CAT 2021 DILR Questions — Slot 3

Reading the schedule. Each bar is one person on one project; the number inside is the % of work they completed, and the bar's length is how many months they worked. Reading all the bars gives two master tables.

Per employee — total months worked, and the month-weighted average completion (the "annual completion index"):

Per project — total employee-months, and the calendar span (start of earliest bar to end of latest):

This question. Check the two statements against the tables.

• Statement I — total project-months the same for all four employees. From the top table: Abani 9, Bahni 9, Danni 9, Tinni 9. All equal → I is true.
• Statement II — total employee-months the same for all five projects. From the bottom table: 6, 5, 9, 10, 6. Clearly not equal → II is false.

Only I holds.

Answer: Only I

Reading the schedule. Each bar is one person on one project; the number inside is the % of work they completed, and the bar's length is how many months they worked. Reading all the bars gives two master tables.

Per employee — total months worked, and the month-weighted average completion (the "annual completion index"):

Per project — total employee-months, and the calendar span (start of earliest bar to end of latest):

This question. "Did not work on multiple projects in any month" means the person's bars never overlap on the calendar. Lay each person's months side by side:

• Abani: Feb–Mar, then May–Jun, then Jul–Nov. No overlap.
• Bahni: Feb–Mar, then Apr–Jul, then Oct–Dec. No overlap.
• Danni: Feb–Apr, then Jun–Aug, then Sep–Oct, then Dec. No overlap.
• Tinni: Jan–Feb, Mar–Apr, then Jul–Sep and Sep–Oct — both include September, so Tinni is on two projects at once that month.

So everyone except Tinni stayed on one project per month.

Answer: Only Abani, Bahni and Danni

Reading the schedule. Each bar is one person on one project; the number inside is the % of work they completed, and the bar's length is how many months they worked. Reading all the bars gives two master tables.

Per employee — total months worked, and the month-weighted average completion (the "annual completion index"):

Per project — total employee-months, and the calendar span (start of earliest bar to end of latest):

This question. A project's duration is the number of months from its first start to its last finish. From the project table:

P1 = 3, P2 = 3, P3 = 5, P4 = 5, P5 = 4.

The equal pairs are P1 & P2 (both 3) and P3 & P4 (both 5). Only P3 & P4 appears among the options.

Answer: Project 3, Project 4

Reading the schedule. Each bar is one person on one project; the number inside is the % of work they completed, and the bar's length is how many months they worked. Reading all the bars gives two master tables.

Per employee — total months worked, and the month-weighted average completion (the "annual completion index"):

Per project — total employee-months, and the calendar span (start of earliest bar to end of latest):

This question. The annual completion index is the month-weighted average already computed in the top table:

Largest to smallest: Danni, Tinni, Abani, Bahni.

Answer: Danni, Tinni, Abani, Bahni

The one rule that matters. The device says "impurity present" exactly when the impurity in the tested sample is 10% or more. If you mix $v_1$ ml that is a fraction $f_1$ impure with $v_2$ ml that is a fraction $f_2$ impure, the mixture's impurity fraction is

$\frac{v_1 f_1 + v_2 f_2}{v_1 + v_2}.$

Every bottle holds 50 ml.

This question. We take 5 ml from A (pure, so $f_A = 0$) and 5 ml from B (impurity fraction $f_B$). The mixture's impurity fraction is

$\frac{5\cdot 0 + 5\cdot f_B}{5 + 5} = \frac{f_B}{2}.$

The device did detect impurity, so

$\frac{f_B}{2} \ge 0.10 \;\Rightarrow\; f_B \ge 0.20.$

Bottle B holds 50 ml, so the amount of impurity in it is $50 \times f_B \ge 50 \times 0.20 = 10$ ml. That is only a floor — B could hold much more. So all we can safely conclude is 10 ml or more.

Answer: 10 ml or more

The one rule that matters. The device says "impurity present" exactly when the impurity in the tested sample is 10% or more. If you mix $v_1$ ml that is a fraction $f_1$ impure with $v_2$ ml that is a fraction $f_2$ impure, the mixture's impurity fraction is

$\frac{v_1 f_1 + v_2 f_2}{v_1 + v_2}.$

Every bottle holds 50 ml.

This question. Four bottles, each entirely pure or entirely impure. We only need a yes/no: are all four pure?

Do it in a single test: take equal amounts from all four and mix them together.

• If all four are pure, the mix is 0% impure → nothing detected → they are ready.
• If even one bottle is fully impure, that bottle alone makes the mix at least $\tfrac14 = 25\%$ impure — well past the 10% line → detected → not ready.

The two situations give opposite readings, so one test settles it.

