CAT 2019 DILR Questions — Slot 2

Setting up the store-wide totals

Reading the radar charts. Each chart is a triangle with one corner per department. The dotted triangles are the percentage levels, from 10% (innermost) out to 50% (the outer edge), and each year's three shares add to 100%. Revenue shares (Clothing / Produce / Electronics): 2016 = 20 / 30 / 50, 2017 = 30 / 40 / 30, 2018 = 40 / 40 / 20. Cost shares: 2016 = 30 / 40 / 30, 2017 = 30 / 30 / 40, 2018 = 20 / 50 / 30.

The radar charts give us percentage shares of revenue and cost for each department in each year. To find actual profit percentages, we first need the total revenue and total cost for each year.

Step 1 — Anchor 2016.

Let the store's total cost in 2016 be $100$. The percentage profit in 2016 was $100\%$, and percentage profit is $\frac{\text{Profit}}{\text{Cost}}\times 100\%$. So:

$\frac{R_{2016}-C_{2016}}{C_{2016}}\times 100 = 100 \implies R_{2016}-C_{2016}=C_{2016} \implies R_{2016}=200$

Step 2 — Use the doubling clues.

• Revenue doubled from 2016 to 2017: $R_{2017}=2\times 200=400$.
• Cost doubled from 2016 to 2018: $C_{2018}=2\times 100=200$.

Step 3 — Use the Electronics-in-2017 clue.

"No profit from Electronics in 2017" means Electronics revenue equals Electronics cost in 2017.

$120 = 0.40\times C_{2017} \implies C_{2017}=300$

Step 4 — Use the Clothing-revenue = Produce-cost (2018) clue.

$0.40\times R_{2018}=100 \implies R_{2018}=250$

Solved totals:

Answering the question

For 2018:

$\text{Profit\%}_{2018}=\frac{250-200}{200}\times 100=\frac{50}{200}\times 100=25\%$

Answer: 25

Setting up the store-wide totals

Reading the radar charts. Each chart is a triangle with one corner per department. The dotted triangles are the percentage levels, from 10% (innermost) out to 50% (the outer edge), and each year's three shares add to 100%. Revenue shares (Clothing / Produce / Electronics): 2016 = 20 / 30 / 50, 2017 = 30 / 40 / 30, 2018 = 40 / 40 / 20. Cost shares: 2016 = 30 / 40 / 30, 2017 = 30 / 30 / 40, 2018 = 20 / 50 / 30.

The radar charts give us percentage shares of revenue and cost for each department. We need the actual totals first.

Step 1 — Anchor 2016. Let 2016 cost $=100$. Since 2016 profit percentage is $100\%$:

$\frac{R_{2016}-C_{2016}}{C_{2016}}\times 100 = 100 \implies R_{2016}=200$

Step 2 — Apply the doubling clues.

• Revenue doubled 2016→2017: $R_{2017}=400$.
• Cost doubled 2016→2018: $C_{2018}=200$.

Step 3 — Electronics in 2017 broke even (revenue = cost):

$0.30\times 400 = 0.40\times C_{2017} \implies 120 = 0.40\times C_{2017} \implies C_{2017}=300$

Step 4 — In 2018, Clothing revenue = Produce cost:

$0.40\times R_{2018} = 0.50\times 200 = 100 \implies R_{2018}=250$

Solved totals:

Answering the question

We need the ratio of Produce-department revenue in 2017 to that in 2018.

From the revenue-share table, Produce's share is $40\%$ in both years — but the total revenue differs.

$\text{Ratio} = 160: 100 = 8: 5$

💡 Teacher tip: Equal percentage shares do not mean equal amounts — always multiply the percentage by that year's total.

Answer: 8: 5 (Option D)

Setting up the store-wide totals

Reading the radar charts. Each chart is a triangle with one corner per department. The dotted triangles are the percentage levels, from 10% (innermost) out to 50% (the outer edge), and each year's three shares add to 100%. Revenue shares (Clothing / Produce / Electronics): 2016 = 20 / 30 / 50, 2017 = 30 / 40 / 30, 2018 = 40 / 40 / 20. Cost shares: 2016 = 30 / 40 / 30, 2017 = 30 / 30 / 40, 2018 = 20 / 50 / 30.

The radar charts give percentage shares of revenue and cost by department. We determine the actual totals step by step.

Step 1 — Anchor 2016. Let 2016 cost $=100$. With a $100\%$ profit percentage:

$R_{2016}=200$

Step 2 — Doubling clues.

• $R_{2017}=2\times 200=400$
• $C_{2018}=2\times 100=200$

Step 3 — Electronics broke even in 2017 (revenue = cost):

$0.30\times 400 = 0.40\times C_{2017} \implies C_{2017}=300$

Step 4 — In 2018, Clothing revenue = Produce cost:

$0.40\times R_{2018}=0.50\times 200=100 \implies R_{2018}=250$

Solved totals:

Answering the question

We need: what percentage of the store's 2016 total profit came from Electronics?

The store's total 2016 profit is $200-100=100$.

Now compute Electronics' own profit in 2016 using its revenue and cost shares:

$\text{Electronics' share of total profit}=\frac{70}{100}\times 100=70\%$

💡 Teacher tip: Department profit = (department's revenue share × total revenue) − (department's cost share × total cost). The percentages apply to different bases (revenue vs. cost), so compute each separately.

Answer: 70

Setting up the store-wide totals

Reading the radar charts. Each chart is a triangle with one corner per department. The dotted triangles are the percentage levels, from 10% (innermost) out to 50% (the outer edge), and each year's three shares add to 100%. Revenue shares (Clothing / Produce / Electronics): 2016 = 20 / 30 / 50, 2017 = 30 / 40 / 30, 2018 = 40 / 40 / 20. Cost shares: 2016 = 30 / 40 / 30, 2017 = 30 / 30 / 40, 2018 = 20 / 50 / 30.

The radar charts give percentage shares of revenue and cost by department. We find the actual totals using the four clues.

Step 1 — Anchor 2016. Let 2016 cost $=100$. With a $100\%$ profit percentage:

$\frac{R_{2016}-100}{100}\times 100 = 100 \implies R_{2016}=200$

Step 2 — Doubling clues.

• Revenue doubled 2016→2017: $R_{2017}=400$.
• Cost doubled 2016→2018: $C_{2018}=200$.

Step 3 — Electronics broke even in 2017 (no profit ⇒ revenue = cost):

$0.30\times 400 = 0.40\times C_{2017} \implies 120 = 0.40\times C_{2017} \implies C_{2017}=300$

Step 4 — In 2018, Clothing revenue = Produce cost:

$0.40\times R_{2018} = 0.50\times 200 = 100 \implies R_{2018}=250$

Solved totals:

Answering the question

We need the difference in the store's profit percentages for 2017 and 2018.

$\text{Profit\%}_{2017}=\frac{400-300}{300}\times 100=\frac{100}{300}\times 100=33.3\%$

$\text{Profit\%}_{2018}=\frac{250-200}{200}\times 100=\frac{50}{200}\times 100=25\%$

$\text{Difference}=33.3-25=8.3$

Answer: 8.3 (Option C)

What were the scores of Chen, David, and Eric after Round 3?

We need to first work out who scored in each of Rounds 1–3. Let us build the full scoring table step by step.

Understanding the round structure

Each round features 6 players in a sliding window:

Each round awards 7, 3, 1 points to its top three (no ties).

