CAT 2020 DILR Questions — Slot 3

We need the total number of mango trees (row X). The fastest route is through the teak row: clue 3 gives us the teak column relationships, and clue 9 converts teak to mango.

Step 1 — Parameterize the teak row (clue 3). Clue 3 says: teak in column 3 = 2 × teak in column 2, and teak in column 3 = ½ × teak in column 4.

Let teak in column 2 = $a$. Then:

• Column 3 teak = $2a$
• Column 4 teak = $4a$

The figure gives $Y_1 = 21$, so teak total $= 21 + a + 2a + 4a = 21 + 7a$.

Step 2 — Relate the rows (clue 9). Clue 9: mango = 2 × teak, so mango total $= 2(21 + 7a) = 42 + 14a$. Also, pine = $205 - \text{mango} - \text{teak} = 205 - 3(21 + 7a) = 142 - 21a$.

Step 3 — Pin down $a$. The pine row already shows $Z_3 = 9$ and $Z_4 = 28$ (37 trees accounted for), so: $Z_1 + Z_2 = 142 - 21a - 37 = 105 - 21a > 0 \implies a < 5$

Every plot must contain a non-zero multiple of 3 or 4, so $a \in \{3, 4\}$.

• $a = 3$: teak in column 4 = $4 \times 3 = 12$. But $X_1 = 12$ already (from the figure), and no two plots may share the same count. ✗
• $a = 4$: teak row = $(21, 4, 8, 16)$ — all distinct, all valid. ✓

Step 4 — Compute mango. Teak total = $21 + 4 + 8 + 16 = 49$. Mango total = $2 \times 49 = \mathbf{98}$.

💡 Teacher tip: When a clue gives ratios between columns (double, half), introduce a single variable and let the ratios propagate — then use the "all values distinct" and "multiples of 3 or 4" constraints to eliminate candidates.

Answer: 98 (Choice C).

This question asks for every daughter's tree total, so we need to solve the full grid — both the tree counts and who owns each plot.

Part A — Tree counts per row. Following the same logic as the mango-tree question:

• Let teak in column 2 = $a$; then columns 3 and 4 have $2a$ and $4a$ teak trees.
• Teak total = $21 + 7a$; mango = $2 \times$ teak; pine = $205 - 3 \times$ teak.
• Pine already has $Z_3 = 9$, $Z_4 = 28$, forcing $a < 5$ with $a$ a multiple of 3 or 4.
• $a = 3$ is rejected (gives $Y_4 = 12 = X_1$, a duplicate). So $a = 4$.

This gives: teak total = 49, mango total = 98, pine total = 58, with teak row = $(21, 4, 8, 16)$.

Part B — Who owns which plot?

1. Plot counts (clue 4 + even counts): 12 plots total, each daughter gets an even number, and Abha & Bina each get more plots than Dipti. The only split is Abha 4, Bina 4, Chitra 2, Dipti 2.

2. Chitra's plots: From the figure, Chitra owns X1 and Z2 — that's her 2 plots.

3. Dipti's plots (clues 5, 6, 8): Dipti needs two adjoining plots in the same row (clue 6), not adjacent — by row, column, or diagonal — to Chitra's plots X1 or Z2 (clue 8). The plots adjacent to X1 are X2, Y1, Y2; those adjacent to Z2 are Z1, Z3, Y1, Y2, Y3. After eliminating all of these, the only pair of adjoining same-row plots left is X3 & X4. So Dipti = {X3, X4}.

4. Corner Z4 (clue 5): Only Bina, Chitra, Dipti got corner plots. X1 = Chitra, X4 = Dipti, Z1 = Bina. Dipti and Chitra are already full (2 plots each), so Z4 = Bina.

5. Bina's remaining plots (clue 7): Bina must have one plot in each row and each column. She already has Z1 (col 1, row Z) and Z4 (col 4, row Z). She needs columns 2 & 3 and rows X & Y. The only fit is X2 (col 2, row X) and Y3 (col 3, row Y). So Bina = {Z1, X2, Y3, Z4}.

6. Abha fills the rest: The remaining plots are {Y1, Y2, Y4, Z3} = {21, 4, 16, 9} → Abha = 50.

Part C — Daughter totals (clue 1).

• Abha = Chitra + 20 → Chitra = 50 − 20 = 30
• Abha = Dipti − 6 → Dipti = 50 + 6 = 56
• Bina = 205 − 50 − 30 − 56 = 69

Part D — Remaining tree counts.

