CAT 2021 DILR Questions — Slot 2

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — object o8. Steps 5 and 6 show that every possible 16-bundle for Charles contains o8, and the distribution places it with him.

Answer: o8 was given to Charles

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — who envies someone. Only the two people below the top value of 16 can possibly envy, namely Amar (12) and Disha (13); the three at 16 envy no one. Check Amar against the 16-bundles:

• Barat's {o9, o7}: Amar values it at 9 + 8 = 17 > 12 → Amar envies Barat.
• Charles's {o3, o8}: Amar values it at 9 + 7 = 16 > 12 → Amar envies Charles.

So Amar envies. Among the four names offered, only Amar is one of the enviers.

Answer: Amar

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — Amar's value for his own bundle. Amar's bundle is {o2, o6}, which he values at

$9 + 3 = 12.$

Answer: 12

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — object o4. In the final split of {o1, o2, o4, o6}, Disha's bundle is {o1, o4}. So o4 went to Disha.

Answer: Disha

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — object o5. Elise's bundle is {o10, o5} (Step 7), so o5 went to Elise.

Answer: Elise

The valuation table (how much each person values each object):

Each person receives a bundle of exactly two objects; a bundle's worth to a person is the sum of that person's two values.

Working out the whole distribution (once).

1. The two 10's are forced. If someone values an object at 10, they must get it. The only 10's are Barat → o9 and Elise → o10. So Barat gets o9, Elise gets o10.
2. Barat is pinned to a total of 16. Barat already holds o9 (worth 10 to him). His total must be even and at most 16, so his second object must be worth an even number no more than 6 to him. The only such value in his row is o7 = 6. So Barat = {o9, o7} = 16.
3. Elise also totals 16. She holds o10 (worth 10). Her second object must be even and at most 6 to her: the choices are o1, o5, o7 (each 6). o7 is taken, so Elise = {o10, o1} or {o10, o5}, worth 16.
4. The third 16 is Charles, not Amar. Amar would value Barat's bundle {o9, o7} at 9 + 8 = 17. If Amar's own bundle were also 16, he would envy Barat — but a person on the top value of 16 is not allowed to envy anyone. So Amar is below 16, and the third person at 16 is Charles, who reaches 16 only with two objects each worth 8 to him — two of {o1, o2, o3, o8}.
5. Charles must hold o8. Objects o1, o2, o3 go to three different people, so Charles can't take two of those; and o1 and o8 go to different people, so {o1, o8} is out. The only pairs left are {o2, o8} and {o3, o8} — both contain o8, so o8 → Charles.
6. Charles = {o3, o8}. If Charles held {o2, o8}, Barat would value it at 9 + 8 = 17 and would envy him — not allowed. So Charles = {o3, o8} = 16, leaving o2 free.
7. Elise = {o10, o5}. The other choice, {o10, o1}, leaves no way to finish without breaking a rule, so Elise = {o10, o5} = 16.
8. The last four objects {o1, o2, o4, o6} split between Amar and Disha. Disha's total must be odd, Amar's even. The only split that works is Amar = {o2, o6} = 9 + 3 = 12 and Disha = {o1, o4} = 8 + 5 = 13.

The complete result:

This question — object o1. The only valid split of the last four objects gives Disha {o1, o4}, so o1 is forced to Disha — nowhere else works.

Answer: o1 was given to Disha

We need to find how many days between Sep 1 and Sep 14 (both inclusive) had no order booked by this client across all categories.

Step 1 — Read every booking day off the chart

Each bar's left endpoint is the booking day. Scanning all 35 bars across the 14 categories, here is the complete data with the booking day highlighted:

Step 2 — Collect the distinct booking days

Looking down the Booked column, the distinct days on which at least one order was booked are:

Sep 1, Sep 2, Sep 3, Sep 4, Sep 5, Sep 6, Sep 7, Sep 13

That is 8 distinct days with at least one booking.

Step 3 — Count days with no booking in Sep 1–14

The window Sep 1 to Sep 14 inclusive contains 14 days. Of these, 8 had at least one booking, so the number of days with no booking is:

$14 - 8 = 6$

The six booking-free days are: Sep 8, Sep 9, Sep 10, Sep 11, Sep 12, Sep 14.

