CAT 2025 DILR Questions — Slot 2

Setting up the full Guru schedule

Let us abbreviate the musicians as A (Ananya), B (Bhaskar), C (Charu), D (Devendra), and the Gurus as M (Pandit Meghnath), S (Ustad Samiran), R (Acharya Raghunath).

Step 1 — Fix Raghunath's span

Clue 2 says Raghunath trained nobody during 2015–2018 or 2021–2024, so his only active years are $\{2013, 2014, 2019, 2020\}$. The longest consecutive run inside these years is 2, and each Guru's span is a single value from $\{2,3,4\}$ (all different). Therefore Raghunath's span is 2, and every student's block under Raghunath is either 2013–14 or 2019–20. This leaves Meghnath and Samiran with spans $\{3,4\}$.

Step 2 — Place students under Raghunath

Clue 4 tells us Ananya started under Meghnath in 2013 and Bhaskar started under Samiran in 2013. Since a student cannot train under two Gurus in the same year, neither Ananya nor Bhaskar can have Raghunath in 2013–14. Therefore Ananya and Bhaskar both train under Raghunath in 2019–20, which forces Charu and Devendra to train under Raghunath in 2013–14.

So the Raghunath-Gurubhai pairs are: A–B (2019–20) and C–D (2013–14).

Step 3 — Determine Meghnath's span and place the remaining Gurubhai pairs

From Clue 3, the pairs that must be Gurubhai for exactly 2 years are A–C and B–D (since A–D and B–C are explicitly never Gurubhai, and A–B / C–D are already covered by Raghunath). Clue 1 says Samiran never trains more than one student at a time, so Samiran creates no Gurubhai pairs. Hence A–C and B–D must overlap under Meghnath.

If Meghnath's span were 3, Ananya's block (starting 2013) would be 2013–15, and Charu would need a 2-year overlap — but any 3-year block overlapping Ananya's by 2 years would collide with Charu's Raghunath block (2013–14). So Meghnath's span is 4 and Samiran's span is 3.

• Ananya under Meghnath: 2013–2016.
• Charu under Meghnath: 2015–2018 (overlaps Ananya in 2015–16 = 2 years, avoids her Raghunath block).
• Bhaskar under Meghnath: 2021–2024 (his only free 4-year run after S = 2013–15 and R = 2019–20).
• Devendra under Meghnath: 2019–2022 (overlaps Bhaskar in 2021–22 = 2 years).

Step 4 — Place Samiran blocks (span 3, one student at a time)

• Bhaskar under Samiran: 2013–2015 (started 2013 per Clue 4).
• Devendra under Samiran: 2016–2018 (only 3-year run avoiding R = 2013–14 and M = 2019–22).
• Charu under Samiran: 2019–2021 (only 3-year run avoiding R = 2013–14 and M = 2015–18).
• Ananya under Samiran: 2022–2024 (only 3-year run avoiding M = 2013–16 and R = 2019–20).

The complete schedule

Answering this question

Ananya and Bhaskar are Gurubhai only under Raghunath, in 2019 and 2020. Among the given options — 2020, 2018, 2021, 2014 — only 2020 appears.

Answer: 2020 (Choice A)

Recalling the solved schedule

Through the deductions for this set (Raghunath's span is 2, Meghnath's span is 4, Samiran's span is 3), the complete Guru schedule is:

Why Charu starts Meghnath in 2015

Meghnath's span is 4 (Raghunath's span of 2 forces $\{\text{M-span}, \text{S-span}\} = \{3,4\}$, and span 3 is ruled out because Charu's required 2-year overlap with Ananya under Meghnath would collide with Charu's Raghunath block of 2013–14).

• Ananya's Meghnath block is 2013–2016 (she started in 2013 per Clue 4).
• Charu's Raghunath block is 2013–2014, so her Meghnath block must avoid those years yet overlap Ananya's block for exactly 2 years.
• The only 4-year consecutive block satisfying both conditions is 2015–2018 (overlap with Ananya = 2015, 2016).

Therefore Charu began her training under Pandit Meghnath in 2015.

Answer: 2015 (Choice B)

Recalling the solved schedule

The complete Guru schedule derived from the clues is:

Reasoning for Bhaskar and Devendra

From Clue 3, Bhaskar and Devendra must be Gurubhai for exactly 2 years. Let us check which Guru can create this overlap:

• Raghunath: Bhaskar's R-block is 2019–20; Devendra's R-block is 2013–14. No overlap.
• Samiran: Clue 1 says Samiran never trains more than one student at a time, so no Gurubhai pair can come from Samiran.
• Meghnath: This is the only possibility.