• Chitra = X1 + Z2 = 12 + Z2 = 30 → Z2 = 18
• Pine total 58 = Z1 + 18 + 9 + 28 → Z1 = 3
• Bina = 3 + X2 + 8 + 28 = 69 → X2 = 30
• Dipti = X3 + X4 = 56; both multiples of 3 or 4, distinct, ≤ 32 → {24, 32}

Fully solved grid:

Trees received: Abha = 50, Bina = 69, Chitra = 30, Dipti = 56.

💡 Teacher tip: The ownership logic chains neatly: fix Chitra first (she's given 2 plots), then Dipti (adjoining + not adjacent to Chitra), then Bina (one per row/column), and Abha takes whatever is left.

Answer: 50, 69, 30, 56 (Choice A).

We need to find how many pine trees Chitra received. From the figure, Chitra owns plot Z2 (row Z = pine), so we need the tree count in Z2.

Key deductions leading to Z2:

1. Teak row: Using clue 3 (column ratios) and clue 9 (mango = 2 × teak), we find teak row = $(21, 4, 8, 16)$, teak total = 49, mango = 98, pine = 58.

2. Plot ownership: Each daughter gets an even number of plots; Abha and Bina each get more than Dipti (clue 4). The only split is Abha 4, Bina 4, Chitra 2, Dipti 2. From the figure, Chitra's two plots are X1 and Z2.

3. Abha's total: Abha's plots are the remaining ones — {Y1, Y2, Y4, Z3} = {21, 4, 16, 9} = 50.

4. Chitra's total (clue 1): Abha = Chitra + 20, so Chitra = 50 − 20 = 30.

5. Solving for Z2: Chitra's total = X1 + Z2 = 12 + Z2 = 30, so Z2 = 18.

Solved grid (showing Chitra's plots):

Chitra's only pine plot is Z2, which contains 18 trees.

Answer: 18 (Choice A).

We need to identify the plot with the smallest number of trees and who owns it. Once the full grid is solved, this is a straightforward scan of all twelve values.

How the grid is solved (brief summary):

1. Clue 3 + clue 9 fix the teak row as $(21, 4, 8, 16)$, giving teak = 49, mango = 98, pine = 58.
2. Plot ownership is determined by clues 4–8: Abha gets 4 plots, Bina gets 4, Chitra gets 2, Dipti gets 2.
3. Clue 1 fixes daughter totals: Abha = 50, Chitra = 30, Dipti = 56, Bina = 69.
4. Individual plot values follow from these totals and the row/column constraints.

Fully solved grid:

The twelve plot values are: 12, 30, 24, 32, 21, 4, 8, 16, 3, 18, 9, 28.

The smallest is 3 trees, in plot Z1 (row Z, column 1), which belongs to Bina.

💡 Teacher tip: Once the grid is fully solved, "smallest plot" questions are a simple scan — but make sure every value is correctly determined, since one wrong entry can change the minimum.

Answer: Bina, 3 trees (Choice B).

We need to find which statement is NOT true. Let's check each one against the solved grid.

How the grid is solved (brief summary):

1. Clue 3 + clue 9 fix the teak row as $(21, 4, 8, 16)$, giving teak = 49, mango = 98, pine = 58.
2. Plot ownership: Abha = {Y1, Y2, Y4, Z3}, Bina = {Z1, X2, Y3, Z4}, Chitra = {X1, Z2}, Dipti = {X3, X4}.
3. Clue 1 gives totals: Abha = 50, Chitra = 30, Dipti = 56, Bina = 69.
4. Individual values: Z2 = 18, Z1 = 3, X2 = 30, X3/X4 = {24, 32}.

Fully solved grid:

Checking each statement:

• A. "Chitra got 12 mango trees." Chitra's mango plot is X1 = 12. ✓ True.
• B. "Bina got 32 pine trees." Bina's pine (row Z) plots are Z1 = 3 and Z4 = 28, totaling $3 + 28 = 31$, not 32. ✗ False.
• C. "Abha got 41 teak trees." Abha's teak (row Y) plots are Y1 = 21, Y2 = 4, Y4 = 16, totaling $21 + 4 + 16 = 41$. ✓ True.
• D. "Dipti got 56 mango trees." Dipti's plots X3 and X4 are both in row X (mango), totaling $24 + 32 = 56$. ✓ True.

Statement B is the one that is NOT true.

💡 Teacher tip: In "which is NOT true" questions, check the arithmetic statements first — summing a daughter's plots in a specific row is quick to verify and often eliminates most options fast.

Answer: "Bina got 32 pine trees" (Choice B).