Answer: 6

We need the average processing time of all orders belonging to categories that used only one type of ship mode during the period.

Step 1 — Identify categories that use only one ship mode

For each category, check whether all its orders share the same ship mode. Here is the relevant slice:

Only Accessories and Envelopes qualify — each has a single order, both shipped Standard Class.

Step 2 — Find the processing time for each qualifying order

Step 3 — Compute the average

$\text{Average} = \frac{18 + 4}{2} = \frac{22}{2} = 11 \text{ days}$

💡 Teacher tip: The question says "all orders in the categories which had only one type of ship mode" — so we average over the orders (2 orders here), not over the categories. Since each qualifying category has exactly one order, both interpretations give the same answer here, but it is good practice to confirm.

Answer: 11

We need to arrange Art, Binders, Paper, and Phones in decreasing order of average processing time of their orders.

Step 1 — Extract processing times for the four categories

From the chart, each bar's processing time = dispatch day − booking day. Here are the orders for the four categories in question:

Step 2 — Compute the average for each category

Step 3 — Rank in decreasing order

Comparing the averages:

$\text{Phones }(5) > \text{Art }(4) > \text{Binders }(3.75) > \text{Paper }(3.33)$

💡 Teacher tip: Phones has a single very long order (12 days) that pulls its average up despite having two short orders. Always check whether an outlier is driving the average.

The decreasing order is Phones, Art, Binders, Paper, which matches Choice B.

Answer: Choice B — Phones, Art, Binders, Paper

We need to find approximately what percentage of all orders had a processing time of exactly 1 day during Sep 1 to Sep 22.

Step 1 — Confirm the total number of orders

Every bar in the chart represents one order, and all 35 bars lie wholly within Sep 1–22. So the total number of orders is 35.

Step 2 — Identify all orders with a processing time of exactly 1 day

Processing time = dispatch day − booking day. Scanning every bar, the orders where this difference equals 1 are:

That gives 7 orders with a processing time of exactly 1 day.

💡 Teacher tip: A quick way to spot these on the chart: a 1-day processing time means the bar is just one day wide — its left and right endpoints are on consecutive gridlines. Scan for the shortest bars in each category row.

Step 3 — Compute the percentage

$\frac{7}{35} \times 100\% = \frac{1}{5} \times 100\% = 20\%$

Answer: Choice C — 20%

Setting up the championship structure

Before answering this specific question, we need to solve the tournament. Here is the key mechanic:

• Within each group: Rank 2 plays Rank 3 in Game 1. The winner of Game 1 plays Rank 1 in Game 2. The winner of Game 2 is the group winner.
• So Rank 1 plays only 1 group game (Game 2).
• A Rank 2/3 player who wins Game 1 and then plays Game 2 plays 2 group games.
• Knockouts: SF1 = Group A winner vs Group B winner; SF2 = Group C winner vs Group D winner; Final = SF1 winner vs SF2 winner.

Step 1 — Total games check

There are 8 group games + 2 semi-finals + 1 final = 11 games, so 22 player-game appearances. From the clue: $4 \times 3 + 2 \times 2 + 6 \times 1 = 22$ ✓.

Step 2 — Which groups' winners reached the final?

Each group stage always accounts for 4 player-games (2 games × 2 players). The group winner then plays 1 semi-final, plus 1 more game only if they reach the final.

So the final was Group A winner vs Group C winner.

Step 3 — Identify the finalists and their ranks

Chitra (Group C's representative) played 3 games and reached the final.

• If Chitra were Rank 2 or 3: she would play Game 1 (1 game) + Game 2 (2 games) + SF2 (3 games) + Final (4 games) = 4 games. But she only played 3.
• If Chitra were Rank 1: she plays Game 2 (1) + SF2 (2) + Final (3) = 3 games ✓.

So Chitra is Rank 1 of Group C. Since Chitra did not win the championship (clue 1), she lost the final. The champion is Group A's winner = Aruna.

Aruna (Group A's representative) played 3 games and won the final.

• If Aruna were Rank 2/3: she'd need 2 group games + SF1 + Final (won) = 4 games. Too many.
• If Aruna were Rank 1: Game 2 (1) + SF1 (2) + Final won (3) = 3 games ✓.

So Aruna is Rank 1 of Group A.

