CAT 2023 DILR Questions — Slot 1

Twelve houses sit in two rows and six columns. Block XX is columns A, B, C; Block YY is columns D, E, F. One road runs vertically between column C and column D; another runs horizontally along the bottom of Row-2.

Road-adjacency value = number of a house's sides touching a road:

Neighbour count = number of a house's sides touching occupied houses in the same block (a road or the block boundary never counts). For a vacant house, quoted price $= \text{base} + 5 \times (\text{road value}) + 3 \times (\text{neighbour count})$, where base $= 10$ without parking or $12$ with parking.

Pinning Block XX from "highest price in XX = 24". Testing each house, the only vacant one that can reach exactly 24 is B2: road value 1, no parking, price $= 10 + 5 + 3n = 15 + 3n$, which is 24 only when $n = 3$ — so all three of B2's block-neighbours A2, C2, B1 are occupied. Row-1 has just one occupied house per block, so B1 is XX's occupied Row-1 house and A1, C1 are vacant.

Pinning Block YY from "lowest price = 15, reached in column E". E1 and E2 are both vacant. Block YY has exactly one house with parking, and columns D and F each need an occupied house. Row-1 has one occupied YY house, and since E1 is vacant it is D1 or F1.

• If D1 were the occupied one, F1 would be vacant; then E1 and F1 are both low-road-value vacant houses worth $\text{base} + 3$, and with only one parking to go round, at least one of them is $10 + 3 = 13 < 15$ — breaking the "lowest = 15" rule. So this case fails.
• Hence F1 is occupied, D1 is vacant, and column D's occupied house is D2.

E1's neighbours are D1 (vacant), F1 (occupied), E2 (vacant), so its neighbour count is 1 and its price is $\text{base} + 3$. For the lowest price to be 15, E1 must carry the parking: $12 + 3 = 15$. That uses up YY's single parking, so every other vacant YY house has base 10.

Solved grid (✔ occupied, ∘ vacant; F2 is the one house the clues leave open):

Definitely occupied: B1, A2, C2, F1, D2. Definitely vacant: A1, C1, B2, D1, E1, E2. Only F2 is undetermined.

This question. Block XX is A1, B1, C1, A2, B2, C2. The occupied ones are B1, A2, C2; the vacant ones are A1, C1, B2 — that is 3 vacant houses.

💡 Teacher tip: The single number "24" cracks the whole block: only B2 can reach it, and only when all three of its neighbours are occupied.

Answer: 3

Twelve houses sit in two rows and six columns. Block XX is columns A, B, C; Block YY is columns D, E, F. One road runs vertically between column C and column D; another runs horizontally along the bottom of Row-2.

Road-adjacency value = number of a house's sides touching a road:

Neighbour count = number of a house's sides touching occupied houses in the same block (a road or the block boundary never counts). For a vacant house, quoted price $= \text{base} + 5 \times (\text{road value}) + 3 \times (\text{neighbour count})$, where base $= 10$ without parking or $12$ with parking.

Pinning Block XX from "highest price in XX = 24". Testing each house, the only vacant one that can reach exactly 24 is B2: road value 1, no parking, price $= 10 + 5 + 3n = 15 + 3n$, which is 24 only when $n = 3$ — so all three of B2's block-neighbours A2, C2, B1 are occupied. Row-1 has just one occupied house per block, so B1 is XX's occupied Row-1 house and A1, C1 are vacant.

Pinning Block YY from "lowest price = 15, reached in column E". E1 and E2 are both vacant. Block YY has exactly one house with parking, and columns D and F each need an occupied house. Row-1 has one occupied YY house, and since E1 is vacant it is D1 or F1.

• If D1 were the occupied one, F1 would be vacant; then E1 and F1 are both low-road-value vacant houses worth $\text{base} + 3$, and with only one parking to go round, at least one of them is $10 + 3 = 13 < 15$ — breaking the "lowest = 15" rule. So this case fails.
• Hence F1 is occupied, D1 is vacant, and column D's occupied house is D2.

