Answers to questions

1. How many taps did Clive receive for his question? — 7

From total taps 40 and Alia’s received = 9 plus Badal/Dilshan/Ehsaan equal (call R), we have 9 + C + 3R = 40. The only integer solution in the feasible range is R = 8 and C = 7. So Clive’s column sum (taps received) is 7, matching the final table.

2. Which two people tapped an equal number of times in total? — Alia and Badal (Option 1)

Row totals come from known totals: A = 6, D = 11, E = 9 and B + C = 14 (since rows must sum to 40). With Clue “Clive tapped more than Badal” the unique split is B = 6, C = 8. Thus Alia and Badal each made 6 taps.

3. What was Clive’s response to Ehsaan’s question? — No (Option 2)

In the filled table C→E = 2. That assignment satisfies column/row totals, the reciprocity constraints for A and E (no matching answers), and the requirement that each column contain at least one 1, one 2, and one 3. So Clive answered “No” to Ehsaan.

4. How many “Yes” responses were received across all the questions? — 6

Count all 1-tap entries in the final table: A row has two 1s, B row two 1s, C row one 1, D row zero 1s, E row one 1 → total 2+2+1+0+1 = 6. This satisfies the constraints and the column diversity requirement.

Steps to make table: CAT 2025 DILR Slot 1

Step 1: Total taps = 40 → sum of column totals = 40 → with A = 9 and B,D,E equal (call each R) we get 9 + C + 3R = 40 → solve → R = 8 and C = 7. (Uses: total = 40; Alia received 9; B,D,E equal.)

Step 2: Therefore columns (taps received) are A=9, B=8, C=7, D=8, E=8. (Arithmetic result.)

Step 3: Row totals given: A tapped 6, D tapped 11, E tapped 9; remaining rows B and C must sum to 40 − (6+11+9) = 14 → B+C = 14. (Uses: row totals A,D,E and global total.)

Step 4: Clue “Clive tapped more than Badal” → C > B and B+C = 14 → unique integer solution is B=6, C=8. (Uses: C > B and B+C constraint.)

Step 5: Now row totals are A=6, B=6, C=8, D=11, E=9 — these match the table’s row sums. (Consequence.)

Step 6: Alia responded “Yes” to Clive and Dilshan → cells A→C = 1 and A→D = 1. (Uses: clue 1.)

Step 7: Fill remaining Alia row taps: A row must total 6 with no diagonal; remaining taps sum = 6 − (1+1) = 4 → split as A→B = 2 and A→E = 2. (Uses: row A total and step 6.)

Step 8: Dilshan responded “No” to Badal → cell D→B = 2. (Uses: clue 2.)

Step 9: Reciprocity mismatch for Alia and Ehsaan: for each X ≠ A, X→A ≠ A→X, and similarly for E → for every X ≠ E, X→E ≠ E→X. Use this to eliminate symmetric choices. (Uses: clue 5.)

Step 10: Limit “No one responded ‘Yes’ more than twice” bounds each row to at most two 1s; combine with known row totals to force specific entries (e.g., D must have many 3s to reach 11 while respecting at least one 1, one 2, one 3 in its column). (Uses: clue 4 and column diversity requirement.)

Step 11: Ehsaan’s row (total 9) must include exactly one 1 toward C (to meet column C = 7 and ensure A→C =1 already) → set E→C = 1; remaining E taps become E→A = 3, E→B = 3, E→D = 2 to reach 9 while satisfying reciprocity with A and column targets. (Combines: column sums, reciprocity with A, row total E.)

Step 12: Clive’s row (total 8) then fits as C→A = 2, C→B = 1, C→D = 3, C→E = 2 to meet row total 8, column targets (especially C column = 7), and the A/E reciprocity constraints. (Fills C row consistent with previous steps.)

Step 13: Badal’s row (total 6) becomes B→A = 1, B→C = 2, B→D = 2, B→E = 1 to satisfy B row sum 6, column sums, and the requirement that each question received at least one 1, one 2, and one 3. (Fills B row consistent with clues.)