Step 4 — Fill in each group's internal results

• Group A: Aruna (Rank 1) plays only Game 2. Clue 2 says Aruna did not play Arif, so Arif didn't win Game 1. Azul won Game 1 (vs Arif), then lost Game 2 to Aruna. Azul = 2 games ✓, Arif = 1 game ✓.
• Group C: Chitra (Rank 1) plays only Game 2. Chetan (2 games) won Game 1 (vs Chhavi), then lost Game 2 to Chitra. Chhavi = 1 game ✓.
• Group B: Biju (3 games) is Rank 2/3 — won Game 1, won Game 2, lost SF1 (2+1=3). Brinda and Brij each played 1 game. Clue 2 (Brij didn't play Brinda) means they were not the Game 1 pair, so one is Rank 1 and the other lost Game 1 to Biju. Biju is Rank 2 or 3 — definitely not Rank 1.
• Group D: Dipen (3 games) is Rank 2/3 — won Game 1, won Game 2, lost SF2 (2+1=3). Donna and Deb are Rank 1 and Game-1 loser in some order. Dipen is Rank 2 or 3 — definitely not Rank 1.

Solved championship table

Knockouts: SF1 — Aruna beats Biju · SF2 — Chitra beats Dipen · Final — Aruna beats Chitra.

Answering this question

We need someone who was definitely NOT Rank 1:

• Dipen — Rank 2 or 3 (had to win Game 1 first). Definitely not Rank 1. ✓
• Aruna — Rank 1 of Group A. ✗
• Brij — could be Rank 1 of Group B. ✗
• Chitra — Rank 1 of Group C. ✗

Answer: Choice A — Dipen

Setting up the championship structure

Key mechanic recap:

• Within each group: Rank 2 plays Rank 3 (Game 1); the winner plays Rank 1 (Game 2). The Game 2 winner is the group winner.
• Knockouts: SF1 = Group A winner vs Group B winner; SF2 = Group C winner vs Group D winner; Final = SF1 winner vs SF2 winner.

Step 1 — Which groups' winners reached the final?

Each group stage uses exactly 4 player-games. The group winner adds 1 for the semi-final, and 1 more only if they reach the final.

So SF1 was won by Group A's winner and SF2 by Group C's winner; the final was A's winner vs C's winner.

Step 2 — Identify the finalists

Chitra played 3 games and reached the final. If she were Rank 2/3, she'd need 2 group games + SF2 + Final = 4 games. Since she only played 3, she must be Rank 1 (1 group game + SF2 + Final = 3). Clue 1 says Chitra didn't win the championship, so she lost the final.

Aruna played 3 games and is Group A's representative. If she were Rank 2/3 and won the final, she'd need 4 games. So she must be Rank 1 (1 + 1 + 1 = 3) and is the champion.

Step 3 — Fill in group results

• Group A: Aruna (Rank 1) plays only Game 2. Clue 2 (Aruna didn't play Arif) means Arif didn't win Game 1 — Azul did. Azul lost Game 2 to Aruna.
• Group C: Chitra (Rank 1) plays only Game 2. Chetan (2 games) won Game 1 (vs Chhavi), then lost Game 2 to Chitra.
• Group B: Biju (Rank 2/3, 3 games) won both group games, lost SF1. Brinda/Brij are Rank 1 and Game-1 loser.
• Group D: Dipen (Rank 2/3, 3 games) won both group games, lost SF2. Donna/Deb are Rank 1 and Game-1 loser.

Solved championship table

Knockouts: SF1 — Aruna beats Biju · SF2 — Chitra beats Dipen · Final — Aruna beats Chitra.

Answering this question

Cross-group matches happen only in the knockouts:

• SF1: Aruna vs Biju
• SF2: Chitra vs Dipen
• Final: Aruna vs Chitra

Now checking each option:

• A. Deb, Donna — both in Group D, but they never play each other (Dipen plays each of them separately). ✗
• B. Azul, Biju — Azul never left Group A; Biju never left SF1. They never met. ✗
• C. Donna, Chetan — different groups, neither advanced to knockouts. Never met. ✗
• D. Chitra, Dipen — met in SF2 (Chitra beat Dipen). ✓

Answer: Choice D — Chitra, Dipen

Setting up the championship structure

Key mechanic recap:

• Within each group: Rank 2 plays Rank 3 (Game 1); the winner plays Rank 1 (Game 2). The Game 2 winner is the group winner.
• Knockouts: SF1 = Group A winner vs Group B winner; SF2 = Group C winner vs Group D winner; Final = SF1 winner vs SF2 winner.