Key deductions for Rounds 1–6

1. Joshin (Player 10) plays only Rounds 5 and 6 in this phase, yet has 14 points. The maximum from two rounds is $7+7=14$, so Joshin took 7 in R5 and 7 in R6.

2. Gordon (Player 7) plays Rounds 2–6 and has 17 points. The clue says Gordon never scored in two consecutive rounds. To reach 17 = 7 + 7 + 3, he needs three non-adjacent rounds from {R2, R3, R4, R5, R6}. The only such triple is R2, R4, R6. Since R6's 7 belongs to Joshin, Gordon gets 7 (R2), 7 (R4), 3 (R6).

3. Amita (Player 1) plays R1 and R6, with 8 points = 7 + 1. Since R6's 7 is Joshin's, Amita gets 7 (R1), 1 (R6).

4. Bala (Player 2) plays R1 and R2, with 2 points = 1 + 1, so 1 (R1), 1 (R2).

5. All the 1-point slots in R1, R2, R6 are now taken. Hansa (Player 8) has just 1 point and plays R3–R6; the only remaining 1 is in R3, so Hansa = 1 (R3).

6. Ikea (Player 9) has 2 points and plays R4–R6; the available 1s are in R4 and R5 (R6's 1 is Amita's), so Ikea = 1 (R4), 1 (R5).

7. Fatima (Player 6) plays all six rounds and has 10 points. Every 7 except R3 is taken (R1→Amita, R2→Gordon, R4→Gordon, R5→Joshin, R6→Joshin), so Fatima = 7 (R3) + one 3.

8. Eric (Player 5) has 3 points (a single 3) from Rounds 1–5. The clue says Eric and Fatima both scored in a round. Fatima scored in R3 (her 7) and in one other round (her 3). Eric and Fatima cannot both get the 3-point slot in the same round, so the only round where they can both appear among the top three is R3 — where Fatima got 7 and Eric gets 3. Thus Eric = 3 (R3).

9. In R5, the 7 (Joshin) and 1 (Ikea) are fixed; the remaining eligible scorer for the 3 is Fatima (Eric, Gordon, Hansa are all done). So Fatima = 3 (R5).

10. In R4, the 7 (Gordon) and 1 (Ikea) are fixed; the 3 must go to David (Eric and Fatima are done). So David = 3 (R4).

11. David has 6 = 3 + 3, so he needs one more 3 in R1 or R2. Chen has 3 = a single 3, in the other of R1/R2. The clues leave open which of R1/R2 goes to David vs Chen — but this does not affect any question.

Solved scoring table (Rounds 1–6)

Answering the question

We read cumulative totals through Round 3:

• Chen: scored a single 3 in either R1 or R2. Either way, by the end of R3 he has 3.
• David: scored a 3 in either R1 or R2 (his second 3 is in R4, which hasn't happened yet). Through R3, David has 3.
• Eric: scored his only 3 in R3 itself. Through R3, Eric has 3.

$\text{Chen} = 3,\quad \text{David} = 3,\quad \text{Eric} = 3$

Answer: 3, 3, 3 (Option B)

Which three players were in the last three positions after Round 4?

We first reconstruct the scoring for Rounds 1–4, then sum each player's points.

Building the scoring table

Each round awards 7, 3, 1 to its top three. Here is how we pin down Rounds 1–4:

1. Joshin (Player 10) plays only R5 and R6 in the first phase but already has 14 points after Round 6 — that forces 7 (R5) and 7 (R6).
2. Gordon (Player 7) has 17 from R2–R6, never scoring in consecutive rounds. The only way to fit 7 + 7 + 3 on non-adjacent rounds is R2, R4, R6; since R6's 7 is Joshin's, Gordon gets 7 (R2), 7 (R4), 3 (R6).
3. Amita (Player 1) has 8 from R1 + R6 = 7 + 1; R6's 7 is taken, so 7 (R1), 1 (R6).
4. Bala (Player 2) has 2 from R1 + R2 = 1 (R1), 1 (R2).
5. Hansa (Player 8) has 1 point; all 1s in R1, R2, R6 are taken, so 1 (R3).
6. Ikea (Player 9) has 2 from R4–R6; available 1s are R4 and R5, so 1 (R4), 1 (R5).
7. Fatima (Player 6) has 10 from all six rounds; the only 7 left is R3, so 7 (R3) + one 3.
8. Eric (Player 5) has 3 (a single 3). Since Eric and Fatima both scored in a round, and they cannot share the 3-point slot, Eric's 3 must be in R3 alongside Fatima's 7: Eric = 3 (R3).
9. Fatima's 3 goes to R5 (the only round where she is the remaining eligible scorer after Joshin's 7 and Ikea's 1).
10. David (Player 4) has 6 = 3 + 3. One 3 is in R4 (the only eligible scorer there after Gordon's 7 and Ikea's 1); the other is in R1 or R2.
11. Chen (Player 3) has 3 = a single 3, in whichever of R1/R2 David did not take.

Solved scoring for Rounds 1–4

Cumulative points through Round 4

💡 Teacher tip: Joshin (Player 10) does not appear until Round 5, so after Round 4 he has zero points — always check which players have not yet entered the competition.

The bottom three are Joshin (0), Hansa (1), and Ikea (1).

Answer: Hansa, Ikea, Joshin (Option D)

Which player scored points in the maximum number of rounds?

To answer this, we need the complete scoring record across all 10 rounds. Let us reconstruct it.

Rounds 1–6 — key deductions

• Joshin plays only R5, R6 in this phase; 14 points forces 7 (R5), 7 (R6).
• Gordon (17 pts, no consecutive scoring rounds, plays R2–R6) must use non-adjacent R2, R4, R6: 7 (R2), 7 (R4), 3 (R6).
• Amita (8 from R1, R6) = 7 (R1), 1 (R6).
• Bala (2 from R1, R2) = 1 (R1), 1 (R2).
• Hansa (1 pt) = 1 (R3); Ikea (2 pts) = 1 (R4), 1 (R5).
• Fatima (10 from all six rounds) = 7 (R3) + 3 (R5).
• Eric (3 pts) = 3 (R3) — forced because Eric and Fatima must share a scoring round, and R3 is the only one where Fatima's 7 and Eric's 3 can coexist.
• David (6 = 3+3) = 3 (R4) + 3 in R1 or R2. Chen (3) = 3 in the other of R1/R2.

Rounds 7–10 — key deductions

Points gained in R7–10: Amita +10, Bala +3, Chen +3, Eric +7, Hansa +3, Ikea +15, Joshin +3 (David, Fatima, Gordon gain 0).

• Eric plays only R10 in this phase; +7 means 7 (R10).
• Ikea (+15 over R7–R9) = 7 + 7 + 1, scoring in all three — one of the two players who scored in three consecutive rounds.
• Chen (+3 over R8–R10) = 1 (R8), 1 (R9), 1 (R10) — the other three-consecutive scorer.
• Amita scored in R10 (given); her +10 = 7 + 3 with non-consecutive rounds → 7 (R7), 3 (R10).
• Joshin scored in R7 (given); +3 = 3 (R7).
• Hansa (+3, plays only R7–R8) = 3 (R8).
• Bala (+3) = 3 (R9).

Complete scoring table

Counting rounds scored per player

Ikea scored in five different rounds — more than any other player.

