CAT 2019 DILR Questions — Slot 1

We are given the cryptarithm:

B H A A G F

• A H J F K F

---

A A F G C A F

Each letter stands for a distinct digit (0–9). Let us work column by column from right to left, tracking carries.

Key observation — the sum has 7 digits, but each addend has only 6.

The leftmost digit of the sum (A) is produced entirely by a carry out of the hundred-thousands column. When we add two 6-digit numbers, the only possible carry into a new (7th) digit is 1. Therefore:

$A = 1$

We can verify this is consistent by working through all the columns:

Everything is consistent with $A = 1$.

💡 Teacher tip: In any cryptarithm where the sum has more digits than the addends, the extra leading digit is always 1 — it can only come from a final carry.

Answer: 1

We are given the cryptarithm:

B H A A G F

• A H J F K F

---

A A F G C A F

Each letter stands for a distinct digit (0–9). Let us work column by column from right to left, tracking carries $c_1, c_2, \ldots$

Step 1 — Determine A.

The sum has 7 digits while each addend has 6, so the leading A is a carry of 1: $A = 1$

Step 2 — Work through each column.

Step 3 — Focus on the hundred-thousands column for B.

In the leftmost addition column of the two addends, we have: $B + A + c_5 = A + 10 \cdot c_6$

We already know $A = 1$ and the carry from the ten-thousands column is $c_5 = 1$ (since $2H = 2 \times 5 = 10$, which produces a carry of 1). Substituting: $B + 1 + 1 = 1 + 10 \cdot c_6$ $B + 1 = 10 \cdot c_6$

Since B is a single digit (0–9) and B ≠ 0 (because F = 0 already), the only solution is $c_6 = 1$ and: $B = 9$

💡 Teacher tip: The carry into the hundred-thousands column ($c_5 = 1$) comes from the H + H column. Since $H = 5$, we get $2 \times 5 = 10$, which writes 0 (matching F = 0) and carries 1.

Answer: 9

We are given the cryptarithm:

B H A A G F

• A H J F K F

---

A A F G C A F

Each letter stands for a distinct digit (0–9). Working column by column from right to left:

So the fixed assignments are: A = 1, F = 0, C = 2, H = 5, B = 9.

The remaining digits are {3, 4, 6, 7, 8}, to be shared among G, J, K and the two unused letters D and E.

Step 1 — Apply the two constraints on G, J, K.

From the columns above:

• $G = J + 1$ (thousands column)
• $G + K = 11$ (tens column), so $K = 11 - G$

Step 2 — Enumerate all valid assignments.

J must be from {3, 4, 6, 7, 8}, and both $G = J + 1$ and $K = 11 - G$ must also be unused digits from the same set:

(J = 4 gives G = 5, but H = 5 already; J = 8 gives G = 9, but B = 9 already — both ruled out.)

Step 3 — Determine which digits D can take.

The letters D and E never appear in the sum, so they simply take whatever two digits are left over in each case. Across the three valid cases, the leftover sets are:

• {6, 8}
• {3, 8}
• {4, 6}

So D can be 3, 4, 6, or 8 — but never 7. In every valid case, 7 is always assigned to G, J, or K.

💡 Teacher tip: When a letter doesn't appear in the puzzle, its possible values are exactly the union of all leftover digits across every valid assignment. The digit that never appears in any leftover set is the answer.

Answer: 7

We are given the cryptarithm:

B H A A G F

• A H J F K F

---

A A F G C A F

Each letter stands for a distinct digit (0–9). Working column by column from right to left:

So the fixed assignments are: A = 1, F = 0, C = 2, H = 5, B = 9.

The remaining digits are {3, 4, 6, 7, 8}, to be shared among G, J, K and the two unused letters D and E.

Step 1 — Apply the two constraints on G, J, K.

From the columns above:

• $G = J + 1$ (thousands column)
• $G + K = 11$ (tens column), so $K = 11 - G$

Step 2 — Enumerate all valid assignments.

J must be from {3, 4, 6, 7, 8}, and both $G = J + 1$ and $K = 11 - G$ must also be unused digits from the same set:

(J = 4 gives G = 5, but H = 5 already; J = 8 gives G = 9, but B = 9 already — both ruled out.)

Step 3 — Determine which digits G can take.

Looking at the G column across all three valid cases, G takes the values 4, 7, or 8. It is never 6.

💡 Teacher tip: To check whether a particular digit can be assigned to a letter, enumerate every valid solution and collect all values that letter takes. If a digit never appears in that collection, it cannot be that letter.

Answer: 6

Finding the highest total score

Key rule: Every bull's eye (score of 5) in Rounds 1–3 earns the player one bonus round (Rounds 4–6). So the number of bonus rounds a player actually played tells us exactly how many 5s they scored in the first three rounds.

Step 1 — Count each player's bonus rounds from the table.

Step 2 — Pin down what we can immediately.

• Xyla scored 5 in all three compulsory rounds, then bonus rounds of 1, 5, and an unknown R6. Her total so far: $5+5+5+1+5+R_6 = 21 + R_6$.
• Wangdu has no 5s at all and R2 = 4, so R1, R3 ∈ {1, 2, 3, 4}. His total = $R_1 + 4 + R_3$, which is at most 12.
• Tanzi has exactly one 5 in R1–R3, and R2 = 4 (not 5), so her 5 is in R1 or R3. Her total = $R_1 + 4 + R_3 + 5$.
• Yonita has exactly one 5 in R1–R3, and R3 = 3 (not 5), so her 5 is in R1 or R2. Her total = $R_1 + R_2 + 3 + 5$.

Step 3 — Use Clue 4 (bull's eyes in R2 = 2 × bull's eyes in R3).

Let $B_2$ = number of players scoring 5 in Round 2, and $B_3$ = number scoring 5 in Round 3. We know $B_2 = 2B_3$.

• Xyla contributes to both $B_2$ and $B_3$.
• Wangdu (R2 = 4, no 5s) contributes to neither.
• Tanzi (R2 = 4) cannot contribute to $B_2$.
• Yonita (R3 = 3) cannot contribute to $B_3$.

Since $B_3 \geq 1$ (Xyla), try $B_3 = 1$: only Xyla has R3 = 5. Then $B_2 = 2$, meaning exactly one other player has R2 = 5. But Umeza (2 bull's eyes, R3 ≠ 5) would need both 5s in R1 and R2, and Zeneca (2 bull's eyes, R3 ≠ 5) would also need both 5s in R1 and R2 — giving $B_2 \geq 3$. Contradiction.

So $B_3 = 2$ and $B_2 = 4$. Besides Xyla, exactly one more player has R3 = 5, and all three of Umeza, Yonita, Zeneca must have R2 = 5 (since Tanzi and Wangdu cannot).

Step 4 — Identify who has the second R3 = 5.