Bhaskar's Meghnath block is 2021–2024 (his other blocks — Samiran 2013–15 and Raghunath 2019–20 — leave only 2021–24 as a free 4-year run). Devendra's Meghnath block is 2019–2022. They share Meghnath in 2021 and 2022.

Among the options — 2022, 2015, 2020, 2018 — only 2022 is listed.

Answer: 2022 (Choice A)

Recalling the solved schedule

The complete Guru schedule is:

Checking each option

• A: Charu under Samiran in 2018 — In 2018, Charu is under Meghnath, not Samiran. ✗
• B: Ananya under Samiran in 2015 — In 2015, Ananya is under Meghnath, not Samiran. ✗
• C: Ananya under Samiran in 2018 — In 2018, Ananya is not training under any Guru (her Meghnath block ended in 2016 and her Raghunath block starts in 2019). ✗
• D: Charu under Samiran in 2019 — Charu's Samiran block is 2019–2021, so she is indeed under Samiran in 2019. ✓

💡 Teacher tip: Charu's Samiran block (2019–21) is forced because her Raghunath (2013–14) and Meghnath (2015–18) blocks are already placed, and Samiran's three single-student slots must be packed without overlap into the remaining years alongside the other students' Samiran blocks.

Answer: Charu was training under Ustad Samiran in 2019 (Choice D)

Recalling the solved schedule with counts

The complete Guru schedule, with the number of musicians training each year, is:

Counting the years with exactly two musicians training

Reading the last column, exactly two of the four musicians are training in:

1. 2017 — Charu (M) and Devendra (S)
2. 2018 — Charu (M) and Devendra (S)
3. 2023 — Ananya (S) and Bhaskar (M)
4. 2024 — Ananya (S) and Bhaskar (M)

That gives 4 years.

Answer: 4

Setting up the size logic

A ball pings on a hoop exactly when the ball's diameter is not larger than the hoop's diameter — i.e. ball-size $\le$ hoop-size. So pings tell us about relative sizes.

Let us decode each clue:

1. H1 (clue 3): Every ball except B3 pings on H1. Since a ball fails only when it is too big for the hoop, B3 is the largest ball, and H1 is big enough for every other ball — so H1 is the largest hoop.
2. H2 (clue 4): Only B2 pings on H2. So B2 is the smallest ball, and H2 is the smallest hoop (every other ball is too big for it).
3. H4 (clue 1): B1 and B6 ping on H4, but B5 does not. So H4 is at least as big as B1 and B6, but smaller than B5: $B1 \le H4 < B5$ and $B6 \le H4 < B5$.
4. H3 (clue 2): B4 pings on H3, but B1 does not. So H3 is at least as big as B4 but smaller than B1: $B4 \le H3 < B1$.

Putting the ball sizes together: $B2 < B4 < B1 < B5 < B3$ (B6 sits somewhere between B2 and B5; its exact spot is not fixed.)

Ping table

A ✓ means the ball fits through that hoop.

Answering this question

We need the total pings made by B1, B2, and B3:

• B1 pings on H1 and H4 → 2
• B2 (the smallest ball) pings on all four hoops → 4
• B3 (the largest ball) pings on none → 0

Total $= 2 + 4 + 0 = 6$.

Answer: 6

What we already know about ball sizes

From the four clues (a ball pings on a hoop iff ball-size $\le$ hoop-size):

• B3 is the largest ball (only B3 fails to ping on H1, the biggest hoop).
• B2 is the smallest ball (only B2 pings on H2, the smallest hoop).
• From H4: B1 and B6 ping but B5 does not → $B1 \le H4 < B5$ and $B6 \le H4 < B5$, so B1 < B5 and B6 < B5.
• From H3: B4 pings but B1 does not → $B4 \le H3 < B1$, so B4 < B1.

So the fixed chain is: $B2 < B4 < B1 < B5 < B3$

The only ball whose exact position is not pinned down is B6. We only know $B2 < B6 < B5$; B6 could be smaller than B4, between B4 and B1, between B1 and B5 — the data does not say.

Testing each option

• A. $B4 < B5 < B3$ — follows from $B4 < B1 < B5$ and $B5 < B3$. Necessarily true.
• B. $B2 < B1 < B5$ — B2 is the smallest ball, and $B1 < B5$. Necessarily true.
• C. $B1 < B6 < B3$ — this requires $B1 < B6$. But B6 only has to be bigger than B2 and smaller than B5; it could well be smaller than B1. Not necessarily true.
• D. $B1 < B5 < B3$ — directly from the fixed chain. Necessarily true.