Answer: Ikea (Option D)

Which players scored points in the last round (Round 10)?

We need to determine who received the 7, 3, and 1 points in Round 10. Let us reconstruct the Rounds 7–10 scoring.

Points gained in Rounds 7–10

Subtracting each player's Round 6 total from their Round 10 total:

David, Fatima, and Gordon gained nothing — they did not score in Rounds 7–10.

Round 10 participants

Round 10 features players 10, 1, 2, 3, 4, 5: Joshin, Amita, Bala, Chen, David, Eric.

Of these, David gained 0, so the three scorers are among {Joshin, Amita, Bala, Chen, Eric}.

Deductions for Rounds 7–10

1. Eric plays only R10 in this phase and gained +7, so Eric = 7 (R10).

2. Ikea gained +15 over R7–R9 (she does not play R10). That equals 7 + 7 + 1, so she scored in all three rounds — one of the two players who scored in three consecutive rounds.

3. Chen is the other three-consecutive scorer (given). He gained +3 over R8–R10 = 1 (R8), 1 (R9), 1 (R10).

4. Amita scored in R10 (given) and gained +10 = 7 + 3. Since she cannot score in consecutive rounds (only Ikea and Chen may), her 7 must be in R7 (non-adjacent to R10). So Amita = 7 (R7), 3 (R10).

5. Joshin scored in R7 (given) and gained +3 = 3 (R7).

6. Hansa gained +3 and plays only R7–R8; with R7's slots taken (Amita 7, Joshin 3, Ikea 1), she gets 3 (R8).

7. Bala gained +3; the remaining 3-point slot is in R9, so Bala = 3 (R9).

Round 10 scoring

The three players who scored in Round 10 are Eric (7), Amita (3), and Chen (1).

Answer: Amita, Chen, Eric (Option D)

Setting up the full arrangement

Before answering any specific question, we need to pin down the structure of both papers.

Step 1 — Find the marks distribution for each paper.

Each paper is worth 100 marks using questions worth 5, 10, or 15 marks. If a paper has $a$ questions worth 5 marks, $b$ worth 10, and $c$ worth 15, then:

$5a + 10b + 15c = 100 \quad\Longrightarrow\quad a + 2b + 3c = 20$

with the constraints $a \ge 4$, $b \ge 3$, $c \ge 2$ (from the set-up), and all 5-mark questions come first, then 10-mark, then 15-mark.

Step 2 — Pin down MT.

Annie prepared the 5th question of MT and it carried 5 marks. Since all 5-mark questions come first, MT must have at least five 5-mark questions ($a \ge 5$). Also, Annie made exactly 1 MT question while every other faculty member made more than 1 (at least 2), so MT has at least $1 + 5 \times 2 = 11$ questions.

The only valid splits $(a, b, c)$ with $a \ge 5$ are:

Step 3 — Use the "ET has more questions" and "equal total" constraints.

ET has more questions than MT, and the maximum any paper can be is 13, so MT must be less than 13. That leaves MT = 11 or 12.

Since each of the six faculty prepared the same total number of questions across both papers, (MT + ET) must be divisible by 6:

• MT = 11, ET = 13 → $24 \div 6 = 4$ ✓ (each faculty prepared 4 questions total)
• MT = 11, ET = 12 → 23, not divisible by 6 ✗
• MT = 12, ET = 13 → 25, not divisible by 6 ✗

So MT has 11 questions (split 5, 3, 3) and ET has 13 questions (split 8, 3, 2). Each faculty prepared 4 questions in total.

In MT: Annie = 1 question, each of the other five = 2 questions. In ET: Annie = 3 questions, each of the other five = 2 questions.

Step 4 — Place everyone's blocks (consecutive in each paper).

MT (11 questions): Fakir is 1st → positions 1–2; Chetan is 3rd → positions 3–4; Annie is 5th (single) → position 5; Dave is last → positions 10–11; Esha is 8th → positions 8–9; the remaining spots 6–7 go to Beti.

ET (13 questions): Dave is 1st → positions 1–2; Chetan is 3rd → positions 3–4; Annie is 5th with 3 questions → positions 5–7; Esha is 8th → positions 8–9; Fakir is last → positions 12–13; the remaining spots 10–11 go to Beti.

💡 Teacher tip: The key breakthrough is realising that the marks-ordering rule (5s, then 10s, then 15s) combined with Annie's 5th question being 5 marks forces $a \ge 5$ for MT, which dramatically narrows the possible paper sizes.

Answering this question

We need the second question in ET. From the table, ET positions 1 and 2 both belong to Dave (he prepared the first question of ET, and since each faculty's questions are consecutive, his block covers positions 1–2).

Answer: D. Dave

Setting up the full arrangement

Step 1 — Find the marks distribution for each paper.

Each paper is worth 100 marks using questions worth 5, 10, or 15 marks. If a paper has $a$ questions worth 5 marks, $b$ worth 10, and $c$ worth 15, then:

$5a + 10b + 15c = 100 \quad\Longrightarrow\quad a + 2b + 3c = 20$

with $a \ge 4$, $b \ge 3$, $c \ge 2$, and all 5-mark questions come first, then 10-mark, then 15-mark.

Step 2 — Pin down MT.

Annie prepared the 5th question of MT and it was worth 5 marks. Since all 5-mark questions precede the 10-mark ones, MT must have at least five 5-mark questions ($a \ge 5$). Also, Annie made exactly 1 MT question while every other faculty member made more than 1 (at least 2), so MT has at least $1 + 5 \times 2 = 11$ questions.

The only valid splits with $a \ge 5$ are: $(5,3,3)=11$ questions, $(6,4,2)=12$ questions, $(8,3,2)=13$ questions.

Step 3 — Use the "ET has more questions" and "equal total" constraints.

ET has more questions than MT, and the maximum is 13, so MT < 13 (MT is 11 or 12). Since each of the six faculty prepared the same total number of questions, (MT + ET) must be divisible by 6:

• MT = 11, ET = 13 → $24 \div 6 = 4$ ✓
• MT = 11, ET = 12 → 23 ✗
• MT = 12, ET = 13 → 25 ✗

So MT has 11 questions with split $(5, 3, 3)$ and ET has 13 questions with split $(8, 3, 2)$.

Step 4 — Place everyone's blocks.

💡 Teacher tip: The marks-ordering rule means we can read off the mark value directly from the position once we know the split. For MT (split 5, 3, 3): positions 1–5 are 5 marks, 6–8 are 10 marks, 9–11 are 15 marks. For ET (split 8, 3, 2): positions 1–8 are 5 marks, 9–11 are 10 marks, 12–13 are 15 marks.

Answering this question

We need the total number of 5-mark questions across both papers.

• MT has five 5-mark questions (positions 1–5).
• ET has eight 5-mark questions (positions 1–8).

$5 + 8 = 13$

Answer: A. 13

Setting up the full arrangement

Step 1 — Marks distribution.

Each paper is worth 100 marks with questions worth 5, 10, or 15 marks. With $a$ five-mark, $b$ ten-mark, and $c$ fifteen-mark questions:

$a + 2b + 3c = 20, \quad a \ge 4,\; b \ge 3,\; c \ge 2$

All 5-mark questions come first, then 10-mark, then 15-mark.

Step 2 — Pin down MT.