• If Zeneca had R3 = 5: her two 5s are R2 and R3, so R1 ≠ 5. But Clue 5 says Tanzi and Zeneca share the same R1. Tanzi's single 5 must be in R1 (since R3 = 5 is taken by Xyla and Zeneca, and R2 = 4), giving Tanzi R1 = 5. Then Zeneca R1 = 5 too — but we just said Zeneca R1 ≠ 5. Contradiction.
• If Tanzi had R3 = 5: her single 5 is R3, so R1 ≠ 5. Clue 5 requires Zeneca R1 = Tanzi R1 ≠ 5. But Zeneca's two 5s are R2 and (since R3 ≠ 5) R1, forcing R1 = 5. Contradiction.
• So Umeza has R3 = 5. Her two 5s are R2 and R3, so R1 ≠ 5.

Step 5 — Apply Clue 5 to fix Tanzi and Zeneca.

With Umeza having the second R3 = 5, Tanzi's R3 ≠ 5, so Tanzi's single 5 is in R1: Tanzi R1 = 5. Clue 5 says Zeneca R1 = Tanzi R1 = 5. Zeneca's two 5s are R1 and R2 (R3 ≠ 5), so Zeneca R3 ∈ {1, 2, 3, 4} and differs from Tanzi's R3.

Step 6 — Use Clue 1 (Tanzi = Umeza = Yonita) to find the common total.

• Tanzi: $5 + 4 + R_3^T + 5 = 14 + R_3^T$
• Umeza: $R_1^U + 5 + 5 + 1 + 2 = 13 + R_1^U$ (R1 ≠ 5, so $R_1^U \in \{1,2,3,4\}$)
• Yonita: $R_1^Y + 5 + 3 + 5 = 13 + R_1^Y$ (R1 ≠ 5, so $R_1^Y \in \{1,2,3,4\}$)

Setting them equal: $14 + R_3^T = 13 + R_1^U = 13 + R_1^Y$, so $R_1^U = R_1^Y = 1 + R_3^T$.

Since $R_1^U \leq 4$, we get $R_3^T \leq 3$. Testing each:

• $R_3^T = 1$: common total = 15 (multiple of 3 ✓)
• $R_3^T = 2$: common total = 16 (not a multiple of 3 — but Clue 2 allows only one non-multiple, and three players share this total, so three non-multiples ✗)
• $R_3^T = 3$: common total = 17 (same problem ✗)

So $R_3^T = 1$, common total = 15, $R_1^U = R_1^Y = 2$.

Step 7 — Use Clue 3 (highest = 2 × lowest + 1) to fix remaining values.

Wangdu's total (at most 12) is the lowest. The highest is either Xyla ($21 + R_6$) or Zeneca ($20 + R_3^Z$).

• If lowest = 12 (Wangdu R1 = R3 = 4): highest = $2 \times 12 + 1 = 25$. Xyla = 25 → $R_6 = 4$. Zeneca = $20 + R_3^Z \leq 24 < 25$ ✓.
• Check Clue 2: totals are 15, 15, 12, 25, 15, 24. Only 25 is not a multiple of 3 ✓.
• Check Clue 5: Tanzi R3 = 1, Zeneca R3 = 4 (different ✓); both R1 = 5 (same ✓).

Completed table:

Xyla has the highest total: $5+5+5+1+5+4 = 25$.

Answer: 25 (Choice A)

Finding Zeneca's total score

Key rule: Every bull's eye (score of 5) in Rounds 1–3 earns the player one bonus round (Rounds 4–6). So the number of bonus rounds a player actually played tells us exactly how many 5s they scored in the first three rounds.

Step 1 — Count each player's bonus rounds from the table.

Step 2 — Use Clue 4 to determine who scored 5 in Rounds 2 and 3.

Clue 4 says the number of bull's eyes in Round 2 was double that in Round 3. Let $B_2$ and $B_3$ be those counts.

• Xyla (R2 = 5, R3 = 5) contributes to both.
• Wangdu (R2 = 4, no 5s) contributes to neither.
• Tanzi (R2 = 4) cannot contribute to $B_2$.
• Yonita (R3 = 3) cannot contribute to $B_3$.

Testing $B_3 = 1$ leads to a contradiction (Umeza and Zeneca would both need R2 = 5, pushing $B_2$ too high). So $B_3 = 2$, $B_2 = 4$, meaning Umeza, Yonita, and Zeneca all have R2 = 5, and besides Xyla, exactly one more player has R3 = 5.

Step 3 — Identify the second R3 = 5 and apply Clue 5.

Checking each candidate:

• Zeneca R3 = 5: her two 5s are R2, R3 → R1 ≠ 5. But Clue 5 requires Zeneca R1 = Tanzi R1, and Tanzi's single 5 must be R1 (since R2 = 4 and R3 is taken) → Tanzi R1 = 5 → Zeneca R1 = 5. Contradiction.
• Tanzi R3 = 5: her single 5 is R3 → R1 ≠ 5. Clue 5 requires Zeneca R1 = Tanzi R1 ≠ 5. But Zeneca's two 5s (R2 and R1, since R3 ≠ 5) force R1 = 5. Contradiction.
• Umeza R3 = 5: her two 5s are R2, R3 → R1 ≠ 5. This works.

So Umeza has R3 = 5. Tanzi's single 5 is in R1 → Tanzi R1 = 5. By Clue 5, Zeneca R1 = 5 as well. Zeneca's two 5s are R1 and R2, so Zeneca R3 ∈ {1, 2, 3, 4} (and differs from Tanzi's R3).

Step 4 — Use Clue 1 (Tanzi = Umeza = Yonita) to find the common total.

• Tanzi: $5 + 4 + R_3^T + 5 = 14 + R_3^T$
• Umeza: $R_1^U + 5 + 5 + 1 + 2 = 13 + R_1^U$
• Yonita: $R_1^Y + 5 + 3 + 5 = 13 + R_1^Y$

Setting equal: $R_1^U = R_1^Y = 1 + R_3^T$. Since $R_1^U \leq 4$, we get $R_3^T \leq 3$. Clue 2 (all but one total is a multiple of 3) eliminates $R_3^T = 2$ (total 16) and $R_3^T = 3$ (total 17) because three players would share a non-multiple-of-3 total. So $R_3^T = 1$, common total = 15, $R_1^U = R_1^Y = 2$.

Step 5 — Use Clue 3 to fix Zeneca's R3 and Xyla's R6.

Wangdu (no 5s, R2 = 4) has the lowest total, at most 12. The highest is Xyla ($21 + R_6$) or Zeneca ($20 + R_3^Z$).

• If Wangdu = 12 (R1 = R3 = 4): highest = $2 \times 12 + 1 = 25$ → Xyla = 25 ($R_6 = 4$), Zeneca = $20 + R_3^Z \leq 24$.
• Clue 2 check: totals 15, 15, 12, 25, 15, 24 — only 25 is not a multiple of 3 ✓.
• Clue 5 check: Tanzi R3 = 1, Zeneca R3 = 4 (different ✓); both R1 = 5 (same ✓).