Answer: 1

The one rule that matters. The device says "impurity present" exactly when the impurity in the tested sample is 10% or more. If you mix $v_1$ ml that is a fraction $f_1$ impure with $v_2$ ml that is a fraction $f_2$ impure, the mixture's impurity fraction is

$\frac{v_1 f_1 + v_2 f_2}{v_1 + v_2}.$

Every bottle holds 50 ml.

This question. Four bottles: exactly one is 80% pure / 20% impure, the other three are pure. We must say which bottle is the impure one.

Test 1 — split into two pairs. Mix equal amounts from bottles 1 and 2.

• If the impure bottle is 1 or 2, the mix is $0.20 \times \tfrac12 = 10\%$ impure → detected.
• If the impure bottle is 3 or 4, the mix is pure → nothing detected.

Either way we now know which pair holds the impure bottle.

Test 2 — check one bottle from that pair alone. A lone impure bottle reads 20% (detected); a pure one reads 0%. That single reading names the culprit.

Why not one test? A single test only gives a yes/no, which can separate at most two cases — but here there are four possible culprits. So 2 is the fewest possible, and the plan above achieves it.

Answer: 2

The one rule that matters. The device says "impurity present" exactly when the impurity in the tested sample is 10% or more. If you mix $v_1$ ml that is a fraction $f_1$ impure with $v_2$ ml that is a fraction $f_2$ impure, the mixture's impurity fraction is

$\frac{v_1 f_1 + v_2 f_2}{v_1 + v_2}.$

Every bottle holds 50 ml.

This question. Four bottles: either 1 or 2 of them are pure and the rest are 85% pure / 15% impure. We must find the exact number of pure bottles — equivalently, whether 2 or 3 bottles are impure.

Do it in one test: mix equal amounts from all four. If $k$ bottles are impure, the mixture's impurity is $\dfrac{k \times 0.15}{4}$.

The two cases land on opposite sides of the 10% line, so a single combined test tells us the exact count.

Answer: 1

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — the two doubles. From the build:

• P10 throws first in round 3 (rank 1 after round 2) and had thrown last in round 2 → double.
• P8 throws first in round 4 (rank 6 after round 3) and had thrown last among the qualifiers in round 3 → double.

Answer: P8, P10

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — the silver medallist. The final standings give rank 2 (silver) to P1, at 88.6 m.

Answer: P1

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — the last throw of the event. In round 6 the order is rank 6 first up to rank 1 last, so the very last throw is by whoever is rank 1 after round 5. After round 5, P7 has just jumped to 89.6 and stands first. So P7 throws the last javelin.

Answer: P7

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — the silver medallist's final score. Silver is P1, who never improves in, so his score stays at his round-3 best of 88.6 m.

Answer: 88.6

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — a possible final score for P8. P8 does not improve in, so its final score is just $d_8$, its single valid round-2 throw. Two conditions pin down the range: $d_8$ must beat P6's 82.5 (so P8 qualifies) yet stay below P1's 82.9 (so P8 is the lowest-ranked qualifier, which is what makes its round-4 double happen). So $82.5 < d_8 < 82.9$.

Among the choices only 82.7 lies in that window. (81.9 is too low to qualify, 85.1 is too high and breaks the double, and 0 would mean no valid throw at all.)

Answer: 82.7

How scoring and order work. A player's score after any round is the longest valid throw he has made so far (an invalid throw counts as 0). After every round players are re-ranked by score, ties broken in favour of the previously higher rank. Throwing order: in (rounds 1–3) the current rank-1 throws first, then rank-2, and so on; in (rounds 4–6) the order reverses — current rank-6 throws first, up to rank-1 last. The top six after round 3 go through to.

Building the whole event once.

Given throws. Round 1: P1 82.9, P3 81.5, P5 86.4, P6 82.5, P7 87.2, P9 84.1 (everyone else invalid → 0). Round 3: P1 88.6, P3 79.0, P9 81.4.

After round 1 the ranking by score is P7 (87.2), P5 (86.4), P9 (84.1), P1 (82.9), P6 (82.5), P3 (81.5), then P2, P4, P8, P10 (all 0). So in round 2 the throwing order runs P7 first … P10 last.

Round 2. Only the last two throws were valid — those belong to the last two throwers, P8 and P10. Call their distances $d_8$ and $d_{10}$. Both qualify for, and neither wins a medal.

After round 3, P1 improves to 88.6; P3 and P9 do not beat their earlier bests. The six qualifiers must be P1, P7, P5, P9, P8, P10 — which means $d_8$ and $d_{10}$ are both above P6's 82.5.

The two "doubles" (a player who throws first in a round and was last to throw in the previous round):

• P10: to throw first in round 3 it must be rank 1 after round 2, so $d_{10}$ is above 87.2. Since P10 threw last in round 2, it gets a double in round 3. This puts $d_{10}$ between 87.2 and 87.6.
• P8: it is the weakest qualifier, so it ends round 3 ranked 6 and throws first in round 4; among the six qualifiers it threw last in round 3, so it gets a double in round 4. This forces $d_8$ between 82.5 and 82.9.