Annie's 5th MT question is worth 5 marks, so MT has at least five 5-mark questions ($a \ge 5$). Annie made exactly 1 MT question; everyone else made at least 2, so MT $\ge 1 + 5 \times 2 = 11$ questions.

Valid splits with $a \ge 5$: $(5,3,3)=11$, $(6,4,2)=12$, $(8,3,2)=13$ questions.

Step 3 — Use the two key constraints.

ET has more questions than MT (so MT < 13), and (MT + ET) must be divisible by 6 (equal total per faculty). Testing:

• MT = 11, ET = 13 → 24 ÷ 6 = 4 ✓
• MT = 11, ET = 12 → 23 ✗
• MT = 12, ET = 13 → 25 ✗

So MT = 11 (split 5, 3, 3) and ET = 13 (split 8, 3, 2).

Step 4 — Place the blocks and assign marks.

💡 Teacher tip: Once you know the split, the marks for each position are fixed by the ordering rule. For MT (5, 3, 3): positions 1–5 → 5 marks, 6–8 → 10 marks, 9–11 → 15 marks. For ET (8, 3, 2): positions 1–8 → 5 marks, 9–11 → 10 marks, 12–13 → 15 marks.

Answering this question

We need to identify every faculty member who prepared at least one 15-mark question across MT and ET.

MT 15-mark questions (positions 9–11):

• Position 9: Esha
• Positions 10–11: Dave

ET 15-mark questions (positions 12–13):

• Positions 12–13: Fakir

No other faculty member appears in a 15-mark position. So the 15-mark questions were prepared by Dave, Esha, and Fakir only.

Answer: D. Only Dave, Esha and Fakir

Setting up the full arrangement

Step 1 — Marks distribution.

Each paper is worth 100 marks with questions worth 5, 10, or 15 marks. With $a$ five-mark, $b$ ten-mark, and $c$ fifteen-mark questions:

$a + 2b + 3c = 20, \quad a \ge 4,\; b \ge 3,\; c \ge 2$

All 5-mark questions come first, then 10-mark, then 15-mark.

Step 2 — Pin down MT.

Annie's 5th MT question is worth 5 marks, so MT has at least five 5-mark questions ($a \ge 5$). Annie made exactly 1 MT question; everyone else made at least 2, so MT $\ge 1 + 5 \times 2 = 11$ questions.

Valid splits with $a \ge 5$: $(5,3,3)=11$, $(6,4,2)=12$, $(8,3,2)=13$ questions.

Step 3 — Use the two key constraints.

ET has more questions than MT (so MT < 13), and (MT + ET) must be divisible by 6 (equal total per faculty). Testing:

• MT = 11, ET = 13 → 24 ÷ 6 = 4 ✓
• MT = 11, ET = 12 → 23 ✗
• MT = 12, ET = 13 → 25 ✗

So MT = 11 (split 5, 3, 3) and ET = 13 (split 8, 3, 2). Each faculty prepared 4 questions total. In ET: Annie = 3, each of the other five = 2.

Step 4 — Place the blocks in ET.

Using the clues: Dave is 1st (positions 1–2), Chetan is 3rd (positions 3–4), Annie is 5th with 3 consecutive questions (positions 5–7), Esha is 8th (positions 8–9), Fakir is last (positions 12–13). The only remaining consecutive pair is positions 10–11, which must be Beti.

💡 Teacher tip: When placing consecutive blocks, work from the fixed anchors (positions given in the clues) inward. Once Dave, Chetan, Annie, Esha, and Fakir are placed, the only gap left determines Beti's position automatically.

Answering this question

From the ET table, Beti's consecutive block occupies positions 10 and 11. Among the given options — seventh, fourth, ninth, and tenth — the one that falls within Beti's block is the tenth question.

Answer: D. Tenth question

Finding the percentage of Heavy Monsoon States with negative deviation

Step 1 — Read the scatter plot. Each state is a point whose horizontal position gives its actual June–August 2019 rainfall (mm) and whose vertical position gives its percentage deviation from LPA.

Step 2 — Identify the Heavy Monsoon States (actual rainfall ≥ 900 mm).

That is 7 Heavy Monsoon States.

Step 3 — Count how many of them have a negative deviation.

Three states — Arunachal (−10%), Kerala (−10%), Meghalaya (−15%) — have negative deviations.

Step 4 — Compute the percentage.

$\frac{3}{7}\times 100\% \approx 42.86\%$

Answer: 42.86 (Option A)

Finding the median deviation of Low Monsoon States

Step 1 — Read the scatter plot. Each state is a point whose horizontal position gives its actual June–August 2019 rainfall (mm) and whose vertical position gives its percentage deviation from LPA.

Step 2 — Identify the Low Monsoon States (actual rainfall ≤ 750 mm) and list their deviations.

That is 9 Low Monsoon States.

Step 3 — Sort the deviations and find the median.

With 9 values, the median is the 5th value when sorted from lowest to highest:

$-60,\; -40,\; -35,\; -30,\; \boxed{-10},\; +10,\; +15,\; +20,\; +25$

The 5th value is −10%.

Answer: −10% (Option A)

Finding the average rainfall of states with ≤ 600 mm rainfall and negative deviation

Step 1 — Read the scatter plot. Each state is a point whose horizontal position gives its actual June–August 2019 rainfall (mm) and whose vertical position gives its percentage deviation from LPA.

Step 2 — Apply both filters: rainfall ≤ 600 mm and negative deviation.

First, the states with rainfall ≤ 600 mm are: Gujarat (600, +25%), Karnataka (600, +20%), Rajasthan (300, +15%), Madhya Pradesh (600, +10%), Assam (600, −10%), West Bengal (600, −30%), Jharkhand (400, −35%), Delhi (300, −40%), Manipur (400, −60%).

Now keep only those with a negative deviation:

💡 Teacher tip: Gujarat, Karnataka, Rajasthan, and MP all have rainfall ≤ 600 mm but positive deviations, so they are excluded by the second condition.

Step 3 — Compute the average.

$\text{Average} = \frac{600 + 600 + 400 + 300 + 400}{5} = \frac{2300}{5} = 460 \text{ mm}$

Answer: 460 mm (Option D)

Finding the LPA of Gujarat in 2019

Step 1 — Read the scatter plot for Gujarat.

The setup tells us the Y-axis shows the percentage deviation from LPA in 2018. So Gujarat's actual 2019 rainfall of 600 mm is 25% above its LPA for 2018.

Step 2 — Compute LPA(2018) for Gujarat.

$\text{Actual 2019} = \text{LPA(2018)} \times (1 + 0.25)$

$600 = \text{LPA(2018)} \times 1.25 \implies \text{LPA(2018)} = \frac{600}{1.25} = 480 \text{ mm}$

Step 3 — Use the LPA definition to find LPA(2019).

The LPA of a year is the average rainfall of the preceding 10 years:

• LPA(2018) = average of 2009–2018
• LPA(2019) = average of 2010–2019

The only difference between these two 10-year windows is that 2019 enters and 2009 leaves:

$\text{LPA(2019)} - \text{LPA(2018)} = \frac{R_{2019} - R_{2009}}{10}$

Step 4 — Find $R_{2009}$ using the given clue.

We are told Gujarat's actual 2019 rainfall is 20% more than the rainfall in 2009:

$R_{2019} = 1.2 \times R_{2009} \implies 600 = 1.2 \times R_{2009} \implies R_{2009} = 500 \text{ mm}$

Step 5 — Compute LPA(2019).