💡 Teacher tip: Whenever a ball's position is only loosely bounded (here B6 is merely "between B2 and B5"), any statement claiming a specific neighbour for it is suspect.

The statement that is not necessarily true is $B1 < B6 < B3$.

Answer: B1 < B6 < B3 (Choice C)

Reading hoop sizes from the pings

A ball pings on a hoop iff ball-size $\le$ hoop-size. So each clue places a hoop relative to the balls that do and do not fit through it.

1. H1 (clue 3): Every ball except B3 pings on H1. So H1 is big enough for all balls except the largest (B3), meaning H1 is the largest hoop.
2. H2 (clue 4): Only B2 pings on H2. So H2 is big enough only for the smallest ball, meaning H2 is the smallest hoop.
3. H3 (clue 2): B4 pings on H3 but B1 does not. So H3 sits between B4 and B1: $B4 \le H3 < B1$.
4. H4 (clue 1): B1 (and B6) ping on H4 but B5 does not. So H4 sits between B1 and B5: $B1 \le H4 < B5$.

Ordering the hoops

From the ball chain $B2 < B4 < B1 < B5 < B3$, the hoops slot in as: $H2 < B4 \le H3 < B1 \le H4 < B5 \le H1 < B3$

Stripping out the balls, the hoop order is: $H2 < H3 < H4 < H1$

Matching against the options, this is Choice A.

Answer: H2 < H3 < H4 < H1 (Choice A)

Solved sizes

From the clues (a ball pings on a hoop iff ball-size $\le$ hoop-size):

• Ball chain: $B2 < B4 < B1 < B5 < B3$, with B6 somewhere between B2 and B5.
• Hoop chain: $H2 < H3 < H4 < H1$.

A ball's ping count equals the number of hoops at least as big as it.

Ping table

Why B6 is the only uncertain one

• B6 definitely pings on H1 (clue 3: all except B3 ping on H1) and on H4 (clue 1).
• B6 definitely does not ping on H2 (clue 4: only B2 pings on H2).
• On H3, it depends on whether B6 is small enough to fit: the data does not fix B6's size relative to H3. So B6 contributes 2 or 3 pings.

Total

$\text{Total} = (2+4+0+3+1) + (2\text{ or }3) = 10 + (2\text{ or }3) = 12\text{ or }13.$

💡 Teacher tip: Only B6's H3 result is undetermined, so the total can swing by exactly 1 — giving a two-value answer, not a wide range.

Answer: 12 or 13 (Choice A)

Setting up the full assignment

We have nine PMs — the distinct multiples of 10 from 10 to 90 — split into 3 NURs and 6 cities. The six cities appear in the fixed increasing order:

Blusterburg < Noodleton < Splutterville < Quackford < Mumpypore < Zingaloo

The PI formula is:

$\text{PI} = 0.5 \times \text{NUR} + 0.25 \times (\text{city}_1 + \text{city}_2)$

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Step 1 — Use the "only one NUR > city" clue to pin down the NURs

The clue says: across all NURs and all cities, there is exactly one pair where the NUR's PM exceeds the city's PM, and both belong to Humbleset.

Let the six city PMs in increasing order be $c_1 < c_2 < c_3 < c_4 < c_5 < c_6$. For exactly one NUR-to-city comparison to have NUR > city:

• Two NURs must be smaller than every city, i.e. smaller than $c_1$.
• Humbleset's NUR must be larger than exactly one city ($c_1$) and smaller than $c_2$.

So we need four distinct PMs arranged as:

$\text{NUR}_\text{small},\; \text{NUR}_\text{small},\; c_1,\; \text{NUR}_H \quad \text{all below } c_2$

That means at least four PMs are below $c_2$, so $c_2 \geq 50$.

• If $c_2 = 50$: the four PMs below 50 are exactly $\{10, 20, 30, 40\}$. We need $c_1 < \text{NUR}_H < 50$, and two NURs below $c_1$. The only fit is $c_1 = 30$, $\text{NUR}_H = 40$, and the two small NURs are $\{10, 20\}$.
• If $c_2 \geq 60$: there would be 5+ PMs below $c_2$, but only 3 NURs and 1 city ($c_1$) to fill those slots — too many leftover PMs. Impossible.

So the NURs are $\{10, 20, 40\}$ and the cities are $\{30, 50, 60, 70, 80, 90\}$:

Humbleset has NUR = 40 (the only NUR that exceeds a city — 40 > 30).

