CAT 2024 DILR Questions — Slot 1

Each of the six surfers hands out exactly 30 stars (all in multiples of 5) among bloggers A, B, C and D. The chart fixes how many went to A and B; each surfer's stars to C and D together must therefore be $30-(\text{A}+\text{B})$.

Totals must be equal. The six surfers give $6\times 30 = 180$ stars in all, and every blogger ends with the same total, so each blogger got $180/4 = 45$. Adding the chart columns, A already has $10+25+0+5+0+5 = 45$ and B has $0+0+0+25+0+20 = 45$ — so C and D must also reach 45 each.

Now pin the remaining cells:

1. Surfer Y: Y gave 5 to A and 20 to B, leaving 5 for C and D. We're told D got more than C, so $C=0,\ D=5$.
2. Surfer M: the clue says M gave a different amount to each blogger. M gave 10 to A and 0 to B, and C+D must be 20 with two different values (not 0 or 10) — so they are 5 and 15.
3. Two surfers gave everything to one blogger. Only O and X gave nothing to A or B, so they are the two who dump all 30 on a single blogger. C and D each still need 45, and O and X supply 60 between them — so one of them goes entirely to C and the other entirely to D (which one is not fixed).
4. Balancing C: C needs 45. One of O/X supplies 30, Y supplies 0, so M and N must supply the other 15. That forces M gives C = 15 (so D = 5) and N gives C = 0 (so D = 5).

This gives the full grid (the only open point is whether O→C & X→D, or the reverse):

This question — total stars received by D. Every blogger's total is the same, and $180 \div 4 = 45$. Reading D's column confirms it: $5+5+30+0+0+5 = 45$ (whichever of O/X feeds D).

Answer: 45

Each of the six surfers hands out exactly 30 stars (all in multiples of 5) among bloggers A, B, C and D. The chart fixes how many went to A and B; each surfer's stars to C and D together must therefore be $30-(\text{A}+\text{B})$.

Totals must be equal. The six surfers give $6\times 30 = 180$ stars in all, and every blogger ends with the same total, so each blogger got $180/4 = 45$. Adding the chart columns, A already has $10+25+0+5+0+5 = 45$ and B has $0+0+0+25+0+20 = 45$ — so C and D must also reach 45 each.

Now pin the remaining cells:

1. Surfer Y: Y gave 5 to A and 20 to B, leaving 5 for C and D. We're told D got more than C, so $C=0,\ D=5$.
2. Surfer M: the clue says M gave a different amount to each blogger. M gave 10 to A and 0 to B, and C+D must be 20 with two different values (not 0 or 10) — so they are 5 and 15.
3. Two surfers gave everything to one blogger. Only O and X gave nothing to A or B, so they are the two who dump all 30 on a single blogger. C and D each still need 45, and O and X supply 60 between them — so one of them goes entirely to C and the other entirely to D (which one is not fixed).
4. Balancing C: C needs 45. One of O/X supplies 30, Y supplies 0, so M and N must supply the other 15. That forces M gives C = 15 (so D = 5) and N gives C = 0 (so D = 5).

This gives the full grid (the only open point is whether O→C & X→D, or the reverse):

This question — stars D got from Y. Y gave 5 to A and 20 to B, so only $30-25 = 5$ stars are left for C and D together. Since D received more than C, the split must be $C=0,\ D=5$.

Answer: 5

Each of the six surfers hands out exactly 30 stars (all in multiples of 5) among bloggers A, B, C and D. The chart fixes how many went to A and B; each surfer's stars to C and D together must therefore be $30-(\text{A}+\text{B})$.

Totals must be equal. The six surfers give $6\times 30 = 180$ stars in all, and every blogger ends with the same total, so each blogger got $180/4 = 45$. Adding the chart columns, A already has $10+25+0+5+0+5 = 45$ and B has $0+0+0+25+0+20 = 45$ — so C and D must also reach 45 each.

