CAT 2022 DILR Questions — Slot 1

Solving the whole set once. Four players score 12 goals across 8 matches; each scores at least one, and all four totals are different. Every even-numbered match (2, 4, 6, 8) has exactly one goal, so those four matches hold 4 goals and the four odd matches share the remaining $12-4=8$.

1. Let the odd-match goal counts be $M_1, M_3, M_5, M_7$ with $M_1+M_3+M_5+M_7 = 8$.
2. We're told $M_3 = M_7$, and this shared value differs from both $M_1$ and $M_5$.
3. The single highest-scoring match is unique and is not Match 5. Two equal matches ($M_3, M_7$) can't be the unique top, and each even match has only 1 goal — so the top match is Match 1.
4. Put $M_3 = M_7 = x$, giving $M_1 + 2x + M_5 = 8$ with $M_1$ strictly largest. The smallest repeat $x=1$ leaves $M_1 + M_5 = 6$ with $M_1 > M_5$ and $M_5 \neq 1$, forcing $M_5 = 2$, $M_1 = 4$. (Any $x \ge 2$ leaves no room for a strictly larger unique $M_1$.)

Goals per match:

Player totals. Bimla scores in Match 1 plus one each in three other consecutive matches — 4 goals. The four distinct totals sum to 12 with Harita > Bimla, so Harita = 5, Bimla = 4, and the remaining two totals are 1 and 2. Harita (top scorer) plays exactly 3 matches including Match 4 and Match 8 (one goal each), so her other 3 goals land in the only large match, Match 1. Bimla takes one of Match 1's goals, then three consecutive matches that avoid Match 4 and Match 8 — only M5, M6, M7 fit. Amla and Sarita (totals 1 and 2) supply the leftovers: Match 2, Match 3, and the second goal of Match 5.

Solved grid:

The three ● goals (Match 2, Match 3, and the extra goal in Match 5) are shared between Amla and Sarita — one of them takes two of these, the other takes one.

This question — goals in Match 7. Straight from the per-match row, $M_7 = 1$.

Answer: 1

Solving the whole set once. Four players score 12 goals across 8 matches; each scores at least one, and all four totals are different. Every even-numbered match (2, 4, 6, 8) has exactly one goal, so those four matches hold 4 goals and the four odd matches share the remaining $12-4=8$.

1. Let the odd-match goal counts be $M_1, M_3, M_5, M_7$ with $M_1+M_3+M_5+M_7 = 8$.
2. We're told $M_3 = M_7$, and this shared value differs from both $M_1$ and $M_5$.
3. The single highest-scoring match is unique and is not Match 5. Two equal matches ($M_3, M_7$) can't be the unique top, and each even match has only 1 goal — so the top match is Match 1.
4. Put $M_3 = M_7 = x$, giving $M_1 + 2x + M_5 = 8$ with $M_1$ strictly largest. The smallest repeat $x=1$ leaves $M_1 + M_5 = 6$ with $M_1 > M_5$ and $M_5 \neq 1$, forcing $M_5 = 2$, $M_1 = 4$. (Any $x \ge 2$ leaves no room for a strictly larger unique $M_1$.)

Goals per match:

Player totals. Bimla scores in Match 1 plus one each in three other consecutive matches — 4 goals. The four distinct totals sum to 12 with Harita > Bimla, so Harita = 5, Bimla = 4, and the remaining two totals are 1 and 2. Harita (top scorer) plays exactly 3 matches including Match 4 and Match 8 (one goal each), so her other 3 goals land in the only large match, Match 1. Bimla takes one of Match 1's goals, then three consecutive matches that avoid Match 4 and Match 8 — only M5, M6, M7 fit. Amla and Sarita (totals 1 and 2) supply the leftovers: Match 2, Match 3, and the second goal of Match 5.

Solved grid:

The three ● goals (Match 2, Match 3, and the extra goal in Match 5) are shared between Amla and Sarita — one of them takes two of these, the other takes one.

This question — the sequence of goals in Matches 1, 3, 5, 7. Reading the per-match row: $M_1 = 4$, $M_3 = 1$, $M_5 = 2$, $M_7 = 1$.

Answer: 4, 1, 2, 1

Solving the whole set once. Four players score 12 goals across 8 matches; each scores at least one, and all four totals are different. Every even-numbered match (2, 4, 6, 8) has exactly one goal, so those four matches hold 4 goals and the four odd matches share the remaining $12-4=8$.

1. Let the odd-match goal counts be $M_1, M_3, M_5, M_7$ with $M_1+M_3+M_5+M_7 = 8$.
2. We're told $M_3 = M_7$, and this shared value differs from both $M_1$ and $M_5$.
3. The single highest-scoring match is unique and is not Match 5. Two equal matches ($M_3, M_7$) can't be the unique top, and each even match has only 1 goal — so the top match is Match 1.
4. Put $M_3 = M_7 = x$, giving $M_1 + 2x + M_5 = 8$ with $M_1$ strictly largest. The smallest repeat $x=1$ leaves $M_1 + M_5 = 6$ with $M_1 > M_5$ and $M_5 \neq 1$, forcing $M_5 = 2$, $M_1 = 4$. (Any $x \ge 2$ leaves no room for a strictly larger unique $M_1$.)

