CAT 2024 DILR Questions — Slot 2

Setting up the staircase grid

Let us label the ten slots by their row–column position:

Rows increase left → right; columns decrease top → bottom.

Key deductions

1. Place 10. The cell d = (R1, C4) is simultaneously the rightmost cell of Row 1 and the topmost cell of Column 4 — so it is the largest in its row and its column. No other cell has this property. Hence d = 10.

2. Place 1 (partially). By the same logic, the smallest number must sit at a cell that is the leftmost of its row and the bottommost of its column. These "valley" cells are the main diagonal a, e, h, j. Clue 3 says 1 shares a row or column with 10 (at d), so 1 is in Row 1 or Column 4. On the diagonal, only a (Row 1) and j (Column 4) qualify — so 1 is at a or j.

3. Clue 4: 2 and 3 avoid both Row 1 and Column 4 (where 10 sits), so {2, 3} must occupy two of {e, f, h}.

4. Place 9. The only cells that can be as large as 9 are c (directly left of 10) and g (directly below 10), since every other cell has at least two cells forced to be larger than it. Testing 9 at g leads to a contradiction (7 and 8 would be squeezed into {a, b, c, h}, and no valid arrangement survives the row/column ordering plus the {2,3} requirement). Hence c = 9.

5. Place 7 and 8. Clue 5 says 7 and 8 avoid Row 1 and Column 3 (where 9 sits), so {7, 8} ⊆ {e, g, i, j}. The only cell among these that can hold 8 is g (since e is bounded above by b < 9, i is bounded above by g, and j is bounded above by i). So g = 8, and then i = 7 (the only remaining option for 7 in {e, i, j}).

6. Clue 6 (4 and 6 in the same row). With c = 9, g = 8, i = 7, d = 10 fixed, the numbers left for {a, b, e, f, h, j} are {1, 2, 3, 4, 5, 6}. The only row that can hold both 4 and 6 is Row 1 (Row 2's open slots e, f must include two of {2, 3, 5}, and Row 3 has only one open slot). So a = 4, b = 6. Since a = 4, the 1 cannot be at a, so j = 1.

7. Finish {e, f, h}. They receive {2, 3, 5}. Since e < f and h < f, the largest (5) must go to f = 5, leaving {e, h} = {2, 3} in either order.

Solved grid

Answering this question — least row-sum

Compute each row's sum:

Row 4, with its single entry of 1, has by far the smallest sum regardless of how 2 and 3 are placed.

Answer: 4

Setting up the staircase grid

Let us label the ten slots by their row–column position:

Rows increase left → right; columns decrease top → bottom.

Key deductions

1. Place 10. Cell d = (R1, C4) is the rightmost of its row and the topmost of its column — the unique "peak." So d = 10.

2. Place 1 (partially). The smallest number must sit at a "valley" cell (leftmost of its row and bottommost of its column): the diagonal a, e, h, j. Clue 3 forces 1 into Row 1 or Column 4, so on the diagonal only a or j qualify.

3. Clue 4: {2, 3} ⊆ {e, f, h} (they avoid Row 1 and Column 4).

4. Place 9. Only c or g can hold 9. Testing 9 at g creates a contradiction, so c = 9. Clue 5 then forces 7 and 8 into {e, g, i, j}; the only consistent assignment is g = 8, i = 7.

5. Clue 6: 4 and 6 must share a row. With c = 9, g = 8, i = 7, d = 10 fixed, the only row that can hold both 4 and 6 is Row 1, giving a = 4, b = 6. Since a = 4, we get j = 1.

6. Finish: {e, f, h} get {2, 3, 5}; f is the largest, so f = 5, and {e, h} = {2, 3} in either order.

Solved grid

Answering this question — which statements MUST be true?

• Statement I: 10 is placed in a slot in Row 1. 10 is fixed at (R1, C4), which is in Row 1. Must be true.

• Statement II: 1 is placed in a slot in Row 4. 1 is fixed at (R4, C4), which is in Row 4. Must be true.

Both statements hold in every valid arrangement.

Answer: Both I and II

Setting up the staircase grid

Let us label the ten slots by their row–column position:

Rows increase left → right; columns decrease top → bottom.

