CAT 2023 DILR Questions — Slot 3

Build the whole month-by-month table from the summary figures, then read off April.

Summary table given (across the five months): Online min/max/median = 40/100/80, Offline = 30/80/50, Total = 110/130/120. Every monthly value is a multiple of 10.

1. January: Offline = 2 × Online. Online is at least 40 (its minimum) and Offline is at most 80 (its maximum), which forces Online = 40, Offline = 80, Total = 120.
2. April: Online = 2 × Offline, so Total = 3 × Offline. Keeping Total ≥ 110 gives Offline = 40, Online = 80, Total = 120.
3. May: Online is largest here, so Online = 100 (its maximum). Total ≤ 130 forces Offline = 30 (also the offline minimum), Total = 130.
4. The remaining months (February, March) are fixed using the medians and the fact that March's online equals February's offline: February = 80/50/130 and March = 50/60/110.

Solved registration table (all values are multiples of 10; Total = Online + Offline):

Checks — Online {40,80,50,80,100}: min 40, max 100, median 80 ✓ · Offline {80,50,60,40,30}: min 30, max 80, median 50 ✓ · Total {110,120,130}: min 110, max 130, median 120 ✓.

This question — April's total. From the table, April = 120.

Answer: 120

The January entry is pinned down first from the "offline = twice online" clue.

Summary table given (across the five months): Online min/max/median = 40/100/80, Offline = 30/80/50, Total = 110/130/120. Every monthly value is a multiple of 10.

1. January: Offline = 2 × Online. Online is at least 40 (its minimum) and Offline is at most 80 (its maximum), which forces Online = 40, Offline = 80, Total = 120.
2. April: Online = 2 × Offline, so Total = 3 × Offline. Keeping Total ≥ 110 gives Offline = 40, Online = 80, Total = 120.
3. May: Online is largest here, so Online = 100 (its maximum). Total ≤ 130 forces Offline = 30 (also the offline minimum), Total = 130.
4. The remaining months (February, March) are fixed using the medians and the fact that March's online equals February's offline: February = 80/50/130 and March = 50/60/110.

Solved registration table (all values are multiples of 10; Total = Online + Offline):

Checks — Online {40,80,50,80,100}: min 40, max 100, median 80 ✓ · Offline {80,50,60,40,30}: min 30, max 80, median 50 ✓ · Total {110,120,130}: min 110, max 130, median 120 ✓.

This question — January's online registrations. Online is at least 40 and Offline (= twice Online) is at most 80, so Online = 40.

Answer: 40

Both statements are checked against the completed table.

Summary table given (across the five months): Online min/max/median = 40/100/80, Offline = 30/80/50, Total = 110/130/120. Every monthly value is a multiple of 10.

1. January: Offline = 2 × Online. Online is at least 40 (its minimum) and Offline is at most 80 (its maximum), which forces Online = 40, Offline = 80, Total = 120.
2. April: Online = 2 × Offline, so Total = 3 × Offline. Keeping Total ≥ 110 gives Offline = 40, Online = 80, Total = 120.
3. May: Online is largest here, so Online = 100 (its maximum). Total ≤ 130 forces Offline = 30 (also the offline minimum), Total = 130.
4. The remaining months (February, March) are fixed using the medians and the fact that March's online equals February's offline: February = 80/50/130 and March = 50/60/110.

Solved registration table (all values are multiples of 10; Total = Online + Offline):

Checks — Online {40,80,50,80,100}: min 40, max 100, median 80 ✓ · Offline {80,50,60,40,30}: min 30, max 80, median 50 ✓ · Total {110,120,130}: min 110, max 130, median 120 ✓.

This question — which statements can be true.

• I. Offline was smallest in May. Offline minimum is 30, which is May's. True.
• II. Total was smallest in February. Total minimum is 110, which is March's, not February (130). False.

Answer: D — Only I

February's offline figure comes out to a single fixed value.

Summary table given (across the five months): Online min/max/median = 40/100/80, Offline = 30/80/50, Total = 110/130/120. Every monthly value is a multiple of 10.