Step 14: Dilshan’s row (total 11) completes as D→A = 3, D→C = 3, D→E = 3 (plus D→B = 2 set earlier) to reach 11 and provide the necessary 3s to meet column diversity and totals. (Fills D row consistent with step 8 and column sums.)

Step 15: Final verification → all off-diagonal cells are 1/2/3; each column has at least one 1, one 2, one 3; row/column totals equal the values found in Steps 2 and 5; reciprocity constraints for A and E are satisfied; Clive tapped more than Badal. (Global check — table validated.)

Working table — Tapping game CAT 2025 DILR Slot 1

Key: Rows = who tapped (respondents). Columns = whose question they answered (questions asked).

1 tap = Yes, 2 taps = No, 3 taps = Maybe.

Quick Answers

What was Eklavya’s score at the end of Quarter 2? — 4

Eklavya started Q1 in Novice and, because Dolly = 2 and Falguni = 3 are constant, the remaining Q1 Novice rating 1 goes to Eklavya. After end-Q1 swaps Eklavya remains Novice and in Q2 the Novice ratings become {1,2,3} with Bunty = 1 given, so Eklavya must take 3 in Q2. Therefore his cumulative after Q2 = 1 (Q1) + 3 (Q2) = 4.

2. How many employees changed groups more than once up to the beginning of Quarter 4? — 0

Only one swap (one promotion + one demotion) happens at the end of each quarter, and the forced sequence (end-Q1, end-Q2, end-Q3) produces at most one move per person. Tracking each employee shows everyone either stayed or moved exactly once, so no one moved more than once — zero employees.

3. What was Bunty’s score at the end of Quarter 3? — 5

Bunty was lowest in Elite in Q1 and thus had Q1 = 1; Q2 = 1 is given; in Q3 Novice (A, B, D) Dolly = 2 and Asha = 1 are fixed, so Bunty must be 3 in Q3. Summing gives 1 + 1 + 3 = 5.

4. For how many employees can the scores at the end of Quarter 3 be determined with certainty? — 4

Dolly and Falguni are constant raters so their Q3 totals are fixed (6 and 9). Bunty and Eklavya’s quarter-by-quarter ratings are uniquely forced, so their end-Q3 totals are fixed (Bunty 5, Eklavya 6). Asha and Chintu’s Q1 ratings can be swapped (2↔3) without violating clues, making their end-Q3 totals ambiguous; hence exactly 4 employees have determinable totals.

5. Which statement(s) is/are NECESSARILY true? — Only II (Asha got 1 in Q2)

To ensure Asha is demoted by end-Q2 (so A, B, C are all Novice by start-Q4), Asha must be the lowest cumulative after Q2; given Falguni’s constant 3 and distinct ratings rule, the only assignment that enforces Asha’s demotion across all valid scenarios is Asha = 1 in Q2. Her Q1 rating, however, can be either 2 or 3, so statement I is not necessary. Therefore Only II is necessarily true.

TableSteps — InnovateX Puzzle

Step 1: Initial groups given → Elite = A, B, C and Novice = D, E, F at start of Q1. (Clue 1)
Step 2: Dolly and Falguni constant ratings → D = 2 always, F = 3 always. (Clue 2)
Step 3: Bunty gets 1 in Q2 → Bunty must be in Novice in Q2 → Bunty was demoted at end Q1. (Clue 3)
Step 4: In Q1 Elite, the lowest rating is 1 → assign Bunty = 1 in Q1 Elite. (Uses promotion rule)
Step 5: Dolly is 2 and Falguni is 3 in Q1 Novice → remaining Novice rating 1 goes to Eklavya. (Clue 2 + Q1 Novice set)
Step 6: At end of Q1, highest in Novice is Falguni (3) → promoted; lowest in Elite is Bunty (1) → demoted. (Rule)
Step 7: Start Q2 Elite becomes A, C, F and Novice becomes B, D, E. (From Step 6)
Step 8: Asha receives 1 in Q3 and Dolly 2 in Q3 → Q3 Novice must contain both Asha and Dolly. (Clue 3)
Step 9: All of A, B, C must be in Novice at start of Q4 → Asha must be demoted by end Q2. (Clue 1 + swap constraints)
Step 10: To demote Asha at end Q2, Asha must have the lowest cumulative score in Q2 → Asha must receive rating 1 in Q2. (Rule + necessity)
Step 11: In Q2 Elite, ratings are {1,2,3} with F = 3 constant → remaining {1,2} assigned as Asha = 1, Chintu = 2. (Step 10 + distinct ratings)
Step 12: After scoring Q1 + Q2, Asha is uniquely the lowest → Asha demoted; Eklavya highest in Novice → promoted. (Rule)