Key deductions

1. Place 10. Cell d = (R1, C4) is the unique "peak" (rightmost of its row, topmost of its column). So d = 10.

2. Place 1 (partially). The smallest number must sit at a "valley" cell on the diagonal a, e, h, j. Clue 3 restricts 1 to Row 1 or Column 4, so 1 is at a or j.

3. Clue 4: {2, 3} ⊆ {e, f, h}.

4. Place 9. Only c or g can hold 9; 9 at g leads to a contradiction, so c = 9. Clue 5 then forces g = 8, i = 7.

5. Clue 6: 4 and 6 must share a row; only Row 1 works, giving a = 4, b = 6, and therefore j = 1.

6. Finish: {e, f, h} get {2, 3, 5}; f is largest, so f = 5, and {e, h} = {2, 3} in either order.

Solved grid

💡 Teacher tip: When two numbers can swap between two cells without violating any clue, neither placement is forced — so any statement asserting one specific arrangement is not a "must be true."

Answering this question — must 2 be in Column 2 and 3 in Column 3?

The numbers 2 and 3 occupy cells (R2, C2) and (R3, C3), but in either order:

• Arrangement A: e = 2 (Column 2) and h = 3 (Column 3) — both statements true.
• Arrangement B: e = 3 (Column 2) and h = 2 (Column 3) — both statements false.

Both arrangements satisfy every clue, so neither statement is forced.

• Statement I (2 in Column 2): Not necessarily — 2 could be in Column 3.
• Statement II (3 in Column 3): Not necessarily — 3 could be in Column 2.

Answer: Neither I nor II

Setting up the staircase grid

Let us label the ten slots by their row–column position:

Rows increase left → right; columns decrease top → bottom.

Key deductions

1. Place 10. Cell d = (R1, C4) is the unique "peak." So d = 10.

2. Place 1 (partially). The smallest number sits on the diagonal {a, e, h, j}; Clue 3 restricts it to a or j.

3. Clue 4: {2, 3} ⊆ {e, f, h}.

4. Place 9. Only c or g can hold 9; 9 at g fails, so c = 9. Clue 5 then gives g = 8, i = 7.

5. Clue 6: 4 and 6 share Row 1: a = 4, b = 6, so j = 1.

6. Finish: {e, f, h} get {2, 3, 5}; f = 5, and {e, h} = {2, 3} in either order.

Solved grid

Answering this question — how many slots are undetermined?

Let us check each slot:

Eight slots are fixed with certainty. Only (R2, C2) and (R3, C3) — which hold 2 and 3 between them in either order — remain ambiguous. No clue distinguishes which is which.

Answer: 2

Setting up the staircase grid

Let us label the ten slots by their row–column position:

Rows increase left → right; columns decrease top → bottom.

Key deductions

1. Place 10. Cell d = (R1, C4) is the unique "peak" (rightmost of its row, topmost of its column). So d = 10.

2. Place 1 (partially). The smallest number sits on the diagonal {a, e, h, j}; Clue 3 restricts it to a or j.

3. Clue 4: {2, 3} ⊆ {e, f, h} (they avoid Row 1 and Column 4).

4. Place 9. Only c or g can hold 9; 9 at g leads to a contradiction, so c = 9. Clue 5 then forces 7 and 8 into {e, g, i, j}; the only consistent assignment is g = 8, i = 7.

5. Clue 6: 4 and 6 must share a row; only Row 1 works, giving a = 4, b = 6, and therefore j = 1.

6. Finish: {e, f, h} get {2, 3, 5}; f is largest, so f = 5, and {e, h} = {2, 3} in either order.

Solved grid

Answering this question — sum of Column 4

Column 4 contains the entries in slots d, g, i, and j:

All four of these are fixed with certainty (the only ambiguity in the entire grid is between 2 and 3, neither of which is in Column 4). So the sum is:

$10 + 8 + 7 + 1 = 26$

Answer: 26

Finding how many players Zara coached

Let's first work out the coach assignments step by step.

Who can coach whom?

• Yuki → only even-numbered players: {2, 4, 6, 8}
• Zara → only odd-numbered players: {1, 3, 5, 7}
• Xena → any player

Step 1 — Use clue 2 (P2, P3, P5 had three different coaches).