Now pin the remaining cells:

1. Surfer Y: Y gave 5 to A and 20 to B, leaving 5 for C and D. We're told D got more than C, so $C=0,\ D=5$.
2. Surfer M: the clue says M gave a different amount to each blogger. M gave 10 to A and 0 to B, and C+D must be 20 with two different values (not 0 or 10) — so they are 5 and 15.
3. Two surfers gave everything to one blogger. Only O and X gave nothing to A or B, so they are the two who dump all 30 on a single blogger. C and D each still need 45, and O and X supply 60 between them — so one of them goes entirely to C and the other entirely to D (which one is not fixed).
4. Balancing C: C needs 45. One of O/X supplies 30, Y supplies 0, so M and N must supply the other 15. That forces M gives C = 15 (so D = 5) and N gives C = 0 (so D = 5).

This gives the full grid (the only open point is whether O→C & X→D, or the reverse):

This question — how many surfers split their stars among exactly two bloggers. Count how many bloggers each surfer gave a non-zero number to:

Exactly two bloggers were used by N and P — so 2 surfers.

Answer: 2

Each of the six surfers hands out exactly 30 stars (all in multiples of 5) among bloggers A, B, C and D. The chart fixes how many went to A and B; each surfer's stars to C and D together must therefore be $30-(\text{A}+\text{B})$.

Totals must be equal. The six surfers give $6\times 30 = 180$ stars in all, and every blogger ends with the same total, so each blogger got $180/4 = 45$. Adding the chart columns, A already has $10+25+0+5+0+5 = 45$ and B has $0+0+0+25+0+20 = 45$ — so C and D must also reach 45 each.

Now pin the remaining cells:

1. Surfer Y: Y gave 5 to A and 20 to B, leaving 5 for C and D. We're told D got more than C, so $C=0,\ D=5$.
2. Surfer M: the clue says M gave a different amount to each blogger. M gave 10 to A and 0 to B, and C+D must be 20 with two different values (not 0 or 10) — so they are 5 and 15.
3. Two surfers gave everything to one blogger. Only O and X gave nothing to A or B, so they are the two who dump all 30 on a single blogger. C and D each still need 45, and O and X supply 60 between them — so one of them goes entirely to C and the other entirely to D (which one is not fixed).
4. Balancing C: C needs 45. One of O/X supplies 30, Y supplies 0, so M and N must supply the other 15. That forces M gives C = 15 (so D = 5) and N gives C = 0 (so D = 5).

This gives the full grid (the only open point is whether O→C & X→D, or the reverse):

This question — which statement is certain?

• I. Stars C got from M. M's four amounts must all differ: A = 10, B = 0, and C + D = 20 split as two distinct values. Balancing C's total to 45 forces M → C = 15. This is fixed — certain.
• II. Stars D got from O. O gives all 30 stars to a single blogger, but that blogger could be C or D — O and X are interchangeable. So D-from-O is either 30 or 0 — not certain.

Only statement I can be determined.

Answer: Only I

Setting up the key formulas

Let Amiya's campaign level be $a$ and Ramya's be $r$, where each is either 1 (staid) or 2 (vigorous).

When both run issue-focused campaigns, the turnout is $20(a+r)\%$ of all students, and votes split proportionally to campaign levels. This means:

$\text{Amiya's base votes} = 20(a+r)\% \times \frac{a}{a+r} = 20a\%$

$\text{Ramya's base votes} = 20(a+r)\% \times \frac{r}{a+r} = 20r\%$

So each candidate's base vote share (as a percentage of all students) is simply 20 × their own level.

When both run attacking campaigns, the rule says: 10% of each candidate's base voters (the voters they would have had under issue campaigns) will not vote at all. So each candidate keeps 90% of their base:

Applying to this question

Both run staid campaigns ($a = 1,\ r = 1$) and both attack.