So Zeneca R3 = 4.

Completed table:

Zeneca's total = $5 + 5 + 4 + 5 + 5 = 24$.

Answer: 24 (Choice D)

Determining which statement is true

Key rule: Every bull's eye (score of 5) in Rounds 1–3 earns the player one bonus round (Rounds 4–6). So the number of bonus rounds a player actually played tells us exactly how many 5s they scored in the first three rounds.

Step 1 — Count each player's bonus rounds from the table.

Step 2 — Use Clue 4 to determine who scored 5 in Rounds 2 and 3.

Clue 4: bull's eyes in R2 = 2 × bull's eyes in R3. After testing possibilities, the only consistent solution is $B_3 = 2$, $B_2 = 4$, meaning Umeza, Yonita, and Zeneca all have R2 = 5, and besides Xyla, Umeza has R3 = 5.

Step 3 — Apply Clue 5 and Clue 1 to fix remaining scores.

• Tanzi's single 5 is in R1 (R2 = 4, R3 ≠ 5) → Tanzi R1 = 5.
• Clue 5: Zeneca R1 = Tanzi R1 = 5. Zeneca's two 5s are R1, R2 → Zeneca R3 ∈ {1, 2, 3, 4}.
• Clue 1 (Tanzi = Umeza = Yonita) combined with Clue 2 (all but one total is a multiple of 3) forces the common total to 15, with Tanzi R3 = 1, Umeza R1 = 2, Yonita R1 = 2.

Step 4 — Use Clue 3 to fix the last unknowns.

Wangdu has the lowest total (at most 12). If Wangdu = 12 (R1 = R3 = 4), then highest = $2 \times 12 + 1 = 25$ → Xyla R6 = 4, Zeneca R3 = 4. All clues check out.

Completed table:

Step 5 — Evaluate each option.

• A. Xyla's score was 23 → Xyla scored 25, not 23. ✗
• B. Zeneca's score was 23 → Zeneca scored 24, not 23. ✗
• C. Zeneca was the highest scorer → Zeneca scored 24, but Xyla scored 25. ✗
• D. Xyla was the highest scorer → Xyla scored 25, the highest of all six players. ✓

Answer: Xyla was the highest scorer (Choice D)

Finding Tanzi's Round 3 score

Key rule: Every bull's eye (score of 5) in Rounds 1–3 earns the player one bonus round (Rounds 4–6). So the number of bonus rounds a player actually played tells us exactly how many 5s they scored in the first three rounds.

Step 1 — Count each player's bonus rounds from the table.

Step 2 — Focus on Tanzi.

Tanzi played exactly one bonus round (R4), so she scored exactly one 5 in Rounds 1–3. From the table, her R2 = 4 (not a 5), so her single bull's eye is in R1 or R3.

Her total = $R_1 + 4 + R_3 + 5 = R_1 + R_3 + 9$, where exactly one of $R_1, R_3$ equals 5 and the other is in {1, 2, 3, 4}.

Step 3 — Use Clue 4 to determine the R2/R3 bull's-eye distribution.

Clue 4: bull's eyes in R2 = 2 × bull's eyes in R3. After testing, the only consistent solution is $B_3 = 2$ (Xyla and Umeza have R3 = 5) and $B_2 = 4$ (Xyla, Umeza, Yonita, Zeneca have R2 = 5).

Since Umeza (not Tanzi) has the second R3 = 5, Tanzi's R3 ≠ 5. Therefore Tanzi's single bull's eye must be in R1: Tanzi R1 = 5.

Step 4 — Apply Clue 5 to confirm.

Clue 5 says Tanzi and Zeneca have the same R1 score. Zeneca's two 5s are in R1 and R2 (R3 ≠ 5), so Zeneca R1 = 5. This matches Tanzi R1 = 5 ✓. Their R3 scores must differ — Tanzi R3 ≠ Zeneca R3 ✓ (we'll confirm values below).

Step 5 — Use Clue 1 and Clue 2 to find Tanzi's total and R3.

• Tanzi: $5 + 4 + R_3^T + 5 = 14 + R_3^T$
• Umeza: $R_1^U + 5 + 5 + 1 + 2 = 13 + R_1^U$ (R1 ≠ 5)
• Yonita: $R_1^Y + 5 + 3 + 5 = 13 + R_1^Y$ (R1 ≠ 5)

Clue 1 sets these equal: $14 + R_3^T = 13 + R_1^U = 13 + R_1^Y$, so $R_1^U = R_1^Y = 1 + R_3^T$.

Since $R_1^U \leq 4$, we get $R_3^T \leq 3$. Testing:

• $R_3^T = 1$ → common total = 15 (multiple of 3 ✓)
• $R_3^T = 2$ → common total = 16 (not a multiple of 3; three players share it, violating Clue 2 ✗)
• $R_3^T = 3$ → common total = 17 (same problem ✗)

So Tanzi R3 = 1.

Verification with the completed table:

All five clues are satisfied: Tanzi's total = $5 + 4 + 1 + 5 = 15$ ✓, and her R3 = 1 differs from Zeneca's R3 = 4 ✓.

Answer: 1 (Choice D)

We need to find which aspect has the smallest median score across the five vendors. With five vendors, the median is simply the third value when the five scores are arranged in ascending order.

Reading the radar chart: the rings are 10 points apart and the centre is 0, so each score is read against the nearest ring. The set's own examples calibrate the read — Vendor 1 Reliability 52, Vendor 2 Features 45, Vendor 3 Cost 90. The readings:

Step 1 — Sort each aspect's five scores and pick the middle (3rd) value.

Step 2 — Identify the smallest median.

The medians are: Reliability 52, Reach 63, Quality 62, Features 56, Customer Service 50, Cost 77.

The lowest median is 50, which belongs to Customer Service.

💡 Teacher tip: With an odd number of values (here 5), the median is the middle value after sorting — no averaging needed. Just sort and pick the 3rd.

Answer: A. Customer Service

A vendor's final score is the average of their six aspect scores. Since every vendor has the same number of aspects (six), the vendor with the highest total will also have the highest average — so we can simply sum each vendor's row and compare.

Reading the radar chart: the rings are 10 points apart and the centre is 0, so each score is read against the nearest ring. The set's own examples calibrate the read — Vendor 1 Reliability 52, Vendor 2 Features 45, Vendor 3 Cost 90. The readings:

Step 1 — Sum each vendor's six scores.

• Vendor 1: $52 + 80 + 72 + 40 + 55 + 77 = 376$
• Vendor 2: $40 + 58 + 69 + 45 + 41 + 81 = 334$
• Vendor 3: $75 + 63 + 62 + 56 + 50 + 90 = 396$
• Vendor 4: $26 + 46 + 40 + 90 + 70 + 71 = 343$
• Vendor 5: $60 + 70 + 48 + 75 + 28 + 50 = 331$

Step 2 — Divide by 6 to get the average (final score).