• P2 is even, so P2's coach is Yuki or Xena.
• P3 and P5 are odd, so each is coached by Zara or Xena.
• For all three to have different coaches, P2 must be coached by Yuki (the only coach available to P2 but not to P3/P5). Then {P3, P5} must be split between Xena and Zara (one each).

Step 2 — Use clue 2 (P1 and P4 share a coach).

• P4 is even → coach is Yuki or Xena.
• P1 is odd → coach is Zara or Xena.
• The only coach common to both lists is Xena. So P1 and P4 are both coached by Xena.

Step 3 — Use clue 3 (P5 and P7 share a coach).

• Suppose P5 = Xena. Then P3 = Zara (from Step 1), and P7 = Xena (shared with P5). But then Zara would coach only P3 — just 1 player, violating the rule that each coach trains at least 2. Impossible.
• Therefore P5 = Zara, which forces P3 = Xena (from Step 1), and P7 = Zara (shared with P5).

So Zara coaches exactly P5 and P7 — that is 2 players.

Step 4 — Confirm no other odd player can go to Zara.

• P1 is already Xena (Step 2), P3 is already Xena (Step 3), P5 and P7 are Zara. All odd players are assigned. No room for Zara to take a third.

Answer: B. Exactly 2

Finding Player-7's rating

We need to determine the shared rating of P5 and P7. Let's build the full picture.

Coach assignments (derived from the clues):

(P6 and P8 are split between Yuki and Xena; the exact split is resolved below using ratings.)

Key deduction — the shared rating of P5 and P7.

Let P5 = P7 = $v$. Clue 3 says every other player got a unique rating, meaning no other player shares a rating with anyone else — and in particular, none of the other six has rating $v$.

So the six unique players use six distinct values from $\{1,2,3,4,5,6,7\} \setminus \{v\}$, and the total sum of all eight ratings is: $2v + (1+2+3+4+5+6+7 - v) = 2v + (28 - v) = 28 + v$

Clue 4 says the overall average is 4, so the total sum is $8 \times 4 = 32$: $28 + v = 32 \implies v = 4$

Therefore P5 = P7 = 4.

Answer: 4

Finding Player-6's rating

Let's work through the ratings using the solved coach assignments.

Coach assignments:

Step 1 — Shared rating of P5 and P7. Let $v$ = P5 = P7. The other six players each have a unique rating, none equal to $v$. So the total sum is $28 + v$. With average 4 (sum = 32): $v = 4$. So P5 = P7 = 4.

Step 2 — Coach averages (clue 6).

• $\text{avg}(\text{Zara}) = \frac{4+4}{2} = 4$
• Clue 6: avg(Yuki) = avg(Zara) + 2 = 6, and avg(Xena) = avg(Yuki) ÷ 2 = 3.

Step 3 — Yuki's players. Yuki coaches P2 and one of {P6, P8}. Clue 5 says P2 got the highest rating, so P2 = 7.

Yuki's average is 6 over 2 players, so the sum is 12: $7 + r_{\text{other}} = 12 \implies r_{\text{other}} = 5$

Step 4 — Is the other Yuki player P6 or P8? Clue 7 says P4's rating = 2 × P8's rating. If P8 were the Yuki player rated 5, then P4 = 2 × 5 = 10, which exceeds the maximum of 7. Impossible.

So the Yuki player rated 5 must be P6, and P8 goes to Xena.

Answer: 5

Counting how many ratings are determined with certainty

Let's assemble the complete solution and then count.

Coach assignments:

Rating deductions:

1. P5 = P7 = 4. Let their shared rating be $v$. The other six players have unique ratings (all different from $v$), so the total sum is $28 + v$. With average 4 (sum 32): $v = 4$.

2. Coach averages (clue 6): avg(Zara) = 4, avg(Yuki) = 6, avg(Xena) = 3.

3. P2 = 7 (highest rating, clue 5). Yuki's two players sum to 12, so the other Yuki player = 5. If that were P8, then P4 = 2 × 5 = 10 (impossible). So P6 = 5 (Yuki) and P8 goes to Xena.