1. January: Offline = 2 × Online. Online is at least 40 (its minimum) and Offline is at most 80 (its maximum), which forces Online = 40, Offline = 80, Total = 120.
2. April: Online = 2 × Offline, so Total = 3 × Offline. Keeping Total ≥ 110 gives Offline = 40, Online = 80, Total = 120.
3. May: Online is largest here, so Online = 100 (its maximum). Total ≤ 130 forces Offline = 30 (also the offline minimum), Total = 130.
4. The remaining months (February, March) are fixed using the medians and the fact that March's online equals February's offline: February = 80/50/130 and March = 50/60/110.

Solved registration table (all values are multiples of 10; Total = Online + Offline):

Checks — Online {40,80,50,80,100}: min 40, max 100, median 80 ✓ · Offline {80,50,60,40,30}: min 30, max 80, median 50 ✓ · Total {110,120,130}: min 110, max 130, median 120 ✓.

This question — offline registrations in February. With January (80) and April (40) already placed and the offline median equal to 50, the only consistent value for February is 50.

Answer: D — 50

Compare the monthly totals for the two proposed pairs.

Summary table given (across the five months): Online min/max/median = 40/100/80, Offline = 30/80/50, Total = 110/130/120. Every monthly value is a multiple of 10.

1. January: Offline = 2 × Online. Online is at least 40 (its minimum) and Offline is at most 80 (its maximum), which forces Online = 40, Offline = 80, Total = 120.
2. April: Online = 2 × Offline, so Total = 3 × Offline. Keeping Total ≥ 110 gives Offline = 40, Online = 80, Total = 120.
3. May: Online is largest here, so Online = 100 (its maximum). Total ≤ 130 forces Offline = 30 (also the offline minimum), Total = 130.
4. The remaining months (February, March) are fixed using the medians and the fact that March's online equals February's offline: February = 80/50/130 and March = 50/60/110.

Solved registration table (all values are multiples of 10; Total = Online + Offline):

Checks — Online {40,80,50,80,100}: min 40, max 100, median 80 ✓ · Offline {80,50,60,40,30}: min 30, max 80, median 50 ✓ · Total {110,120,130}: min 110, max 130, median 120 ✓.

This question — pairs with equal totals.

• I. January and April: 120 and 120 — equal.
• II. February and May: 130 and 130 — equal.

Both pairs match.

Answer: A — Both I and II

Reconstruct every team's route, then tally visits to each station.

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F. Each street takes 30 minutes, and the 09:00–12:00 window allows exactly six crossings, so every team walks a 6-street loop starting and ending at A. Rule 1 means no two teams share the same street in the same 30-minute interval.

• Team 4 never enters B, D or F, so it uses only A, C, E via streets A–C and A–E. It is at E at 11:30 (back at A at 12:00) and cannot be at E at 10:30 (reserved for T1 and T3), so its loop is A–E–A–C–A–E–A.
• Team 1 is at E at 10:30 and at the dead-end B at 11:30, so it is at A at 11:00. Working the 09:00 interval with the street-sharing rule forces it to start A–B, giving A–B–A–E–A–B–A.
• Team 3 is alone at D at 10:00 and at E at 10:30, crossing D–E then; since it cannot use A–E, it returns via D and C, giving A–C–D–E–D–C–A.
• Team 2 is alone at E at 10:00 but never uses A–E, so it reaches E through F, giving A–F–E–F–(A or E)–F–A.

Solved position timeline (each team makes a 6-street closed loop from A back to A; a team is at a station on every half-hour mark):

Routes: T1 A→B→A→E→A→B→A · T2 A→F→E→F→(A or E)→F→A · T3 A→C→D→E→D→C→A · T4 A→E→A→C→A→E→A.

This question — the most-visited station. Counting appearances across the seven marks: C = 3, D = 2, E = 5, F = 3. Station E is visited most often.

Answer: B — Station E

Only one team can ever reach the dead-end B.