Step 1 — Which groups' winners reached the final?

Each group stage uses exactly 4 player-games. The group winner adds 1 for the semi-final, and 1 more only if they reach the final.

So the final was Group A winner vs Group C winner.

Step 2 — Who are the finalists?

Chitra played 3 games and is Group C's representative in the final.

• If Chitra were Rank 2/3: 2 group games + SF2 + Final = 4 games. But she played only 3.
• If Chitra were Rank 1: 1 group game + SF2 + Final = 3 games ✓.

So Chitra is Rank 1 of Group C and reached the final.

💡 Teacher tip: The number of games a group winner plays depends on their rank. A Rank 1 winner plays 1 group game + knockouts; a Rank 2/3 winner plays 2 group games + knockouts. This difference is what lets us pin down ranks from the games-played counts.

Aruna played 3 games and is Group A's representative in the final.

• If Aruna were Rank 2/3 and won the final: 2 + 1 + 1 = 4 games. Too many.
• If Aruna were Rank 1: 1 + 1 + 1 = 3 games ✓.

So Aruna is Rank 1 of Group A and reached the final.

Step 3 — Who won the final?

Clue 1 states Chitra did not win the championship. Since the final was Aruna vs Chitra, and Chitra didn't win:

Aruna won the final and is the champion.

Verification — full group results

Knockouts: SF1 — Aruna beats Biju · SF2 — Chitra beats Dipen · Final — Aruna beats Chitra.

Games-per-player check: Aruna 3, Azul 2, Arif 1, Biju 3, Brinda 1, Brij 1, Chitra 3, Chetan 2, Chhavi 1, Dipen 3, Donna 1, Deb 1 ✓

Answer: Choice B — Aruna

Setting up the championship structure

Key mechanic recap:

• Within each group: Rank 2 plays Rank 3 (Game 1); the winner plays Rank 1 (Game 2). The Game 2 winner is the group winner.
• Knockouts: SF1 = Group A winner vs Group B winner; SF2 = Group C winner vs Group D winner; Final = SF1 winner vs SF2 winner.

Step 1 — Which groups' winners reached the final?

Each group stage uses exactly 4 player-games. The group winner adds 1 for the semi-final, and 1 more only if they reach the final.

So the final was Group A winner vs Group C winner.

Step 2 — Identify Chitra's rank and games

Chitra played 3 games and is Group C's representative in the final.

• If Rank 2/3: 2 group games + SF2 + Final = 4 games. Too many.
• If Rank 1: 1 group game (Game 2) + SF2 + Final = 3 games ✓.

So Chitra is Rank 1 of Group C. As Rank 1, she plays only Game 2 in the group stage.

Step 3 — Fill in Group C's internal results

Chitra (Rank 1) plays only Game 2. The two other players (Chetan, Chhavi) play Game 1. Chetan played 2 games (Game 1 + Game 2), so Chetan won Game 1 (vs Chhavi) and then lost Game 2 to Chitra. Chhavi played 1 game (lost Game 1).

Step 4 — Full solved championship

Knockouts: SF1 — Aruna beats Biju · SF2 — Chitra beats Dipen · Final — Aruna beats Chitra.

Answering this question

Chitra played exactly 3 games. Let's list her opponents:

1. Game 2 of Group C — Chitra vs Chetan (Chitra won)
2. SF2 — Chitra vs Dipen (Chitra won)
3. Final — Chitra vs Aruna (Chitra lost)

So Chitra played against Chetan, Dipen, and Aruna.

Now checking the options:

• A. Aruna — played Chitra in the final. ✗
• B. Chetan — played Chitra in Group C Game 2. ✗
• C. Dipen — played Chitra in SF2. ✗
• D. Biju — Biju only reached SF1 (lost to Aruna) and never crossed into Chitra's side of the draw. Biju never played Chitra. ✓

Answer: Choice D — Biju

We need to fill in the entire rating grid before we can identify the single smallest rating across all customers and parameters. Let us build it step by step from the clues.