4. Xena's players {P1, P3, P4, P8} sum to $4 \times 3 = 12$. Clue 7: $r_4 = 2 \times r_8$ and $r_4 < r_5 = 4$. The available ratings for Xena's players are $\{1, 2, 3, 6\}$ (since 4, 5, 7 are used). Testing $r_4 = 2r_8$ with $r_4 < 4$:

• $r_8 = 1,\ r_4 = 2$ ✓ (both available, $2 < 4$)
• Any other option either gives $r_4 \geq 4$ or uses unavailable ratings.

So P8 = 1, P4 = 2, leaving {P1, P3} = {3, 6} — but we cannot tell which is 3 and which is 6.

Final arrangement:

Counting certain ratings:

• Fixed: P2 = 7, P4 = 2, P5 = 4, P6 = 5, P7 = 4, P8 = 1 → 6 players
• Ambiguous: P1 and P3 (could be 3 and 6 in either order) → 2 players

Answer: 6

Identifying Xena's players

Let's determine the full coach assignment from the clues.

Who can coach whom?

• Yuki → only even-numbered: {2, 4, 6, 8}
• Zara → only odd-numbered: {1, 3, 5, 7}
• Xena → any player

Step 1 — Clue 2: P2, P3, P5 had three different coaches.

• P2 (even) → Yuki or Xena
• P3 (odd) → Zara or Xena
• P5 (odd) → Zara or Xena
• For three different coaches, P2 must be Yuki, and {P3, P5} = {Xena, Zara} (one each).

Step 2 — Clue 2: P1 and P4 share a coach.

• P4 (even) → Yuki or Xena; P1 (odd) → Zara or Xena.
• Only common option: Xena. So P1 = P4 = Xena.

Step 3 — Clue 3: P5 and P7 share a coach.

• If P5 = Xena, then P3 = Zara and P7 = Xena. Zara would have only P3 (1 player) — below the minimum of 2. Impossible.
• So P5 = Zara, P3 = Xena, P7 = Zara.

Step 4 — Remaining evens P6, P8.

• Currently: Xena has {P1, P3, P4} = 3; Yuki has {P2} = 1; Zara has {P5, P7} = 2.
• Yuki needs at least 2, so takes one of P6/P8.
• Clue 1: Xena trained more than Yuki. If Yuki takes 1 more (total 2), Xena takes the other (total 4): 4 > 2 ✓. If Yuki takes both (total 3), Xena stays at 3: 3 > 3 ✗.
• So one of P6/P8 goes to Yuki, the other to Xena.

Step 5 — Use ratings to split P6 and P8.

• P5 = P7 = 4 (shared rating; total sum = 32 forces this).
• avg(Zara) = 4 → avg(Yuki) = 6, avg(Xena) = 3 (clue 6).
• P2 = 7 (highest). Yuki's sum = 12, so other Yuki player = 5.
• If P8 = 5 (Yuki), then P4 = 2 × 5 = 10 (clue 7) — impossible. So P6 = Yuki (rated 5) and P8 = Xena.

Final coach assignment:

Xena's players: P1, P3, P4, P8.

Answer: B. Player-1, Player-3, Player-4, Player-8

Finding the number of buyers on Day 1

We need to connect Day 1 and Day 2 through their cumulative averages.

Step 1 — Read Day 2 from the chart.

Total buyers on Day 2 = $5 + 10 + 5 + 20 + 10 = 50$.

Sum of all Day-2 ratings: $1(5) + 2(10) + 3(5) + 4(20) + 5(10) = 5 + 20 + 15 + 80 + 50 = 170$

Step 2 — Set up the cumulative-average link.

Let $n_1$ = number of buyers on Day 1 and $S_1$ = sum of their ratings.

• Cumulative average after Day 1 is 3, so $S_1 = 3n_1$.
• Cumulative average after Day 2 is 3.1, meaning the average of all ratings through Day 2 is 3.1:

$\frac{S_1 + 170}{n_1 + 50} = 3.1$

Step 3 — Solve for $n_1$.

Substitute $S_1 = 3n_1$:

$\frac{3n_1 + 170}{n_1 + 50} = 3.1$

$3n_1 + 170 = 3.1(n_1 + 50)$

$3n_1 + 170 = 3.1n_1 + 155$

$170 - 155 = 3.1n_1 - 3n_1$

$15 = 0.1\,n_1$

$n_1 = 150$

💡 Teacher tip: The cumulative average is a running average — it pools every rating from Day 1 onward. So the Day-2 cumulative average combines Day 1's sum and Day 2's sum in a single fraction.