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F. Each street takes 30 minutes, and the 09:00–12:00 window allows exactly six crossings, so every team walks a 6-street loop starting and ending at A. Rule 1 means no two teams share the same street in the same 30-minute interval.

• Team 4 never enters B, D or F, so it uses only A, C, E via streets A–C and A–E. It is at E at 11:30 (back at A at 12:00) and cannot be at E at 10:30 (reserved for T1 and T3), so its loop is A–E–A–C–A–E–A.
• Team 1 is at E at 10:30 and at the dead-end B at 11:30, so it is at A at 11:00. Working the 09:00 interval with the street-sharing rule forces it to start A–B, giving A–B–A–E–A–B–A.
• Team 3 is alone at D at 10:00 and at E at 10:30, crossing D–E then; since it cannot use A–E, it returns via D and C, giving A–C–D–E–D–C–A.
• Team 2 is alone at E at 10:00 but never uses A–E, so it reaches E through F, giving A–F–E–F–(A or E)–F–A.

Solved position timeline (each team makes a 6-street closed loop from A back to A; a team is at a station on every half-hour mark):

Routes: T1 A→B→A→E→A→B→A · T2 A→F→E→F→(A or E)→F→A · T3 A→C→D→E→D→C→A · T4 A→E→A→C→A→E→A.

This question — times any team passes through B. B connects only to A. Team 4 is barred from B, and Teams 2 and 3's loops never touch it. Only Team 1 visits B — at 09:30 and 11:30, so 2 times.

Answer: 2

10:15 falls inside the 10:00–10:30 interval, so we ask who is crossing D–E then.

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F. Each street takes 30 minutes, and the 09:00–12:00 window allows exactly six crossings, so every team walks a 6-street loop starting and ending at A. Rule 1 means no two teams share the same street in the same 30-minute interval.

• Team 4 never enters B, D or F, so it uses only A, C, E via streets A–C and A–E. It is at E at 11:30 (back at A at 12:00) and cannot be at E at 10:30 (reserved for T1 and T3), so its loop is A–E–A–C–A–E–A.
• Team 1 is at E at 10:30 and at the dead-end B at 11:30, so it is at A at 11:00. Working the 09:00 interval with the street-sharing rule forces it to start A–B, giving A–B–A–E–A–B–A.
• Team 3 is alone at D at 10:00 and at E at 10:30, crossing D–E then; since it cannot use A–E, it returns via D and C, giving A–C–D–E–D–C–A.
• Team 2 is alone at E at 10:00 but never uses A–E, so it reaches E through F, giving A–F–E–F–(A or E)–F–A.

Solved position timeline (each team makes a 6-street closed loop from A back to A; a team is at a station on every half-hour mark):

Routes: T1 A→B→A→E→A→B→A · T2 A→F→E→F→(A or E)→F→A · T3 A→C→D→E→D→C→A · T4 A→E→A→C→A→E→A.

This question — who patrols street D–E at 10:15. In the 10:00–10:30 interval, Team 3 crosses from D to E; by the street-sharing rule no other team is on D–E then.

Answer: D — Team 3

Read Team 4's route and count its stops at E.

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F. Each street takes 30 minutes, and the 09:00–12:00 window allows exactly six crossings, so every team walks a 6-street loop starting and ending at A. Rule 1 means no two teams share the same street in the same 30-minute interval.

• Team 4 never enters B, D or F, so it uses only A, C, E via streets A–C and A–E. It is at E at 11:30 (back at A at 12:00) and cannot be at E at 10:30 (reserved for T1 and T3), so its loop is A–E–A–C–A–E–A.
• Team 1 is at E at 10:30 and at the dead-end B at 11:30, so it is at A at 11:00. Working the 09:00 interval with the street-sharing rule forces it to start A–B, giving A–B–A–E–A–B–A.
• Team 3 is alone at D at 10:00 and at E at 10:30, crossing D–E then; since it cannot use A–E, it returns via D and C, giving A–C–D–E–D–C–A.
• Team 2 is alone at E at 10:00 but never uses A–E, so it reaches E through F, giving A–F–E–F–(A or E)–F–A.