Goals per match:

Player totals. Bimla scores in Match 1 plus one each in three other consecutive matches — 4 goals. The four distinct totals sum to 12 with Harita > Bimla, so Harita = 5, Bimla = 4, and the remaining two totals are 1 and 2. Harita (top scorer) plays exactly 3 matches including Match 4 and Match 8 (one goal each), so her other 3 goals land in the only large match, Match 1. Bimla takes one of Match 1's goals, then three consecutive matches that avoid Match 4 and Match 8 — only M5, M6, M7 fit. Amla and Sarita (totals 1 and 2) supply the leftovers: Match 2, Match 3, and the second goal of Match 5.

Solved grid:

The three ● goals (Match 2, Match 3, and the extra goal in Match 5) are shared between Amla and Sarita — one of them takes two of these, the other takes one.

This question — which statement(s) is/are true?

• Statement 1 (Amla and Sarita never scored in the same match): their goals live only in Match 2, Match 3 and the extra goal of Match 5. Each of these is a single goal in its match, so at most one of the two is present — they never share a match. True.
• Statement 2 (Harita and Sarita never scored in the same match): Harita scores only in Matches 1, 4, 8; Sarita only among Matches 2, 3, 5. No overlap. True.

Answer: Both the statements

Solving the whole set once. Four players score 12 goals across 8 matches; each scores at least one, and all four totals are different. Every even-numbered match (2, 4, 6, 8) has exactly one goal, so those four matches hold 4 goals and the four odd matches share the remaining $12-4=8$.

1. Let the odd-match goal counts be $M_1, M_3, M_5, M_7$ with $M_1+M_3+M_5+M_7 = 8$.
2. We're told $M_3 = M_7$, and this shared value differs from both $M_1$ and $M_5$.
3. The single highest-scoring match is unique and is not Match 5. Two equal matches ($M_3, M_7$) can't be the unique top, and each even match has only 1 goal — so the top match is Match 1.
4. Put $M_3 = M_7 = x$, giving $M_1 + 2x + M_5 = 8$ with $M_1$ strictly largest. The smallest repeat $x=1$ leaves $M_1 + M_5 = 6$ with $M_1 > M_5$ and $M_5 \neq 1$, forcing $M_5 = 2$, $M_1 = 4$. (Any $x \ge 2$ leaves no room for a strictly larger unique $M_1$.)

Goals per match:

Player totals. Bimla scores in Match 1 plus one each in three other consecutive matches — 4 goals. The four distinct totals sum to 12 with Harita > Bimla, so Harita = 5, Bimla = 4, and the remaining two totals are 1 and 2. Harita (top scorer) plays exactly 3 matches including Match 4 and Match 8 (one goal each), so her other 3 goals land in the only large match, Match 1. Bimla takes one of Match 1's goals, then three consecutive matches that avoid Match 4 and Match 8 — only M5, M6, M7 fit. Amla and Sarita (totals 1 and 2) supply the leftovers: Match 2, Match 3, and the second goal of Match 5.

Solved grid:

The three ● goals (Match 2, Match 3, and the extra goal in Match 5) are shared between Amla and Sarita — one of them takes two of these, the other takes one.

This question — which statement(s) is/are false?

• Statement 1 (every match had at least one goal): the per-match totals are 4, 1, 1, 1, 2, 1, 1, 1 — all at least 1. True, so not false.
• Statement 2 (no two players scored in the same number of matches): Harita scored in 3 matches, Bimla in 4; Amla and Sarita, with totals 1 and 2, scored in 1 and 2 matches. The counts 3, 4, 1, 2 are all different. True, so not false.

Neither statement is false.

Answer: None of the statements

Solving the whole set once. Four players score 12 goals across 8 matches; each scores at least one, and all four totals are different. Every even-numbered match (2, 4, 6, 8) has exactly one goal, so those four matches hold 4 goals and the four odd matches share the remaining $12-4=8$.

1. Let the odd-match goal counts be $M_1, M_3, M_5, M_7$ with $M_1+M_3+M_5+M_7 = 8$.
2. We're told $M_3 = M_7$, and this shared value differs from both $M_1$ and $M_5$.
3. The single highest-scoring match is unique and is not Match 5. Two equal matches ($M_3, M_7$) can't be the unique top, and each even match has only 1 goal — so the top match is Match 1.
4. Put $M_3 = M_7 = x$, giving $M_1 + 2x + M_5 = 8$ with $M_1$ strictly largest. The smallest repeat $x=1$ leaves $M_1 + M_5 = 6$ with $M_1 > M_5$ and $M_5 \neq 1$, forcing $M_5 = 2$, $M_1 = 4$. (Any $x \ge 2$ leaves no room for a strictly larger unique $M_1$.)

