

Which is/are true? A: Non-candidates from M&S voted Qureshi. B: Non-candidates from F&A voted Qureshi.

  • (1) Only B
  • (2) Only A
  • (3) Both
  • (4) Neither
Show solution
(1) Only B. M&S non-candidates voted for Ramaswamy (not Qureshi); F&A non-candidates voted Qureshi. So only B is true.

CAT 2024 & 2025, recent

Logical-reasoning sets from the actual CAT 2024 and CAT 2025 papers, distributed here from the year-paper pages. Each set's setup is summarised; full data tables are at the source.

CAT 2024 · Slot 1

SET, Election Campaigns. Amiya and Ramya each run a campaign at one of two intensity levels, staid = level 1, vigorous = level 2, and each either focuses on issues or attacks the opponent. Base rule (both focus on issues): voter turnout = 20 × (sum of the two levels) %, and vote share is split in proportion to the two campaign levels. Attack modifications: if Amiya attacks & Ramya stays on issues, 10% of Amiya's voters switch to Ramya and 10% don't vote; if Ramya attacks & Amiya stays on issues, 20% of Ramya's voters switch to Amiya and 5% don't vote; if both attack, 10% of each candidate's voters don't vote.

ModerateCAT 2024 · Slot 1

If both of them run staid campaigns attacking the other, then what percentage of students will vote in the election?

  • (A) 40%
  • (B) 64%
  • (C) 60%
  • (D) 36%
Show solution
(D) 36%. Both staid ⇒ levels sum to 2 ⇒ base turnout = 20 × 2 = 40%. Mutual attack drops 10% of each side's voters ⇒ 40% × 0.9 = 36%.
ModerateCAT 2024 · Slot 1

What is the minimum percentage of students who will vote in the election?

  • (A) 32%
  • (B) 40%
  • (C) 38%
  • (D) 36%
Show solution
(D) 36%. The lowest base (both staid) is 40%; the largest proportional loss applies under mutual attack, giving 36% as the floor.
HardCAT 2024 · Slot 1

If Amiya runs a campaign focusing on issues, then what is the maximum percentage of votes that she can get?

  • (A) 48%
  • (B) 44%
  • (C) 40%
  • (D) 36%
Show solution
(A) 48%. Amiya vigorous (level 2) on issues, Ramya vigorous but attacking ⇒ 80% turnout, with the 20% transfer from Ramya's voters boosting Amiya to 48%.
HardCAT 2024 · Slot 1

If Ramya runs a campaign attacking Amiya, then what is the minimum percentage of votes that she is guaranteed to get?

  • (A) 12%
  • (B) 15%
  • (C) 30%
  • (D) 18%
Show solution
(B) 15%. Ramya staid (level 1) attacking loses 20% + 5% = 25% from her baseline share of 20% ⇒ 15%.
HardCAT 2024 · Slot 1

What is the maximum possible voting margin with which one of the candidates can win?

  • (A) 20%
  • (B) 29%
  • (C) 28%
  • (D) 26%
Show solution
(B) 29%. Amiya vigorous on issues (≈44%) against Ramya staid attacking (15%) gives a margin of 44 − 15 = 29%.

SET, Countries Visited. The chart below provides complete information about the number of countries visited by Dheeraj, Samantha and Nitesh, in Asia, Europe and the rest of the world (ROW). The following additional facts are also known: (1) 32 countries were visited by at least one of them. (2) USA (in ROW) is the only country that was visited by all three of them. (3) China (in Asia) is the only country that was visited by both Dheeraj and Nitesh, but not by Samantha. (4) France (in Europe) is the only country outside Asia, which was visited by both Dheeraj and Samantha, but not by Nitesh. (5) Half of the countries visited by both Samantha and Nitesh are in Europe.

(The source data is a bar chart of countries-visited-per-region per person; the chart image cannot be transcribed verbatim here. The official answers below are confirmed.)

ModerateCAT 2024 · Slot 1 TITA

How many distinct countries in Asia were visited by at least one of the three?