E1's neighbours are D1 (vacant), F1 (occupied), E2 (vacant), so its neighbour count is 1 and its price is $\text{base} + 3$. For the lowest price to be 15, E1 must carry the parking: $12 + 3 = 15$. That uses up YY's single parking, so every other vacant YY house has base 10.

Solved grid (✔ occupied, ∘ vacant; F2 is the one house the clues leave open):

Definitely occupied: B1, A2, C2, F1, D2. Definitely vacant: A1, C1, B2, D1, E1, E2. Only F2 is undetermined.

This question. Check the four options against the grid: A1 is vacant, F2 is undetermined, D2 is occupied and B1 is occupied. Of the listed options, B1 is definitely occupied (it is the occupied neighbour that lets B2 reach 24, and it is Row-1's XX occupant).

Answer: B1

Twelve houses sit in two rows and six columns. Block XX is columns A, B, C; Block YY is columns D, E, F. One road runs vertically between column C and column D; another runs horizontally along the bottom of Row-2.

Road-adjacency value = number of a house's sides touching a road:

Neighbour count = number of a house's sides touching occupied houses in the same block (a road or the block boundary never counts). For a vacant house, quoted price $= \text{base} + 5 \times (\text{road value}) + 3 \times (\text{neighbour count})$, where base $= 10$ without parking or $12$ with parking.

Pinning Block XX from "highest price in XX = 24". Testing each house, the only vacant one that can reach exactly 24 is B2: road value 1, no parking, price $= 10 + 5 + 3n = 15 + 3n$, which is 24 only when $n = 3$ — so all three of B2's block-neighbours A2, C2, B1 are occupied. Row-1 has just one occupied house per block, so B1 is XX's occupied Row-1 house and A1, C1 are vacant.

Pinning Block YY from "lowest price = 15, reached in column E". E1 and E2 are both vacant. Block YY has exactly one house with parking, and columns D and F each need an occupied house. Row-1 has one occupied YY house, and since E1 is vacant it is D1 or F1.

• If D1 were the occupied one, F1 would be vacant; then E1 and F1 are both low-road-value vacant houses worth $\text{base} + 3$, and with only one parking to go round, at least one of them is $10 + 3 = 13 < 15$ — breaking the "lowest = 15" rule. So this case fails.
• Hence F1 is occupied, D1 is vacant, and column D's occupied house is D2.

E1's neighbours are D1 (vacant), F1 (occupied), E2 (vacant), so its neighbour count is 1 and its price is $\text{base} + 3$. For the lowest price to be 15, E1 must carry the parking: $12 + 3 = 15$. That uses up YY's single parking, so every other vacant YY house has base 10.

Solved grid (✔ occupied, ∘ vacant; F2 is the one house the clues leave open):

Definitely occupied: B1, A2, C2, F1, D2. Definitely vacant: A1, C1, B2, D1, E1, E2. Only F2 is undetermined.

This question. Row-2 is A2, B2, C2, D2, E2, F2. From the grid A2, C2, D2 are occupied; B2 and E2 are vacant; F2 is the one open house.

• If F2 is occupied, the vacant Row-2 houses are B2, E2 — 2.
• If F2 is vacant, they are B2, E2, F2 — 3.

So the count is either 2 or 3.

Answer: Either 2 or 3

Twelve houses sit in two rows and six columns. Block XX is columns A, B, C; Block YY is columns D, E, F. One road runs vertically between column C and column D; another runs horizontally along the bottom of Row-2.

Road-adjacency value = number of a house's sides touching a road:

Neighbour count = number of a house's sides touching occupied houses in the same block (a road or the block boundary never counts). For a vacant house, quoted price $= \text{base} + 5 \times (\text{road value}) + 3 \times (\text{neighbour count})$, where base $= 10$ without parking or $12$ with parking.