We need to find which column has the highest total number of trees. From the solved grid

Setting up the decoded operations

Before answering any question, we need to translate the cumulative table into per-day figures. Let's define, for orders booked on day $d$:

• $x_d$ = orders delivered the next day ($d+1$)
• $y_d$ = orders delivered the second day ($d+2$)
• $l_d$ = orders lost (declared lost at the end of day $d+2$)

Then $B_d = x_d + y_d + l_d$ (total booked on day $d$).

Key relationships from the delivery rules:

• Orders delivered on day $t$ = $x_{t-1} + y_{t-2}$ (next-day deliveries from yesterday + second-day deliveries from two days ago).
• Orders lost on day $t$ = $l_{t-2}$ (orders booked two days ago that were never delivered).

Step 1 — Daily bookings (differences of cumulative booked):

Step 2 — Lost per booking day (differences of cumulative lost, shifted by 2 days):

(e.g., cumulative lost rose from 92 on the 14th to 94 on the 15th, so 2 orders were lost on the 15th — these were booked on the 13th, hence $l_{13}=2$.)

Step 3 — Use the given 2-day delivery figures $y_{11}=4,\ y_{12}=6,\ y_{13}=8$ and the delivered-on-day equation:

• Day 13 delivered: $x_{12}+y_{11}=11 \Rightarrow x_{12}=7$
• Day 14 delivered: $x_{13}+y_{12}=27 \Rightarrow x_{13}=21$
• Day 15 delivered: $x_{14}+y_{13}=23 \Rightarrow x_{14}=15$

Step 4 — Chain through using $B_d = x_d+y_d+l_d$ and the delivered-on-day equation for subsequent days:

• $B_{12}=7+6+1=14$, $B_{13}=21+8+2=31$
• $B_{14}=15+y_{14}+12=30 \Rightarrow y_{14}=3$
• Day 16 delivered: $x_{15}+y_{14}=11 \Rightarrow x_{15}=8$; then $y_{15}=28-8-12=8$
• Day 17 delivered: $x_{16}+y_{15}=21 \Rightarrow x_{16}=13$; then $y_{16}=25-13-2=10$
• Day 18 delivered: $x_{17}+y_{16}=13 \Rightarrow x_{17}=3$; then $y_{17}=25-3-9=13$
• Day 19 delivered: $x_{18}+y_{17}=14 \Rightarrow x_{18}=1$

Fully decoded operations:

Solving Question 1

We need the fraction of that day's bookings that were lost, i.e. $l_d / B_d$, for each of the four listed days.

💡 Teacher tip: When comparing fractions with different denominators, either cross-multiply or convert to decimals. Here $12/28$ vs $12/30$ is an easy call — same numerator, smaller denominator wins.

The largest fraction is on the 15th ($\approx 0.429$), edging out the 14th ($0.40$).

Answer: 15th (Choice A)

Setting up the decoded operations

Before answering any question, we need to translate the cumulative table into per-day figures. Let's define, for orders booked on day $d$:

• $x_d$ = orders delivered the next day ($d+1$)
• $y_d$ = orders delivered the second day ($d+2$)
• $l_d$ = orders lost (declared lost at the end of day $d+2$)

Then $B_d = x_d + y_d + l_d$ (total booked on day $d$).

Key relationships from the delivery rules:

• Orders delivered on day $t$ = $x_{t-1} + y_{t-2}$
• Orders lost on day $t$ = $l_{t-2}$

Step 1 — Daily bookings (differences of cumulative booked):

Step 2 — Lost per booking day (differences of cumulative lost, shifted by 2 days):

Step 3 — Use the given 2-day delivery figures $y_{11}=4,\ y_{12}=6,\ y_{13}=8$:

• Day 13 delivered: $x_{12}+y_{11}=11 \Rightarrow x_{12}=7$
• Day 14 delivered: $x_{13}+y_{12}=27 \Rightarrow x_{13}=21$
• Day 15 delivered: $x_{14}+y_{13}=23 \Rightarrow x_{14}=15$

Step 4 — Chain through using $B_d = x_d+y_d+l_d$ and the delivered-on-day equation:

• $B_{12}=7+6+1=14$, $B_{13}=21+8+2=31$
• $B_{14}=15+y_{14}+12=30 \Rightarrow y_{14}=3$
• Day 16: $x_{15}+y_{14}=11 \Rightarrow x_{15}=8$; then $y_{15}=28-8-12=8$
• Day 17: $x_{16}+y_{15}=21 \Rightarrow x_{16}=13$; then $y_{16}=25-13-2=10$
• Day 18: $x_{17}+y_{16}=13 \Rightarrow x_{17}=3$; then $y_{17}=25-3-9=13$
• Day 19: $x_{18}+y_{17}=14 \Rightarrow x_{18}=1$

Fully decoded operations:

Solving Question 2

We need the day with the highest number of orders booked ($B_d$) among the four listed options.