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Step 2 — Parity constraint on city pairs

For a state's PI to be an integer, $0.25 \times (\text{city}_1 + \text{city}_2)$ must be an integer. Since each city PM is a multiple of 10, this means $\text{city}_1/10 + \text{city}_2/10$ must be even — the two cities must share the same parity (both even or both odd when divided by 10).

• Even (÷10): 60, 80 → Splutterville, Mumpypore
• Odd (÷10): 30, 50, 70, 90 → Blusterburg, Noodleton, Quackford, Zingaloo

The two even-valued cities (60, 80) must be paired together. The four odd-valued cities form two pairs.

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Step 3 — Humbleset must contain Blusterburg (30)

Since Humbleset's NUR (40) exceeds only Blusterburg (30), Humbleset must contain Blusterburg. So Humbleset's city pair includes 30. The possible pairs with 30 are: (30, 50), (30, 70), or (30, 90).

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Step 4 — Test each case

Humbleset's NUR = 40, so $\text{PI}_H = 20 + 0.25 \times (30 + \text{city}_2)$.

The even pair (60, 80) always goes to one of the other two states, with NUR ∈ {10, 20}.

• (30, 50), PI_H = 40: The (60, 80) pair gives PI = 5+35 = 40 or 10+35 = 45. The (70, 90) pair gives PI = 5+40 = 45 or 10+40 = 45. Either way, we get a duplicate (40 or 45). ✗

• (30, 70), PI_H = 45: The (60, 80) pair gives PI = 40 or 45. The (50, 90) pair gives PI = 5+35 = 40 or 10+35 = 45. Always a duplicate with Humbleset's 45. ✗

• (30, 90), PI_H = 50: The (60, 80) pair gives PI = 40 (NUR=10) or 45 (NUR=20). The (50, 70) pair gives PI = 35 (NUR=10) or 40 (NUR=20). For distinct PIs we need NUR=20 with (60, 80) → PI = 45, and NUR=10 with (50, 70) → PI = 35. This gives PIs 35, 45, 50 — all distinct! ✓

Since Fogglia is lowest (35) and Humbleset is highest (50):

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Answering the question

Whimshire's PI:

$\text{PI}_W = 0.5 \times 20 + 0.25 \times (60 + 80) = 10 + 35 = 45$

Answer: 45

Recalling the solved assignment

Working through the clues (as detailed in the set's full solution), the unique assignment is:

How we got here (key steps):

1. The "only one NUR > city" clue forces the NURs to be $\{10, 20, 40\}$ and the cities to be $\{30, 50, 60, 70, 80, 90\}$ (with Blusterburg = 30 being the only city exceeded by an NUR — Humbleset's 40).
2. Integer-PI requirements force same-parity city pairs, so 60 and 80 must share a state.
3. Testing all valid pairings, only one arrangement yields three distinct integer PIs with Humbleset highest and Fogglia lowest.

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Computing Fogglia's PI

Fogglia has NUR = 10 and cities Noodleton (50) and Quackford (70):

$\text{PI}_F = 0.5 \times 10 + 0.25 \times (50 + 70) = 5 + 30 = 35$

This is indeed the lowest PI among the three states (35 < 45 < 50), matching the clue.

Answer: 35

Recalling the solved assignment

Working through all the clues, the unique assignment of PMs to states is:

Key deductions that fix this arrangement:

1. The "only one NUR > city" clue forces NURs = $\{10, 20, 40\}$ and cities = $\{30, 50, 60, 70, 80, 90\}$. Humbleset's NUR (40) is the only NUR exceeding a city (40 > 30 = Blusterburg), so Humbleset must contain Blusterburg.
2. Integer-PI requirements force same-parity city pairs: 60 and 80 must be together; the odd-valued cities pair up from {30, 50, 70, 90}.
3. Testing all pairings, only Humbleset = (30, 90) with NUR = 40 yields three distinct PIs (35, 45, 50) with the correct ranking.

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Computing Humbleset's PI

Humbleset has NUR = 40 and cities Blusterburg (30) and Zingaloo (90):

$\text{PI}_H = 0.5 \times 40 + 0.25 \times (30 + 90) = 20 + 30 = 50$

This is the highest PI (50 > 45 > 35), exactly as the clue requires. Note also that Humbleset is the state where the NUR (40) exceeds one of its cities (Blusterburg = 30) — the unique such pair across all states.

Answer: 50

Recalling the solved assignment

After working through all clues, the complete city-to-state assignment is:

The city PMs in increasing order are: Blusterburg (30) < Noodleton (50) < Splutterville (60) < Quackford (70) < Mumpypore (80) < Zingaloo (90).