Pinning Block XX from "highest price in XX = 24". Testing each house, the only vacant one that can reach exactly 24 is B2: road value 1, no parking, price $= 10 + 5 + 3n = 15 + 3n$, which is 24 only when $n = 3$ — so all three of B2's block-neighbours A2, C2, B1 are occupied. Row-1 has just one occupied house per block, so B1 is XX's occupied Row-1 house and A1, C1 are vacant.

Pinning Block YY from "lowest price = 15, reached in column E". E1 and E2 are both vacant. Block YY has exactly one house with parking, and columns D and F each need an occupied house. Row-1 has one occupied YY house, and since E1 is vacant it is D1 or F1.

• If D1 were the occupied one, F1 would be vacant; then E1 and F1 are both low-road-value vacant houses worth $\text{base} + 3$, and with only one parking to go round, at least one of them is $10 + 3 = 13 < 15$ — breaking the "lowest = 15" rule. So this case fails.
• Hence F1 is occupied, D1 is vacant, and column D's occupied house is D2.

E1's neighbours are D1 (vacant), F1 (occupied), E2 (vacant), so its neighbour count is 1 and its price is $\text{base} + 3$. For the lowest price to be 15, E1 must carry the parking: $12 + 3 = 15$. That uses up YY's single parking, so every other vacant YY house has base 10.

Solved grid (✔ occupied, ∘ vacant; F2 is the one house the clues leave open):

Definitely occupied: B1, A2, C2, F1, D2. Definitely vacant: A1, C1, B2, D1, E1, E2. Only F2 is undetermined.

This question. Column E has E1 and E2, both vacant.

• E1: road value 0, parking (base 12), neighbour count 1 (only F1 occupied) — price $= 12 + 0 + 3 = 15$, fixed.
• E2: road value 1, base 10 (parking is used up by E1), neighbours D2 (occupied), F2 (open), E1 (vacant). If F2 is occupied the neighbour count is 2, giving $10 + 5 + 6 = 21$; if F2 is vacant it is $18$.

The largest possible is E2 = 21 (when F2 is occupied).

Answer: 21

Twelve houses sit in two rows and six columns. Block XX is columns A, B, C; Block YY is columns D, E, F. One road runs vertically between column C and column D; another runs horizontally along the bottom of Row-2.

Road-adjacency value = number of a house's sides touching a road:

Neighbour count = number of a house's sides touching occupied houses in the same block (a road or the block boundary never counts). For a vacant house, quoted price $= \text{base} + 5 \times (\text{road value}) + 3 \times (\text{neighbour count})$, where base $= 10$ without parking or $12$ with parking.

Pinning Block XX from "highest price in XX = 24". Testing each house, the only vacant one that can reach exactly 24 is B2: road value 1, no parking, price $= 10 + 5 + 3n = 15 + 3n$, which is 24 only when $n = 3$ — so all three of B2's block-neighbours A2, C2, B1 are occupied. Row-1 has just one occupied house per block, so B1 is XX's occupied Row-1 house and A1, C1 are vacant.

Pinning Block YY from "lowest price = 15, reached in column E". E1 and E2 are both vacant. Block YY has exactly one house with parking, and columns D and F each need an occupied house. Row-1 has one occupied YY house, and since E1 is vacant it is D1 or F1.

• If D1 were the occupied one, F1 would be vacant; then E1 and F1 are both low-road-value vacant houses worth $\text{base} + 3$, and with only one parking to go round, at least one of them is $10 + 3 = 13 < 15$ — breaking the "lowest = 15" rule. So this case fails.
• Hence F1 is occupied, D1 is vacant, and column D's occupied house is D2.

E1's neighbours are D1 (vacant), F1 (occupied), E2 (vacant), so its neighbour count is 1 and its price is $\text{base} + 3$. For the lowest price to be 15, E1 must carry the parking: $12 + 3 = 15$. That uses up YY's single parking, so every other vacant YY house has base 10.