Answer: 150

Finding the daily average rating of Day 3

First we must deduce the exact distribution of ratings on Day 3, then compute its average.

Step 1 — Summarise the Day-3 clues.

1. Total buyers = 100.
2. The modes are 4 and 5 (tied), so ratings 4 and 5 have the highest counts, and those two counts are equal.
3. Every rating count is a non-zero multiple of 10.
4. Rating-1 count = Rating-2 count, and that number is half the Rating-3 count.

Step 2 — Translate into variables.

Let the Rating-1 and Rating-2 counts each be $x$ (they are equal). Then the Rating-3 count is $2x$.

Since 4 and 5 are tied modes, let each of their counts be $m$, with $m \ge 2x$ (they must be at least as large as every other count).

Step 3 — Use the total.

$x + x + 2x + m + m = 100$

$4x + 2m = 100 \;\Longrightarrow\; 2x + m = 50$

Step 4 — Apply the constraints.

Since $m \ge 2x$: $50 - 2x \ge 2x \;\Longrightarrow\; 50 \ge 4x \;\Longrightarrow\; x \le 12.5$

Since $x$ must be a non-zero multiple of 10, the only value satisfying $x \le 12.5$ is $x = 10$.

Then $m = 50 - 2(10) = 30$.

Step 5 — Write the Day-3 distribution.

Check: $10 + 10 + 20 + 30 + 30 = 100$ ✓ Modes are 4 and 5 (each with 30) ✓ All counts are non-zero multiples of 10 ✓

Step 6 — Compute the daily average.

Sum of ratings: $1(10) + 2(10) + 3(20) + 4(30) + 5(30) = 10 + 20 + 60 + 120 + 150 = 360$

$\text{Daily average} = \frac{360}{100} = 3.6$

💡 Teacher tip: When modes are tied, the tied values share the largest count. Here, saying 4 and 5 are both modes forces their counts to be equal and at least as large as every other rating's count — that inequality is what pins down the unique solution.

Answer: A. 3.6

Finding the median of all Day-3 ratings

We first need the exact Day-3 distribution. (If you've already worked it out, skip to Step 3.)

Step 1 — Deduce the Day-3 distribution.

From the clues: 100 buyers total, all counts are non-zero multiples of 10, Rating-1 count = Rating-2 count = $x$ and equals half the Rating-3 count (= $2x$), and ratings 4 and 5 are tied modes with equal count $m \ge 2x$.

$x + x + 2x + m + m = 100 \;\Longrightarrow\; 2x + m = 50$

With $m \ge 2x$ we get $x \le 12.5$, so the only valid multiple of 10 is $x = 10$, giving $m = 30$.

Step 2 — Write the distribution.

Step 3 — Find the median of 100 values.

With 100 ratings (an even number), the median is the average of the 50th and 51st values when all ratings are sorted in order.

Build the cumulative count:

The 50th value falls in the rating-4 block (positions 41–70), and so does the 51st value.

$\text{Median} = \frac{4 + 4}{2} = 4$

💡 Teacher tip: For a sorted list of 100 items, you only need items #50 and #51. Build a running cumulative count and locate which rating band each position lands in.

Answer: 4

Comparing cumulative averages: Day 2 vs Day 3

We need the cumulative average at the end of Day 2 and at the end of Day 3, then the percentage change.

Step 1 — Cumulative average after Day 2.

This is given directly: 3.1.

(For completeness: Day 1 had 150 buyers with sum $3 \times 150 = 450$; Day 2 had 50 buyers with sum 170. Cumulative average $= \frac{450 + 170}{150 + 50} = \frac{620}{200} = 3.1$.)

Step 2 — Cumulative average after Day 3.

We need the Day-3 distribution. From the clues (100 buyers, non-zero multiples of 10, Rating-1 = Rating-2 = $x$ = half of Rating-3 = $2x$, and ratings 4 and 5 are tied modes with count $m$):

$2x + m = 50,\quad m \ge 2x \;\Longrightarrow\; x = 10,\; m = 30$

Day-3 sum $= 10 + 20 + 60 + 120 + 150 = 360$.