Solved position timeline (each team makes a 6-street closed loop from A back to A; a team is at a station on every half-hour mark):

Routes: T1 A→B→A→E→A→B→A · T2 A→F→E→F→(A or E)→F→A · T3 A→C→D→E→D→C→A · T4 A→E→A→C→A→E→A.

This question — times Team 4 is at E. Team 4's loop is A–E–A–C–A–E–A, placing it at E at 09:30 and 11:30 — 2 times.

Answer: 2

We check which teams' routes include station C.

Streets: A–B, A–C, A–E, A–F, C–D, D–E, E–F. Each street takes 30 minutes, and the 09:00–12:00 window allows exactly six crossings, so every team walks a 6-street loop starting and ending at A. Rule 1 means no two teams share the same street in the same 30-minute interval.

• Team 4 never enters B, D or F, so it uses only A, C, E via streets A–C and A–E. It is at E at 11:30 (back at A at 12:00) and cannot be at E at 10:30 (reserved for T1 and T3), so its loop is A–E–A–C–A–E–A.
• Team 1 is at E at 10:30 and at the dead-end B at 11:30, so it is at A at 11:00. Working the 09:00 interval with the street-sharing rule forces it to start A–B, giving A–B–A–E–A–B–A.
• Team 3 is alone at D at 10:00 and at E at 10:30, crossing D–E then; since it cannot use A–E, it returns via D and C, giving A–C–D–E–D–C–A.
• Team 2 is alone at E at 10:00 but never uses A–E, so it reaches E through F, giving A–F–E–F–(A or E)–F–A.

Solved position timeline (each team makes a 6-street closed loop from A back to A; a team is at a station on every half-hour mark):

Routes: T1 A→B→A→E→A→B→A · T2 A→F→E→F→(A or E)→F→A · T3 A→C→D→E→D→C→A · T4 A→E→A→C→A→E→A.

This question — how many teams pass through C. Team 3 reaches C (09:30, 11:30) and Team 4 reaches C (10:30). Teams 1 and 2 never do. So 2 teams pass through C.

Answer: D — 2

Reconstruct every project and test score, then read Rini's project figure.

Three project teams (each one female + one male share a project score); the test is individual; aggregate = (1−t)·Project + t·Test.

1. Project: min 40, max 80, average 60 over six scores summing to 360; as three team-scores counted twice, the team scores are {40, 60, 80}.
2. Test: min 40, max 80, average 60, four distinct values plus two at the average — the multiset is {40, 50, 60, 60, 70, 80}. Shyamal (2nd highest) = 70; Biman (2nd lowest) = 50.
3. Amala–Koli: Amala's project = 2 × Koli's forces Koli = 40, Amala = 80. Koli's test = Amala's + 20; combined with the remaining clues this gives Amala-test = 60, Koli-test = 80, then Rini-test = 60 and Mathew-test = 40 (Mathew < Rini).
4. Teams: Amala (project 80) pairs with the 80-male Mathew; the others are (Koli, Biman) = 40 and (Rini, Shyamal) = 60.
5. Weights: Shyamal's aggregate = Koli's + 2 = Amala's − 2 solves to test weight t = 0.6 (project weight 0.4).

Solved score table (aggregate = 0.4 × Project + 0.6 × Test):

Project teams: (Amala, Mathew) = 80 · (Koli, Biman) = 40 · (Rini, Shyamal) = 60. Weights: Project 0.4, Test 0.6.

This question — Rini's project score. Rini is on the team scoring 60, so her project score is 60.

Answer: 60

The weight follows from lining up three aggregates that differ by 2.

Three project teams (each one female + one male share a project score); the test is individual; aggregate = (1−t)·Project + t·Test.