Step 1 — Packaging

Fact 2 says the four Packaging ratings are distinct integers from 1–9 summing to 29. The largest possible sum of four distinct values from 1–9 is $9+8+7+6=30$. To get 29 we must drop the sum by exactly 1, which means replacing 6 with 5. So the Packaging set is {5, 7, 8, 9}.

Fact 4 says Bihari gave the highest Packaging rating, so Bihari = 9. The remaining three — Atal, Chirag, Deepak — share {5, 7, 8} in some order.

Step 2 — Hygiene

Fact 2 says the four Hygiene ratings are distinct and sum to 26. Fact 3 says Chirag's Hygiene equals his Packaging, so Chirag's common value is one of {5, 7, 8}. Fact 4 says Chirag gave the highest Hygiene rating, so his value must be the maximum of the Hygiene set.

We list the four-distinct-value sets from 1–9 that sum to 26:

• {2, 7, 8, 9} — max 9 (not in {5,7,8})
• {3, 6, 8, 9} — max 9 (not in {5,7,8})
• {4, 5, 8, 9} — max 9 (not in {5,7,8})
• {4, 6, 7, 9} — max 9 (not in {5,7,8})
• {5, 6, 7, 8} — max 8 ✓

Only {5, 6, 7, 8} has a maximum (8) that Chirag can also hold in Packaging. So Hygiene = {5, 6, 7, 8}, Chirag = 8 in Hygiene, and therefore Chirag = 8 in Packaging as well.

Step 3 — Behaviour

Fact 5 says every Behaviour rating is between 5 and 7 (inclusive), with Atal giving the unique maximum and Deepak the unique minimum. The only way to place four values in {5, 6, 7} with a unique max and unique min is:

• Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6

Step 4 — Timeliness (partially)

Fact 1 says the Timeliness total is 21, three customers gave the same rating, and Atal rated higher than both Bihari and Chirag. Since Atal outranks Bihari and Chirag, he cannot be part of the equal trio. So the three equal raters are Bihari, Chirag, and Deepak, and Atal is the odd one.

Let the common value be $x$ and Atal's value be $y$, with $y > x$: $3x + y = 21$ The valid pairs are $(x, y) = (5, 6)$ or $(4, 9)$. We resolve this using the money total in Step 6.

Step 5 — Splitting Packaging and Hygiene among Atal, Deepak

After Step 2, Packaging remaining for Atal and Deepak is {5, 7} (since Chirag = 8, Bihari = 9). Hygiene remaining for Atal, Bihari, Deepak is {5, 6, 7} (since Chirag = 8).

Fact 6 says Atal's rank in Packaging equals his rank in Hygiene, and the same holds for Deepak.

Test: Atal Packaging = 5, Deepak Packaging = 7.

• In Packaging, Atal's 5 is the lowest → rank 4. Deepak's 7 is third-highest → rank 3.
• For rank-matching, Atal needs Hygiene rank 4 (lowest = 5) and Deepak needs Hygiene rank 3. Hygiene values left are {5, 6, 7}. If Atal = 5 (rank 4) and Deepak = 6 (rank 3), then Bihari = 7 (rank 2). This is consistent. ✓

The other assignment (Atal Packaging = 7, Deepak Packaging = 5) does not produce matching ranks, so we take:

• Atal Packaging = 5, Deepak Packaging = 7
• Atal Hygiene = 5, Deepak Hygiene = 6, Bihari Hygiene = 7

Step 6 — Resolving Timeliness via the ₹120 total

Summing Behaviour + Packaging + Hygiene for each customer:

Now test the two Timeliness options:

Option A — (Atal, Bihari, Chirag, Deepak) = (9, 4, 4, 4): Totals become Atal 26, Bihari 26, Chirag 26, Deepak 22. Three customers cross 25 → three bonuses = ₹60. Only Deepak (total 22) can tip, and the max tip is ₹50. Maximum money = ₹60 + ₹50 = ₹110 ≠ ₹120. ✗

Option B — (6, 5, 5, 5): Totals become Atal 23, Bihari 27, Chirag 27, Deepak 23. Two customers (Bihari, Chirag) cross 25 → two bonuses = ₹40. Atal and Deepak (no bonus) must tip ₹80 together, which is ₹50 + ₹30. Total = ₹40 + ₹80 = ₹120. ✓

So Timeliness: Atal = 6, Bihari = Chirag = Deepak = 5.