Vendor 3 has the highest final score at 66.0.

💡 Teacher tip: When comparing averages across groups of equal size, you can skip the division and just compare the sums — the ranking is identical.

Answer: D. Vendor 3

We need to find which vendors appear among the top two scorers on the maximum number of aspects. The approach: for each aspect, identify the two vendors with the highest scores, then count how many times each vendor appears in these top-two lists.

Reading the radar chart: the rings are 10 points apart and the centre is 0, so each score is read against the nearest ring. The set's own examples calibrate the read — Vendor 1 Reliability 52, Vendor 2 Features 45, Vendor 3 Cost 90. The readings:

Step 1 — Identify the top two scorers on each aspect.

Step 2 — Tally appearances.

Vendor 1 and Vendor 5 each appear in the top two on 3 aspects — more than any other vendor.

💡 Teacher tip: Read each row carefully to spot the two largest values. A quick scan from highest to lowest per row keeps you from missing a tie or a close call.

Answer: B. Vendor 1 and Vendor 5

We need to find every vendor who is among the top three scorers on all six aspects — meaning the vendor must appear in the top three of every single row. If a vendor falls outside the top three on even one aspect, they are eliminated.

Reading the radar chart: the rings are 10 points apart and the centre is 0, so each score is read against the nearest ring. The set's own examples calibrate the read — Vendor 1 Reliability 52, Vendor 2 Features 45, Vendor 3 Cost 90. The readings:

Step 1 — Identify the top three scorers on each aspect.

Step 2 — Check which vendor appears in every row.

• Vendor 1: appears in Reliability ✓, Reach ✓, Quality ✓, Features ✗ (scored only 40, ranked 4th), Customer Service ✓, Cost ✓ → eliminated (misses Features).
• Vendor 3: appears in Reliability ✓, Reach ✓, Quality ✓, Features ✓, Customer Service ✓, Cost ✓ → appears in all six.
• Vendor 2: misses Reliability, Reach, Features, Customer Service → eliminated.
• Vendor 4: misses Reliability, Reach, Quality, Cost → eliminated.
• Vendor 5: misses Quality, Customer Service, Cost → eliminated.

Only Vendor 3 is among the top three on all six aspects.

💡 Teacher tip: The fastest way is to pick the vendor who looks strongest overall (here Vendor 3, who had the highest average in Question 2) and verify them first. Then check whether any other vendor also clears all six rows.

Answer: C. Vendor 3

We need to place 12 items on 16 shelves, so 4 shelves stay empty. The type counts are 5 biscuits, 3 candies, and 4 savouries (since $12 - 5 - 3 = 4$).

Step 1 — Fix the savoury block

Clue 3 says D, E, F are savouries placed after all biscuits and candies, so the savoury block is the last block. Clue 4 puts K at shelf 16. Since K sits after all biscuits and candies, K must also be a savoury. Therefore the 4 savouries are D, E, F, K occupying shelves 13–16:

A gap is required between different types, so shelf 12 is empty.

Step 2 — Identify the types of every item

• Clue 5: L and J are the same type; H is a different type.
• If L and J were candies, then I (consecutive to J by clue 2) would also be a candy — giving 4 candies, which is too many. So L and J are biscuits, and I (adjacent to J) is a biscuit too.
• H differs from L/J, so H is not a biscuit → H is a candy.
• Clue 6 says C is a candy. Clue 1 says A and B are adjacent, hence the same type. If A and B were candies we would have C, H, A, B = 4 candies — too many. So A and B are biscuits, and the third candy is G.

Types: Biscuits = A, B, I, J, L · Candies = C, G, H · Savouries = D, E, F, K.

Step 3 — Fix the internal order of each block

• Clue 7: L has exactly one empty shelf before it, so L starts the biscuit block. Inside the block: L, A, B, then I, J (A–B increasing; I–J both higher than A, B). The order of I and J is not fixed → 2 choices.
• Clue 6: C has two empty shelves before it, so C starts the candy block: C, then G, H. The order of G and H is not fixed → 2 choices.

Step 4 — Place the two blocks before the savouries

The biscuit block and candy block can appear in either order before the savoury block → 2 choices. All 4 empty shelves are accounted for: two before C, one before L, and shelf 12.

Family 1 — candies first:

Family 2 — biscuits first:

Step 5 — Count

Total arrangements = 2 (block order) × 2 (G/H swap) × 2 (I/J swap): $2 \times 2 \times 2 = 8.$

Answer: 8 (Choice A)

We need to determine each item's type. There are 12 items on 16 shelves (4 empty), split as 5 biscuits, 3 candies, 4 savouries.

Step 1 — Savouries

Clue 3 puts D, E, F (savouries) after all biscuits and candies. Clue 4 puts K at shelf 16, which is after all biscuits and candies, so K is also a savoury. Thus savouries = D, E, F, K at shelves 13–16, with shelf 12 empty.

Step 2 — L, J, I, H

Clue 5: L and J are the same type; H is different. If L and J were candies, then I (consecutive to J, clue 2) would also be a candy — 4 candies is too many. So L, J, I are biscuits. H differs from them, so H is a candy.

Step 3 — A, B, G, C

Clue 6: C is a candy. Clue 1: A and B are adjacent, hence the same type. If A and B were candies, we would have C, H, A, B = 4 candies — too many. So A and B are biscuits, and the remaining candy is G.

Type summary

Among the options L, A, B, G, the item G is a candy, not a biscuit.

Answer: G (Choice D)

We first build the full arrangement. There are 12 items on 16 shelves (4 empty), with types 5 biscuits, 3 candies, 4 savouries.

Deductions

1. Savouries: D, E, F are savouries after all biscuits/candies (clue 3); K at shelf 16 (clue 4) is also a savoury. So D=13, E=14, F=15, K=16, and shelf 12 is empty.
2. Types: L, J, I are biscuits; H is a candy; A, B are biscuits; C and G are candies. So Biscuits = A, B, I, J, L; Candies = C, G, H; Savouries = D, E, F, K.
3. Block starts: L has one empty before it (clue 7) → L starts the biscuit block. C has two empties before it (clue 6) → C starts the candy block.
4. The biscuit and candy blocks can come in either order before the savouries. The 4 empty shelves are always: the two before C, the one before L, and shelf 12.

The two families

Candies first — empties at 1, 2, 6, 12:

Biscuits first — empties at 1, 7, 8, 12:

Checking the options

The only possible empty-shelf sets are {1, 2, 6, 12} and {1, 7, 8, 12}.

• A. 1, 7, 11, 12 — does not match either family.
• B. 1, 5, 6, 12 — does not match either family.
• C. 1, 2, 6, 12 — matches the candies-first arrangement. ✓
• D. 1, 2, 8, 12 — does not match either family.