1. Project: min 40, max 80, average 60 over six scores summing to 360; as three team-scores counted twice, the team scores are {40, 60, 80}.
2. Test: min 40, max 80, average 60, four distinct values plus two at the average — the multiset is {40, 50, 60, 60, 70, 80}. Shyamal (2nd highest) = 70; Biman (2nd lowest) = 50.
3. Amala–Koli: Amala's project = 2 × Koli's forces Koli = 40, Amala = 80. Koli's test = Amala's + 20; combined with the remaining clues this gives Amala-test = 60, Koli-test = 80, then Rini-test = 60 and Mathew-test = 40 (Mathew < Rini).
4. Teams: Amala (project 80) pairs with the 80-male Mathew; the others are (Koli, Biman) = 40 and (Rini, Shyamal) = 60.
5. Weights: Shyamal's aggregate = Koli's + 2 = Amala's − 2 solves to test weight t = 0.6 (project weight 0.4).

Solved score table (aggregate = 0.4 × Project + 0.6 × Test):

Project teams: (Amala, Mathew) = 80 · (Koli, Biman) = 40 · (Rini, Shyamal) = 60. Weights: Project 0.4, Test 0.6.

This question — the test weight. Setting Shyamal's aggregate = Koli's + 2 = Amala's − 2 gives $t = 0.6$.

Answer: A — 0.60

With all scores and the weights known, we find the top aggregate.

Three project teams (each one female + one male share a project score); the test is individual; aggregate = (1−t)·Project + t·Test.

1. Project: min 40, max 80, average 60 over six scores summing to 360; as three team-scores counted twice, the team scores are {40, 60, 80}.
2. Test: min 40, max 80, average 60, four distinct values plus two at the average — the multiset is {40, 50, 60, 60, 70, 80}. Shyamal (2nd highest) = 70; Biman (2nd lowest) = 50.
3. Amala–Koli: Amala's project = 2 × Koli's forces Koli = 40, Amala = 80. Koli's test = Amala's + 20; combined with the remaining clues this gives Amala-test = 60, Koli-test = 80, then Rini-test = 60 and Mathew-test = 40 (Mathew < Rini).
4. Teams: Amala (project 80) pairs with the 80-male Mathew; the others are (Koli, Biman) = 40 and (Rini, Shyamal) = 60.
5. Weights: Shyamal's aggregate = Koli's + 2 = Amala's − 2 solves to test weight t = 0.6 (project weight 0.4).

Solved score table (aggregate = 0.4 × Project + 0.6 × Test):

Project teams: (Amala, Mathew) = 80 · (Koli, Biman) = 40 · (Rini, Shyamal) = 60. Weights: Project 0.4, Test 0.6.

This question — the maximum aggregate. Amala's aggregate = $0.4 \times 80 + 0.6 \times 60 = 32 + 36 = 68$, the highest (Shyamal is next at 66).

Answer: A — 68

Mathew's test score is fixed once the female test scores are placed.

Three project teams (each one female + one male share a project score); the test is individual; aggregate = (1−t)·Project + t·Test.

1. Project: min 40, max 80, average 60 over six scores summing to 360; as three team-scores counted twice, the team scores are {40, 60, 80}.
2. Test: min 40, max 80, average 60, four distinct values plus two at the average — the multiset is {40, 50, 60, 60, 70, 80}. Shyamal (2nd highest) = 70; Biman (2nd lowest) = 50.
3. Amala–Koli: Amala's project = 2 × Koli's forces Koli = 40, Amala = 80. Koli's test = Amala's + 20; combined with the remaining clues this gives Amala-test = 60, Koli-test = 80, then Rini-test = 60 and Mathew-test = 40 (Mathew < Rini).
4. Teams: Amala (project 80) pairs with the 80-male Mathew; the others are (Koli, Biman) = 40 and (Rini, Shyamal) = 60.
5. Weights: Shyamal's aggregate = Koli's + 2 = Amala's − 2 solves to test weight t = 0.6 (project weight 0.4).

Solved score table (aggregate = 0.4 × Project + 0.6 × Test):

Project teams: (Amala, Mathew) = 80 · (Koli, Biman) = 40 · (Rini, Shyamal) = 60. Weights: Project 0.4, Test 0.6.

This question — Mathew's test score. From the table, Mathew's test score is 40.

Answer: 40

Check the two proposed pairs against the actual project teams.