Final rating grid

Answering this question

We scan every cell of the final grid for the smallest value. The value 5 appears in Deepak's Behaviour, Atal's Packaging, Atal's Hygiene, and the Timeliness ratings of Bihari, Chirag, and Deepak. No rating anywhere is below 5.

Answer: 5

We need the complete rating grid to find who gave the highest total. Let us build it from the clues.

Step 1 — Packaging

Four distinct values from 1–9 summing to 29. The maximum possible sum is $9+8+7+6=30$, so to reach 29 we replace 6 with 5: the set is {5, 7, 8, 9}. Fact 4 (Bihari highest in Packaging) gives Bihari = 9; Atal, Chirag, Deepak share {5, 7, 8}.

Step 2 — Hygiene

Four distinct values summing to 26, with Chirag highest (Fact 4) and Chirag's Hygiene = Chirag's Packaging (Fact 3, so his value is in {5, 7, 8}). Checking all four-distinct sets summing to 26, only {5, 6, 7, 8} has a maximum of 8 — which Chirag can hold. So Hygiene = {5, 6, 7, 8}, Chirag = 8 in both Hygiene and Packaging.

Step 3 — Behaviour

All values in {5, 6, 7} (Fact 5), with Atal the unique max and Deepak the unique min. The only fit: Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6.

Step 4 — Timeliness (partially)

Total 21, three equal, Atal > Bihari and Atal > Chirag (Fact 1). Atal cannot be in the equal trio, so Bihari, Chirag, Deepak share value $x$ and Atal has $y > x$: $3x + y = 21$, giving $(x,y) = (5,6)$ or $(4,9)$. Resolved in Step 6.

Step 5 — Assigning Packaging and Hygiene to Atal and Deepak

Packaging left for Atal and Deepak: {5, 7}. Hygiene left for Atal, Bihari, Deepak: {5, 6, 7}.

Fact 6 requires Atal's Packaging rank = his Hygiene rank, and Deepak's Packaging rank = his Hygiene rank.

• Atal Packaging = 5 (rank 4 in Packaging) → Atal Hygiene must also be rank 4 → Atal Hygiene = 5. Deepak Packaging = 7 (rank 3) → Deepak Hygiene must be rank 3 → Deepak Hygiene = 6. Then Bihari Hygiene = 7. ✓

The reverse assignment breaks the rank-matching, so we keep this one.

Step 6 — Resolving Timeliness via ₹120

Behaviour + Packaging + Hygiene subtotals: Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

• (9, 4, 4, 4): Totals → 26, 26, 26, 22. Three bonuses (₹60) + max one tip (₹50) = ₹110 ≠ ₹120. ✗
• (6, 5, 5, 5): Totals → 23, 27, 27, 23. Two bonuses (₹40) + two tips (₹50 + ₹30 = ₹80) = ₹120. ✓

Final rating grid

Answering this question

From the Total column: Atal = 23, Bihari = 27, Chirag = 27, Deepak = 23. The maximum total is 27, achieved by both Bihari and Chirag.

Answer: Choice C — Bihari and Chirag

To find Atal's Timeliness rating, we need to resolve the full grid — particularly the Timeliness column, which has two candidates that can only be distinguished using the ₹120 money total.

Step 1 — Packaging

Four distinct values from 1–9 summing to 29. Since $9+8+7+6=30$, the only set summing to 29 is {5, 7, 8, 9}. Bihari is highest (Fact 4) → Bihari = 9. Atal, Chirag, Deepak share {5, 7, 8}.

Step 2 — Hygiene

Four distinct values summing to 26. Chirag is highest (Fact 4) and his Hygiene equals his Packaging (Fact 3), so his value is in {5, 7, 8}. Of all four-distinct sets summing to 26, only {5, 6, 7, 8} has a maximum (8) within {5, 7, 8}. So Hygiene = {5, 6, 7, 8}, Chirag = 8 in both Hygiene and Packaging.

Step 3 — Behaviour

All values in {5, 6, 7} (Fact 5), Atal unique max, Deepak unique min → Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6.