Show solution
3. Only Dheeraj and Nitesh visited Asia, overlapping solely on China (Samantha visited 0 Asian countries); the distinct Asian countries resolve to 3.
HardCAT 2024 · Slot 1 TITA

How many countries in Europe were visited only by Nitesh?

Show solution
2. After accounting for France (Dheeraj + Samantha) and the Samantha-Nitesh shared European countries, Nitesh's Europe-only count is 2.
ModerateCAT 2024 · Slot 1 TITA

How many ROW countries were visited by both Nitesh and Samantha?

Show solution
4. Three shared ROW countries plus the USA (visited by all three) = 4.
HardCAT 2024 · Slot 1

How many countries in Europe were visited by exactly one person?

  • (A) 10
  • (B) 5
  • (C) 14
  • (D) 12
Show solution
(D) 12. From the European visits, remove the shared (Samantha-Nitesh) overlap and the France pair (Dheeraj + Samantha); 12 European countries were visited by exactly one person.

CAT 2025 · Slot 1

SET, Employee ratings & promotions. At InnovateX, six employees (Asha, Bunty, Chintu, Dolly, Eklavya, Falguni) are split into two groups of three: Elite (managed by Kuku) and Novice (managed by Lalu). Each quarter, the members of a group receive distinct integer ratings 1-3; an employee's score at the end of a quarter is their cumulative rating from the start of the year. After each quarter, the highest-scoring Novice is promoted to Elite and the lowest-scoring Elite is demoted to Novice (ties broken by the higher rating in the latest quarter). Facts: Asha, Bunty and Chintu were in Elite at the beginning of Quarter 1, and all of them were in Novice at the beginning of Quarter 4. Dolly and Falguni were the only employees who got the same rating across all the quarters. In Quarter 3, under Lalu, Asha and Dolly received ratings of 1 and 2 respectively. Bunty received a rating of 1 in Quarter 2.

HardCAT 2025 · Slot 1 · Set 2

Q1: What was Eklavya's score at the end of Quarter 2?   Q2: How many employees changed groups more than once up to the beginning of Quarter 4?   Q3: What was Bunty's score at the end of Quarter 3?   Q4: For how many employees can the scores at the end of Quarter 3 be determined with certainty?   Q5: Which of the following is/are NECESSARILY true? I. Asha received a rating of 2 in Quarter 1. II. Asha received a rating of 1 in Quarter 2.

  • Q5: (A) Neither I nor II (B) Both I and II (C) Only I (D) Only II
Show solution
4 · 0 · 5 · 4 · (D) Only II (Q1-Q5).

SET, Round-table seating. Seven chairs (1-7 clockwise). Four friends, Aslam, Bashir, Chhavi, Davies. Initially Aslam-Chhavi are adjacent, and Bashir and Davies each have empty chairs on both sides. On each turn a person moves to the first empty chair (clockwise or anti-clockwise). Turn order: Aslam, Bashir, Chhavi, Davies, Aslam, Bashir, Chhavi. Friends sit in adjacent chairs only after Turn 2 and Turn 6. Davies: chair 2 after Turn 1, chair 4 after Turn 5. Chhavi in chair 7 after Turn 2.

HardCAT 2025 · Slot 1 · Set 3

Q1: What is the number of the chair initially occupied by Bashir?   Q2: Who sits on the chair numbered 4 at the end of Turn 3?   Q3: Which of the chairs are occupied at the end of Turn 6?   Q4: Which of the following BEST describes the friends sitting on chairs adjacent to Bashir at the end of Turn 7?

  • Q2: (A) Bashir (B) Chhavi (C) Davies (D) No one
  • Q3: (A) Chairs 4,5,6,7 (B) Chairs 1,2,3,4 (C) Chairs 2,3,4,5 (D) Chairs 1,2,6,7
  • Q4: (A) Chhavi only (B) Davies only (C) Chhavi and Davies (D) Aslam and Chhavi
Show solution
4 · (D) No one · (A) Chairs 4,5,6,7 · (B) Davies only (Q1-Q4).