Three project teams (each one female + one male share a project score); the test is individual; aggregate = (1−t)·Project + t·Test.

1. Project: min 40, max 80, average 60 over six scores summing to 360; as three team-scores counted twice, the team scores are {40, 60, 80}.
2. Test: min 40, max 80, average 60, four distinct values plus two at the average — the multiset is {40, 50, 60, 60, 70, 80}. Shyamal (2nd highest) = 70; Biman (2nd lowest) = 50.
3. Amala–Koli: Amala's project = 2 × Koli's forces Koli = 40, Amala = 80. Koli's test = Amala's + 20; combined with the remaining clues this gives Amala-test = 60, Koli-test = 80, then Rini-test = 60 and Mathew-test = 40 (Mathew < Rini).
4. Teams: Amala (project 80) pairs with the 80-male Mathew; the others are (Koli, Biman) = 40 and (Rini, Shyamal) = 60.
5. Weights: Shyamal's aggregate = Koli's + 2 = Amala's − 2 solves to test weight t = 0.6 (project weight 0.4).

Solved score table (aggregate = 0.4 × Project + 0.6 × Test):

Project teams: (Amala, Mathew) = 80 · (Koli, Biman) = 40 · (Rini, Shyamal) = 60. Weights: Project 0.4, Test 0.6.

This question — which pairs shared a project team. The teams are (Amala, Mathew), (Koli, Biman), (Rini, Shyamal).

• i) Amala and Biman — not a team (Amala is with Mathew).
• ii) Koli and Mathew — not a team (Koli is with Biman).

Neither pair shared a team.

Answer: C — Neither i) nor ii)

Build the four-way grid of AC types, then split each type across the dealers.

Sort every AC into four buckets: Window-Inverter, Window-Non-inverter, Split-Inverter, Split-Non-inverter.

1. Window is 25% of the total and Split is 75%. Among Inverters, 20% are Window, so Split-Inverter = 4 × Window-Inverter. Given Window-Non-inverter = 6 and Split-Inverter = 36, we get Window-Inverter = 36 ÷ 4 = 9. So Window total = 9 + 6 = 15, and since that is 25% of everything, the grand total = 60, Split = 45 and Split-Non-inverter = 45 − 36 = 9.
2. Window by dealer: only D2 and D3 sold Window-Non-inverter, with D2 = 2 × D3, so D3 = 2 and D2 = 4. D3 and D4 sold equal Window numbers, each one-third of D2's Window; with every dealer selling at least 2 Window ACs, this forces D1 = 5, D2 = 6, D3 = 2, D4 = 2 Window ACs.
3. Split-Inverter by dealer: D1 sold 13 Inverters, so its Split-Inverter = 13 − 5 = 8. D3 and D4 sold equal Split-Inverters, each half of D2's; the total 36 then gives D2 = 14, D3 = D4 = 7.
4. Non-inverter by dealer: D3 sold 5 Non-inverters, so its Split-Non = 3. D1's Split = 2 × its Window = 10, so its Split-Non = 2. That leaves 4 Split-Non-inverter ACs shared between D2 and D4.

Category totals (Window = 25% of total; among Inverters 20% are Window, so Split-Inverter = 4 × Window-Inverter; given 6 Window-Non-inverter and 36 Split-Inverter):

Dealer-wise grid:

where $a + b = 4$ (the split of the last 4 Split-Non-inverter ACs between D2 and D4 is not individually fixed).

This question — D2's Split-Inverter count. This value is fully fixed by the grid: 14.

Answer: 14

We need the city-wide share of Non-inverter ACs.

Sort every AC into four buckets: Window-Inverter, Window-Non-inverter, Split-Inverter, Split-Non-inverter.