$\text{Amiya's votes} = 18 \times 1 = 18\%$ $\text{Ramya's votes} = 18 \times 1 = 18\%$ $\text{Total turnout} = 18 + 18 = \mathbf{36\%}$

Answer: D. 36%

Setting up the key formulas

Let Amiya's campaign level be $a$ and Ramya's be $r$, where each is either 1 (staid) or 2 (vigorous).

When both run issue-focused campaigns, each candidate's base vote share (as a percentage of all students) is:

$\text{Amiya's base} = 20a\%, \qquad \text{Ramya's base} = 20r\%$

The attack rules then shift portions of these bases. Here is the complete turnout formula for every scenario:

💡 Teacher tip: Turnout always increases when campaign levels ($a$ or $r$) increase, because every formula has a positive coefficient on both $a$ and $r$. So to minimise turnout, set both levels to their lowest value: $a = 1,\ r = 1$.

Checking all four scenarios with $a = 1,\ r = 1$

The lowest turnout across all scenarios is 36%, which occurs when both candidates run staid campaigns and attack each other.

Answer: D. 36%

Setting up the key formulas

Let Amiya's campaign level be $a$ and Ramya's be $r$, where each is either 1 (staid) or 2 (vigorous).

When both run issue-focused campaigns, each candidate's base vote share (as a percentage of all students) is:

$\text{Amiya's base} = 20a\%, \qquad \text{Ramya's base} = 20r\%$

The attack rules then shift portions of these bases:

Applying to this question

Amiya runs an issue-focused campaign. So the possible scenarios are:

1. Both issues → Amiya gets $20a$. Maximum at $a = 2$: $20 \times 2 = 40\%$.
2. Ramya attacks, Amiya issues → Amiya gets $20a + 4r$ (she gains 20% of Ramya's base). Maximum at $a = 2,\ r = 2$: $40 + 8 = 48\%$.

Between these two, the larger value is 48%.

Let's verify the calculation for the winning case ($a = 2,\ r = 2$, Ramya attacks):

• Amiya's base = $20 \times 2 = 40\%$
• Ramya's base = $20 \times 2 = 40\%$
• 20% of Ramya's base switches to Amiya: $0.20 \times 40 = 8\%$
• Amiya's total = $40 + 8 = 48\%$ ✓

Answer: A. 48%

Setting up the key formulas

Let Amiya's campaign level be $a$ and Ramya's be $r$, where each is either 1 (staid) or 2 (vigorous).

When both run issue-focused campaigns, each candidate's base vote share (as a percentage of all students) is:

$\text{Amiya's base} = 20a\%, \qquad \text{Ramya's base} = 20r\%$

The attack rules then shift portions of these bases:

Applying to this question

Ramya runs an attacking campaign. Two sub-cases are possible depending on what Amiya does:

Case 1: Amiya runs issues (only Ramya attacks)

From the rule: 20% of Ramya's base voters go to Amiya, and another 5% don't vote. So Ramya keeps only 75% of her base: $\text{Ramya's votes} = 20r \times 0.75 = 15r$

Case 2: Amiya also attacks (both attack)

From the rule: 10% of each candidate's base voters don't vote. Ramya keeps 90%: $\text{Ramya's votes} = 20r \times 0.9 = 18r$

Finding the minimum guaranteed

Ramya cannot control what Amiya does, so we must consider the worst case for Ramya — which is Case 1 (she gets fewer votes when Amiya stays on issues).

$\text{Ramya's votes (worst case)} = 15r$

To minimise this, set $r = 1$ (staid campaign): $15 \times 1 = \mathbf{15\%}$

💡 Teacher tip: "Guaranteed" means the minimum across all scenarios the candidate doesn't control. Ramya controls her own level and that she attacks, but she does not control Amiya's choice — so we take the worst of the two sub-cases.

Answer: B. 15%

Setting up the key formulas

Let Amiya's campaign level be $a$ and Ramya's be $r$, where each is either 1 (staid) or 2 (vigorous).