Answer: 1, 2, 6, 12 (Choice C)

We need the statement that holds in every valid arrangement. Let us first build the two possible families.

Setup

12 items on 16 shelves (4 empty); types are 5 biscuits, 3 candies, 4 savouries.

• Savouries D, E, F, K occupy shelves 13–16 (clues 3, 4); shelf 12 is empty.
• Biscuits = A, B, I, J, L; Candies = C, G, H.
• L starts the biscuit block (one empty before it, clue 7); C starts the candy block (two empties before it, clue 6).
• The biscuit and candy blocks may come in either order before the savouries.

The two families

Candies first:

Here B is at shelf 9 and C is at shelf 3.

Biscuits first:

Here B is at shelf 4 and C is at shelf 9.

Testing each option

• A. All biscuits before candies — true only in the biscuits-first family, false in the other. Not necessary.
• B. Two empty shelves between biscuits and candies — in the candies-first family there is only one empty shelf (shelf 6) between the candy and biscuit blocks. Not necessary.
• C. All candies before biscuits — true only in the candies-first family. Not necessary.
• D. At least four shelves between B and C.
• Candies first: B at 9, C at 3 → shelves between = 4, 5, 6, 7, 8 = 5 shelves.
• Biscuits first: B at 4, C at 9 → shelves between = 5, 6, 7, 8 = 4 shelves.

In both families there are at least four shelves between B and C. (The G↔H and I↔J swaps do not move B or C, so this holds in all 8 arrangements.)

Answer: There are at least four shelves between items B and C. (Choice D)

Finding Kerala's IPC rank

Reading the stacked bars. Each bar is split into IPC (blue), SLL (red) and Other (green) crimes, and a segment's length is its count. Read each boundary off the axis (gridlines every 10) and subtract — e.g. Kerala's blue segment ends near 8, the red runs from 8 to about 23 (≈15 SLL), and the green from 23 to about 34 (≈11 Other). Reads are good to about ±1, which is why some counts below are ranges. West Bengal's 520 SLL cases come from the set-up, not the chart.

We need to order the ten states/UTs by their IPC crime counts and use the known ranks to pin down Kerala's position.

Step 1 — List IPC counts (from the bar chart and West Bengal data):

Step 2 — Use the given ranks to anchor the ordering.

From the rank table:

• Karnataka → IPC rank 3
• Maharashtra → IPC rank 3
• Telangana → IPC rank 6

Since Karnataka and Maharashtra share rank 3, exactly 2 states have more IPC cases than them. Looking at the counts, those two must be Delhi (63–64) and Goa (25–26). So:

• Delhi = rank 1
• Goa = rank 2
• Karnataka & Maharashtra = rank 3 (tied)

Step 3 — Determine Kerala's position.

After a two-way tie at rank 3, the next available rank is 5 (ranks 1, 2, 3, 3 are taken, so the next state gets rank 5).

Telangana has rank 6, meaning exactly 5 states exceed its IPC count of 3–4. Those five are: Delhi, Goa, Karnataka, Maharashtra, and Kerala (7–8). Since Kerala's IPC count (7–8) is the next highest after the Karnataka/Maharashtra pair, Kerala takes rank 5.

💡 Teacher tip: When two states tie at rank r, the next state gets rank r + 2 (not r + 1), because the tie "uses up" two positions. Here, the tie at rank 3 means the next rank is 5.

Answer: 5

IPC: SLL ratio in the two states with the highest total cases

Reading the stacked bars. Each bar is split into IPC (blue), SLL (red) and Other (green) crimes, and a segment's length is its count. Read each boundary off the axis (gridlines every 10) and subtract — e.g. Kerala's blue segment ends near 8, the red runs from 8 to about 23 (≈15 SLL), and the green from 23 to about 34 (≈11 Other). Reads are good to about ±1, which is why some counts below are ranges. West Bengal's 520 SLL cases come from the set-up, not the chart.

Step 1 — Identify the two states with the highest totals.

The two highest are West Bengal (520) and Delhi (~142) — far ahead of the rest.

Step 2 — Extract their IPC and SLL counts.

West Bengal's cases are entirely SLL (given in the set-up), so its IPC contribution is 0.

Step 3 — Compute the combined IPC and SLL totals.

$\text{Total IPC} = 0 + 64 \approx 64$

$\text{Total SLL} = 520 + 36 \approx 556$

Step 4 — Find the ratio.

$\frac{\text{IPC}}{\text{SLL}} = \frac{64}{556} \approx 0.115 \approx \frac{1}{9}$

This is overwhelmingly dominated by West Bengal's 520 SLL cases, making the ratio very small — close to $1:9$.

💡 Teacher tip: When one state contributes 520 out of ~556 total SLL cases, the ratio is driven almost entirely by that one state. The small IPC contribution from Delhi (~64) against this huge SLL number points directly to $1:9$.

Answer: D. 1: 9

Definite ranks in the 'Other Crimes' category

Reading the stacked bars. Each bar is split into IPC (blue), SLL (red) and Other (green) crimes, and a segment's length is its count. Read each boundary off the axis (gridlines every 10) and subtract — e.g. Kerala's blue segment ends near 8, the red runs from 8 to about 23 (≈15 SLL), and the green from 23 to about 34 (≈11 Other). Reads are good to about ±1, which is why some counts below are ranges. West Bengal's 520 SLL cases come from the set-up, not the chart.

We need to check whether Tamil Nadu is definitely rank 2 and Puducherry is definitely rank 3 in 'Other Crimes'.

Step 1 — List 'Other Crimes' counts for all ten states/UTs.

Step 2 — Use the known rank-8 information as an anchor.

From the rank table, both Maharashtra and Telangana have Other-crimes rank 8. By the ranking rule, rank 8 means exactly 7 states/UTs have a larger count of other crimes.

Step 3 — Count how many of the ten listed states exceed Telangana's other-crimes count (~6–7).

Going down the list:

1. Delhi (42–43) ✓
2. Tamil Nadu (35–36) ✓
3. Puducherry (30) ✓
4. Karnataka (25–26) ✓
5. Goa (18–19) ✓
6. Kerala (10–11) ✓
7. Haryana (9–10) ✓

That is exactly 7 states — matching the requirement for rank 8.

💡 Teacher tip: This is the critical deduction. Since rank 8 allows exactly 7 states above, and we've found exactly 7 from the listed states, no unlisted state can sit among the top 7. The entire top-8 ordering is locked.

Step 4 — Assign the ranks.

Both statements are forced:

• (i) Tamil Nadu = rank 2 ✓
• (ii) Puducherry = rank 3 ✓

Answer: A. Both i) and ii)

Sum of Delhi's ranks across all three crime categories

Reading the stacked bars. Each bar is split into IPC (blue), SLL (red) and Other (green) crimes, and a segment's length is its count. Read each boundary off the axis (gridlines every 10) and subtract — e.g. Kerala's blue segment ends near 8, the red runs from 8 to about 23 (≈15 SLL), and the green from 23 to about 34 (≈11 Other). Reads are good to about ±1, which is why some counts below are ranges. West Bengal's 520 SLL cases come from the set-up, not the chart.