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Checking each option

Only Noodleton and Quackford are both in Fogglia.

Answer: Noodleton, Quackford (Choice D)

Recalling the fully solved assignment

Every clue in the set converges to a single unique arrangement. Here is the complete identification of all nine entities:

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Why every entity is identifiable

1. NUR values are fixed by the "only one NUR > city" clue: the NURs must be $\{10, 20, 40\}$ (two NURs below the smallest city, one between the smallest and second-smallest city).
2. City values are fixed by the increasing-order clue: once the NURs are known, the remaining six PMs $\{30, 50, 60, 70, 80, 90\}$ map to the six cities in order.
3. City pairings are fixed by the integer-PI parity rule: 60 and 80 must share a state; the odd-valued cities pair up, and only one pairing yields distinct PIs.
4. State assignments are fixed by the PI ranking (Humbleset highest, Fogglia lowest) and the NUR > city clue (Humbleset must contain Blusterburg).

Since the entire arrangement is forced, all $6 + 3 = 9$ entities can be identified.

Answer: 9

Setting up the relationship from the plot

The plot gives each country two percentage changes:

• X-axis: % increase in SI from 2016 to 2020, i.e. $X = \frac{\text{SI}_{2020} - \text{SI}_{2016}}{\text{SI}_{2016}}$
• Y-axis: % increase in SI from 2020 to 2024, i.e. $Y = \frac{\text{SI}_{2024} - \text{SI}_{2020}}{\text{SI}_{2020}}$

So if a country's 2016 SI is $s$: $\text{SI}_{2020} = s(1+X), \qquad \text{SI}_{2024} = s(1+X)(1+Y)$

Pinning E's 2016 value directly

E's plotted point is (25%, 20%):

• $\text{E}_{2020} = 1.25 \times s_E$
• $\text{E}_{2024} = 1.25 \times 1.20 \times s_E = 1.50 \times s_E$

Clue 3 says E was the only country with SI = 90 in 2024, so: $1.50 \times s_E = 90 \implies s_E = 60$

That already answers the question, but let us verify the full table is consistent.

Deriving the complete SI table

2016 rank order (Clues 1 & 2): B(1) > C(2) > E(3) > A(4) > D(5) > F(6), since F is lowest in every year.

2016 range = 60 (Clue 4): B is highest, F is lowest, so $s_B - s_F = 60$.

Since $s_E = 60$, we need $s_C > 60$ and $s_B > s_C$.

Integrality constraints (all SI values are integers):

• C at (−20%, 40%): $\text{C}_{2020} = 0.8\,s_C = \frac{4}{5}s_C$ and $\text{C}_{2024} = 1.12\,s_C = \frac{28}{25}s_C$, so $s_C$ must be divisible by 25. With $s_C > 60$ and $s_C < s_B \le 100$, the only option is $s_C = 75$.
• B at (−25%, −25%): $\text{B}_{2020} = 0.75\,s_B = \frac{3}{4}s_B$ and $\text{B}_{2024} = 0.5625\,s_B = \frac{9}{16}s_B$, so $s_B$ must be divisible by 16. With $s_B > 75$ and $s_B \le 100$, we get $s_B \in \{80, 96\}$.
• A at (25%, 50%): $\text{A}_{2020} = 1.25\,s_A = \frac{5}{4}s_A$ and $\text{A}_{2024} = 1.875\,s_A = \frac{15}{8}s_A$, so $s_A$ must be divisible by 8.

Testing $s_B = 96$ (so $s_F = 36$): Then $\text{F}_{2020} = 2 \times 36 = 72$. For A to stay above F in 2020: $1.25\,s_A > 72 \Rightarrow s_A > 57.6$. But $s_A < s_E = 60$ and $s_A$ divisible by 8 gives $s_A \in \{40, 48, 56\}$ — none exceeds 57.6. Impossible.

So $s_B = 80$, $s_F = 20$. Then $\text{F}_{2020} = 40$, and we need $\text{A}_{2020} = 1.25\,s_A > 40 \Rightarrow s_A > 32$, giving $s_A \in \{40, 48, 56\}$.

Clue 3 (only E has SI 90 in 2024): $\text{A}_{2024} = 1.875\,s_A$. If $s_A = 48$, then $\text{A}_{2024} = 90$, conflicting with E being the only country at 90. If $s_A = 56$, then $\text{A}_{2024} = 105 > 100$, invalid. So $s_A = 40$.