Solved grid (✔ occupied, ∘ vacant; F2 is the one house the clues leave open):

Definitely occupied: B1, A2, C2, F1, D2. Definitely vacant: A1, C1, B2, D1, E1, E2. Only F2 is undetermined.

This question. To keep Block YY's lowest price at 15 (reached in column E), E1 must hold the parking: its neighbour count is 1, so without parking it would be $10 + 3 = 13$, dropping the minimum below 15. Giving E1 parking makes it $12 + 3 = 15$, and since YY has exactly one parking house, no other YY house has it.

Answer: E1

Setting up. Department sizes are F&A 9, M&S 7, O&Q 5, B&H 3 (24 faculty in all). The four candidates are Pakrasi (P), Qureshi (Q), Ramaswamy (R) and Samuel (S). Rules: everyone votes; within a department all the non-candidates vote for the same person; nobody votes for themselves or for a candidate from their own department; exactly one candidate is from O&Q; at most two candidates per department.

The tally: P 3, Q 14, R 6, S 1 (adds to 24 ✓). The candidates' own votes are given: P→R, Q→S, R→Q, S→P.

Separate each candidate's total into "one candidate ballot" plus "department blocks":

The 20 non-candidates split into department blocks. Since exactly one candidate is in O&Q, its block is $5-1=4$. Matching the needed numbers to possible block sizes: the "2" can only be the B&H block (size 2), the "5" must be the M&S block (size 5, so M&S has 2 candidates), which leaves F&A with no candidate — its whole block of 9 goes to Q, and with the O&Q block of 4 that gives Q his 13. S gets no block at all.

Placing the candidates. M&S holds two candidates; R can't be one (its own block votes R). Q and S can't share a department (Q votes S), so the M&S pair must be Pakrasi and Qureshi. That leaves Ramaswamy and Samuel for O&Q and B&H — one each, and either arrangement obeys every rule.

The solved picture:

Check: Q $=9+4+1=14$, R $=5+1=6$, P $=2+1=3$, S $=1$. ✓

This question — which two candidates can be in the same department?

The only department that holds two candidates is M&S, and that pair is Pakrasi and Qureshi. No other pair can share a department: R is barred from M&S, and Samuel cannot sit with Pakrasi (he voted for Pakrasi) or with Qureshi (Qureshi voted for him).

Answer: Prof. Pakrasi and Prof. Qureshi

Recall the solved picture (from splitting each candidate's votes into one candidate ballot plus whole-department blocks):

This question — how many votes could Qureshi have got from a single department?

Qureshi's 14 votes come from: the F&A block — all 9 faculty; the O&Q block of 4 (plus Ramaswamy's own vote, 1, if Ramaswamy sits in O&Q, so 4 or 5 from there); and 0 or 1 from B&H depending on where Ramaswamy sits. M&S gives him nothing.

So a single department can contribute 9 (F&A), 4 or 5 (O&Q), or 0/1 (B&H). Of the options 7, 6, 8, 9, only 9 is achievable.

Answer: 9

Recall the solved picture:

Ramaswamy and Samuel take O&Q and B&H, one each — that is the only piece left open.

This question — if Samuel belongs to B&H

If Samuel is in B&H, then Ramaswamy must be the O&Q candidate. Now test both statements:

• Statement A: "Pakrasi belongs to M&S." Pakrasi is always in the M&S pair — true.
• Statement B: "Ramaswamy belongs to O&Q." With Samuel in B&H, Ramaswamy is forced into O&Q — true.

Both hold.

Answer: Both statements A and B

Recall the solved picture:

This question — what can we conclude about the O&Q candidate?

The M&S pair is Pakrasi and Qureshi, and F&A has no candidate, so the single O&Q candidate must be Ramaswamy or Samuel. Both arrangements (Ramaswamy in O&Q, or Samuel in O&Q) fit every clue, so we cannot pin it down further.

Answer: It was either Prof. Ramaswamy or Prof. Samuel.