Delhi's ranks are all missing (marked *), so we must deduce each one from the data.

Step 1 — Delhi's IPC rank.

Delhi has the highest IPC count among all ten states (63–64, far ahead of Goa at 25–26). No unlisted state can exceed it either, since Delhi is among the top-ten states by total cases and its IPC count dwarfs every other visible state. → IPC rank = 1.

Step 2 — Delhi's Other Crimes rank.

Delhi has the highest Other-crimes count (42–43). As established in the previous question, Delhi is rank 1 in Other crimes. → Other rank = 1.

Step 3 — Delhi's SLL rank.

From the rank table:

• Karnataka → SLL rank 2 (given)
• Goa → SLL rank 4 (given)
• Maharashtra → SLL rank 4 (given)

West Bengal (520) is clearly rank 1, and Karnataka (48–49) is rank 2 — both confirmed.

Goa and Maharashtra share rank 4, meaning exactly 3 states have more SLL cases than them. Two of those three are West Bengal and Karnataka. The third must be a state with SLL cases ≥ 35–36 that is strictly greater than Goa and Maharashtra.

Delhi's SLL count (35–36) is at the same approximate level as Goa and Maharashtra. But since Goa and Maharashtra are rank 4 (not rank 3), Delhi must have slightly more SLL cases than them — a difference too small for the bar chart to resolve, but pinned down by the rank table. Therefore Delhi sits alone at rank 3.

→ SLL rank = 3

💡 Teacher tip: When the bar chart gives approximate values that look equal (all ~35–36), use the rank table to break the tie. Goa and Maharashtra being rank 4 (not 3) proves that exactly one more state sits above them — and Delhi is the only candidate.

Step 4 — Sum the three ranks.

$\text{Sum} = 1\;(\text{IPC}) + 3\;(\text{SLL}) + 1\;(\text{Other}) = \mathbf{5}$

Answer: 5

Setting up the core constraint

Let the number of items of each type be $n_a, n_b, n_c, \dots$ We are told:

1. There is exactly one type-a item (the diamond ring), so $n_a = 1$.
2. Each type has at least twice as many items as the previous type: $n_b \ge 2n_a$, $n_c \ge 2n_b$, and so on.
3. The total across all types is $100$.

Finding the minimum number of types

We want the fewest types possible.

• One type alone is impossible: the only type would have to be type a, but there is exactly one type-a item — not 100.
• Two types: set $n_a = 1$ and $n_b = 99$. Check the doubling rule: $99 \ge 2 \times 1 = 2$ ✓. The total is $1 + 99 = 100$ ✓.

So two types are achievable, and one type is not. The minimum is therefore 2.

Answer: 2

Setting up the core constraint

Let the number of items of each type be $n_a, n_b, n_c, \dots$ with:

• $n_a = 1$ (exactly one diamond ring),
• each next type at least double the previous: $n_{i+1} \ge 2\,n_i$,
• total $= 100$.

Finding the maximum number of types

To fit as many types as possible, we make each type as small as the rules allow — the tightest chain is

$1,\ 2,\ 4,\ 8,\ 16,\ 32,\ 64,\ \dots$

whose running totals are:

• Seven types would need at least $1+2+4+8+16+32+64 = 127$ items, but we only have 100 — impossible.
• Six types need only $63$, which is well within 100. The remaining $100 - 63 = 37$ items can be added to the last type, giving for example $1, 2, 4, 8, 16, 69$ (each count is still at least double the previous, and the sum is 100). ✓

💡 Teacher tip: To maximise the number of types, shrink each type to its minimum allowed size; to minimise the number of types, dump as many items as possible into one type.

Answer: 6

Recalling the constraint

Each type has at least twice as many items as the previous type, with $n_a = 1$ and total $= 100$.

We test each option by trying to build a valid chain that includes the stated count and sums to exactly 100.

Checking each option

Option A — 75 items of type e. Try $1, 2, 4, 18, 75$:

• $2 \ge 2$ ✓, $4 \ge 4$ ✓, $18 \ge 8$ ✓, $75 \ge 36$ ✓
• Sum $= 1+2+4+18+75 = 100$ ✓

Possible.

Option B — 30 items of type b. With $n_a = 1,\ n_b = 30$, the remaining $69$ items go into type c (or later). We need $n_c \ge 2 \times 30 = 60$, and $69 \ge 60$ ✓. Take $1, 30, 69$:

• Sum $= 100$ ✓

Possible.

Option D — 60 items of type d. We need $n_c \le 60/2 = 30$ and $n_c \ge 2\,n_b$. Try $n_b = 12,\ n_c = 27$ (so $27 \ge 24$ ✓ and $27 \le 30$ ✓), then $n_d = 60$ (and $60 \ge 54$ ✓):

• $1, 12, 27, 60$ sums to $100$ ✓

Possible.

Option C — 45 items of type c. Here $n_c = 45$, so $n_b \le 45/2 = 22.5$, i.e. $n_b \le 22$. The first three types can then total at most

$1 + 22 + 45 = 68.$

That leaves $100 - 68 = 32$ items for type d and beyond. But type d must have at least $2 \times 45 = 90$ items — far more than 32. There is no way to absorb the remaining items, so the total can never reach exactly 100.

💡 Teacher tip: When a middle type is large, the next type's minimum (double of it) can exceed what's left — that's the trap here.

Impossible.

Answer: C — There are exactly 45 items of type c.

What we learn about box 45's type

Box 45's type appears:

• $31$ times in boxes 1–44,
• $43$ times in boxes 46–100,
• and once in box 45 itself.

So the total number of items of this type is

$31 + 43 + 1 = 75.$

Where can a 75-item type sit in the chain?

Each subsequent type must be at least double the previous. If any type came after the 75-item type, it would need at least $2 \times 75 = 150$ items — but there are only 100 boxes total. So the 75-item type must be the last (largest) type.

That leaves $100 - 75 = 25$ items to be distributed among the types before it.

Maximising the number of smaller types

To fit as many smaller types as possible into those 25 items, use the tightest chain $1, 2, 4, 8, \dots$ and check running totals:

• Four smaller types need only $15 \le 25$ ✓.
• Five smaller types would need $31 > 25$ ✗ — too many.

So at most 4 smaller types can precede the 75-item type.

Total types

$4 \text{ (smaller types)} + 1 \text{ (the 75-item type)} = 5.$

A concrete example: $1, 2, 4, 18, 75$, which sums to $100$ and satisfies every doubling rule.

💡 Teacher tip: Once you pin a type's count, check whether it must be the last type (if doubling it exceeds the remaining total). That locks the structure and turns the problem into a smaller version of itself.