Goals per match:

Player totals. Bimla scores in Match 1 plus one each in three other consecutive matches — 4 goals. The four distinct totals sum to 12 with Harita > Bimla, so Harita = 5, Bimla = 4, and the remaining two totals are 1 and 2. Harita (top scorer) plays exactly 3 matches including Match 4 and Match 8 (one goal each), so her other 3 goals land in the only large match, Match 1. Bimla takes one of Match 1's goals, then three consecutive matches that avoid Match 4 and Match 8 — only M5, M6, M7 fit. Amla and Sarita (totals 1 and 2) supply the leftovers: Match 2, Match 3, and the second goal of Match 5.

Solved grid:

The three ● goals (Match 2, Match 3, and the extra goal in Match 5) are shared between Amla and Sarita — one of them takes two of these, the other takes one.

This question — the extra condition: Harita scored in one more match than Sarita. Harita scored in 3 matches, so Sarita scored in 2 (making Sarita's total 2 and Amla's 1). Sarita then fills two of the three ● slots {Match 2, Match 3, extra goal of Match 5}, and Amla the remaining one.

• Statement 1 (Amla scored in consecutive matches): Amla scores in only one match, so "consecutive matches" cannot hold. Not necessarily true.
• Statement 2 (Sarita scored in consecutive matches): Sarita's two matches are two of {2, 3, 5}. That could be {2, 3} (consecutive) or {2, 5} / {3, 5} (not consecutive). Not necessarily true.

Neither is necessarily true.

Answer: None of the statements

Building the timing toolkit once. A train halts 2 min at a junction (R, S, T, V) and 1 min at any ordinary station. A hop between neighbouring stations takes 2 min east–west and 3 min north–south. The first train leaves every terminal at 6:00 am; north–south trains run every 15 min, east–west every 10 min; after a full trip a train rests at least 15 min before returning.

Counting off the map. The dots between stations are: M–R 3, R–S 9, S–N 5 on the northern east–west line, and the same on P–T–V–Q; A–R 2, R–T 3, T–B 2 on the western north–south line, and the same on C–S–V–D. So each east–west line has 17 ordinary stations, 2 junctions and 20 hops; each north–south line has 7 ordinary stations, 2 junctions and 10 hops.

One-way run times (travel + all halts in between):

C–D mirrors A–B (41 min); P–Q mirrors M–N (61 min).

Handy partial times (you don't count the halt at your starting station):

• $M \to R = 11$ min; $R \to S$ (or $S \to R$) $= 29$ min.
• $B \to T$ (north) $= 11$ min; $T \to R$ (north) $= 15$ min.
• $A \to R$ (south) $= 11$ min; $R \to B$ (south) $= 28$ min.

This question — M to N, ready to board at 8:05 am. Trains leave M every 10 minutes from 6:00 (…, 7:50, 8:00, 8:10, …). At 8:05 the next one is 8:10. The full M→N journey is 61 minutes, so he arrives at $8{:}10 + 61 =$ 9:11 am.

Answer: 9:11 am

Building the timing toolkit once. A train halts 2 min at a junction (R, S, T, V) and 1 min at any ordinary station. A hop between neighbouring stations takes 2 min east–west and 3 min north–south. The first train leaves every terminal at 6:00 am; north–south trains run every 15 min, east–west every 10 min; after a full trip a train rests at least 15 min before returning.

Counting off the map. The dots between stations are: M–R 3, R–S 9, S–N 5 on the northern east–west line, and the same on P–T–V–Q; A–R 2, R–T 3, T–B 2 on the western north–south line, and the same on C–S–V–D. So each east–west line has 17 ordinary stations, 2 junctions and 20 hops; each north–south line has 7 ordinary stations, 2 junctions and 10 hops.

One-way run times (travel + all halts in between):

C–D mirrors A–B (41 min); P–Q mirrors M–N (61 min).

Handy partial times (you don't count the halt at your starting station):

• $M \to R = 11$ min; $R \to S$ (or $S \to R$) $= 29$ min.
• $B \to T$ (north) $= 11$ min; $T \to R$ (north) $= 15$ min.
• $A \to R$ (south) $= 11$ min; $R \to B$ (south) $= 28$ min.

This question — T to S, ready to board at 10:25 am. The route is T → R (north on the A–B line), then R → S (east on the M–N line).

1. Northbound A–B trains reach T at …, 10:11, 10:26, … and leave 2 minutes later. At 10:25 she catches the one leaving T at 10:28.
2. $T \to R$ takes 15 min → she reaches R at 10:43.
3. Eastbound M–N trains leave R at …, 10:33, 10:43, 10:53, … (they arrive at:41 and leave at:43 after the junction halt). Arriving exactly at 10:43, she just catches the 10:43 departure.
4. $R \to S$ takes 29 min → arrival $10{:}43 + 29 =$ 11:12 am.

💡 Teacher tip: "Arrives exactly as it leaves" counts as catching it — that single minute is what makes 11:12 possible instead of the next train's 11:22.

Answer: 11:12 am

Building the timing toolkit once. A train halts 2 min at a junction (R, S, T, V) and 1 min at any ordinary station. A hop between neighbouring stations takes 2 min east–west and 3 min north–south. The first train leaves every terminal at 6:00 am; north–south trains run every 15 min, east–west every 10 min; after a full trip a train rests at least 15 min before returning.