$\text{LPA(2019)} = \text{LPA(2018)} + \frac{R_{2019} - R_{2009}}{10} = 480 + \frac{600 - 500}{10} = 480 + 10 = 490 \text{ mm}$

💡 Teacher tip: The key insight is that two consecutive 10-year LPAs differ by exactly $\frac{R_{\text{new}} - R_{\text{old}}}{10}$, because 9 of the 10 years overlap. Here the shift is only 10 mm, so LPA(2019) sits just 10 mm above LPA(2018).

Answer: 490 mm (Option C)

Setting up the full arrangement

First, let us note the languages each person speaks:

Step 1 — Convert language-team counts into person-team counts.

A team "speaks" a language if any member speaks it. So if a language is spoken by exactly $k$ teams, the only people who speak that language must be spread across exactly $k$ teams.

• Arabic → 1 team: Only Robert speaks Arabic, and Robert is in Team 1 (Fact 5). So Robert is in exactly 1 team.
• Dutch → 1 team: Only Quentin speaks Dutch, and Quentin is in Team 3 (Fact 5). So Quentin is in exactly 1 team.
• Basque → 2 teams: Only Sally speaks Basque, so Sally is in exactly 2 teams.
• French → 2 teams: Robert (Team 1) covers one team; Terence is the only other French speaker, so Terence must cover the second team — Terence is in at least 1 team.
• Chinese → 3 teams: Only Paula and Terence speak Chinese, so between them they must appear in all three teams.
• English → 3 teams: Paula, Quentin, and Sally must collectively cover all three teams.

Step 2 — Determine team size.

All three teams have the same size $k$ (either 2 or 3), so total memberships $= 3k$.

We already know $Q = 1$, $R = 1$, $S = 2$. Fact 4 says Paula and Sally share exactly two teams, so Paula is in at least 2 teams.

$P + Q + R + S + T = 3k \implies P + 1 + 1 + 2 + T = 3k \implies P + T = 3k - 4$

• If $k = 2$: $P + T = 2$. With $P \geq 2$, we get $P = 2, T = 0$. But $T = 0$ means no one provides French to a second team — contradiction.
• If $k = 3$: $P + T = 5$. With $P \geq 2$ and $T \geq 1$, the viable option (after checking all constraints) is $P = 3, T = 2$.

So each team has 3 members, with membership counts: Paula 3, Sally 2, Terence 2, Quentin 1, Robert 1.

Step 3 — Build the teams.

Paula is in every team. Robert is only in Team 1.

• Team 1: Paula + Robert already give Chinese, English, Arabic, French — that's exactly 4 languages. The third member must add no new language. Sally would add Basque (a 5th language), so the third member must be Terence (Chinese and French, both already covered).
• Terence needs a second team. If it were Team 3, Sally couldn't reach her required 2 teams (she can't be in Team 1). So Terence's second team is Team 2.
• Sally fills the remaining seats in Teams 2 and 3.

Verification: English & Chinese appear in all three teams ✓; Basque (T2, T3) and French (T1, T2) in exactly two ✓; Arabic (T1) and Dutch (T3) in exactly one ✓; Paula & Sally share exactly Teams 2 and 3 ✓; Quentin and Robert never share a team ✓.

---

Answering the question

Team 2 = {Paula, Sally, Terence}. Among the four options — Paula, Terence, Quentin, Sally — the one not in Team 2 is Quentin (he is only in Team 3).

Answer: Quentin

Setting up the full arrangement

First, let us note the languages each person speaks:

Step 1 — Convert language-team counts into person-team counts.

A team "speaks" a language if any member speaks it. So if a language is spoken by exactly $k$ teams, the only people who speak that language must be spread across exactly $k$ teams.

• Arabic → 1 team: Only Robert speaks Arabic, and Robert is in Team 1 (Fact 5). So Robert is in exactly 1 team.
• Dutch → 1 team: Only Quentin speaks Dutch, and Quentin is in Team 3 (Fact 5). So Quentin is in exactly 1 team.
• Basque → 2 teams: Only Sally speaks Basque, so Sally is in exactly 2 teams.
• French → 2 teams: Robert (Team 1) covers one team; Terence is the only other French speaker, so Terence must cover the second team — Terence is in at least 1 team.
• Chinese → 3 teams: Only Paula and Terence speak Chinese, so between them they must appear in all three teams.
• English → 3 teams: Paula, Quentin, and Sally must collectively cover all three teams.

Step 2 — Determine team size.

All three teams have the same size $k$ (either 2 or 3), so total memberships $= 3k$.

We already know $Q = 1$, $R = 1$, $S = 2$. Fact 4 says Paula and Sally share exactly two teams, so Paula is in at least 2 teams.

$P + Q + R + S + T = 3k \implies P + 1 + 1 + 2 + T = 3k \implies P + T = 3k - 4$

• If $k = 2$: $P + T = 2$. With $P \geq 2$, we get $P = 2, T = 0$. But $T = 0$ means no one provides French to a second team — contradiction.
• If $k = 3$: $P + T = 5$. With $P \geq 2$ and $T \geq 1$, the viable option (after checking all constraints) is $P = 3, T = 2$.

So each team has 3 members, with membership counts: Paula 3, Sally 2, Terence 2, Quentin 1, Robert 1.

Step 3 — Build the teams.

Paula is in every team. Robert is only in Team 1.

• Team 1: Paula + Robert already give Chinese, English, Arabic, French — exactly 4 languages. The third member must add no new language. Sally would add Basque (a 5th language), so the third member must be Terence.
• Terence needs a second team. If it were Team 3, Sally couldn't reach her required 2 teams. So Terence's second team is Team 2.
• Sally fills the remaining seats in Teams 2 and 3.

Verification: English & Chinese in all three ✓; Basque (T2, T3) and French (T1, T2) in exactly two ✓; Arabic (T1) and Dutch (T3) in exactly one ✓; Paula & Sally share exactly Teams 2 and 3 ✓; Quentin and Robert never share a team ✓.

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Answering the question

From the membership counts derived above:

Among the four options — Paula (3), Quentin (1), Sally (2), Robert (1) — the person in exactly two teams is Sally.

Answer: Sally

Setting up the full arrangement

First, let us note the languages each person speaks:

Step 1 — Convert language-team counts into person-team counts.

A team "speaks" a language if any member speaks it. So if a language is spoken by exactly $k$ teams, the only people who speak that language must be spread across exactly $k$ teams.

• Arabic → 1 team: Only Robert speaks Arabic, and Robert is in Team 1 (Fact 5). So Robert is in exactly 1 team.
• Dutch → 1 team: Only Quentin speaks Dutch, and Quentin is in Team 3 (Fact 5). So Quentin is in exactly 1 team.
• Basque → 2 teams: Only Sally speaks Basque, so Sally is in exactly 2 teams.
• French → 2 teams: Robert (Team 1) covers one team; Terence is the only other French speaker, so Terence must cover the second team — Terence is in at least 1 team.
• Chinese → 3 teams: Only Paula and Terence speak Chinese, so between them they must appear in all three teams.
• English → 3 teams: Paula, Quentin, and Sally must collectively cover all three teams.

Step 2 — Determine team size.

All three teams have the same size $k$ (either 2 or 3), so total memberships $= 3k$.