SET, Tapping game. Alia, Badal, Clive, Dilshan and Ehsaan played a game in which each asks a unique question to all the others, and they respond by tapping their feet, either once or twice or thrice. One tap means "Yes", two taps mean "No", and three taps mean "Maybe". A total of 40 taps were heard across the five questions, and each question received at least one "Yes", one "No" and one "Maybe". Also: (1) Alia tapped a total of 6 times and received 9 taps to her question; she responded "Yes" to the questions asked by both Clive and Dilshan. (2) Dilshan and Ehsaan tapped a total of 11 and 9 times respectively; Dilshan responded "No" to Badal. (3) Badal, Dilshan and Ehsaan received an equal number of taps to their respective questions. (4) No one responded "Yes" more than twice. (5) No one's answer to Alia's question matched the answer that Alia gave to that person's question. (6) Clive tapped more times in total than Badal.

HardCAT 2025 · Slot 1 · Set 5

Q1: How many taps did Clive receive for his question?   Q2: Which two people tapped an equal number of times in total?   Q3: What was Clive's response to Ehsaan's question?   Q4: How many "Yes" responses were received across all the questions?

  • Q2: (A) Badal and Dilshan (B) Clive and Ehsaan (C) Dilshan and Clive (D) Alia and Badal
  • Q3: (A) No (B) Maybe (C) Cannot be determined (D) Yes
Show solution
7 · (D) Alia and Badal · (B) Maybe · 6 (Q1-Q4).

CAT 2025 · Slot 2

SET, Balls & hoops. Six balls (B1-B6) tested on four hoops (H1-H4). A ball "pings" if its diameter ≤ hoop diameter, else it gets stuck. Facts: (1) B1 and B6 pinged H4 but B5 didn't; (2) B4 pinged H3 but B1 didn't; (3) all balls except B3 pinged H1; (4) only B2 pinged H2.

HardCAT 2025 · Slot 2 · Set 2

Q6: Total pings by B1, B2 and B3?   Q7: Which statement about ball sizes is NOT necessarily true?   Q8: Which statement about hoop sizes is true?   Q9: What best describes total pings across all tests?

  • Q7: (A) B4<B5<B3 (B) B1<B6<B3 (C) B1<B5<B3 (D) B2<B1<B5
  • Q8: (A) H2<H4<H3<H1 (B) H1<H3<H4<H2 (C) H2<H3<H4<H1 (D) H1<H4<H3<H2
  • Q9: (A) 13 or 14 (B) 12 or 13 (C) at least 9 (D) 12, 13 or 14
Show solution
6 · (B) · (C) · (B) 12 or 13 (Q6-Q9).

SET, Musicians & gurus. Four musicians (Ananya, Bhaskar, Charu, Devendra) trained under three gurus (Pandit Meghnath, Ustad Samiran, Acharya Raghunath), 2013-2024. Each guru trains for a different consecutive span (2, 3 or 4 years) and never more than two students at once. "Gurubhai" pairs train under the same guru in the same year. Facts: Samiran trained at most one student per year; Raghunath inactive 2015-18 and 2021-24; Ananya-Devendra and Bhaskar-Charu were never gurubhai, all other pairs were gurubhai exactly 2 years; in 2013 Ananya started under Meghnath, Bhaskar under Samiran.

HardCAT 2025 · Slot 2 · Set 3

Q10: In which year were Ananya and Bhaskar gurubhai?   Q11: In which year did Charu begin under Pandit Meghnath?   Q12: In which year were Bhaskar and Devendra gurubhai?   Q13: Which statement is true?

  • Q10: (A) 2018 (B) 2014 (C) 2021 (D) 2020
  • Q11: (A) 2021 (B) 2015 (C) 2017 (D) 2016
  • Q12: (A) 2022 (B) 2015 (C) 2018 (D) 2020
  • Q13: (A) Ananya under Samiran in 2015 (B) Ananya under Samiran in 2018 (C) Charu under Samiran in 2019 (D) Charu under Samiran in 2018
Show solution
(D) 2020 · (B) 2015 · (A) 2022 · (C) Charu under Samiran in 2019 (Q10-Q13). Q14 (years with only two musicians training): 4.