Counting off the map. The dots between stations are: M–R 3, R–S 9, S–N 5 on the northern east–west line, and the same on P–T–V–Q; A–R 2, R–T 3, T–B 2 on the western north–south line, and the same on C–S–V–D. So each east–west line has 17 ordinary stations, 2 junctions and 20 hops; each north–south line has 7 ordinary stations, 2 junctions and 10 hops.

One-way run times (travel + all halts in between):

C–D mirrors A–B (41 min); P–Q mirrors M–N (61 min).

Handy partial times (you don't count the halt at your starting station):

• $M \to R = 11$ min; $R \to S$ (or $S \to R$) $= 29$ min.
• $B \to T$ (north) $= 11$ min; $T \to R$ (north) $= 15$ min.
• $A \to R$ (south) $= 11$ min; $R \to B$ (south) $= 28$ min.

This question — the latest she can be ready at S to reach B before 1 am, going via R. Route: S → R (west on M–N), then R → B (south on A–B). Work backwards from B.

1. The last southbound A–B train leaves terminal A at midnight. $A \to R$ is 11 min, so it reaches R at 00:11 and, after the halt, leaves R at 00:13 — the last train through R. From there $R \to B$ is 28 min, reaching B at 00:41 (before 1 am ✓). So she must be at R by 00:13.
2. Westbound $S \to R$ takes 29 min, so she must leave S no later than $00{:}13 - 29 = 23{:}44$.
3. Westbound M–N trains leave S at …, 23:39, 23:49, … (arriving at:37, leaving at:39). The latest departure that is 23:44 or earlier is 23:39; it reaches R at 00:08, in time for the 00:13 southbound. The 23:49 train would reach R at 00:18 — too late.

So she must be ready to board at S by 11:39 pm.

Answer: 11:39 pm

Building the timing toolkit once. A train halts 2 min at a junction (R, S, T, V) and 1 min at any ordinary station. A hop between neighbouring stations takes 2 min east–west and 3 min north–south. The first train leaves every terminal at 6:00 am; north–south trains run every 15 min, east–west every 10 min; after a full trip a train rests at least 15 min before returning.

Counting off the map. The dots between stations are: M–R 3, R–S 9, S–N 5 on the northern east–west line, and the same on P–T–V–Q; A–R 2, R–T 3, T–B 2 on the western north–south line, and the same on C–S–V–D. So each east–west line has 17 ordinary stations, 2 junctions and 20 hops; each north–south line has 7 ordinary stations, 2 junctions and 10 hops.

One-way run times (travel + all halts in between):

C–D mirrors A–B (41 min); P–Q mirrors M–N (61 min).

Handy partial times (you don't count the halt at your starting station):

• $M \to R = 11$ min; $R \to S$ (or $S \to R$) $= 29$ min.
• $B \to T$ (north) $= 11$ min; $T \to R$ (north) $= 15$ min.
• $A \to R$ (south) $= 11$ min; $R \to B$ (south) $= 28$ min.

This question — minimum trains to run the A–B line (both directions). One way A→B takes 41 min, a train must rest at least 15 min at a terminal, and a departure is needed every 15 min.

Follow one train from A at 6:00: it reaches B at 6:41, rests at least to 6:56, and must leave B at a scheduled slot (:00,:15,:30,:45) → 7:00. Back to A at 7:41, rest to ≥ 7:56, next A slot → 8:00. So it returns to "leaving A" after $8{:}00 - 6{:}00 = 120$ minutes.

With a train needed to depart every 15 minutes, the number required is $120 \div 15 =$ 8.

Answer: 8

Building the timing toolkit once. A train halts 2 min at a junction (R, S, T, V) and 1 min at any ordinary station. A hop between neighbouring stations takes 2 min east–west and 3 min north–south. The first train leaves every terminal at 6:00 am; north–south trains run every 15 min, east–west every 10 min; after a full trip a train rests at least 15 min before returning.

Counting off the map. The dots between stations are: M–R 3, R–S 9, S–N 5 on the northern east–west line, and the same on P–T–V–Q; A–R 2, R–T 3, T–B 2 on the western north–south line, and the same on C–S–V–D. So each east–west line has 17 ordinary stations, 2 junctions and 20 hops; each north–south line has 7 ordinary stations, 2 junctions and 10 hops.

One-way run times (travel + all halts in between):

C–D mirrors A–B (41 min); P–Q mirrors M–N (61 min).

Handy partial times (you don't count the halt at your starting station):

• $M \to R = 11$ min; $R \to S$ (or $S \to R$) $= 29$ min.
• $B \to T$ (north) $= 11$ min; $T \to R$ (north) $= 15$ min.
• $A \to R$ (south) $= 11$ min; $R \to B$ (south) $= 28$ min.

This question — minimum trains for the whole city. Find each line's full cycle (round trip + the two rests, rounded up to the next scheduled departure) and divide by that line's gap between trains.