We already know $Q = 1$, $R = 1$, $S = 2$. Fact 4 says Paula and Sally share exactly two teams, so Paula is in at least 2 teams.

$P + Q + R + S + T = 3k \implies P + 1 + 1 + 2 + T = 3k \implies P + T = 3k - 4$

• If $k = 2$: $P + T = 2$. With $P \geq 2$, we get $P = 2, T = 0$. But $T = 0$ means no one provides French to a second team — contradiction.
• If $k = 3$: $P + T = 5$. With $P \geq 2$ and $T \geq 1$, the viable option (after checking all constraints) is $P = 3, T = 2$.

💡 Teacher tip: The key insight is that Chinese must be spoken by all three teams, but only Paula and Terence speak Chinese. If Paula is in only 2 teams, Terence must cover the third — but then the French constraint and the team-size equation force a contradiction. Paula must be in all three teams.

So each team has 3 members, with membership counts: Paula 3, Sally 2, Terence 2, Quentin 1, Robert 1.

Step 3 — Build the teams.

Paula is in every team. Robert is only in Team 1.

• Team 1: Paula + Robert already give Chinese, English, Arabic, French — exactly 4 languages. The third member must add no new language. Sally would add Basque (a 5th language), so the third member must be Terence.
• Terence needs a second team. If it were Team 3, Sally couldn't reach her required 2 teams. So Terence's second team is Team 2.
• Sally fills the remaining seats in Teams 2 and 3.

Verification: English & Chinese in all three ✓; Basque (T2, T3) and French (T1, T2) in exactly two ✓; Arabic (T1) and Dutch (T3) in exactly one ✓; Paula & Sally share exactly Teams 2 and 3 ✓; Quentin and Robert never share a team ✓.

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Answering the question

Looking at the solved table, Paula appears in Team 1, Team 2, and Team 3 — she is the only person in all three teams. This makes sense: she is the only person who speaks both Chinese and English, and since every team must speak both of those languages, she is needed everywhere.

Answer: Paula

Setting up the full arrangement

First, let us note the languages each person speaks:

Step 1 — Convert language-team counts into person-team counts.

A team "speaks" a language if any member speaks it. So if a language is spoken by exactly $k$ teams, the only people who speak that language must be spread across exactly $k$ teams.

• Arabic → 1 team: Only Robert speaks Arabic, and Robert is in Team 1 (Fact 5). So Robert is in exactly 1 team.
• Dutch → 1 team: Only Quentin speaks Dutch, and Quentin is in Team 3 (Fact 5). So Quentin is in exactly 1 team.
• Basque → 2 teams: Only Sally speaks Basque, so Sally is in exactly 2 teams.
• French → 2 teams: Robert (Team 1) covers one team; Terence is the only other French speaker, so Terence must cover the second team — Terence is in at least 1 team.
• Chinese → 3 teams: Only Paula and Terence speak Chinese, so between them they must appear in all three teams.
• English → 3 teams: Paula, Quentin, and Sally must collectively cover all three teams.

Step 2 — Determine team size.

All three teams have the same size $k$ (either 2 or 3), so total memberships $= 3k$.

We already know $Q = 1$, $R = 1$, $S = 2$. Fact 4 says Paula and Sally share exactly two teams, so Paula is in at least 2 teams.

$P + Q + R + S + T = 3k \implies P + 1 + 1 + 2 + T = 3k \implies P + T = 3k - 4$

• If $k = 2$: $P + T = 2$. With $P \geq 2$, we get $P = 2, T = 0$. But $T = 0$ means no one provides French to a second team — contradiction.
• If $k = 3$: $P + T = 5$. With $P \geq 2$ and $T \geq 1$, the viable option (after checking all constraints) is $P = 3, T = 2$.

So each team has 3 members, with membership counts: Paula 3, Sally 2, Terence 2, Quentin 1, Robert 1.

Step 3 — Build the teams.

Paula is in every team. Robert is only in Team 1.

• Team 1: Paula + Robert already give Chinese, English, Arabic, French — exactly 4 languages. The third member must add no new language. Sally would add Basque (a 5th language), so the third member must be Terence.
• Terence needs a second team. If it were Team 3, Sally couldn't reach her required 2 teams. So Terence's second team is Team 2.
• Sally fills the remaining seats in Teams 2 and 3.

Verification: English & Chinese in all three ✓; Basque (T2, T3) and French (T1, T2) in exactly two ✓; Arabic (T1) and Dutch (T3) in exactly one ✓; Paula & Sally share exactly Teams 2 and 3 ✓; Quentin and Robert never share a team ✓.

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Answering the question

Team 1 = {Paula, Robert, Terence}.

• Paula contributes Chinese and English.
• Robert contributes Arabic and French.
• Terence contributes Chinese and French (both already covered).

So the four languages spoken by Team 1 are: Chinese, English, Arabic, French. Apart from Chinese and English, the remaining two are Arabic and French.

Answer: Arabic and French

We need to find, among students who supported proposal A, what percentage preferred Sunita. To do that, we first build the complete survey grid from the seven clues.

Step 1 — Establish the two big totals

• Total surveyed = 500.
• Every student prefers exactly one of Sunita or Ragini.
• 250 support proposal A; 250 support proposal B.

Step 2 — Split each candidate's supporters using the percentage clues

1. 200 prefer Sunita, and 80% of them support A: $0.80 \times 200 = 160$ So 160 Sunita-preferrers support A.

2. 300 prefer Ragini, and 30% of them support A: $0.30 \times 300 = 90$ So 90 Ragini-preferrers support A.

Check: $160 + 90 = 250$, which matches the total A-supporters. ✓

1. 20% of B-supporters prefer Sunita: $0.20 \times 250 = 50$ So among B-supporters, 50 prefer Sunita and 200 prefer Ragini.

2. Students who did not support B = $500 - 250 = 250$. Of these, 40% prefer Ragini: $0.40 \times 250 = 100$ So among non-B students, 100 prefer Ragini and 150 prefer Sunita.

3. Every Sunita-preferrer who supports B also supports A. So the cell "Sunita, B, not-A" is 0.

4. Among Ragini-preferrers, 20% support neither proposal: $0.20 \times 300 = 60$ So the Ragini "not-A, not-B" cell = 60.

Step 3 — Fill the grid

How the Sunita row fills: She has 50 B-supporters, and all of them also support A, so A&B = 50 and not-A&B = 0. Her total A-supporters are 160, leaving $160 - 50 = 110$ in A&not-B. Her row total is 200, so the remaining not-A&not-B cell = $200 - 50 - 110 - 0 = 40$.

How the Ragini row fills: not-A&not-B = 60 (from clue 7). Her non-B total is 100, so A&not-B = $100 - 60 = 40$. Her total A-supporters are 90, so A&B = $90 - 40 = 50$. Her B-total is 200, so not-A&B = $200 - 50 = 150$.

Column totals confirm A = $100 + 150 = 250$ and B = $100 + 150 = 250$. ✓

Step 4 — Answer the question

Among A-supporters, those who prefer Sunita are in the two A-columns of Sunita's row: $50 + 110 = 160$

Total A-supporters = 250, so $\frac{160}{250} \times 100 = 64\%$

Answer: 64

We need the percentage of students who did not support proposal A and preferred Ragini. We first build the full survey grid from the clues.

Step 1 — Establish the two big totals

• Total surveyed = 500.
• Every student prefers exactly one of Sunita or Ragini.
• 250 support proposal A; 250 support proposal B.