1. Window is 25% of the total and Split is 75%. Among Inverters, 20% are Window, so Split-Inverter = 4 × Window-Inverter. Given Window-Non-inverter = 6 and Split-Inverter = 36, we get Window-Inverter = 36 ÷ 4 = 9. So Window total = 9 + 6 = 15, and since that is 25% of everything, the grand total = 60, Split = 45 and Split-Non-inverter = 45 − 36 = 9.
2. Window by dealer: only D2 and D3 sold Window-Non-inverter, with D2 = 2 × D3, so D3 = 2 and D2 = 4. D3 and D4 sold equal Window numbers, each one-third of D2's Window; with every dealer selling at least 2 Window ACs, this forces D1 = 5, D2 = 6, D3 = 2, D4 = 2 Window ACs.
3. Split-Inverter by dealer: D1 sold 13 Inverters, so its Split-Inverter = 13 − 5 = 8. D3 and D4 sold equal Split-Inverters, each half of D2's; the total 36 then gives D2 = 14, D3 = D4 = 7.
4. Non-inverter by dealer: D3 sold 5 Non-inverters, so its Split-Non = 3. D1's Split = 2 × its Window = 10, so its Split-Non = 2. That leaves 4 Split-Non-inverter ACs shared between D2 and D4.

Category totals (Window = 25% of total; among Inverters 20% are Window, so Split-Inverter = 4 × Window-Inverter; given 6 Window-Non-inverter and 36 Split-Inverter):

Dealer-wise grid:

where $a + b = 4$ (the split of the last 4 Split-Non-inverter ACs between D2 and D4 is not individually fixed).

This question — percentage of Non-inverter ACs. Non-inverter = Window-Non + Split-Non = 6 + 9 = 15 out of 60:

$\frac{15}{60} \times 100 = 25\%$

Answer: C — 25.00%

Even though D2 and D4's Split-Non-inverter counts are not individually fixed, their sum is — which is all this question needs.

Sort every AC into four buckets: Window-Inverter, Window-Non-inverter, Split-Inverter, Split-Non-inverter.

1. Window is 25% of the total and Split is 75%. Among Inverters, 20% are Window, so Split-Inverter = 4 × Window-Inverter. Given Window-Non-inverter = 6 and Split-Inverter = 36, we get Window-Inverter = 36 ÷ 4 = 9. So Window total = 9 + 6 = 15, and since that is 25% of everything, the grand total = 60, Split = 45 and Split-Non-inverter = 45 − 36 = 9.
2. Window by dealer: only D2 and D3 sold Window-Non-inverter, with D2 = 2 × D3, so D3 = 2 and D2 = 4. D3 and D4 sold equal Window numbers, each one-third of D2's Window; with every dealer selling at least 2 Window ACs, this forces D1 = 5, D2 = 6, D3 = 2, D4 = 2 Window ACs.
3. Split-Inverter by dealer: D1 sold 13 Inverters, so its Split-Inverter = 13 − 5 = 8. D3 and D4 sold equal Split-Inverters, each half of D2's; the total 36 then gives D2 = 14, D3 = D4 = 7.
4. Non-inverter by dealer: D3 sold 5 Non-inverters, so its Split-Non = 3. D1's Split = 2 × its Window = 10, so its Split-Non = 2. That leaves 4 Split-Non-inverter ACs shared between D2 and D4.

Category totals (Window = 25% of total; among Inverters 20% are Window, so Split-Inverter = 4 × Window-Inverter; given 6 Window-Non-inverter and 36 Split-Inverter):

Dealer-wise grid:

where $a + b = 4$ (the split of the last 4 Split-Non-inverter ACs between D2 and D4 is not individually fixed).

This question — total ACs sold by D2 and D4. D2 = 20 + a and D4 = 9 + b, so their sum is $29 + (a + b) = 29 + 4 = 33$.

💡 Teacher tip: When two unknowns always add to a fixed number, any quantity that depends only on their sum is determined — no need to pin down each one.

Answer: 33

We test each statement against the grid, remembering only $a + b = 4$ is free.

Sort every AC into four buckets: Window-Inverter, Window-Non-inverter, Split-Inverter, Split-Non-inverter.