Answer: A — 5

Setting up the grid and solving the placement

The intersections form a 4-row × 3-column grid:

A person at any intersection can see along three types of straight lines: their row, their column, and the single diagonal b–g. For example, a person at g sees along row 3 (c, k), column 2 (e, f, h), and the diagonal (b) — giving sightlines to b, c, e, f, h, k.

Step-by-step deduction

1. Clue 1 — Triangle: X, U, Z stand at the three corners of a triangle formed by street segments. The only triangles in this grid use the diagonal b–g: either b–c–g or b–f–g. So X, U, Z occupy {b, c, g} or {b, f, g}.

2. Clue 2 — X sees only U and Z: X must be at a position where exactly two of the six people are visible. Testing X at b: b's sightlines reach a, c, d, f, g, j. If U and Z are at g and f respectively, X sees only them. ✓

3. Clue 4 — U sees V behind Z: U must be positioned so that V is the next intersection beyond Z along a straight line from U. If U is at g and Z is at f, then looking up column 2 (e–f–g–h), U at g sees Z at f, and V at e is the next intersection beyond Z. ✓

4. Clue 3 — Y sees only U and W: Y at k sees along row 3 (c, g) and column 3 (i, j, l). With U at g and W at l, Y sees exactly U and W. ✓

5. Clue 5 — W sees neither V nor Z: W at l sees d, h, i, j, k — none of which is e (V) or f (Z). ✓

6. Clue 6 — Nobody at d: Confirmed; d is unoccupied. ✓

Final placement

Answering this question

The six people occupy intersections g, e, l, b, k, f. Intersection a is not among them, so no one is standing there.

Answer: No one (Choice C)

Setting up the grid and solving the placement

The intersections form a 4-row × 3-column grid:

A person at any intersection can see along their row, their column, and the diagonal b–g.

Step-by-step deduction

1. Clue 1: X, U, Z form a triangle of street segments — the only such triangles use the b–g diagonal: b–c–g or b–f–g.

2. Clue 2 (X sees only U and Z): Placing X at b, whose sightlines reach a, c, d, f, g, j. With U at g and Z at f, X sees exactly U and Z. ✓

3. Clue 4 (U sees V behind Z): U at g looks up column 2 (e–f–g–h), sees Z at f, and V at e is the next intersection beyond. ✓

4. Clue 3 (Y sees only U and W): Y at k sees c, g, i, j, l. With U at g and W at l, Y sees exactly U and W. ✓

5. Clue 5 (W sees neither V nor Z): W at l sees d, h, i, j, k — neither e nor f. ✓

6. Clue 6: Nobody at d. ✓

Final placement

Answering this question — whom can V see?

V is at e. Let's trace e's sightlines:

• Row 1 (a–e–i): sees a and i
• Column 2 (e–f–g–h): sees f, g, h

So from e, the visible intersections are: a, i, f, g, h.

Now check which of these have people:

• a → no one
• i → no one
• f → Z
• g → U
• h → no one

💡 Teacher tip: Always list all visible intersections first, then filter for occupied ones. Don't shortcut by guessing — trace every sightline.

V sees U and Z only.

Answer: U and Z only (Choice C)

Setting up the grid and solving the placement

The intersections form a 4-row × 3-column grid:

A person at any intersection can see along their row, their column, and the diagonal b–g.

Step-by-step deduction

1. Clue 1: X, U, Z form a triangle of street segments — the only such triangles use the b–g diagonal: b–c–g or b–f–g.

2. Clue 2 (X sees only U and Z): X at b sees a, c, d, f, g, j. With U at g and Z at f, X sees exactly U and Z. ✓

3. Clue 4 (U sees V behind Z): U at g looks up column 2 (e–f–g–h), sees Z at f, and V at e is beyond. ✓

4. Clue 3 (Y sees only U and W): Y at k sees c, g, i, j, l. With U at g and W at l, Y sees exactly U and W. ✓

5. Clue 5 (W sees neither V nor Z): W at l sees d, h, i, j, k — neither e nor f. ✓

6. Clue 6: Nobody at d. ✓

Final placement

Answering this question — minimum street segments from X to Y

X is at b and Y is at k. We need the shortest path along street segments.

Let's examine the available routes:

💡 Teacher tip: The diagonal b–g is a street segment, so it counts as just one segment — this is what makes the 2-segment route possible. Don't overlook diagonals when computing shortest paths.

The shortest route is b → g → k, using the diagonal b–g and then row 3 segment g–k, for a total of 2 street segments.

Answer: 2 (Choice C)

Setting up the grid and solving the placement

The intersections form a 4-row × 3-column grid:

A person at any intersection can see along their row, their column, and the diagonal b–g.

Step-by-step deduction

1. Clue 1: X, U, Z form a triangle of street segments — the only such triangles use the b–g diagonal: b–c–g or b–f–g.

2. Clue 2 (X sees only U and Z): X at b sees a, c, d, f, g, j. With U at g and Z at f, X sees exactly U and Z. ✓

3. Clue 4 (U sees V behind Z): U at g looks up column 2 (e–f–g–h), sees Z at f, and V at e is beyond. ✓

4. Clue 3 (Y sees only U and W): Y at k sees c, g, i, j, l. With U at g and W at l, Y sees exactly U and W. ✓

5. Clue 5 (W sees neither V nor Z): W at l sees d, h, i, j, k — neither e nor f. ✓

6. Clue 6: Nobody at d. ✓

Final placement

Answering this question — whom would a person at d see?

A new person stands at d. Let's trace d's sightlines:

• Row 4 (d–h–l): sees h and l
• Column 1 (a–b–c–d): sees a, b, c

So from d, the visible intersections are: a, b, c, h, l.

Now check which of these have people:

• a → no one
• b → X
• c → no one
• h → no one
• l → W

💡 Teacher tip: Note that d is at the bottom of column 1 and the left end of row 4 — it has no diagonal connection, so only two sightlines (row and column) apply.

A person at d would see W (at l) and X (at b) only.

Answer: W and X only (Choice A)

Setting up the full schedule

There are $4 \text{ dancers} \times 2 \text{ items} = 8$ performances in total.

Key structural deductions:

1. Since no dancer performs her second item before every dancer has performed her first, the four first items occupy performances 1–4 and the four second items occupy performances 5–8.
2. The second items follow the same dancer order as the first items.
3. Each composer assigns exactly $8 \div 4 = 2$ items, and since the four first items use four different composers, each composer has one first item (in 1–4) and one second item (in 5–8).

Anchors from clues (iii) and (iv):

• Performance 1 = Princess, composed by Badal.
• Performance 8 = Rani, composed by Gagan.

So the dancer order is: Princess,?,?, Rani.

Applying clue (v) — Ashman's items are consecutive: Since Ashman has one item in 1–4 and one in 5–8, the only way they can be consecutive is performances 4 and 5.