Step 2 — Split each candidate's supporters using the percentage clues

1. 200 prefer Sunita, and 80% of them support A: $0.80 \times 200 = 160$ So 160 Sunita-preferrers support A.

2. 300 prefer Ragini, and 30% of them support A: $0.30 \times 300 = 90$ So 90 Ragini-preferrers support A.

Check: $160 + 90 = 250$, matching total A-supporters. ✓

1. 20% of B-supporters prefer Sunita: $0.20 \times 250 = 50$ So among B-supporters, 50 prefer Sunita and 200 prefer Ragini.

2. Non-B students = $500 - 250 = 250$. Of these, 40% prefer Ragini: $0.40 \times 250 = 100$ So among non-B students, 100 prefer Ragini and 150 prefer Sunita.

3. Every Sunita-preferrer who supports B also supports A, so the "Sunita, B, not-A" cell = 0.

4. Among Ragini-preferrers, 20% support neither proposal: $0.20 \times 300 = 60$ So the Ragini "not-A, not-B" cell = 60.

Step 3 — Fill the grid

Sunita row: 50 B-supporters all support A (A&B = 50, not-A&B = 0); A-total 160 leaves 110 in A&not-B; row total 200 leaves 40 in not-A&not-B.

Ragini row: not-A&not-B = 60; non-B total 100 leaves 40 in A&not-B; A-total 90 leaves 50 in A&B; B-total 200 leaves 150 in not-A&B.

Column totals confirm A = 250 and B = 250. ✓

Step 4 — Answer the question

Students who did not support A = $500 - 250 = 250$.

Among these non-A students, those who prefer Ragini sit in the two not-A columns of Ragini's row: $150 + 60 = 210$

So the required percentage is $\frac{210}{250} \times 100 = 84\%$

Answer: 84

We need the percentage of students who supported both proposals A and B and preferred Sunita. We first build the full survey grid from the clues.

Step 1 — Establish the two big totals

• Total surveyed = 500.
• Every student prefers exactly one of Sunita or Ragini.
• 250 support proposal A; 250 support proposal B.

Step 2 — Split each candidate's supporters using the percentage clues

1. 200 prefer Sunita, and 80% of them support A: $0.80 \times 200 = 160$ So 160 Sunita-preferrers support A.

2. 300 prefer Ragini, and 30% of them support A: $0.30 \times 300 = 90$ So 90 Ragini-preferrers support A.

Check: $160 + 90 = 250$, matching total A-supporters. ✓

1. 20% of B-supporters prefer Sunita: $0.20 \times 250 = 50$ So among B-supporters, 50 prefer Sunita and 200 prefer Ragini.

2. Non-B students = $500 - 250 = 250$. Of these, 40% prefer Ragini: $0.40 \times 250 = 100$ So among non-B students, 100 prefer Ragini and 150 prefer Sunita.

3. Every Sunita-preferrer who supports B also supports A, so the "Sunita, B, not-A" cell = 0.

4. Among Ragini-preferrers, 20% support neither proposal: $0.20 \times 300 = 60$ So the Ragini "not-A, not-B" cell = 60.

Step 3 — Fill the grid

Sunita row: 50 B-supporters all support A (A&B = 50, not-A&B = 0); A-total 160 leaves 110 in A&not-B; row total 200 leaves 40 in not-A&not-B.

Ragini row: not-A&not-B = 60; non-B total 100 leaves 40 in A&not-B; A-total 90 leaves 50 in A&B; B-total 200 leaves 150 in not-A&B.

Column totals confirm A = 250 and B = 250. ✓

Step 4 — Answer the question

Students who supported both A and B are in the "A & B" column, which totals 100.

Of these, those who prefer Sunita = 50 (the Sunita entry in the A&B column).

$\frac{50}{100} \times 100 = 50\%$

Answer: D. 50

We need the count of students who supported proposal B, did not support proposal A, and preferred Ragini. We first build the full survey grid from the clues.

Step 1 — Establish the two big totals

• Total surveyed = 500.
• Every student prefers exactly one of Sunita or Ragini.
• 250 support proposal A; 250 support proposal B.

Step 2 — Split each candidate's supporters using the percentage clues

1. 200 prefer Sunita, and 80% of them support A: $0.80 \times 200 = 160$ So 160 Sunita-preferrers support A.

2. 300 prefer Ragini, and 30% of them support A: $0.30 \times 300 = 90$ So 90 Ragini-preferrers support A.

Check: $160 + 90 = 250$, matching total A-supporters. ✓

1. 20% of B-supporters prefer Sunita: $0.20 \times 250 = 50$ So among B-supporters, 50 prefer Sunita and 200 prefer Ragini.

2. Non-B students = $500 - 250 = 250$. Of these, 40% prefer Ragini: $0.40 \times 250 = 100$ So among non-B students, 100 prefer Ragini and 150 prefer Sunita.

3. Every Sunita-preferrer who supports B also supports A, so the "Sunita, B, not-A" cell = 0.

4. Among Ragini-preferrers, 20% support neither proposal: $0.20 \times 300 = 60$ So the Ragini "not-A, not-B" cell = 60.

Step 3 — Fill the grid

Sunita row: 50 B-supporters all support A (A&B = 50, not-A&B = 0); A-total 160 leaves 110 in A&not-B; row total 200 leaves 40 in not-A&not-B.

Ragini row: not-A&not-B = 60; non-B total 100 leaves 40 in A&not-B; A-total 90 leaves 50 in A&B; B-total 200 leaves 150 in not-A&B.

Column totals confirm A = 250 and B = 250. ✓

Step 4 — Answer the question

We need students who are: B-supporter + not-A + prefer Ragini.

That is exactly the Ragini entry in the "not-A & B" column, which is 150.

💡 Teacher tip: When a question specifies three conditions at once (support B, not support A, prefer Ragini), it is pointing at a single cell of the grid — read it directly rather than re-deriving.

Answer: A. 150

Finding the maximum number of patients on a single day

Key setup. The clinic operates from 9:00 a.m. to 11:30 a.m., giving a working window of 150 minutes each day. Three doctors work simultaneously in three rooms:

The room-to-doctor assignment changes by day, but every day all three doctors are present:

Reasoning. To maximise the number of patients seen, we assume the queue is never empty — every doctor is busy for the entire 150-minute window, seeing patients back-to-back with no idle time. The number of patients each doctor can see depends only on their consultation time, not on which room they occupy:

$\text{Ben}: \frac{150}{10} = 15 \text{ patients}$

$\text{Kane}: \frac{150}{15} = 10 \text{ patients}$

$\text{Wayne}: \frac{150}{25} = 6 \text{ patients}$

Since all three doctors work in parallel, the daily total is the sum:

$15 + 10 + 6 = 31$

💡 Teacher tip: When the queue never empties, each doctor acts like an independent pipeline. The room number only affects which patient goes to which doctor — it does not change how many patients a doctor can see in 150 minutes.

Answer: C. 31

Who earns the most on a busy Saturday?

Key setup. On Saturday the room assignments are:

The clinic window is 9:00 a.m. – 11:30 a.m. = 150 minutes. Since the queue is never empty, every doctor works the full 150 minutes without any idle time.

Step 1 — Patients seen by each doctor.

Step 2 — Earnings.