• A–B and C–D (north–south, 41 min each way, trains every 15 min): the cycle works out to 120 min → $120 \div 15 = 8$ trains each.
• M–N and P–Q (east–west, 61 min each way, trains every 10 min): leave M 6:00 → N 7:01, rest to 7:16, next 10-min slot 7:20 → M 8:21, rest to 8:36, next slot 8:40. Cycle = 160 min → $160 \div 10 = 16$ trains each.

Total $= 8 + 8 + 16 + 16 =$ 48 trains.

Answer: 48

Solving the whole set once. Each interviewer always awards the same face value from $\{2,3,5,7,11,13\}$, and a candidate's funding $\div 1000$ equals the product of the face values she received. Factor the five fundings:

Every value 2–13 appears somewhere, so each interviewer matches exactly one face value.

1. Fathima gives to everyone except Qahira — 4 of the 5 candidates. Only the value 3 appears in exactly 4 products (all but 77). So Fathima = 3 and Qahira = 77 = {7, 11} (2 tokens).
2. Adhara gives only to Pragnyaa, so her value sits in just one product. Only 13 is unique (to 390). So Adhara = 13 and Pragnyaa = 390 = {2, 3, 5, 13}.
3. Rasheeda got the most tokens (the maximum is 4); of the rest only 210 has 4 factors, so Rasheeda = 210 = {2, 3, 5, 7}. She got nothing from Esther, whose value is therefore not 2, 3, 5 or 7 → Esther = 11.
4. Qahira = {7, 11}: Esther (11) supplies one; Dhanavi gives to Qahira, so Dhanavi = 7.
5. That leaves {2, 5} for Bithi and Chhaya. Bithi gives to Smera, and the last two candidates, Smera and Tantra, hold 165 and 66 in some order.

Solved grid (✓ = a token was awarded):

Smera and Tantra hold 165 and 66 in some order; Bithi and Chhaya split {2, 5} so that Bithi's value lies in Smera's product. Everything else is fixed.

This question — how many tokens did Qahira receive? Her funding of 77,000 means a product of $77 = 7 \times 11$ — two face values, from Dhanavi and Esther. So 2 tokens.

Answer: 2

Solving the whole set once. Each interviewer always awards the same face value from $\{2,3,5,7,11,13\}$, and a candidate's funding $\div 1000$ equals the product of the face values she received. Factor the five fundings:

Every value 2–13 appears somewhere, so each interviewer matches exactly one face value.

1. Fathima gives to everyone except Qahira — 4 of the 5 candidates. Only the value 3 appears in exactly 4 products (all but 77). So Fathima = 3 and Qahira = 77 = {7, 11} (2 tokens).
2. Adhara gives only to Pragnyaa, so her value sits in just one product. Only 13 is unique (to 390). So Adhara = 13 and Pragnyaa = 390 = {2, 3, 5, 13}.
3. Rasheeda got the most tokens (the maximum is 4); of the rest only 210 has 4 factors, so Rasheeda = 210 = {2, 3, 5, 7}. She got nothing from Esther, whose value is therefore not 2, 3, 5 or 7 → Esther = 11.
4. Qahira = {7, 11}: Esther (11) supplies one; Dhanavi gives to Qahira, so Dhanavi = 7.
5. That leaves {2, 5} for Bithi and Chhaya. Bithi gives to Smera, and the last two candidates, Smera and Tantra, hold 165 and 66 in some order.

Solved grid (✓ = a token was awarded):

Smera and Tantra hold 165 and 66 in some order; Bithi and Chhaya split {2, 5} so that Bithi's value lies in Smera's product. Everything else is fixed.

This question — who definitely got a token from Bithi but not from Dhanavi? Dhanavi (value 7) gives to the candidates whose product contains 7: Rasheeda (210) and Qahira (77). Bithi (value 2 or 5) gives to Pragnyaa and Rasheeda (both 2 and 5 divide 390 and 210) and to Smera (Bithi's value lies in Smera's product).

Check the options: Rasheeda and Qahira both got Dhanavi (excluded); Tantra never gets Bithi (Tantra's product lacks Bithi's value either way). Pragnyaa gets a Bithi token in every case and, with no 7 in 390, never a Dhanavi token.

Answer: Pragnyaa

Solving the whole set once. Each interviewer always awards the same face value from $\{2,3,5,7,11,13\}$, and a candidate's funding $\div 1000$ equals the product of the face values she received. Factor the five fundings:

Every value 2–13 appears somewhere, so each interviewer matches exactly one face value.

1. Fathima gives to everyone except Qahira — 4 of the 5 candidates. Only the value 3 appears in exactly 4 products (all but 77). So Fathima = 3 and Qahira = 77 = {7, 11} (2 tokens).
2. Adhara gives only to Pragnyaa, so her value sits in just one product. Only 13 is unique (to 390). So Adhara = 13 and Pragnyaa = 390 = {2, 3, 5, 13}.
3. Rasheeda got the most tokens (the maximum is 4); of the rest only 210 has 4 factors, so Rasheeda = 210 = {2, 3, 5, 7}. She got nothing from Esther, whose value is therefore not 2, 3, 5 or 7 → Esther = 11.
4. Qahira = {7, 11}: Esther (11) supplies one; Dhanavi gives to Qahira, so Dhanavi = 7.
5. That leaves {2, 5} for Bithi and Chhaya. Bithi gives to Smera, and the last two candidates, Smera and Tantra, hold 165 and 66 in some order.