When both run issue-focused campaigns, each candidate's base vote share (as a percentage of all students) is:

$\text{Amiya's base} = 20a\%, \qquad \text{Ramya's base} = 20r\%$

The attack rules then shift portions of these bases. Here is the complete table of vote shares:

Finding the maximum margin

We need the largest absolute value of the margin across all scenarios and all choices of $a, r \in \{1, 2\}$.

Scenario 1 — Both issues: $|20(a-r)|$, max = $20 \times 1 = 20\%$.

Scenario 2 — Amiya attacks, Ramya issues: $|14a - 20r|$

• $(a=2, r=1)$: $|28 - 20| = 8\%$
• $(a=1, r=2)$: $|14 - 40| = 26\%$ ← Ramya wins by 26%
• $(a=2, r=2)$: $|28 - 40| = 12\%$
• $(a=1, r=1)$: $|14 - 20| = 6\%$

Scenario 3 — Ramya attacks, Amiya issues: $|20a - 11r|$

• $(a=2, r=1)$: $|40 - 11| = \mathbf{29\%}$ ← Amiya wins by 29%
• $(a=1, r=2)$: $|20 - 22| = 2\%$
• $(a=2, r=2)$: $|40 - 22| = 18\%$
• $(a=1, r=1)$: $|20 - 11| = 9\%$

Scenario 4 — Both attack: $|18(a-r)|$, max = $18 \times 1 = 18\%$.

Result

The largest margin across all scenarios is 29%, achieved when Amiya runs a vigorous issue-focused campaign ($a = 2$) and Ramya runs a staid attacking campaign ($r = 1$).

Verification:

• Amiya's base = $20 \times 2 = 40\%$
• Ramya's base = $20 \times 1 = 20\%$
• 20% of Ramya's base switches to Amiya: $0.20 \times 20 = 4\%$
• 5% of Ramya's base doesn't vote: $0.05 \times 20 = 1\%$
• Amiya = $40 + 4 = 44\%$, Ramya = $20 \times 0.75 = 15\%$
• Margin = $44 - 15 = 29\%$ ✓

Answer: B. 29%

Reading the candlesticks. Gridlines are Rs. 200 apart. A red (bearish) candle has its Open at the top of the box and its Close at the bottom; a green (bullish) candle is the other way round. The top of the wick is the day's High and the bottom is the day's Low — B has no upper wick (High = Open = 2000), and C and E have no lower wick (Low = 800 and 1100).

Written out as four prices per share:

The Share Price Variability is $\text{SPV}=\frac{\text{High}-\text{Low}}{\tfrac12(\text{Open}+\text{Close})}.$ We only need the four shares asked about — A, C, D and F.

1. A: $\dfrac{2400-1200}{(2200+1800)/2}=\dfrac{1200}{2000}=0.60$.
2. C: $\dfrac{1400-800}{(800+1200)/2}=\dfrac{600}{1000}=0.60$.
3. D: $\dfrac{1200-300}{(500+1000)/2}=\dfrac{900}{750}=1.20$.
4. F: $\dfrac{2000-1200}{(1800+1600)/2}=\dfrac{800}{1700}\approx 0.47$.

The largest value is D's, at 1.20.

💡 Teacher tip: D has both a wide High−Low swing (900) and the smallest average price (750), so the ratio shoots up. A big numerator over a small denominator wins.

Answer: D

Reading the candlesticks. Gridlines are Rs. 200 apart. A red (bearish) candle has its Open at the top of the box and its Close at the bottom; a green (bullish) candle is the other way round. The top of the wick is the day's High and the bottom is the day's Low — B has no upper wick (High = Open = 2000), and C and E have no lower wick (Low = 800 and 1100).

Written out as four prices per share, then the Share Price Variability for every share: $\text{SPV}=\frac{\text{High}-\text{Low}}{\tfrac12(\text{Open}+\text{Close})}.$

The shares with SPV greater than 0.5 are A, C, D and G — that is 4 shares.