Now pool all three days:

$\text{Cumulative average after Day 3} = \frac{980}{300} \approx 3.267$

Step 3 — Compute the percentage increase from Day 2 to Day 3.

$\text{Percentage increase} = \frac{3.267 - 3.1}{3.1} \times 100 = \frac{0.167}{3.1} \times 100 \approx 5.4\%$

Step 4 — Match to the options.

• A: less than 5% → No (5.4% > 5%)
• B: decreased → No (it increased)
• C: between 5% and 8% → Yes (5.4% lies in this range)
• D: more than 8% → No

💡 Teacher tip: The cumulative average changes slowly when the existing pool is large. Here, 200 ratings already exist before Day 3, so even though Day 3's daily average (3.6) is much higher than 3.1, the cumulative average only moves from 3.1 to about 3.27 — roughly a 5.4% rise.

Answer: C. The cumulative average of Day 3 increased by a percentage between 5% and 8% from Day 2.

Finding the firm with the highest annual rate of growth (ARG) in PAT

First, let us consolidate the data from both bubble plots. The position of each dot gives (ES, PAT), and the bubble area is proportional to PRD (the percentage of PAT spent on R&D), with the same area scale in both years.

Key insight: We are told the annual rate of growth in PAT remained constant over the years for each firm. The period 2019 → 2023 spans 4 years. If the constant annual growth rate is $r$, then:

$\text{PAT}_{2023} = \text{PAT}_{2019} \times (1+r)^4$

This means the firm with the largest ratio $\text{PAT}_{2023}/\text{PAT}_{2019}$ also has the largest $r$. So we simply need to compare PAT ratios for firms A, B, C, and E.

💡 Teacher tip: When a growth rate is constant over a fixed number of years, you never need to compute the actual rate — just compare the overall growth ratios. The firm with the biggest total ratio automatically has the biggest annual rate.

Firm E's ratio (≈ 1.458) is the highest among the four, so Firm E had the highest constant annual rate of growth in PAT.

Answer: Firm E

Finding the ratio of Firm C's R&D spending: 2019 to 2023

Let us first consolidate the relevant data from the two bubble plots. The dot position gives (ES, PAT), and the bubble area is proportional to PRD (the percentage of PAT spent on R&D), with the same area scale across both years.

Key relationship: PRD is the percentage of PAT spent on R&D. Therefore:

$\text{R\&D amount} = \text{PAT} \times \text{PRD}$

Since PRD is given as a relative index (proportional to bubble area), the actual R&D amount is proportional to PAT × PRD. For a ratio between two years, the constant of proportionality cancels out, so we can use the index directly.

Step 1 — Firm C's R&D spend in 2019:

$\text{R\&D}_{2019} \propto 2400 \times 8.33 = 19992$

Step 2 — Firm C's R&D spend in 2023:

$\text{R\&D}_{2023} \propto 3000 \times 3.37 = 10110$

Step 3 — Take the ratio (2019: 2023):

$\frac{\text{R\&D}_{2019}}{\text{R\&D}_{2023}} = \frac{19992}{10110} \approx 1.98 \approx \frac{2}{1}$

Step 4 — Compare with the options:

💡 Teacher tip: The computed value ≈ 1.98 sits between 9:4 (= 2.25) and 9:5 (= 1.80). Check which is closer: |1.98 − 2.25| = 0.27, while |1.98 − 1.80| = 0.18. So 9:5 is the closest match.

The ratio closest to 1.98 is 9: 5.

Answer: 9: 5

Finding the firm with the maximum PAT per employee in 2023

Let us first consolidate the data from the two bubble plots. The dot position gives (ES, PAT), and the bubble area is proportional to PRD, with the same area scale across both years.

What we need: PAT per employee in 2023 for firms A, C, E, and F. This is simply:

$\text{PAT per employee} = \frac{\text{PAT}}{\text{ES}}$

Note that PRD (the bubble size) is irrelevant here — we only need the dot's coordinates.

Step 1 — Compute PAT/ES for each candidate firm in 2023:

💡 Teacher tip: PAT per employee is read entirely from the dot's position — the bubble size (PRD) plays no role. Don't let the extra information distract you.