Step 13: Start Q3 Elite becomes C, F, E and Novice becomes A, B, D. (From Step 12)
Step 14: Q3 Novice ratings forced: Asha = 1, Dolly = 2 → Bunty must take 3. (Clue + distinct ratings)
Step 15: Q3 Elite ratings forced: F = 3, Eklavya = 2, Chintu = 1. (Constants + distinct ratings)
Step 16: End Q3 swap: highest in Novice = Bunty (3) but cannot be promoted (final Elite must be D,E,F per Clue 1) → Dolly (2) is promoted instead; lowest in Elite = Chintu (1) → demoted. (Constraint + rule)
Step 17: Start Q4 Elite becomes D, E, F as required; Novice becomes A, B, C. (Matches Clue 1)

Working Table:  

Legend:

• Groups: Elite (E), Novice (N) — shown as starting group each quarter.
• Ratings in each quarter are integers 1 (lowest), 2, 3 (highest) within each group.
• “(amb)” marks cells that have two possible values (ambiguous between two scenarios).

Converting Graphs to table for ease of reading

Bar Graph: Import Tarrifs in billion USD charged by each country on other Countries

Table A: Import Tariffs Collected (in Billion USD) (Row = Exporter, Column = Importer)

Radar chart: Import Tariff percentage changed by each country on other countries

Table B: Tariff Percentage Rates (%) (Row = Exporter, Column = Importer)

Tariff amount=Tariff %×Import value (in billion USD)

Import value (Imports by X from Y) = Tariff amount (X on Y)/ Tariff % (X on Y)

1. Correct Option: 2 Rationale: To determine the highest value, we calculate the trade value for each option using the formula: Trade Value = Tariff Collected / Tariff Percentage.

For option 1, Exports by France to Japan: 3.0 billion / 0.30 = 10 billion.

For option 2, Imports by US from France (which is Exports by France to US): 6.0 billion / 0.20 = 30 billion.

For option 3, Exports by Japan to UK: 6.0 billion / 0.40 = 15 billion.

For option 4, Imports by France from India (which is Exports by India to France): 6.5 billion / 0.40 = 16.25 billion. Difficulty: Moderate

2. Correct Option: 4 Rationale: Trade Surplus or Deficit is calculated as Exports minus Imports.

First, calculate India’s exports to UK: Tariff collected is 3.0 billion and the rate is 30%. Value = 3.0 / 0.30 = 10 billion.

Second, calculate India’s imports from UK: Tariff collected is 5.0 billion and the rate is 20%. Value = 5.0 / 0.20 = 25 billion.

Trade Balance = 10 billion – 25 billion = -15 billion.

A negative balance indicates a trade deficit of 15.0 billion USD.

Difficulty: Moderate

3. Correct Option: 4 Rationale: Identify the row for the exporter (Japan) and the column for the importer (India).

From Table A (Tariffs Collected), the value for Row Japan, Column India is 3.5 billion.

From Table B (Tariff Rates), the value for Row Japan, Column India is 50% (or 0.50).

Calculate the export value: 3.5 / 0.50 = 7.0 billion USD.