SET, Research papers by authors. Four authors (Arman, Brajen, Chintan, Devon) write single-, two-, three- and four-author papers. Charts show paper counts by type and author-contribution totals. Facts: each author wrote ≥1 of each type; all four wrote different numbers of single-author papers; both Chintan and Devon wrote more three-author papers than Brajen; Brajen's single- and two-author counts were equal.

HardCAT 2025 · Slot 2 · Set 5

Q19: Total two- and three-author papers by Brajen?   Q20: Which statements are necessarily true (i: Chintan wrote exactly three two-author papers; ii: Chintan wrote more single-author papers than Devon)?   Q21: Which are necessarily true (i: Arman's three-author papers only with Chintan & Devon; ii: Brajen's three-author papers only with Chintan & Devon)?   Q22: If Devon wrote more than one two-author paper, how many did Chintan write?

  • Q20: (A) Neither (B) Only ii (C) Both (D) Only i
  • Q21: (A) Neither (B) Only ii (C) Only i (D) Both
Show solution
4 · (A) Neither · (D) Both · 3 (Q19-Q22).

CAT 2025 · Slot 3

SET, International travel. Aurevia, Brelosia, Cyrenia and Zerathania are four countries, with currencies Aurels, Brins, Crowns and Zentars respectively. One Crown is worth 0.5 Zentars. Three travellers, Jano, Kira and Lian, set out from Zerathania, each visiting exactly two of the other countries, and each country is visited by exactly two travellers. Each traveller's Flight Cost is the total airfare to both countries and back; Jano's Flight Cost was 4000 Zentars and the other two were 5000 and 6000 Zentars (in some order). When visiting a country a traveller spent 1000, 2000 or 3000 in the local currency, with a different spend in each of the two countries; across all visits there were exactly two spends of 1000 and exactly one spend of 3000. Observations: Aurevia citizen, "Jano and Kira visited our country, and their Travel Costs were 3500 and 8000, respectively." Brelosia citizen, "Kira and Lian visited our country, spending 2000 and 3000, respectively. Kira's Travel Cost was 4000." Cyrenia citizen, "Lian visited our country and her Travel Cost was 36000."

HardCAT 2025 · Slot 3 · Set 2

Q5: What is the sum of the Travel Costs of all three travellers (in Zentars)?   Q6: How much did Lian spend in total in the two countries she visited (in Zentars)?   Q7: What was Jano's total spend in Aurels?   Q8: One Brin equals how many Crowns?   Q9: Which of the following statements is NOT true about money spent in the local currency?

  • Q8: (A) 8 (B) 0.125 (C) 0.5 (D) 4
  • Q9: (A) Jano spent 2000 in Aurevia (B) Lian spent 2000 in Cyrenia (C) Jano spent 2000 in Cyrenia (D) Kira spent 1000 in Aurevia
Show solution
41000 · 13000 · 1500 · (A) 8 · (A) Jano spent 2000 in Aurevia (Q5-Q9).

SET, Passing the Buck. Seven children, Aarav, Bina, Chirag, Divya, Eshan, Farhan and Gaurav, sit in a circle facing inward (not necessarily in that order) and play 'Passing the Buck' over 10 rounds. In each round the child holding the Buck must pass it directly to a child sitting immediately to the left, immediately to the right, second to the left, or second to the right. The game starts with Bina passing the Buck and ends with Chirag receiving the Buck. A table gives some of the pass types and the child receiving the Buck, with some entries missing (marked '?').

HardCAT 2025 · Slot 3 · Set 4

Q14: Who is sitting immediately to the right of Bina?   Q15: Who is sitting third to the left of Eshan?   Q16: For which pass type can the total number of occurrences be uniquely determined?   Q17: For which of the following children is it possible to determine how many times they received the Buck?

  • Q14: (A) Aarav (B) Eshan (C) Farhan (D) Chirag
  • Q15: (A) Gaurav (B) Divya (C) Chirag (D) Aarav
  • Q16: (A) Immediately to the left (B) Second to the right (C) Immediately to the right (D) Second to the left
  • Q17: (A) Farhan (B) Eshan (C) Bina (D) Gaurav
Show solution
(B) Eshan · (C) Chirag · (C) Immediately to the right · (D) Gaurav (Q14-Q17).