D at (100%, 20%): $\text{D}_{2024} = 2.4\,s_D = \frac{12}{5}s_D$, so $s_D$ divisible by 5. With $20 < s_D < 40$: $s_D \in \{25, 30, 35\}$ (not uniquely determined, but irrelevant for this question).

2024 range check: Max = $\text{E}_{2024} = 90$, Min = $\text{F}_{2024} = 30$, range $= 60$ ✓

Complete solved table

💡 Teacher tip: Clue 3 alone — combined with E's plotted coordinates — is enough to pin E's 2016 SI. The rest of the table is derived to confirm consistency.

Answer: 60

Recap: how the plot translates to SI values

For a country with 2016 SI $= s$ and plotted point $(X, Y)$: $\text{SI}_{2020} = s(1+X), \qquad \text{SI}_{2024} = s(1+X)(1+Y)$

F's coordinates and formulas

F's plotted point is (100%, −25%):

• $\text{F}_{2020} = s_F \times 2.00 = 2\,s_F$
• $\text{F}_{2024} = 2\,s_F \times 0.75 = 1.5\,s_F$

Pinning $s_F$ using the clues

Clue 2 says F has the lowest SI in 2016, 2020, and 2024.

Clue 4 says the 2016 range is 60. From Clue 1, B is ranked 1 (highest) in 2016, and from Clue 2, F is lowest, so: $s_B - s_F = 60$

Clue 3 gives $\text{E}_{2024} = 90$, and E's point (25%, 20%) yields $\text{E}_{2024} = 1.5\,s_E = 90$, so $s_E = 60$.

The 2016 rank order is $s_B > s_C > s_E = 60 > s_A > s_D > s_F$, so $s_C > 60$ and $s_B > s_C$.

Integrality forces:

• C at (−20%, 40%): $s_C$ divisible by 25 → $s_C = 75$ (only value between 60 and 100 that keeps $s_B \le 100$).
• B at (−25%, −25%): $s_B$ divisible by 16 → $s_B \in \{80, 96\}$.

Testing $s_B = 96$ ($s_F = 36$): $\text{F}_{2020} = 72$. Then A (at 25%, 50%) needs $\text{A}_{2020} = 1.25\,s_A > 72$, i.e. $s_A > 57.6$. But $s_A < 60$ and $s_A$ divisible by 8 gives $s_A \le 56$. Contradiction — F would not stay lowest in 2020.

Therefore $s_B = 80$ and $s_F = 20$.

Computing F's 2020 SI

$\text{F}_{2020} = 2 \times s_F = 2 \times 20 = 40$

Complete solved table (for reference)

💡 Teacher tip: The key is that F must remain the lowest in 2020. Since F doubles from 2016 to 2020 (100% increase), a large $s_F$ would push F's 2020 value too high to stay below A. The integrality + range constraint together force $s_F = 20$.

Answer: 40

Recap: how the plot translates to SI values

For a country with 2016 SI $= s$ and plotted point $(X, Y)$: $\text{SI}_{2020} = s(1+X), \qquad \text{SI}_{2024} = s(1+X)(1+Y)$

C's coordinates and formulas

C's plotted point is (−20%, 40%):

• $\text{C}_{2020} = 0.80 \times s_C = \frac{4}{5}\,s_C$
• $\text{C}_{2024} = 0.80 \times 1.40 \times s_C = 1.12\,s_C = \frac{28}{25}\,s_C$

Determining $s_C$

From Clue 1, the 2016 rank order is B(1) > C(2) > E(3) > A(4) > D(5) > F(6) (F is lowest by Clue 2).

From Clue 3 and E's point (25%, 20%): $\text{E}_{2024} = 1.5\,s_E = 90 \Rightarrow s_E = 60$.

So $s_C > s_E = 60$, and $s_C < s_B \le 100$.

Integrality: Since $\text{C}_{2020} = \frac{4}{5}s_C$ and $\text{C}_{2024} = \frac{28}{25}s_C$ must both be integers, $s_C$ must be divisible by 25.

The only multiple of 25 strictly between 60 and 100 is 75, so $s_C = 75$.

(This is confirmed by the range constraint: $s_B - s_F = 60$ with $s_B$ divisible by 16 and $s_B > 75$ gives $s_B = 80$, $s_F = 20$.)

Computing C's 2024 SI

$\text{C}_{2024} = 1.12 \times s_C = 1.12 \times 75 = 84$

Complete solved table (for reference)

💡 Teacher tip: C's 2024 value depends on $\frac{28}{25}s_C$ being an integer, which forces $s_C$ to be a multiple of 25. Combined with the rank constraint $60 < s_C < 100$, only $s_C = 75$ works.