Recall the solved picture:

This question — evaluating the two statements

• Statement A: "Non-candidates from M&S voted for Qureshi." The M&S block (5 votes) actually went to Ramaswamy — false.
• Statement B: "Non-candidates from F&A voted for Qureshi." The whole F&A block of 9 went to Qureshi — true.

Answer: Only statement B

The office runs 20 slots of 15 minutes each from 9:00 am to 2:00 pm, with 10 applications booked per slot — 200 applications in all. There are 10 counters: 4 for US, 2 for UK, 2 for Schengen, 2 for Others. Each category has the same number of applications in every slot, and half of all applications are US.

Working out the category split per slot.

1. US per day $= 50\% \times 200 = 100$, so US per slot $= 100/20 = 5$ (filling the 4 US counters, each application taking 10 minutes).
2. That leaves $10 - 5 = 5$ per slot to share, in fixed numbers, among UK, Schengen and Others.
3. Ira, Vijay and Nandini are three Schengen applicants in the 9:15 slot, so Schengen is at least 3 per slot. Mahira and Osman are two Others applicants in the 9:30 slot, so Others is at least 2 per slot.
4. With UK + Schengen + Others = 5, Schengen $\ge 3$ and Others $\ge 2$, the only split is Schengen = 3, Others = 2, UK = 0.

Finding the Schengen processing time. The 9:00 Schengen slot has three applicants; two start at 9:00 on the two counters and the third starts when a counter frees, at time $t$ later, finishing at $2t$ past 9:00. The trio Ira–Vijay–Nandini were the whole 9:15 Schengen slot but arrived at 9:20. Ira takes the counter that is already free at 9:20; Vijay takes the counter that frees when the 9:00 third applicant finishes, at $2t$ past 9:00. Vijay is called at 9:25, so $2t = 25$, giving $t = 12.5$ minutes.

Schengen timeline (2 counters):

The "six busy counters at 9:20" check. At 9:20 all four US counters are working and both Schengen counters are engaged (one on the 9:00 third applicant, the other just picked up Ira) — that is six. So the Others counters must both be idle at 9:20. Each Others slot brings 2 applicants; for the 9:00 pair and the 9:15 pair to both be done by 9:20, an Others application must take at most 5 minutes. Both counters then sit idle until Mahira and Osman start at 9:30.

This question. Each slot already has US = 5, Schengen = 3 and Others = 2, which sums to 10 with no room for UK. So UK is 0 per slot, and across all 20 slots the number of UK applications is $0$.

💡 Teacher tip: "Six counters busy at 9:20" is the pivot clue — it is the only thing that forces the Others time down to 5 minutes and confirms UK is empty.

Answer: 0

The office runs 20 slots of 15 minutes each from 9:00 am to 2:00 pm, with 10 applications booked per slot — 200 applications in all. There are 10 counters: 4 for US, 2 for UK, 2 for Schengen, 2 for Others. Each category has the same number of applications in every slot, and half of all applications are US.

Working out the category split per slot.

1. US per day $= 50\% \times 200 = 100$, so US per slot $= 100/20 = 5$ (filling the 4 US counters, each application taking 10 minutes).
2. That leaves $10 - 5 = 5$ per slot to share, in fixed numbers, among UK, Schengen and Others.
3. Ira, Vijay and Nandini are three Schengen applicants in the 9:15 slot, so Schengen is at least 3 per slot. Mahira and Osman are two Others applicants in the 9:30 slot, so Others is at least 2 per slot.
4. With UK + Schengen + Others = 5, Schengen $\ge 3$ and Others $\ge 2$, the only split is Schengen = 3, Others = 2, UK = 0.

Finding the Schengen processing time. The 9:00 Schengen slot has three applicants; two start at 9:00 on the two counters and the third starts when a counter frees, at time $t$ later, finishing at $2t$ past 9:00. The trio Ira–Vijay–Nandini were the whole 9:15 Schengen slot but arrived at 9:20. Ira takes the counter that is already free at 9:20; Vijay takes the counter that frees when the 9:00 third applicant finishes, at $2t$ past 9:00. Vijay is called at 9:25, so $2t = 25$, giving $t = 12.5$ minutes.