Ranking going into:

. Exactly one player improves in each of rounds 4, 5, 6, always by the same amount. The gold medallist improves in round 5, the bronze in round 6, and one medallist in round 4. The final gaps are Gold − Silver = Silver − Bronze = 1.0 m. Working the arithmetic:

• Bronze = P5: 86.4 → improves once in round 6 by 1.2 → 87.6.
• Silver = P1: never improves → stays 88.6.
• Gold = P7: 87.2 → improves in round 4 (→ 88.4) and round 5 (→ 89.6) → 89.6.

Check: 89.6 − 88.6 = 1.0 and 88.6 − 87.6 = 1.0. The common step is 1.2 m.

Final standings: Gold P7 (89.6), Silver P1 (88.6), Bronze P5 (87.6). P10, P9, P8 do not improve; P8 finishes on $d_8$ (somewhere between 82.5 and 82.9).

This question — the gold medallist's phase-2 improvement. Gold is P7, who went from 87.2 (end of) to 89.6 (final), improving in round 4 and again in round 5 by 1.2 m each time. Total gain:

$89.6 - 87.2 = 2.4 \text{ m}.$

Answer: 2.4

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — how many were DEFINITELY created by Amal? A question is definitely Amal's only if no other creator could produce its reviewer set. From the table those are the ones reviewed by Bimal-then-Komal: Q07, Q09, Q12. Q06 and Q13 are the only other questions Amal could relate to, and both are ambiguous, so they don't count.

Answer: 3

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — how many were DEFINITELY created by Komal? A Komal-created question is reviewed first by Amal alone (and by Bimal only if Amal disapproves). The only question reviewed by Amal alone is Q10 — Amal approved, so it was accepted. Q06 and Q13 (reviewed by Amal and Bimal) could be Komal's, but they could equally be an expert's, so they are not definite.

Answer: 1

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — how many were DEFINITELY created by the experts (SMEs)? The expert signature is a question reviewed by both Amal and Bimal. Three such questions are unambiguous — Q02, Q04, Q08 — because Komal also stepped in (the split that only an expert-created question can force). Q06 and Q13 show the Amal-and-Bimal pattern too, but with no Komal step they could instead be Komal-created, so they are excluded from "definitely".

Answer: 3

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — how many were DEFINITELY disapproved by Bimal? Bimal reviewed Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13. Decide his verdict on each:

Definitely disapproved by Bimal: Q06, Q07, Q09, Q12 → 4.

Answer: 4

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — Amal's approval ratio (approvals ÷ questions he reviewed). Amal reviewed 8 questions. Pin down what we can:

So Amal has 2 certain approvals and 4 uncertain questions.

• Fewest approvals = 2 → ratio $2/8 = 0.25$.
• Most approvals = 6 → ratio $6/8 = 0.75$.

Everything in between is possible, so his ratio lies between 0.25 and 0.75.

Answer: lies between 0.25 and 0.75

Setting up the whole set once. Every question is created either by an outside expert (an SME) or by one of the three reviewers — Amal, Bimal, Komal. A reviewer never reviews a question they themselves created. Who reviews a question, and in what order, depends on who created it:

What we are told after all reviews are finished:

• Rejected: Q02, Q06, Q09, Q11, Q12. Accepted: Q01, Q03, Q04, Q05, Q07, Q08, Q10, Q13.
• Amal reviewed: Q02, Q03, Q04, Q06, Q08, Q10, Q11, Q13.
• Bimal reviewed: Q02, Q04, Q06, Q07, Q08, Q09, Q12, Q13.
• Komal reviewed: Q01, Q02, Q03, Q04, Q05, Q07, Q08, Q09, Q11, Q12.

Reading off each question's creator. For each question, write down exactly which reviewers touched it, then match that to the only creation route that could produce that set of reviewers. The tell-tale signs: a question reviewed by both Amal and Bimal is the expert signature (Komal joins only on a split); a question reviewed by just one person means that first reviewer already settled it.

Definite creators: Amal → Q07, Q09, Q12 (3). Bimal → Q01, Q03, Q05, Q11 (4). Komal → Q10 (1). Expert → Q02, Q04, Q08 (3). Only Q06 and Q13 stay ambiguous — each is either an expert's or Komal's, so they are not definitely anyone's.

This question — of the questions created by Amal or Bimal, how many were disapproved by at least one of the other reviewers? The questions definitely created by Amal or Bimal are Q01, Q03, Q05 (Bimal), Q07, Q09, Q12 (Amal) and Q11 (Bimal). For each, check whether any reviewer other than its creator disapproved it:

Count of "Yes": Q03, Q07, Q09, Q11, Q12 → 5.

Answer: 5