Difficulty: Easy

4. Correct Option: 1 Rationale: We must compare exports to the US against imports from the US for both countries.

For France: Exports to US = 6.0 / 0.20 = 30 billion. Imports from US (US exports to France) = Tariff 5.5 / Rate 30% = 5.5 / 0.30 = 18.33 billion.

Since 30 > 18.33, France has a trade surplus.

For UK: Exports to US = 3.0 / 0.30 = 10 billion.

Imports from US (US exports to UK) = Tariff 2.5 / Rate 20% = 2.5 / 0.20 = 12.5 billion.

Since 10 < 12.5, UK has a trade deficit. Therefore, only France has a trade surplus with the US.

Difficulty: Moderate

1. Answer option 1: 70%

Reasoning: Seats on D–E come only from tickets A→E, B→E and C→E (D→E = 0). Let A→E = x and C→E = y; B→E is given as 30. Clue 3 says exactly four-sevenths of D–E seats were from stations before C, so x + 30 = (4/7)(x + 30 + y). Rearranging gives y = (3x + 90)/4. Counts must be multiples of 10 and x > 30 (clue 4). Trying multiples of 10 yields x = 50 and y = 60. Thus seats on D–E = 50 + 30 + 60 = 140. Capacity is 200, so occupancy = 140/200 = 0.70 = 70%. D–E = A→E (50) + B→E (30) + C→E (60) = 140 seats → 70% occupancy. 

2. Answer: 50 tickets.

Reasoning: This is x from the equation above. From the 4/7 condition and the multiple-of-10 constraint we found x = 50, and clue 4 requires A→C = A→E = x, so A→E = 50.

3. Answer: 80 tickets.

Reasoning: Tickets originating at C are C→D and C→E. From the solved counts C→D = 20 and C→E = 60, so total from C = 20 + 60 = 80.

4. Answer: 40 tickets.

Reasoning: Tickets to station C are A→C and B→C. A→C = 50 (equal to A→E) and B→C = 40, so total to C = 50 + 40 = 90. Tickets to station D are A→D and C→D (B→D = 0). Using the consistent solution A→D = 30 and C→D = 20 gives total to D = 30 + 20 = 50. Difference = 90 − 50 = 40.

5. Answer: 60 tickets.

Reasoning: Exactly-one-segment trips are the adjacent-station bookings: A→B, B→C, C→D, D→E. Given A→B = 0 and D→E = 0, and the solved values B→C = 40 and C→D = 20, the total = 40 + 20 = 60.

Step by Step Table Making

Step 1: Clue 2 gives B→C = 40 and B→E = 30.

Step 2: Clue 3 states that among seats reserved on D–E, exactly 4/7 were from stations before C. Seats on D–E come only from A→E, B→E, C→E (D→E is zero). Let A→E = x and C→E = y. Then seats on D–E = x + 30 + y and seats from before C = x + 30. Condition: x + 30 = (4/7)(x + 30 + y).

Step 3: Clue 4 says A→C = A→E and that value is higher than B→E (30). So x is a multiple of 10 and x > 30. Also A→C = x.

Step 4: Clue 5 sets A→B = 0, B→D = 0, D→E = 0.

Step 5: Clue 6 restricts counts to multiples of 10.

Step 6: From the equation in Step 2 and the multiple-of-10 constraint, solving yields x = 50 and y = 60. (Detailed algebra: x+30 = (4/7)(x+30+y) → 7(x+30)=4(x+30+y) → 7x+210 = 4x+120+4y → 3x +90 = 4y → y = (3x+90)/4. Trying x values (40,50,60,…) and enforcing y integer multiple of 10 gives x=50 → y=(150+90)/4=240/4=60.)

Step 7: Clue 4 also requires A→C = A→E = 50 and A→E > B→E (50>30 satisfied). We still need A→D. Observing segment C–D must reach 190 seats (given 95% occupancy). Using current contributions from B→E (30 contributes to C–D), C→E (60), A→E (50) and unknown A→D and C→D, solving for A→D and C→D with multiples of 10 yields A→D = 30 and C→D = 20 (these satisfy totals and other constraints).