Step 4 — Timeliness candidates

Total 21, three equal, Atal > Bihari and Atal > Chirag (Fact 1). Atal cannot be in the equal trio, so Bihari, Chirag, Deepak each gave $x$ and Atal gave $y > x$: $3x + y = 21 \implies (x, y) = (5, 6) \text{ or } (4, 9)$

Step 5 — Packaging and Hygiene assignments

Packaging left for Atal, Deepak: {5, 7}. Hygiene left for Atal, Bihari, Deepak: {5, 6, 7}.

Fact 6 (rank-matching for Atal and Deepak across Packaging and Hygiene):

• Atal Packaging = 5 (rank 4) ↔ Atal Hygiene = 5 (rank 4) ✓
• Deepak Packaging = 7 (rank 3) ↔ Deepak Hygiene = 6 (rank 3) ✓
• Bihari Hygiene = 7

Step 6 — Resolving Timeliness with the ₹120 total

Subtotals (Behaviour + Packaging + Hygiene): Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

Only (6, 5, 5, 5) reaches ₹120. So Atal's Timeliness = 6.

Final rating grid

Answer: 6

To determine what we can conclude about Deepak's tip, we need the full grid — especially which customers gave bonuses and which gave tips.

Step 1 — Packaging

Four distinct values from 1–9 summing to 29. Since $9+8+7+6=30$, the only set is {5, 7, 8, 9}. Bihari highest (Fact 4) → Bihari = 9. Atal, Chirag, Deepak share {5, 7, 8}.

Step 2 — Hygiene

Four distinct values summing to 26. Chirag is highest (Fact 4) and his Hygiene = his Packaging (Fact 3), so his value is in {5, 7, 8}. Only {5, 6, 7, 8} (max = 8) works. So Chirag = 8 in both Hygiene and Packaging.

Step 3 — Behaviour

All in {5, 6, 7}, Atal unique max, Deepak unique min → Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6.

Step 4 — Timeliness candidates

Total 21, three equal, Atal > Bihari and Atal > Chirag. The equal trio is Bihari, Chirag, Deepak (value $x$); Atal has $y > x$. So $3x + y = 21$, giving $(x,y) = (5,6)$ or $(4,9)$.

Step 5 — Packaging and Hygiene assignments via Fact 6

Packaging left for Atal, Deepak: {5, 7}. Hygiene left for Atal, Bihari, Deepak: {5, 6, 7}.

Rank-matching (Fact 6):

• Atal Packaging = 5 (rank 4) ↔ Atal Hygiene = 5 (rank 4) ✓
• Deepak Packaging = 7 (rank 3) ↔ Deepak Hygiene = 6 (rank 3) ✓
• Bihari Hygiene = 7

Step 6 — Resolving Timeliness and money

Subtotals (B + P + H): Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

• (9, 4, 4, 4): Totals → 26, 26, 26, 22. Three bonuses = ₹60; only Deepak can tip (≤ ₹50). Max = ₹110 ≠ ₹120. ✗
• (6, 5, 5, 5): Totals → 23, 27, 27, 23. Two bonuses (Bihari, Chirag) = ₹40. Atal and Deepak (no bonus) must tip ₹80 = ₹50 + ₹30. Total = ₹120. ✓

Final rating grid

Answering this question

Deepak's total is 23, which is below the 25-point bonus threshold, so he did not give a bonus — he gave a tip instead (since no customer gives both). We know Atal and Deepak together tipped ₹80, split as ₹50 + ₹30. However, nothing in the clues tells us which of the two gave ₹50 and which gave ₹30. Both assignments are equally valid.

💡 Teacher tip: When the data supports two indistinguishable arrangements, the answer is "either value" — not a single fixed value.

Therefore, Deepak's tip is either ₹30 or ₹50.

Answer: Choice B — Either ₹30 or ₹50

We need Atal's four ratings to identify which parameter he rated highest. Let us build the full grid.

Step 1 — Packaging

Four distinct values from 1–9 summing to 29. Since $9+8+7+6=30$, the only set is {5, 7, 8, 9}. Bihari highest (Fact 4) → Bihari = 9. Atal, Chirag, Deepak share {5, 7, 8}.

Step 2 — Hygiene

Four distinct values summing to 26. Chirag is highest (Fact 4) and his Hygiene = his Packaging (Fact 3), so his value is in {5, 7, 8}. Only {5, 6, 7, 8} (max = 8) works. So Chirag = 8 in both Hygiene and Packaging.