Solved grid (✓ = a token was awarded):

Smera and Tantra hold 165 and 66 in some order; Bithi and Chhaya split {2, 5} so that Bithi's value lies in Smera's product. Everything else is fixed.

This question — how many tokens did Chhaya award? Chhaya's value is 2 or 5. Value 2 appears in 390, 210, 66; value 5 appears in 390, 210, 165. Either way, her value sits in exactly three candidates' products (Pragnyaa, Rasheeda, and whichever of Smera/Tantra holds the matching product). So Chhaya awards 3 tokens.

Answer: 3

Solving the whole set once. Each interviewer always awards the same face value from $\{2,3,5,7,11,13\}$, and a candidate's funding $\div 1000$ equals the product of the face values she received. Factor the five fundings:

Every value 2–13 appears somewhere, so each interviewer matches exactly one face value.

1. Fathima gives to everyone except Qahira — 4 of the 5 candidates. Only the value 3 appears in exactly 4 products (all but 77). So Fathima = 3 and Qahira = 77 = {7, 11} (2 tokens).
2. Adhara gives only to Pragnyaa, so her value sits in just one product. Only 13 is unique (to 390). So Adhara = 13 and Pragnyaa = 390 = {2, 3, 5, 13}.
3. Rasheeda got the most tokens (the maximum is 4); of the rest only 210 has 4 factors, so Rasheeda = 210 = {2, 3, 5, 7}. She got nothing from Esther, whose value is therefore not 2, 3, 5 or 7 → Esther = 11.
4. Qahira = {7, 11}: Esther (11) supplies one; Dhanavi gives to Qahira, so Dhanavi = 7.
5. That leaves {2, 5} for Bithi and Chhaya. Bithi gives to Smera, and the last two candidates, Smera and Tantra, hold 165 and 66 in some order.

Solved grid (✓ = a token was awarded):

Smera and Tantra hold 165 and 66 in some order; Bithi and Chhaya split {2, 5} so that Bithi's value lies in Smera's product. Everything else is fixed.

This question — how many tokens did Smera receive? Smera holds either $165 = 3\cdot5\cdot11$ or $66 = 2\cdot3\cdot11$; both are products of three face values. So Smera received 3 tokens, whichever of the two funding amounts is hers.

Answer: 3

Solving the whole set once. Each interviewer always awards the same face value from $\{2,3,5,7,11,13\}$, and a candidate's funding $\div 1000$ equals the product of the face values she received. Factor the five fundings:

Every value 2–13 appears somewhere, so each interviewer matches exactly one face value.

1. Fathima gives to everyone except Qahira — 4 of the 5 candidates. Only the value 3 appears in exactly 4 products (all but 77). So Fathima = 3 and Qahira = 77 = {7, 11} (2 tokens).
2. Adhara gives only to Pragnyaa, so her value sits in just one product. Only 13 is unique (to 390). So Adhara = 13 and Pragnyaa = 390 = {2, 3, 5, 13}.
3. Rasheeda got the most tokens (the maximum is 4); of the rest only 210 has 4 factors, so Rasheeda = 210 = {2, 3, 5, 7}. She got nothing from Esther, whose value is therefore not 2, 3, 5 or 7 → Esther = 11.
4. Qahira = {7, 11}: Esther (11) supplies one; Dhanavi gives to Qahira, so Dhanavi = 7.
5. That leaves {2, 5} for Bithi and Chhaya. Bithi gives to Smera, and the last two candidates, Smera and Tantra, hold 165 and 66 in some order.

Solved grid (✓ = a token was awarded):

Smera and Tantra hold 165 and 66 in some order; Bithi and Chhaya split {2, 5} so that Bithi's value lies in Smera's product. Everything else is fixed.

This question — which amount could Tantra have received? Every candidate except Smera and Tantra is fixed: Pragnyaa 390,000, Rasheeda 210,000, Qahira 77,000. That leaves 165,000 and 66,000 for Smera and Tantra in some order, and the clues allow either assignment (Bithi's value simply follows whichever product Smera holds). So Tantra could have received either Rs. 66,000 or Rs. 165,000.

Answer: Both (a) and (b)

Solving the whole set once. There are 15 girls and $n$ boys. Interest is nested: anyone up for 3 days is up for 2 days, and anyone up for 2 days is up for 1 day. Singers: 6 total, 4 boys → 2 girl singers. Dancers: 10 total, 4 girls → 6 boy dancers. Nobody is both a singer and a dancer.

Finding the number of boys.

• 60% of boys are up for a 2-day event: $0.6n$ boys.
• Of those, 70% are neither singer nor dancer, so the other 30% — that is $0.3 \times 0.6n = 0.18n$ boys — are singers or dancers.
• All 4 male singers are up for the 3-day event, hence for the 2-day event too. So $0.18n = 4 + (\text{male dancers in the 2-day event})$.