Step 2 — Identify the maximum: Comparing 3.00, 3.80, 2.50, and 3.20, the highest value is 3.80, which belongs to Firm C.

Answer: Firm C

Finding the firm with the least R&D spending per employee in 2023

Let us first consolidate the data from the two bubble plots. The dot position gives (ES, PAT), and the bubble area is proportional to PRD (the percentage of PAT spent on R&D), with the same area scale across both years.

Key relationships:

• PRD is the percentage of PAT spent on R&D, so: $\text{R\&D amount} = \text{PAT} \times \text{PRD}$
• R&D spending per employee divides this by ES:

$\text{R\&D per employee} = \frac{\text{PAT} \times \text{PRD}}{\text{ES}}$

Since PRD is a relative index (proportional to bubble area), the R&D per employee values we compute are proportional to the true values. For comparing firms within the same year, this is perfectly sufficient.

Step 1 — Compute (PAT × PRD) / ES for each candidate firm in 2023:

💡 Teacher tip: Notice that C, D, and F all have the same PRD (3.37) in 2023 — their bubbles are identical in size. So among those three, R&D per employee reduces to comparing PAT/ES. Firm D has both the lowest PAT and a moderate ES, pushing it to the bottom. Firm E's much larger PRD (8.95) lifts it well above the others despite a high ES.

Step 2 — Identify the minimum: Comparing 12.80, 10.11, 22.38, and 10.78, the smallest value is 10.11, which belongs to Firm D.

Answer: Firm D

Setting up the walkway network

The gated area is a 4×4 grid of points with two diagonal shortcuts. From the six given lengths, every edge in the grid is determined:

A --150-- B --300-- C --300-- D | | | | 400 400 400 400 | | | | H --150-- G --300-- F --300-- E | | | | 200 200 200 200 | | | | I --150-- J --300-- K --300-- L | | | | 400 400 400 400 | | | | P --150-- O --300-- N --300-- M

Two diagonal walkways:

• I–G = √(150² + 200²) = √62500 = 250 m
• O–K = √(300² + 400²) = √250000 = 500 m

Solving the problem

The resident at L must visit P (post office) and B (bank), then return to L. The shortest such closed trip equals the sum of the three pairwise shortest distances — the visiting order does not matter:

$d(L,P) + d(P,B) + d(B,L)$

1. Shortest path L → P (950 m)

The O–K diagonal is the key shortcut. Instead of walking the full grid perimeter from K down to O, the resident cuts directly across:

(The straight grid route L→K→J→O→P would cost 300+300+400+150 = 1150 m, so the diagonal saves 200 m.)

2. Shortest path P → B (1050 m)

Here the I–G diagonal shortcuts the detour through H:

(The grid route P→I→H→G→B would cost 400+200+150+400 = 1150 m.)

3. Shortest path B → L (1200 m)

Several equal-cost routes exist; one is:

(Equivalently, B→G→F→E→L = 400+300+300+200 = 1200 m.)

Final calculation

$950 + 1050 + 1200 = 3200 \text{ m}$

Answer: B. 3200

Setting up the walkway network

The gated area is a 4×4 grid of points with two diagonal shortcuts. From the six given lengths, every edge is determined:

A --150-- B --300-- C --300-- D | | | | 400 400 400 400 | | | | H --150-- G --300-- F --300-- E | | | | 200 200 200 200 | | | | I --150-- J --300-- K --300-- L | | | | 400 400 400 400 | | | | P --150-- O --300-- N --300-- M

Diagonals: I–G = 250 m, O–K = 500 m.

Identifying the lake-boundary edges

The three rectangular lakes have the following boundary edges:

The walker must stay next to a lake at all times, so only these 12 boundary edges are available. The lakes touch at vertex F (lakes 1 & 2) and vertex K (lakes 2 & 3), but they do not share any edge.

Checking whether all 12 edges can be covered

💡 Teacher tip: A connected graph where every vertex has even degree admits an Euler circuit — a route that uses every edge exactly once and returns to the start. This is exactly what we need: walk every lake-edge once, no repeats.

Let us check the degree of each vertex in the 12-edge lake-boundary subgraph:

Every vertex has even degree (2 or 4), so an Euler circuit exists. The walker can cover all 12 lake-boundary edges in a single trip without repeating any path.