Schengen timeline (2 counters):

The "six busy counters at 9:20" check. At 9:20 all four US counters are working and both Schengen counters are engaged (one on the 9:00 third applicant, the other just picked up Ira) — that is six. So the Others counters must both be idle at 9:20. Each Others slot brings 2 applicants; for the 9:00 pair and the 9:15 pair to both be done by 9:20, an Others application must take at most 5 minutes. Both counters then sit idle until Mahira and Osman start at 9:30.

This question. There are 40 Others applications over the day (2 per slot). Each takes the same time $t_O$, and the six-busy-counters condition forces $t_O \le 5$ minutes. The total processing time is $40 \times t_O$, largest when $t_O = 5$:

$40 \times 5 = 200 \text{ minutes.}$

Answer: 200

The office runs 20 slots of 15 minutes each from 9:00 am to 2:00 pm, with 10 applications booked per slot — 200 applications in all. There are 10 counters: 4 for US, 2 for UK, 2 for Schengen, 2 for Others. Each category has the same number of applications in every slot, and half of all applications are US.

Working out the category split per slot.

1. US per day $= 50\% \times 200 = 100$, so US per slot $= 100/20 = 5$ (filling the 4 US counters, each application taking 10 minutes).
2. That leaves $10 - 5 = 5$ per slot to share, in fixed numbers, among UK, Schengen and Others.
3. Ira, Vijay and Nandini are three Schengen applicants in the 9:15 slot, so Schengen is at least 3 per slot. Mahira and Osman are two Others applicants in the 9:30 slot, so Others is at least 2 per slot.
4. With UK + Schengen + Others = 5, Schengen $\ge 3$ and Others $\ge 2$, the only split is Schengen = 3, Others = 2, UK = 0.

Finding the Schengen processing time. The 9:00 Schengen slot has three applicants; two start at 9:00 on the two counters and the third starts when a counter frees, at time $t$ later, finishing at $2t$ past 9:00. The trio Ira–Vijay–Nandini were the whole 9:15 Schengen slot but arrived at 9:20. Ira takes the counter that is already free at 9:20; Vijay takes the counter that frees when the 9:00 third applicant finishes, at $2t$ past 9:00. Vijay is called at 9:25, so $2t = 25$, giving $t = 12.5$ minutes.

Schengen timeline (2 counters):

The "six busy counters at 9:20" check. At 9:20 all four US counters are working and both Schengen counters are engaged (one on the 9:00 third applicant, the other just picked up Ira) — that is six. So the Others counters must both be idle at 9:20. Each Others slot brings 2 applicants; for the 9:00 pair and the 9:15 pair to both be done by 9:20, an Others application must take at most 5 minutes. Both counters then sit idle until Mahira and Osman start at 9:30.

This question. Nandini is the third of the trio. Ira finishes at 9:32:30 and Vijay at 9:37:30, so the next counter to free is Ira's at 9:32:30. Nandini starts then and takes 12.5 minutes:

$9{:}32{:}30 + 12{:}30 = 9{:}45.$

Among the options this is exactly 9:45 am.

Answer: 9:45 am

The office runs 20 slots of 15 minutes each from 9:00 am to 2:00 pm, with 10 applications booked per slot — 200 applications in all. There are 10 counters: 4 for US, 2 for UK, 2 for Schengen, 2 for Others. Each category has the same number of applications in every slot, and half of all applications are US.

Working out the category split per slot.

1. US per day $= 50\% \times 200 = 100$, so US per slot $= 100/20 = 5$ (filling the 4 US counters, each application taking 10 minutes).
2. That leaves $10 - 5 = 5$ per slot to share, in fixed numbers, among UK, Schengen and Others.
3. Ira, Vijay and Nandini are three Schengen applicants in the 9:15 slot, so Schengen is at least 3 per slot. Mahira and Osman are two Others applicants in the 9:30 slot, so Others is at least 2 per slot.
4. With UK + Schengen + Others = 5, Schengen $\ge 3$ and Others $\ge 2$, the only split is Schengen = 3, Others = 2, UK = 0.