Answer: 84

Recap: how the plot translates to SI values

For a country with 2016 SI $= s$ and plotted point $(X, Y)$: $\text{SI}_{2020} = s(1+X), \qquad \text{SI}_{2024} = s(1+X)(1+Y)$

B's coordinates and formulas

B's plotted point is (−25%, −25%):

• $\text{B}_{2020} = 0.75 \times s_B = \frac{3}{4}\,s_B$
• $\text{B}_{2024} = 0.75 \times 0.75 \times s_B = 0.5625\,s_B = \frac{9}{16}\,s_B$

Determining $s_B$

From Clue 1, B is ranked 1 (highest SI) in 2016. From Clue 2, F is lowest in 2016. From Clue 4, the 2016 range is 60: $s_B - s_F = 60$

From Clue 3 and E's point (25%, 20%): $\text{E}_{2024} = 1.5\,s_E = 90 \Rightarrow s_E = 60$.

The 2016 rank order is $s_B > s_C > 60 > s_A > s_D > s_F$.

Integrality for C (at −20%, 40%): $s_C$ divisible by 25, with $60 < s_C < s_B \le 100$ → $s_C = 75$.

Integrality for B: $\frac{3}{4}s_B$ and $\frac{9}{16}s_B$ must be integers, so $s_B$ divisible by 16. With $s_B > 75$ and $s_B \le 100$: $s_B \in \{80, 96\}$.

Testing $s_B = 96$ ($s_F = 36$): F's 2020 SI $= 2 \times 36 = 72$. A (at 25%, 50%) would need $\text{A}_{2020} = 1.25\,s_A > 72$, i.e. $s_A > 57.6$. But $s_A < 60$ and $s_A$ divisible by 8 gives $s_A \le 56$. F would not remain lowest in 2020 — contradiction.

Therefore $s_B = 80$ (and $s_F = 20$).

Computing B's 2024 SI

$\text{B}_{2024} = 0.5625 \times s_B = 0.5625 \times 80 = \frac{9}{16} \times 80 = 45$

Complete solved table (for reference)

Among the given options (A. 60, B. 54, C. 80, D. 45), the answer is 45.

💡 Teacher tip: B declines by 25% twice — a compounding effect. The factor $0.75 \times 0.75 = 0.5625 = \frac{9}{16}$, and since $s_B = 80 = 5 \times 16$, the arithmetic works out cleanly to 45.

Answer: 45 (Choice D)

We need to build the author × paper-type matrix from the two charts and the four clues, then read off Brajen's two-author and three-author counts.

Step 1 — Set up the matrix. Let each cell be the number of papers of a given type that an author is part of. The four-author papers (there are 2) contain all four authors, so every author has 2 four-author papers.

Step 2 — Column sums (counting author-slots).

Total author-slots = $10+8+9+8 = 35$.

Step 3 — Row sums (papers per author). $5+8+12+10 = 35$ ✓ — this matches, so the matrix is consistent.

Step 4 — Subtract the four-author papers (2 each) from every row.

Each author wrote at least one of each type (fact 1), so every remaining cell is $\ge 1$.

Step 5 — Fix Arman. Arman has only 3 left across three categories, each $\ge 1$, so Arman = (Single 1, Two 1, Three 1).

Step 6 — Fix Brajen. Brajen has 6 left, and his single-count equals his two-count (fact 4). Let Single = Two = $b$; then Three = $6-2b$. With each $\ge 1$ we get $b=1$ or $b=2$.

But the four authors wrote different numbers of single-author papers (fact 2), and Arman already has 1. So Brajen cannot also have 1; hence $b=2$.

Brajen: Single 2, Two 2, Three 2.

💡 Teacher tip: The equality clue (fact 4) plus the distinctness clue (fact 2) together pin Brajen down uniquely — the equality gives two candidates, and distinctness kills the $b=1$ one.

Step 7 — The matrix so far.

(The remaining ambiguity between Chintan and Devon does not affect this question.)

Brajen wrote $2$ two-author papers and $2$ three-author papers, so the total is $2+2=4$.

Answer: 4

We need to determine whether each statement is forced by the data, so we first complete the matrix as far as the clues allow and check whether anything is left open.

Building the matrix (key steps).