Step 8: Populate the tables with these values and compute segment sums.

Train bookings — Working table and calculations

Final route-wise ticket counts (all multiples of 10)

Total tickets booked (sum of routes) = 280.

Seat reservations contributed by each route to each segment

Segments are: A–B, B–C, C–D, D–E. A ticket from X to Y reserves one seat on every segment between X and Y.

1. Correct Answer: 4. Reason: The starting constraints (A and C adjacent; B and D each with empty neighbours) combined with the fixed post-move facts (D at chair 2 after Turn 1, D at chair 4 after Turn 5, and C at chair 7 after Turn 2) produce a unique initial arrangement that satisfies all conditions: A = 6, B = 4, C = 7, D = 2. Any other initial placement for B forces a clash with the “adjacent after Turn 2 / Turn 6” conditions or the specified Davies positions. Hence Bashir starts at chair 4.

2. Correct Answer: No one. Reason: Using the valid move sequence, positions after Turn 3 are A = 1, B = 3, C = 6, D = 2. Davies moves into chair 4 only on Turn 4. Therefore at the end of Turn 3 chair 4 is empty; none of the four friends occupies it.

3. Correct Answer: chairs 4, 5, 6, and 7. Reason: After Turn 6 the positions are A = 7, B = 5, C = 6, D = 4. That set {4,5,6,7} is the contiguous block of four chairs required by the condition “adjacent only at end of Turn 6.” This matches the required adjacency and the computed move sequence.

4. Correct Answer: Davies only. Reason: At end of Turn 7 Bashir is at chair 5 in both valid end-states; his neighbours are chairs 4 and 6. Chair 4 is occupied by Davies, chair 6 is empty in both possible final placements for Chhavi (C ends at 1 or 3). Thus only Davies sits adjacent to Bashir.

Table Steps

Step 1: A and C must start on adjacent chairs, while B and D must each have empty chairs on both sides → forces C at 7 and A at 6 with D at 2 and B at 4. (Uses start-position rules.)

Step 2: Turn 1 requires Aslam to move to the first empty seat and Davies must occupy chair 2 after Turn 1 → A moves from 6 to 1, confirming D stays at 2. (From clue 2.)

Step 3: Turn 2 must end with all four friends in adjacent chairs → B must move from 4 to 3, giving occupied block 1–2–3–7. (Uses adjacency condition at end of Turn 2.)

Step 4: Chhavi must be at chair 7 after Turn 2 and must move on Turn 3 to the first empty chair → C moves from 7 to 6, making chair 4 empty. (From clue 2 and move rule.)

Step 5: Davies must reach chair 4 after Turn 5 → D moves at Turn 4 from 2 to 4 and remains there through Turn 5. (Clue 2.)

Step 6: Turn 6 must again produce four adjacent chairs → B moves from 3 to 5, forming block 4–5–6–7. (From adjacency condition at end of Turn 6.)

Step 7: Turn 7 movement by C can go to chair 1 or 3, but Bashir at 5 keeps only Davies adjacent at chair 4 → final adjacency answer fixed. (Uses Turn 7 placement requirement.)

These steps derive: B starts at 4; chair 4 is empty at end of Turn 3; chairs 4–5–6–7 are occupied at end Turn 6; only Davies is adjacent to Bashir at Turn 7.

Working Table

Chairs are numbered 1 to 7 clockwise.

Aslam = 6, Bashir = 4, Chhavi = 7, Davies = 2

Mover order: Turn1 A, Turn2 B, Turn3 C, Turn4 D, Turn5 A, Turn6 B, Turn7 C.

Each mover moves to FIRST empty seat in chosen direction.

Two valid direction sequences exist, but both produce identical positions at all key turns.

Below is the consolidated movement table (positions after each turn).