Answer: 4

Reading the candlesticks. Gridlines are Rs. 200 apart. A red (bearish) candle has its Open at the top of the box and its Close at the bottom; a green (bullish) candle is the other way round. The top of the wick is the day's High and the bottom is the day's Low — B has no upper wick (High = Open = 2000), and C and E have no lower wick (Low = 800 and 1100).

Daily loss is defined as $\text{Daily loss}=\frac{\text{Open}-\text{Close}}{\text{Open}}.$

A share only makes a loss if it closed below its opening price (a bearish, red share). We are asked about A, B, F and G:

1. A: $\dfrac{2200-1800}{2200}=\dfrac{400}{2200}\approx 18.2\%$.
2. B: $\dfrac{2000-1700}{2000}=\dfrac{300}{2000}=15.0\%$.
3. F: $\dfrac{1800-1600}{1800}=\dfrac{200}{1800}\approx 11.1\%$.
4. G: G closed at 1700, above its open of 1200 — it made a gain, not a loss.

The highest daily loss is A's, at about 18.2%.

Answer: A

Reading the candlesticks. Gridlines are Rs. 200 apart. A red (bearish) candle has its Open at the top of the box and its Close at the bottom; a green (bullish) candle is the other way round. The top of the wick is the day's High and the bottom is the day's Low — B has no upper wick (High = Open = 2000), and C and E have no lower wick (Low = 800 and 1100).

Written out as four prices per share:

A bullish share is one that closed above its opening price (the green shares). From the table those are C, D and G.

The trader buys one of each at the opening price and sells each at the day's high:

1. Total cost (buy at open): $800+500+1200 = 2500$.
2. Total received (sell at high): $1400+1200+1900 = 4500$.
3. Percentage gain: $\dfrac{4500-2500}{2500}\times 100 = \dfrac{2000}{2500}\times 100 = \mathbf{80\%}$.

Answer: 80%

Finding Asian countries visited by at least one person

We first convert the chart and the five clues into a complete Venn-diagram count, broken down by region.

Step 1 — Total visits per person (from the chart).

Sum of all visits $= 11 + 13 + 19 = 43$.

Step 2 — Use the 32 distinct countries (Fact 1).

Let $e_1$ = countries visited by exactly one person, $e_2$ = exactly two, $e_3$ = exactly three.

$e_1 + e_2 + e_3 = 32 \quad\text{(distinct countries)}$ $e_1 + 2e_2 + 3e_3 = 43 \quad\text{(total visits)}$

Subtracting: $e_2 + 2e_3 = 11$.

Fact 2 says USA is the only country visited by all three, so $e_3 = 1$. Hence $e_2 = 9$ and $e_1 = 22$.

Step 3 — Split the 9 "exactly-two" countries.

• D & N only = 1 (China, Fact 3).
• D & S only = 1 (France, Fact 4 — Samantha visited 0 Asian countries, so France is the sole D&S-only country).
• S & N only $= 9 - 1 - 1 = 7$.

So Samantha-and-Nitesh together (including USA) $= 7 + 1 = 8$ countries. Fact 5 says half are in Europe → 4 in Europe, the other 4 outside Europe.

Step 4 — Resolve region by region.

💡 Teacher tip: The key to the whole set is the inclusion–exclusion identity $e_2 + 2e_3 = \text{(sum of visits)} - \text{(distinct countries)}$. Once you know $e_3 = 1$ from the USA clue, everything else cascades.

Answering the question.

Samantha visited 0 Asian countries. Dheeraj's 3 Asian countries include China (shared with Nitesh), and Nitesh's single Asian country is China. So Asia contains:

• Dheeraj-only: 2
• China (D & N): 1

$\text{Asian countries visited by at least one} = 2 + 1 = 3$

Answer: 3

Finding European countries visited only by Nitesh

We first build the full Venn count from the chart and the five clues.

Step 1 — Total visits per person (from the chart).

Sum of all visits $= 11 + 13 + 19 = 43$.

Step 2 — Use the 32 distinct countries (Fact 1).