1. Four-author papers = 2, all four authors each → every author has 2 four-author papers.
2. Column author-slots: Single 10, Two $4\times2=8$, Three $3\times3=9$, Four $2\times4=8$; total 35 = row total $5+8+12+10$ ✓.
3. Subtract 2 (four-author) from each row: Arman has 3, Brajen 6, Chintan 10, Devon 8 left for Single+Two+Three, each cell $\ge 1$.
4. Arman = (1, 1, 1) — only way to split 3 into three positive parts.
5. Brajen: Single = Two = $b$ (fact 4), Three = $6-2b$. Single-counts must be distinct (fact 2); Arman already has 1, so $b\neq 1$, giving $b=2$. Brajen = (2, 2, 2).
6. Three-author column: Arman 1 + Brajen 2 = 3 used; column total is 9, so Chintan + Devon = 6. Fact 3 says both Chintan and Devon wrote more three-author papers than Brajen (2), so each $\ge 3$. Hence Chintan = 3, Devon = 3.
7. Single-author column: total 10; Arman 1 + Brajen 2 = 3 used, so Chintan + Devon = 7. Distinct and $\ge 1$, different from 1 and 2 → {3, 4} in some order.
8. Two-author column: total 8; Arman 1 + Brajen 2 = 3 used, so Chintan + Devon = 5. Each author's row total then locks the pairing:

• Chintan's row: Single + Two = $12-3-2=7$. If Single = 3 then Two = 4; if Single = 4 then Two = 3.
• Devon's row: Single + Two = $10-3-2=5$. If Single = 4 then Two = 1; if Single = 3 then Two = 2.

The two equally valid completions:

Checking the statements.

• i. Chintan wrote exactly three two-author papers. In one valid completion Chintan's two-count is 4 (when his single-count is 3). So this is not necessarily true.
• ii. Chintan wrote more single-author papers than Devon. Chintan could have 3 single-author papers while Devon has 4. So this is not necessarily true either.

Neither statement is forced by the clues.

Answer: Neither i nor ii (Choice A)

Both statements are about who co-authored the three-author papers, so we focus on that column and reconstruct the actual papers.

The three-author column (fully determined).

There are 3 three-author papers, giving $3\times3=9$ author-slots. From the matrix:

(How we get here: Arman has only 3 slots total across Single/Two/Three with each $\ge1$, so Arman = 1 in every column. Brajen's Single = Two = 2 by the equality clue and distinctness, leaving Three = 2. The three-author column sums to 9; Arman + Brajen = 3, so Chintan + Devon = 6, and fact 3 forces each of them above Brajen's 2, i.e. each = 3.)

Reconstructing the three papers.

There are exactly 3 three-author papers.

• Chintan's count is 3 → Chintan appears in all 3 three-author papers.
• Devon's count is 3 → Devon appears in all 3 three-author papers.

So every three-author paper already contains both Chintan and Devon, occupying 2 of the 3 seats. The third seat across the 3 papers is filled by Arman once and Brajen twice ($1+2=3$). The three papers are therefore:

1. {Arman, Chintan, Devon}
2. {Brajen, Chintan, Devon}
3. {Brajen, Chintan, Devon}

💡 Teacher tip: When an author's count in a column equals the total number of papers of that type, that author is in every such paper. Here both Chintan and Devon hit 3 out of 3, which immediately forces them into every three-author paper.

Checking the statements.

• i. Arman wrote three-author papers only with Chintan and Devon. Arman's lone three-author paper is {Arman, Chintan, Devon} — his co-authors are exactly Chintan and Devon. Necessarily true.
• ii. Brajen wrote three-author papers only with Chintan and Devon. Both of Brajen's three-author papers are {Brajen, Chintan, Devon} — his co-authors are exactly Chintan and Devon. Necessarily true.

Both statements are forced.

Answer: Both i and ii (Choice B)

We use the solved matrix and the added condition to pick the correct completion.

The matrix with its one remaining ambiguity.

After fixing Arman, Brajen, and the three-author column, the single- and two-author counts of Chintan and Devon can go two ways (each author's row total locks the pairing):

The two valid completions are:

• Completion 1: Chintan Single = 3, Two = 4; Devon Single = 4, Two = 1.
• Completion 2: Chintan Single = 4, Two = 3; Devon Single = 3, Two = 2.

(Consistency check: in both, the two-author column sums to $1+2+4+1=8$ or $1+2+3+2=8$ ✓, and the single-author column sums to 10 ✓.)

Applying the condition. The condition says Devon wrote more than one two-author paper, i.e. Devon's two-count $>1$.

• In Completion 1, Devon's two-count = 1 — rejected.
• In Completion 2, Devon's two-count = 2 — accepted.

So Completion 2 holds, and in it Chintan's two-author count is 3.

Answer: 3