Finding the Schengen processing time. The 9:00 Schengen slot has three applicants; two start at 9:00 on the two counters and the third starts when a counter frees, at time $t$ later, finishing at $2t$ past 9:00. The trio Ira–Vijay–Nandini were the whole 9:15 Schengen slot but arrived at 9:20. Ira takes the counter that is already free at 9:20; Vijay takes the counter that frees when the 9:00 third applicant finishes, at $2t$ past 9:00. Vijay is called at 9:25, so $2t = 25$, giving $t = 12.5$ minutes.

Schengen timeline (2 counters):

The "six busy counters at 9:20" check. At 9:20 all four US counters are working and both Schengen counters are engaged (one on the 9:00 third applicant, the other just picked up Ira) — that is six. So the Others counters must both be idle at 9:20. Each Others slot brings 2 applicants; for the 9:00 pair and the 9:15 pair to both be done by 9:20, an Others application must take at most 5 minutes. Both counters then sit idle until Mahira and Osman start at 9:30.

This question. Key times: Nandini finishes at 9:45, Vijay at 9:37:30, while Mahira and Osman (Others, 9:30 slot) start at 9:30 and finish by 9:35.

• "Osman completed before 9:45 am." Osman is done by 9:35 — true.
• "Mahira started after Nandini's." Mahira starts at 9:30, Nandini at 9:32:30, so Mahira starts before — false.
• "Osman completed before Vijay's." Osman by 9:35, Vijay at 9:37:30 — true.
• "Mahira completed before Nandini's." Mahira by 9:35, Nandini at 9:45 — true.

The false statement is that Mahira's process started after Nandini's.

Answer: The application process of Mahira started after Nandini's.

The office runs 20 slots of 15 minutes each from 9:00 am to 2:00 pm, with 10 applications booked per slot — 200 applications in all. There are 10 counters: 4 for US, 2 for UK, 2 for Schengen, 2 for Others. Each category has the same number of applications in every slot, and half of all applications are US.

Working out the category split per slot.

1. US per day $= 50\% \times 200 = 100$, so US per slot $= 100/20 = 5$ (filling the 4 US counters, each application taking 10 minutes).
2. That leaves $10 - 5 = 5$ per slot to share, in fixed numbers, among UK, Schengen and Others.
3. Ira, Vijay and Nandini are three Schengen applicants in the 9:15 slot, so Schengen is at least 3 per slot. Mahira and Osman are two Others applicants in the 9:30 slot, so Others is at least 2 per slot.
4. With UK + Schengen + Others = 5, Schengen $\ge 3$ and Others $\ge 2$, the only split is Schengen = 3, Others = 2, UK = 0.

Finding the Schengen processing time. The 9:00 Schengen slot has three applicants; two start at 9:00 on the two counters and the third starts when a counter frees, at time $t$ later, finishing at $2t$ past 9:00. The trio Ira–Vijay–Nandini were the whole 9:15 Schengen slot but arrived at 9:20. Ira takes the counter that is already free at 9:20; Vijay takes the counter that frees when the 9:00 third applicant finishes, at $2t$ past 9:00. Vijay is called at 9:25, so $2t = 25$, giving $t = 12.5$ minutes.

Schengen timeline (2 counters):

The "six busy counters at 9:20" check. At 9:20 all four US counters are working and both Schengen counters are engaged (one on the 9:00 third applicant, the other just picked up Ira) — that is six. So the Others counters must both be idle at 9:20. Each Others slot brings 2 applicants; for the 9:00 pair and the 9:15 pair to both be done by 9:20, an Others application must take at most 5 minutes. Both counters then sit idle until Mahira and Osman start at 9:30.