$0.18n = \frac{9n}{50}$ is a whole number only when $n$ is a multiple of 50. Take $n = 50$: then $0.18n = 9$, so male dancers in the 2-day event $= 9 - 4 = 5$ (fine — there are 6 male dancers). $n = 100$ would need 14 male dancers, impossible. So $n = 50$ boys, class size $50 + 15 = 65$.

Filling in the rest.

• Boys: 80% of 50 = 40 up for 1 day; 60% = 30 up for 2 days; 6 up for 3 days (4 male singers + 2 dancers, all boys since no girl does 3 days).
• The 30 two-day boys = 4 singers + 5 dancers + 21 neither.
• Girls: "singers in the 2-day event = dancers in the 2-day event + 1", with 4 male singers and 5 male dancers, forces 2 girl singers and 0 girl dancers in the 2-day event; the 60%-are-neither rule then makes girls in the 2-day event = 5 (2 singers + 3 neither). All 15 girls do 1 day; none do 3.

Solved counts:

Two-day breakdown — boys (30): 4 singers + 5 dancers + 21 neither. Girls (5): 2 singers + 0 dancers + 3 neither.

This question — how many boys are in the class? The key step is that the count of "singers-or-dancers among the two-day boys", $0.18n$, must be a whole number. That forces $n$ to be a multiple of 50, and only $n = 50$ survives the check on the number of male dancers.

Answer: 50

Solving the whole set once. There are 15 girls and $n$ boys. Interest is nested: anyone up for 3 days is up for 2 days, and anyone up for 2 days is up for 1 day. Singers: 6 total, 4 boys → 2 girl singers. Dancers: 10 total, 4 girls → 6 boy dancers. Nobody is both a singer and a dancer.

Finding the number of boys.

• 60% of boys are up for a 2-day event: $0.6n$ boys.
• Of those, 70% are neither singer nor dancer, so the other 30% — that is $0.3 \times 0.6n = 0.18n$ boys — are singers or dancers.
• All 4 male singers are up for the 3-day event, hence for the 2-day event too. So $0.18n = 4 + (\text{male dancers in the 2-day event})$.

$0.18n = \frac{9n}{50}$ is a whole number only when $n$ is a multiple of 50. Take $n = 50$: then $0.18n = 9$, so male dancers in the 2-day event $= 9 - 4 = 5$ (fine — there are 6 male dancers). $n = 100$ would need 14 male dancers, impossible. So $n = 50$ boys, class size $50 + 15 = 65$.

Filling in the rest.

• Boys: 80% of 50 = 40 up for 1 day; 60% = 30 up for 2 days; 6 up for 3 days (4 male singers + 2 dancers, all boys since no girl does 3 days).
• The 30 two-day boys = 4 singers + 5 dancers + 21 neither.
• Girls: "singers in the 2-day event = dancers in the 2-day event + 1", with 4 male singers and 5 male dancers, forces 2 girl singers and 0 girl dancers in the 2-day event; the 60%-are-neither rule then makes girls in the 2-day event = 5 (2 singers + 3 neither). All 15 girls do 1 day; none do 3.

Solved counts:

Two-day breakdown — boys (30): 4 singers + 5 dancers + 21 neither. Girls (5): 2 singers + 0 dancers + 3 neither.

This question — which quantity can be determined?

• I. Boys up for a 1-day event who are neither dancers nor singers. There are 40 one-day boys. The 4 male singers are surely among them, but the number of male dancers among the one-day boys is 5 or 6 (five are certain via the 2-day event; the sixth may or may not attend anything). So the "neither" count is 31 or 30 — not determined.
• II. Female dancers up for a 1-day event. There are 4 female dancers and all girls are up for a 1-day event, so this is exactly 4 — determined.

Only II can be determined.

Answer: Only II

Solving the whole set once. There are 15 girls and $n$ boys. Interest is nested: anyone up for 3 days is up for 2 days, and anyone up for 2 days is up for 1 day. Singers: 6 total, 4 boys → 2 girl singers. Dancers: 10 total, 4 girls → 6 boy dancers. Nobody is both a singer and a dancer.

Finding the number of boys.

• 60% of boys are up for a 2-day event: $0.6n$ boys.
• Of those, 70% are neither singer nor dancer, so the other 30% — that is $0.3 \times 0.6n = 0.18n$ boys — are singers or dancers.
• All 4 male singers are up for the 3-day event, hence for the 2-day event too. So $0.18n = 4 + (\text{male dancers in the 2-day event})$.

$0.18n = \frac{9n}{50}$ is a whole number only when $n$ is a multiple of 50. Take $n = 50$: then $0.18n = 9$, so male dancers in the 2-day event $= 9 - 4 = 5$ (fine — there are 6 male dancers). $n = 100$ would need 14 male dancers, impossible. So $n = 50$ boys, class size $50 + 15 = 65$.

Filling in the rest.