💡 Teacher tip: A higher fee per patient does not automatically mean higher total earnings. Kane sees fewer patients than Ben but charges double, which tips the balance in his favour. Wayne charges the most per visit but sees so few patients (only 6) that he falls short.

Dr. Kane earns Rs. 2,000 — more than both Ben (Rs. 1,500) and Wayne (Rs. 1,800).

Answer: B. Dr. Kane

On which day was Mr. Singh at the clinic the longest?

Key setup. Mr. Singh arrives at 8:50 a.m. on all three days and holds token 13 each day. Since his arrival time is identical across days, the day he leaves latest is the day with the maximum total duration.

The clinic opens at 9:00 a.m. (t = 0). Patients 1, 2, 3 immediately fill rooms 1, 2, 3 respectively. We simulate the queue forward, always assigning the next token to the lowest-numbered free room.

---

Monday — Rooms: Ben(1), Kane(2), Wayne(3)

Token 13 sees Wayne (25 min): enters at t = 50 (9:50 a.m.), leaves at t = 75 (10:15 a.m.).

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Wednesday — Rooms: Wayne(1), Ben(2), Kane(3)

Token 13 sees Ben (10 min): enters at t = 50 (9:50 a.m.), leaves at t = 60 (10:00 a.m.).

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Friday — Rooms: Kane(1), Wayne(2), Ben(3)

Token 13 sees Ben (10 min): enters at t = 50 (9:50 a.m.), leaves at t = 60 (10:00 a.m.).

---

Summary:

💡 Teacher tip: Token 13 enters at the same clock time (t = 50) on all three days, because the queue dynamics up to that point are symmetric — the same number of patients have been processed. What differs is which doctor token 13 draws, and that depends on the room assignment for that day. On Monday, token 13 lands in Wayne's room (the slowest doctor), making the visit 15 minutes longer.

Mr. Singh leaves at 10:15 a.m. on Monday versus 10:00 a.m. on the other two days, so Monday is the longest.

Answer: C. Monday

Total time all three doctors are simultaneously free on a slow Thursday

Key setup. On Thursday the room assignments are:

At 9:00 a.m. (t = 0) only two patients are waiting, so they take rooms 1 and 2 (the two lowest-numbered rooms). Room 3 starts empty. After that, two patients arrive together every 15 minutes: at t = 15, 30, 45, 60, …

Step 1 — Simulate the first cycle (t = 0 to t = 30).

At t = 30, all three rooms are free for an instant, but two patients arrive at that exact moment and immediately occupy rooms 1 and 2.

Step 2 — Recognise the repeating pattern.

From t = 30 onward, the same 30-minute cycle repeats:

• Two patients take rooms 1 (Wayne, 25 min) and 2 (Ben, 10 min). Room 3 is free.
• At t + 10: Room 2 frees → rooms 2 and 3 free, room 1 busy.
• At t + 15: Two arrive → take rooms 2 (Ben) and 3 (Kane, 15 min). All three busy.
• At t + 25: Rooms 1 and 2 free. Room 3 still busy.
• At t + 30: Room 3 frees and two new patients arrive simultaneously → rooms 1 and 2 reoccupied instantly.

This cycle continues through t = 90, 120, and 150 (clinic close). At every moment when the third room would free up alongside the other two, a fresh pair of patients arrives at that exact instant.

💡 Teacher tip: The key is that Kane's 15-minute consultation time is a perfect divisor of the 15-minute arrival interval. This causes room 3 to free up exactly when the next batch arrives, so the all-free state is always instantaneous — it has zero duration.

Step 3 — Total up.

At no point during the entire clinic session do all three rooms stay empty for any positive length of time.

$\text{Total simultaneously-free time} = 0 \text{ minutes}$

Answer: D. 0

Setting up the grid

Each of the 9 slots holds three pouches. The table gives the minimum and maximum rupee amounts among those three pouches. If the min is $m$ and the max is $M$, the third pouch $x$ satisfies $m \le x \le M$, so each slot's total lies in the range $[2m+M,\; m+2M]$.

The key constraint: every row of 9 pouches and every column of 9 pouches has an integer average, so each row total and each column total must be a multiple of 9.

We are also told the slot at Row 3, Column 1 totals Rs. 4.

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Step 1 — Fix R3C1

For R3C1 the table gives $(\min,\max)=(1,2)$ and the total is 4. The three pouches are $\{1,1,2\}$ (since $1+1+2=4$ and min=1, max=2). ✓

Step 2 — Bound each row and column, then squeeze to a multiple of 9

Computing the total range for each slot:

• Row 1 range: $[8+20+5,\; 10+22+7]=[33,39]$ → only multiple of 9 is 36.
• Row 2 range: $[11+3+32,\; 13+3+46]=[46,62]$ → only multiple of 9 is 54.
• Row 3 range (with R3C1=4): $[4+4+9,\; 4+5+12]=[17,21]$ → only multiple of 9 is 18.
• Column 1 range (with R3C1=4): $[8+11+4,\; 10+13+4]=[23,27]$ → only multiple of 9 is 27.
• Column 2 range: $[20+3+4,\; 22+3+5]=[27,30]$ → only multiple of 9 is 27.
• Grand total $=36+54+18=108$, so Column 3 $=108-27-27=$ 54.

Step 3 — Solve each slot

• Column 1 = 27: $R1C1 + R2C1 + 4 = 27 \Rightarrow R1C1+R2C1=23$. With $R1C1\le 10$ and $R2C1\le 13$, the only option is $R1C1=10,\; R2C1=13$.
• R1C1 = 10 → pouches $\{2,4,4\}$
• R2C1 = 13 → pouches $\{3,5,5\}$

The remaining slots are solved similarly, giving the complete grid:

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Answering the question

The question asks for the total in the three pouches at Row 2, Column 1. From the solved grid, that slot holds $\{3,5,5\}$, which sums to $3+5+5=13$.

Answer: 13

Setting up the grid

Each of the 9 slots holds three pouches. The table gives the minimum and maximum rupee amounts among those three. If the min is $m$ and the max is $M$, the third pouch $x$ satisfies $m \le x \le M$, so each slot's total lies in $[2m+M,\; m+2M]$.

Every row of 9 pouches and every column of 9 pouches has an integer average, so each row total and each column total is a multiple of 9. We are also told R3C1 totals Rs. 4.

Deducing the full grid

R3C1: $(1,2)$ with total 4 → pouches $\{1,1,2\}$.

Row/column totals (squeezing each range to the only multiple of 9 inside it):

Solving slot by slot:

• Col 1 = 27: $R1C1+R2C1=23$ with maxima 10 and 13 → $R1C1=10\;\{2,4,4\}$, $R2C1=13\;\{3,5,5\}$.
• Col 2 = 27: $R1C2+R3C2=24$ with $R3C2\in\{4,5\}$ → $R3C2=4\;\{1,1,2\}$, $R1C2=20\;\{6,6,8\}$.
• Row 3 = 18: $R3C3=18-4-4=10\;\{2,3,5\}$.
• Row 1 = 36: $R1C3=36-10-20=6\;\{1,2,3\}$.
• Col 3 = 54: $R2C3=54-6-10=38\;\{6,12,20\}$; $R2C2=\{1,1,1\}=3$.

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Answering the question

We need the number of pouches containing exactly one coin (Rs. 1). Scanning each slot for 1-rupee pouches:

$1+3+2+2=8$

Answer: 8