Equal-gap condition for Badal, Gagan, Dyu: The remaining slots are:

• In 1–4: performances 2 and 3 (Badal is at 1, Ashman at 4).
• In 5–8: performances 6 and 7 (Ashman at 5, Gagan at 8).

We need the number of performances between each composer's two items to be equal. Testing the two possibilities:

• Gagan at 3 & 8 → gap of 4; Badal at 1 & 6 → gap of 4; Dyu at 2 & 7 → gap of 4. ✅ All equal.

This fixes the composer sequence:

Determining the dancers (clues i and ii):

Each dancer's two composers are:

• Princess: {Badal, Ashman}
• Dancer in slot 2: {Dyu, Badal}
• Dancer in slot 3: {Gagan, Dyu}
• Rani: {Ashman, Gagan}

Clue (i): No composer who assigned to Princess also assigned to Queen. Princess's composers are {Badal, Ashman}, so Queen cannot have Badal or Ashman. The slot-2 dancer has Badal, so Queen must be in slot 3, and Samragni is in slot 2. (Clue (ii) is then satisfied: Rani's {Ashman, Gagan} and Samragni's {Dyu, Badal} share no composer.)

Full schedule:

Answering this question

From the schedule, performance 2 was composed by Dyu (Samragni's first item). Checking the options:

• A: Second performance by Dyu → ✅ True
• B: Third performance by Dyu → False (it's Gagan)
• C: Third performance by Ashman → False (it's Gagan)
• D: Second performance by Gagan → False (it's Dyu)

Answer: A. The second performance was composed by Dyu.

Setting up the full schedule

There are $4 \text{ dancers} \times 2 \text{ items} = 8$ performances in total.

Key structural deductions:

1. Since no dancer performs her second item before every dancer has performed her first, the four first items occupy performances 1–4 and the four second items occupy performances 5–8.
2. The second items follow the same dancer order as the first items.
3. Each composer assigns exactly $8 \div 4 = 2$ items, and since the four first items use four different composers, each composer has one first item (in 1–4) and one second item (in 5–8).

Anchors from clues (iii) and (iv):

• Performance 1 = Princess, composed by Badal.
• Performance 8 = Rani, composed by Gagan.

So the dancer order is: Princess,?,?, Rani.

Applying clue (v) — Ashman's items are consecutive: Since Ashman has one item in 1–4 and one in 5–8, the only way they can be consecutive is performances 4 and 5.

Equal-gap condition for Badal, Gagan, Dyu: The remaining slots are performances 2, 3 (in 1–4) and 6, 7 (in 5–8). We need the gap between each composer's two items to be equal:

• Gagan at 3 & 8 → gap of 4; Badal at 1 & 6 → gap of 4; Dyu at 2 & 7 → gap of 4. ✅

Determining the dancers (clues i and ii):

Each dancer's two composers are:

• Princess: {Badal, Ashman}
• Slot 2 dancer: {Dyu, Badal}
• Slot 3 dancer: {Gagan, Dyu}
• Rani: {Ashman, Gagan}

Clue (i): No composer shared between Princess and Queen. Since Princess has {Badal, Ashman}, Queen cannot have either. The slot-2 dancer has Badal, so Queen = slot 3 and Samragni = slot 2. Clue (ii) is satisfied (Rani's {Ashman, Gagan} ∩ Samragni's {Dyu, Badal} = ∅).

Full schedule:

Answering this question

We need the false statement. Let's check each option against the schedule:

• A. Samragni did not perform in any item by Ashman. → Samragni's composers: Dyu (perf 2) and Badal (perf 6). No Ashman. ✅ True.
• B. Princess did not perform in any item by Dyu. → Princess's composers: Badal (perf 1) and Ashman (perf 5). No Dyu. ✅ True.
• C. Rani did not perform in any item by Badal. → Rani's composers: Ashman (perf 4) and Gagan (perf 8). No Badal. ✅ True.
• D. Queen did not perform in any item by Gagan. → Queen's composers: Gagan (perf 3) and Dyu (perf 7). Queen did perform in an item by Gagan. ❌ False.

Answer: D. Queen did not perform in any item composed by Gagan.

Setting up the full schedule

There are $4 \text{ dancers} \times 2 \text{ items} = 8$ performances in total.

Key structural deductions:

1. Since no dancer performs her second item before every dancer has performed her first, the four first items occupy performances 1–4 and the four second items occupy performances 5–8.
2. The second items follow the same dancer order as the first items.
3. Each composer assigns exactly $8 \div 4 = 2$ items, with one in 1–4 and one in 5–8.

Anchors from clues (iii) and (iv):

• Performance 1 = Princess, composed by Badal.
• Performance 8 = Rani, composed by Gagan.

Dancer order: Princess,?,?, Rani.

Clue (v): Ashman's items are consecutive → performances 4 and 5. For the equal-gap condition, the only arrangement that works is:

• Badal at 1 & 6 (gap 4), Dyu at 2 & 7 (gap 4), Gagan at 3 & 8 (gap 4).

Dancers (clues i and ii): Each dancer's composers: Princess {Badal, Ashman}, slot 2 {Dyu, Badal}, slot 3 {Gagan, Dyu}, Rani {Ashman, Gagan}. Clue (i) forces Queen = slot 3 (since slot 2 shares Badal with Princess), so Samragni = slot 2.

Full schedule:

Answering this question

From the schedule, performance 6 is Samragni's second item, composed by Badal.

Answer: A. Badal

Setting up the full schedule

There are $4 \text{ dancers} \times 2 \text{ items} = 8$ performances in total.

Key structural deductions:

1. Since no dancer performs her second item before every dancer has performed her first, the four first items occupy performances 1–4 and the four second items occupy performances 5–8.
2. The second items follow the same dancer order as the first items.
3. Each composer assigns exactly $8 \div 4 = 2$ items, with one in 1–4 and one in 5–8.

Anchors from clues (iii) and (iv):

• Performance 1 = Princess, composed by Badal.
• Performance 8 = Rani, composed by Gagan.

Dancer order: Princess,?,?, Rani.

Clue (v): Ashman's items are consecutive → performances 4 and 5. For the equal-gap condition, the only arrangement that works is:

• Badal at 1 & 6 (gap 4), Dyu at 2 & 7 (gap 4), Gagan at 3 & 8 (gap 4).

Dancers (clues i and ii): Each dancer's composers: Princess {Badal, Ashman}, slot 2 {Dyu, Badal}, slot 3 {Gagan, Dyu}, Rani {Ashman, Gagan}. Clue (i) forces Queen = slot 3 (since slot 2 shares Badal with Princess), so Samragni = slot 2.

Full schedule:

Answering this question

We need the pair of performances that share the same composer. Checking each option:

• A. 1st and 7th: Badal and Dyu → different ❌
• B. 3rd and 7th: Gagan and Dyu → different ❌
• C. 2nd and 6th: Dyu and Badal → different ❌
• D. 1st and 6th: Badal and Badal → same ✅

Answer: D. The first and the sixth