• Boys: 80% of 50 = 40 up for 1 day; 60% = 30 up for 2 days; 6 up for 3 days (4 male singers + 2 dancers, all boys since no girl does 3 days).
• The 30 two-day boys = 4 singers + 5 dancers + 21 neither.
• Girls: "singers in the 2-day event = dancers in the 2-day event + 1", with 4 male singers and 5 male dancers, forces 2 girl singers and 0 girl dancers in the 2-day event; the 60%-are-neither rule then makes girls in the 2-day event = 5 (2 singers + 3 neither). All 15 girls do 1 day; none do 3.

Solved counts:

Two-day breakdown — boys (30): 4 singers + 5 dancers + 21 neither. Girls (5): 2 singers + 0 dancers + 3 neither.

This question — fraction of the class up for a 2-day event. From the table, two-day attendance is $30 + 5 = 35$ out of a class of $65$:

$\frac{35}{65} = \frac{7}{13}.$

Answer: 7/13

Solving the whole set once. There are 15 girls and $n$ boys. Interest is nested: anyone up for 3 days is up for 2 days, and anyone up for 2 days is up for 1 day. Singers: 6 total, 4 boys → 2 girl singers. Dancers: 10 total, 4 girls → 6 boy dancers. Nobody is both a singer and a dancer.

Finding the number of boys.

• 60% of boys are up for a 2-day event: $0.6n$ boys.
• Of those, 70% are neither singer nor dancer, so the other 30% — that is $0.3 \times 0.6n = 0.18n$ boys — are singers or dancers.
• All 4 male singers are up for the 3-day event, hence for the 2-day event too. So $0.18n = 4 + (\text{male dancers in the 2-day event})$.

$0.18n = \frac{9n}{50}$ is a whole number only when $n$ is a multiple of 50. Take $n = 50$: then $0.18n = 9$, so male dancers in the 2-day event $= 9 - 4 = 5$ (fine — there are 6 male dancers). $n = 100$ would need 14 male dancers, impossible. So $n = 50$ boys, class size $50 + 15 = 65$.

Filling in the rest.

• Boys: 80% of 50 = 40 up for 1 day; 60% = 30 up for 2 days; 6 up for 3 days (4 male singers + 2 dancers, all boys since no girl does 3 days).
• The 30 two-day boys = 4 singers + 5 dancers + 21 neither.
• Girls: "singers in the 2-day event = dancers in the 2-day event + 1", with 4 male singers and 5 male dancers, forces 2 girl singers and 0 girl dancers in the 2-day event; the 60%-are-neither rule then makes girls in the 2-day event = 5 (2 singers + 3 neither). All 15 girls do 1 day; none do 3.

Solved counts:

Two-day breakdown — boys (30): 4 singers + 5 dancers + 21 neither. Girls (5): 2 singers + 0 dancers + 3 neither.

This question — male dancers up for a 1-day event. There are 6 male dancers. Five of them are in the 2-day event, so they are certainly up for the 1-day event. The 6th is not in the 2-day event; since only 40 of the 50 boys attend the 1-day event (10 attend nothing), that 6th dancer may or may not be among the 40. So the count is 5 or 6.

Answer: 5 or 6

Solving the whole set once. There are 15 girls and $n$ boys. Interest is nested: anyone up for 3 days is up for 2 days, and anyone up for 2 days is up for 1 day. Singers: 6 total, 4 boys → 2 girl singers. Dancers: 10 total, 4 girls → 6 boy dancers. Nobody is both a singer and a dancer.

Finding the number of boys.

• 60% of boys are up for a 2-day event: $0.6n$ boys.
• Of those, 70% are neither singer nor dancer, so the other 30% — that is $0.3 \times 0.6n = 0.18n$ boys — are singers or dancers.
• All 4 male singers are up for the 3-day event, hence for the 2-day event too. So $0.18n = 4 + (\text{male dancers in the 2-day event})$.

$0.18n = \frac{9n}{50}$ is a whole number only when $n$ is a multiple of 50. Take $n = 50$: then $0.18n = 9$, so male dancers in the 2-day event $= 9 - 4 = 5$ (fine — there are 6 male dancers). $n = 100$ would need 14 male dancers, impossible. So $n = 50$ boys, class size $50 + 15 = 65$.

Filling in the rest.

• Boys: 80% of 50 = 40 up for 1 day; 60% = 30 up for 2 days; 6 up for 3 days (4 male singers + 2 dancers, all boys since no girl does 3 days).
• The 30 two-day boys = 4 singers + 5 dancers + 21 neither.
• Girls: "singers in the 2-day event = dancers in the 2-day event + 1", with 4 male singers and 5 male dancers, forces 2 girl singers and 0 girl dancers in the 2-day event; the 60%-are-neither rule then makes girls in the 2-day event = 5 (2 singers + 3 neither). All 15 girls do 1 day; none do 3.

Solved counts:

Two-day breakdown — boys (30): 4 singers + 5 dancers + 21 neither. Girls (5): 2 singers + 0 dancers + 3 neither.

This question — female dancers up for a 2-day event. The rule "girl singers in the 2-day event = girl dancers in the 2-day event + 1", together with the boys' figures, gave girl singers = 2 and girl dancers = 0. No female dancer is up for a 2-day event.

Answer: 0