1. Window is 25% of the total and Split is 75%. Among Inverters, 20% are Window, so Split-Inverter = 4 × Window-Inverter. Given Window-Non-inverter = 6 and Split-Inverter = 36, we get Window-Inverter = 36 ÷ 4 = 9. So Window total = 9 + 6 = 15, and since that is 25% of everything, the grand total = 60, Split = 45 and Split-Non-inverter = 45 − 36 = 9.
2. Window by dealer: only D2 and D3 sold Window-Non-inverter, with D2 = 2 × D3, so D3 = 2 and D2 = 4. D3 and D4 sold equal Window numbers, each one-third of D2's Window; with every dealer selling at least 2 Window ACs, this forces D1 = 5, D2 = 6, D3 = 2, D4 = 2 Window ACs.
3. Split-Inverter by dealer: D1 sold 13 Inverters, so its Split-Inverter = 13 − 5 = 8. D3 and D4 sold equal Split-Inverters, each half of D2's; the total 36 then gives D2 = 14, D3 = D4 = 7.
4. Non-inverter by dealer: D3 sold 5 Non-inverters, so its Split-Non = 3. D1's Split = 2 × its Window = 10, so its Split-Non = 2. That leaves 4 Split-Non-inverter ACs shared between D2 and D4.

Category totals (Window = 25% of total; among Inverters 20% are Window, so Split-Inverter = 4 × Window-Inverter; given 6 Window-Non-inverter and 36 Split-Inverter):

Dealer-wise grid:

where $a + b = 4$ (the split of the last 4 Split-Non-inverter ACs between D2 and D4 is not individually fixed).

This question — which statement is necessarily false.

• A. D2 sold the highest number of ACs. D2 = 20 + a ≥ 20, while D1 = 15, D3 = 12, D4 ≤ 13. Always true.
• B. D4 sold more Split ACs than D3. D4-Split = 7 + b, D3-Split = 10; true when b = 4. Can be true.
• C. D1 and D3 sold equal Split ACs. Both = 10. Always true.
• D. D1 and D3 together sold more than D2 and D4 together. D1 + D3 = 27, but D2 + D4 = 33, so 27 > 33 is impossible — necessarily false.

Answer: D

We add one extra condition and see what it pins down.

Sort every AC into four buckets: Window-Inverter, Window-Non-inverter, Split-Inverter, Split-Non-inverter.

1. Window is 25% of the total and Split is 75%. Among Inverters, 20% are Window, so Split-Inverter = 4 × Window-Inverter. Given Window-Non-inverter = 6 and Split-Inverter = 36, we get Window-Inverter = 36 ÷ 4 = 9. So Window total = 9 + 6 = 15, and since that is 25% of everything, the grand total = 60, Split = 45 and Split-Non-inverter = 45 − 36 = 9.
2. Window by dealer: only D2 and D3 sold Window-Non-inverter, with D2 = 2 × D3, so D3 = 2 and D2 = 4. D3 and D4 sold equal Window numbers, each one-third of D2's Window; with every dealer selling at least 2 Window ACs, this forces D1 = 5, D2 = 6, D3 = 2, D4 = 2 Window ACs.
3. Split-Inverter by dealer: D1 sold 13 Inverters, so its Split-Inverter = 13 − 5 = 8. D3 and D4 sold equal Split-Inverters, each half of D2's; the total 36 then gives D2 = 14, D3 = D4 = 7.
4. Non-inverter by dealer: D3 sold 5 Non-inverters, so its Split-Non = 3. D1's Split = 2 × its Window = 10, so its Split-Non = 2. That leaves 4 Split-Non-inverter ACs shared between D2 and D4.

Category totals (Window = 25% of total; among Inverters 20% are Window, so Split-Inverter = 4 × Window-Inverter; given 6 Window-Non-inverter and 36 Split-Inverter):

Dealer-wise grid:

where $a + b = 4$ (the split of the last 4 Split-Non-inverter ACs between D2 and D4 is not individually fixed).

This question — with D3 and D4 selling equal totals. D3 = 12, so D4 = 9 + b = 12 gives b = 3, hence a = 1. D2's Non-inverter ACs = Window-Non + Split-Non = 4 + 1 = 5.

Answer: B — 5