Step 3 — Behaviour

All in {5, 6, 7}, Atal unique max, Deepak unique min → Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6.

Step 4 — Timeliness candidates

Total 21, three equal, Atal > Bihari and Atal > Chirag. Equal trio: Bihari, Chirag, Deepak (value $x$); Atal has $y > x$. So $3x + y = 21$, giving $(x,y) = (5,6)$ or $(4,9)$.

Step 5 — Packaging and Hygiene assignments via Fact 6

Packaging left for Atal, Deepak: {5, 7}. Hygiene left for Atal, Bihari, Deepak: {5, 6, 7}.

Rank-matching (Fact 6):

• Atal Packaging = 5 (rank 4) ↔ Atal Hygiene = 5 (rank 4) ✓
• Deepak Packaging = 7 (rank 3) ↔ Deepak Hygiene = 6 (rank 3) ✓
• Bihari Hygiene = 7

Step 6 — Resolving Timeliness via ₹120

Subtotals (B + P + H): Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

• (9, 4, 4, 4): Totals → 26, 26, 26, 22. Three bonuses = ₹60 + max one tip ₹50 = ₹110 ≠ ₹120. ✗
• (6, 5, 5, 5): Totals → 23, 27, 27, 23. Two bonuses = ₹40 + two tips (₹50 + ₹30) = ₹80 → ₹120. ✓

Final rating grid

Answering this question

Atal's four ratings are:

The highest is 7 in Behaviour.

Answer: Choice B — Behaviour

We need Deepak's Packaging rating. This is determined in the early steps of the deduction, but let us build the full grid to be certain.

Step 1 — Packaging

Four distinct values from 1–9 summing to 29. Since $9+8+7+6=30$, the only set is {5, 7, 8, 9}. Bihari highest (Fact 4) → Bihari = 9. Atal, Chirag, Deepak share {5, 7, 8}.

Step 2 — Hygiene

Four distinct values summing to 26. Chirag is highest (Fact 4) and his Hygiene = his Packaging (Fact 3), so his value is in {5, 7, 8}. Only {5, 6, 7, 8} (max = 8) works. So Chirag = 8 in both Hygiene and Packaging.

Now Packaging remaining for Atal and Deepak: {5, 7}.

Step 3 — Behaviour

All in {5, 6, 7}, Atal unique max, Deepak unique min → Atal = 7, Deepak = 5, Bihari = 6, Chirag = 6.

Step 4 — Timeliness candidates

Total 21, three equal, Atal > Bihari and Atal > Chirag. Equal trio: Bihari, Chirag, Deepak (value $x$); Atal has $y > x$. So $3x + y = 21$, giving $(x,y) = (5,6)$ or $(4,9)$.

Step 5 — Splitting Packaging between Atal and Deepak (Fact 6)

Fact 6 requires Atal's Packaging rank = his Hygiene rank, and Deepak's Packaging rank = his Hygiene rank.

Hygiene remaining for Atal, Bihari, Deepak: {5, 6, 7}.

Test: Atal Packaging = 5, Deepak Packaging = 7.

• Packaging ranks: Atal's 5 is lowest → rank 4; Deepak's 7 is third → rank 3.
• For matching Hygiene ranks: Atal needs rank 4 → Hygiene = 5; Deepak needs rank 3 → Hygiene = 6; Bihari gets Hygiene = 7 (rank 2). ✓ Consistent.

The reverse (Atal = 7, Deepak = 5) fails the rank-matching, so:

💡 Teacher tip: Fact 6 is the key that locks the Packaging and Hygiene assignments. Always test both splits against the rank condition.

Deepak Packaging = 7.

Step 6 — Confirming with Timeliness and money

Subtotals (B + P + H): Atal = 17, Bihari = 22, Chirag = 22, Deepak = 18.

• (9, 4, 4, 4): Three bonuses + one tip ≤ ₹110 ≠ ₹120. ✗
• (6, 5, 5, 5): Two bonuses (₹40) + two tips (₹80) = ₹120. ✓

Final rating grid

Answering this question

From Step 5 and the final grid, Deepak's Packaging rating is 7.

Answer: Choice A — 7