Let $e_1$ = exactly one, $e_2$ = exactly two, $e_3$ = exactly three.

$e_1 + e_2 + e_3 = 32, \qquad e_1 + 2e_2 + 3e_3 = 43$

Subtracting: $e_2 + 2e_3 = 11$. Fact 2 gives $e_3 = 1$ (USA is the only all-three country), so $e_2 = 9$, $e_1 = 22$.

Step 3 — Split the 9 "exactly-two" countries.

• D & N only = 1 (China, Fact 3).
• D & S only = 1 (France, Fact 4 — Samantha has no Asian country, so France is the only D&S-only).
• S & N only $= 9 - 1 - 1 = 7$.

Samantha-and-Nitesh together (including USA) $= 7 + 1 = 8$. Fact 5: half in Europe → 4 in Europe.

Step 4 — Resolve region by region.

Answering the question.

Nitesh visited 6 European countries in total. In Europe, Nitesh shares countries only with Samantha (the 4 S&N countries); he shares nothing with Dheeraj in Europe (China, the only D&N country, is in Asia), and USA (all three) is in ROW.

$\text{Nitesh-only in Europe} = 6 - 4 = 2$

Answer: 2

Finding ROW countries visited by both Nitesh and Samantha

We first build the full Venn count from the chart and the five clues.

Step 1 — Total visits per person (from the chart).

Sum of all visits $= 11 + 13 + 19 = 43$.

Step 2 — Use the 32 distinct countries (Fact 1).

Let $e_1$ = exactly one, $e_2$ = exactly two, $e_3$ = exactly three.

$e_1 + e_2 + e_3 = 32, \qquad e_1 + 2e_2 + 3e_3 = 43$

Subtracting: $e_2 + 2e_3 = 11$. Fact 2 gives $e_3 = 1$ (USA is the only all-three country), so $e_2 = 9$, $e_1 = 22$.

Step 3 — Split the 9 "exactly-two" countries.

• D & N only = 1 (China, Fact 3).
• D & S only = 1 (France, Fact 4 — Samantha has no Asian country, so France is the only D&S-only).
• S & N only $= 9 - 1 - 1 = 7$.

Samantha-and-Nitesh together (including USA) $= 7 + 1 = 8$ countries. Fact 5: half in Europe → 4 in Europe.

Step 4 — Resolve region by region.

💡 Teacher tip: "Visited by both Nitesh and Samantha" includes countries visited by all three — USA counts here. Always check whether the question says "both" (inclusive of all-three) or "exactly two."

Answering the question.

Samantha-and-Nitesh together total 8 countries. Of these, 4 are in Europe (Fact 5). Samantha visited 0 Asian countries, so none of the S&N countries are in Asia. The remaining $8 - 4 = 4$ must all be in ROW.

From the ROW row of the table: S&N-only = 3, plus USA (all three) = 1, giving $3 + 1 = 4$.

Answer: 4

Finding European countries visited by exactly one person

We first build the full Venn count from the chart and the five clues.

Step 1 — Total visits per person (from the chart).

Sum of all visits $= 11 + 13 + 19 = 43$.

Step 2 — Use the 32 distinct countries (Fact 1).

Let $e_1$ = exactly one, $e_2$ = exactly two, $e_3$ = exactly three.

$e_1 + e_2 + e_3 = 32, \qquad e_1 + 2e_2 + 3e_3 = 43$

Subtracting: $e_2 + 2e_3 = 11$. Fact 2 gives $e_3 = 1$ (USA is the only all-three country), so $e_2 = 9$, $e_1 = 22$.

Step 3 — Split the 9 "exactly-two" countries.

• D & N only = 1 (China, Fact 3).
• D & S only = 1 (France, Fact 4 — Samantha has no Asian country, so France is the only D&S-only).
• S & N only $= 9 - 1 - 1 = 7$.

Samantha-and-Nitesh together (including USA) $= 7 + 1 = 8$. Fact 5: half in Europe → 4 in Europe.