This question. US has 5 arrivals per slot across 4 counters at 10 minutes each. In each 15-minute slot the 4 counters can clear $4 \times 1.5 = 6$ applications while only 5 arrive, so a small fixed backlog carries over and the finish time of each slot's last US applicant rises by 15 minutes per slot:

So the last US applicant of the slot starting at $9{:}00 + 15k$ minutes finishes at $9{:}20 + 15k$. The final slot is the 20th ($k = 19$):

$9{:}20 + 15 \times 19 = 9{:}20 + 285 \text{ min} = 2{:}05 \text{ pm.}$

Answer: 2:05 pm

Setting up the grid. Five restaurants (R1–R5) each give an integer rating from 1 to 5 to five workers, so we have a $5\times5$ table to fill. Two facts turn the averages into hard totals:

• Multiply each restaurant's mean by 5 to get its row sum: R1 $=17$, R2 $=11$, R3 $=19$, R4 $=14$, R5 $=17$.
• Multiply each worker's mean by 5 to get their column sum: Ullas $=11$, Vasu $=19$, Waman $=17$, Xavier $=18$, Yusuf $=13$. (Both sets add to 78 — a nice consistency check.)

Fixing each worker's five ratings from their sum, median, mode, range and the known 1's and 5's:

1. Ullas (sum 11, median 2, mode 2, range 3, has a 1): the only set that fits is $\{1,2,2,2,4\}$.
2. Vasu (sum 19, median 4, mode 4, range 3, has a 5): only $\{2,4,4,4,5\}$.
3. Waman (sum 17, median 4, mode 5, range 4, has two 5's and a 1): the rest must be $\{2,4\}$, giving $\{1,2,4,5,5\}$.
4. Xavier (sum 18, median 4, mode 5, range 4, has two 5's): the other three are $\{1,3,4\}$, so $\{1,3,4,5,5\}$.
5. Yusuf (sum 13, median 3, modes 1 and 4, range 3, has two 1's): needs two 4's, so $\{1,1,3,4,4\}$.

Placing the values using the row sums. R3 already has Waman 5, Xavier 5, Yusuf 1, so Ullas + Vasu $=8$, forcing both to 4. Ullas's remaining three cells must then all be 2. Working R2, R1, R4 and R5 the same way, everything lands in exactly one place.

The completed grid:

Every column matches its worker's mean, median, mode and range, so this grid is unique — there is no second way to fill it.

This question — how many ratings cannot be determined

Because the grid above is forced into one and only one arrangement, every single one of the 25 ratings is pinned down. None is left uncertain.

Answer: 0

Recall the fully solved grid (built from the row and column sums plus each worker's median, mode and range — it is the unique fill):

This question — how many workers did R2 give a rating of 4?

Read across R2's row: Ullas 2, Vasu 2, Waman 1, Xavier 5, Yusuf 1. There is no 4 anywhere in that row.

Answer: 0

Recall the fully solved grid (the unique fill from the row/column sums and each worker's summary statistics):

This question — what rating did R1 give to Xavier?

In R1's row, Ullas 1, Waman 5 and Vasu 4 are already fixed, leaving Xavier + Yusuf $=17-10=7$. Xavier's value comes from $\{1,3,4\}$ and Yusuf's from $\{3,4,4\}$; the only pair adding to 7 that also keeps the R4 and R5 rows consistent is Xavier 3, Yusuf 4.

Answer: 3

Recall the fully solved grid:

This question — median of the five ratings R3 gave

R3's row is Ullas 4, Vasu 4, Waman 5, Xavier 5, Yusuf 1. Sort it: $1, 4, 4, 5, 5$. The middle (third) value is the median.

Answer: 4

Recall the fully solved grid:

This question — which restaurant gave its own median rating to exactly one worker?

Find each restaurant's median, then count how many of its five ratings equal that median:

Only R4 gave its median rating (3) to exactly one worker — Yusuf.

Answer: R4