Computing the total distance

$\text{Total} = 1400 + 1000 + 1400 = 3800 \text{ m}$

One valid circuit (starting and ending at C): C → D → E → F → G → J → K → L → M → N → K → F → C.

Answer: C. 3800

Setting up the walkway network

The gated area is a 4×4 grid of points with two diagonal shortcuts. From the six given lengths, every edge is determined:

A --150-- B --300-- C --300-- D | | | | 400 400 400 400 | | | | H --150-- G --300-- F --300-- E | | | | 200 200 200 200 | | | | I --150-- J --300-- K --300-- L | | | | 400 400 400 400 | | | | P --150-- O --300-- N --300-- M

Diagonals: I–G = 250 m, O–K = 500 m.

Structuring the problem

The resident starts and ends at A, visiting no point (other than A) more than once. This is a simple cycle through A.

Point A has only two neighbours: B and H. So every such cycle must look like:

$A \to B \to \cdots \to H \to A$

To maximise the distance, we want the heaviest possible simple path from B to H that passes through as many other vertices as possible, using the longest edges.

Identifying the heaviest edges

The eight edges of weight 400 m are the vertical segments spanning a full row gap:

AH, BG, CF, DE, IP, JO, KN, LM

Using all eight in one cycle gives $8 \times 400 = 3200$ m as a base. We then connect them with the heaviest available remaining edges.

Building the optimal cycle

The cycle that threads all 16 points using every 400 m edge plus the heaviest connectors is:

$A \to B \to G \to F \to C \to D \to E \to L \to M \to N \to K \to J \to O \to P \to I \to H \to A$

Every one of the 16 grid points appears exactly once (a Hamiltonian cycle), so the walk is valid.

Computing the total

$\text{Total} = 8 \times 400 + (300 + 300 + 200 + 300 + 300 + 150 + 200 + 150)$ $= 3200 + 1900 = 5100 \text{ m}$

💡 Teacher tip: The two diagonals (250 m and 500 m) might seem tempting, but including either one in a Hamiltonian cycle forces us to drop multiple 400 m edges, which always results in a lower total. The all-400 m strategy is optimal.

Answer: 5100

Setting up the walkway network

The gated area is a 4×4 grid of points with two diagonal shortcuts. From the six given lengths, every edge is determined:

A --150-- B --300-- C --300-- D | | | | 400 400 400 400 | | | | H --150-- G --300-- F --300-- E | | | | 200 200 200 200 | | | | I --150-- J --300-- K --300-- L | | | | 400 400 400 400 | | | | P --150-- O --300-- N --300-- M

Diagonals: I–G = 250 m, O–K = 500 m.

Identifying the restrictions

The visitor enters and exits at C and must:

1. Not pass by any residence — residences lie on paths AH and ML, with named residence points at A, H, M, L. So the visitor must avoid these four points and all edges touching them.
2. Not cross any point (except C) more than once — the route is a simple cycle through C.

Removed points: A, H, M, L Removed edges: AB, AH, HG, HI, EL, KL, LM, MN (and any other edge touching a removed point)

Available vertices and edges

The remaining vertices are: B, C, D, E, F, G, I, J, K, N, O, P (12 points).

The available edges (with weights) are:

Finding the longest simple cycle through C

We want the heaviest cycle that starts and ends at C, uses only the available edges, and visits no point twice (except C).

💡 Teacher tip: The two diagonals (OK = 500, IG = 250) look attractive, but each creates a bottleneck: inserting either diagonal into a cycle forces us to skip several 400 m edges, which always reduces the total. The optimal cycle avoids both diagonals and instead chains together the maximum number of 400 m segments.

The optimal route is:

$C \to D \to E \to F \to K \to N \to O \to P \to I \to J \to G \to B \to C$

Verification:

• No point among A, H, M, L is visited. ✓
• No point (except C) is repeated. ✓
• The route uses four of the 400 m edges (DE, KN, PI, GB) plus the heaviest available connectors.

Computing the total

$300 + 400 + 300 + 200 + 400 + 300 + 150 + 400 + 150 + 200 + 400 + 300 = 3500 \text{ m}$

Answer: 3500