Step 4 — Resolve region by region.

Answering the question.

"Exactly one" means we read the D-only, S-only, and N-only cells of the Europe row:

• Dheeraj-only in Europe: 6
• Samantha-only in Europe: 4
• Nitesh-only in Europe: 2

$6 + 4 + 2 = 12$

Answer: D. 12

How many games each team plays. Six teams are split into two groups of three. Inside a group each team meets the other two once (2 games); against the other group it meets each of the three teams twice ($3\times 2 = 6$ games). So every team plays $2+6 = 8$ games — one per round — meaning 8 rounds of 3 games each.

Finding the two groups. Same-group pairs meet once; different-group pairs meet twice.

1. Teams 1 and 5 meet only once (clue 4), so they are in the same group.
2. Teams 4 and 6 meet twice — in Rounds 1 and 2 (clue 3) — so they are in different groups.
3. Round 8 is entirely cross-group (clue 1). Its three games are 3–6 and 2–5 (clue 6) plus the leftover pair 1–4. So 1&4, 2&5 and 3&6 are each cross-group.
4. Chaining these: 1 and 5 together; 4 opposite 1; 2 opposite 5; 6 opposite 4; 3 opposite 6. This splits the teams cleanly:

$\textbf{Group X} = \{1,5,6\} \qquad \textbf{Group Y} = \{2,3,4\}$

Building the schedule. Using clue 2 (Round 4 repeats in Round 7, Round 5 repeats in Round 8) together with the fixed games — 4–6 in Rounds 1 & 2, 1–5 in Round 2, 3–4 in Round 3, 1–6 in Round 6 — the whole timetable is forced:

This question — how many rounds. Each team plays 2 games inside its group plus 6 against the other group, i.e. 8 games, and every team plays exactly one game per round. So there are 8 rounds.

Answer: 8

How many games each team plays. Six teams are split into two groups of three. Inside a group each team meets the other two once (2 games); against the other group it meets each of the three teams twice ($3\times 2 = 6$ games). So every team plays $2+6 = 8$ games — one per round — meaning 8 rounds of 3 games each.

Finding the two groups. Same-group pairs meet once; different-group pairs meet twice.

1. Teams 1 and 5 meet only once (clue 4), so they are in the same group.
2. Teams 4 and 6 meet twice — in Rounds 1 and 2 (clue 3) — so they are in different groups.
3. Round 8 is entirely cross-group (clue 1). Its three games are 3–6 and 2–5 (clue 6) plus the leftover pair 1–4. So 1&4, 2&5 and 3&6 are each cross-group.
4. Chaining these: 1 and 5 together; 4 opposite 1; 2 opposite 5; 6 opposite 4; 3 opposite 6. This splits the teams cleanly:

$\textbf{Group X} = \{1,5,6\} \qquad \textbf{Group Y} = \{2,3,4\}$

Building the schedule. Using clue 2 (Round 4 repeats in Round 7, Round 5 repeats in Round 8) together with the fixed games — 4–6 in Rounds 1 & 2, 1–5 in Round 2, 3–4 in Round 3, 1–6 in Round 6 — the whole timetable is forced:

This question — who played Team 1 in Round 5. Round 5's games are 1–4, 2–5 and 3–6. Team 1's opponent is Team 4.

Answer: 4

How many games each team plays. Six teams are split into two groups of three. Inside a group each team meets the other two once (2 games); against the other group it meets each of the three teams twice ($3\times 2 = 6$ games). So every team plays $2+6 = 8$ games — one per round — meaning 8 rounds of 3 games each.

Finding the two groups. Same-group pairs meet once; different-group pairs meet twice.

1. Teams 1 and 5 meet only once (clue 4), so they are in the same group.
2. Teams 4 and 6 meet twice — in Rounds 1 and 2 (clue 3) — so they are in different groups.
3. Round 8 is entirely cross-group (clue 1). Its three games are 3–6 and 2–5 (clue 6) plus the leftover pair 1–4. So 1&4, 2&5 and 3&6 are each cross-group.
4. Chaining these: 1 and 5 together; 4 opposite 1; 2 opposite 5; 6 opposite 4; 3 opposite 6. This splits the teams cleanly:

$\textbf{Group X} = \{1,5,6\} \qquad \textbf{Group Y} = \{2,3,4\}$

Building the schedule. Using clue 2 (Round 4 repeats in Round 7, Round 5 repeats in Round 8) together with the fixed games — 4–6 in Rounds 1 & 2, 1–5 in Round 2, 3–4 in Round 3, 1–6 in Round 6 — the whole timetable is forced:

This question — which of Teams 2, 3, 4, 5 is not in the same group as the others. The groups are X = {1, 5, 6} and Y = {2, 3, 4}. Among the four listed, 2, 3 and 4 all sit in Group Y; only Team 5 belongs to Group X.

Answer: 5

How many games each team plays. Six teams are split into two groups of three. Inside a group each team meets the other two once (2 games); against the other group it meets each of the three teams twice ($3\times 2 = 6$ games). So every team plays $2+6 = 8$ games — one per round — meaning 8 rounds of 3 games each.

Finding the two groups. Same-group pairs meet once; different-group pairs meet twice.

1. Teams 1 and 5 meet only once (clue 4), so they are in the same group.
2. Teams 4 and 6 meet twice — in Rounds 1 and 2 (clue 3) — so they are in different groups.
3. Round 8 is entirely cross-group (clue 1). Its three games are 3–6 and 2–5 (clue 6) plus the leftover pair 1–4. So 1&4, 2&5 and 3&6 are each cross-group.
4. Chaining these: 1 and 5 together; 4 opposite 1; 2 opposite 5; 6 opposite 4; 3 opposite 6. This splits the teams cleanly:

$\textbf{Group X} = \{1,5,6\} \qquad \textbf{Group Y} = \{2,3,4\}$

Building the schedule. Using clue 2 (Round 4 repeats in Round 7, Round 5 repeats in Round 8) together with the fixed games — 4–6 in Rounds 1 & 2, 1–5 in Round 2, 3–4 in Round 3, 1–6 in Round 6 — the whole timetable is forced:

This question — who played Team 1 in Round 7. Round 7 repeats Round 4's match-ups: 1–3, 4–5 and 2–6. Team 1's opponent is Team 3.

Answer: 3

How many games each team plays. Six teams are split into two groups of three. Inside a group each team meets the other two once (2 games); against the other group it meets each of the three teams twice ($3\times 2 = 6$ games). So every team plays $2+6 = 8$ games — one per round — meaning 8 rounds of 3 games each.

Finding the two groups. Same-group pairs meet once; different-group pairs meet twice.

1. Teams 1 and 5 meet only once (clue 4), so they are in the same group.
2. Teams 4 and 6 meet twice — in Rounds 1 and 2 (clue 3) — so they are in different groups.
3. Round 8 is entirely cross-group (clue 1). Its three games are 3–6 and 2–5 (clue 6) plus the leftover pair 1–4. So 1&4, 2&5 and 3&6 are each cross-group.
4. Chaining these: 1 and 5 together; 4 opposite 1; 2 opposite 5; 6 opposite 4; 3 opposite 6. This splits the teams cleanly:

$\textbf{Group X} = \{1,5,6\} \qquad \textbf{Group Y} = \{2,3,4\}$

Building the schedule. Using clue 2 (Round 4 repeats in Round 7, Round 5 repeats in Round 8) together with the fixed games — 4–6 in Rounds 1 & 2, 1–5 in Round 2, 3–4 in Round 3, 1–6 in Round 6 — the whole timetable is forced:

This question — who played Team 6 in Round 3. Round 3's games are 3–4, 1–2 and 5–6. Team 6's opponent is Team 5.

Answer: 5