The 13th had 31 bookings — the highest among the listed days (and indeed the highest in the entire decoded table).

💡 Teacher tip: The 12th and 13th bookings aren't directly readable from the cumulative column (which starts on the 13th), but they are recovered from the decoding: $B_{12}=x_{12}+y_{12}+l_{12}=7+6+1=14$ and $B_{13}=21+8+2=31$.

Answer: 13th (Choice C)

Setting up the decoded operations

Before answering any question, we need to translate the cumulative table into per-day figures. Let's define, for orders booked on day $d$:

• $x_d$ = orders delivered the next day ($d+1$)
• $y_d$ = orders delivered the second day ($d+2$)
• $l_d$ = orders lost (declared lost at the end of day $d+2$)

Then $B_d = x_d + y_d + l_d$ (total booked on day $d$).

Key relationships from the delivery rules:

• Orders delivered on day $t$ = $x_{t-1} + y_{t-2}$
• Orders lost on day $t$ = $l_{t-2}$

Step 1 — Daily bookings (differences of cumulative booked):

Step 2 — Lost per booking day (differences of cumulative lost, shifted by 2 days):

Step 3 — Use the given 2-day delivery figures $y_{11}=4,\ y_{12}=6,\ y_{13}=8$:

• Day 13 delivered: $x_{12}+y_{11}=11 \Rightarrow x_{12}=7$
• Day 14 delivered: $x_{13}+y_{12}=27 \Rightarrow x_{13}=21$
• Day 15 delivered: $x_{14}+y_{13}=23 \Rightarrow x_{14}=15$

Step 4 — Chain through using $B_d = x_d+y_d+l_d$ and the delivered-on-day equation:

• $B_{12}=7+6+1=14$, $B_{13}=21+8+2=31$
• $B_{14}=15+y_{14}+12=30 \Rightarrow y_{14}=3$
• Day 16: $x_{15}+y_{14}=11 \Rightarrow x_{15}=8$; then $y_{15}=28-8-12=8$
• Day 17: $x_{16}+y_{15}=21 \Rightarrow x_{16}=13$; then $y_{16}=25-13-2=10$
• Day 18: $x_{17}+y_{16}=13 \Rightarrow x_{17}=3$; then $y_{17}=25-3-9=13$
• Day 19: $x_{18}+y_{17}=14 \Rightarrow x_{18}=1$

Fully decoded operations:

Solving Question 3

The delivery ratio for a day is defined as next-day deliveries: second-day deliveries, i.e. the value $x_d / y_d$. We compute this for each listed day:

💡 Teacher tip: The 14th stands out because it has a very small number of second-day deliveries ($y_{14}=3$) relative to next-day deliveries ($x_{14}=15$). A small denominator drives the ratio up sharply.

The highest delivery ratio is on the 14th ($= 5$).

Answer: 14th (Choice D)

Setting up the decoded operations

Before answering any question, we need to translate the cumulative table into per-day figures. Let's define, for orders booked on day $d$:

• $x_d$ = orders delivered the next day ($d+1$)
• $y_d$ = orders delivered the second day ($d+2$)
• $l_d$ = orders lost (declared lost at the end of day $d+2$)

Then $B_d = x_d + y_d + l_d$ (total booked on day $d$).

Key relationships from the delivery rules:

• Orders delivered on day $t$ = $x_{t-1} + y_{t-2}$
• Orders lost on day $t$ = $l_{t-2}$

Step 1 — Daily bookings (differences of cumulative booked):

Step 2 — Lost per booking day (differences of cumulative lost, shifted by 2 days):

Step 3 — Use the given 2-day delivery figures $y_{11}=4,\ y_{12}=6,\ y_{13}=8$:

• Day 13 delivered: $x_{12}+y_{11}=11 \Rightarrow x_{12}=7$
• Day 14 delivered: $x_{13}+y_{12}=27 \Rightarrow x_{13}=21$
• Day 15 delivered: $x_{14}+y_{13}=23 \Rightarrow x_{14}=15$

Step 4 — Chain through using $B_d = x_d+y_d+l_d$ and the delivered-on-day equation:

• $B_{12}=7+6+1=14$, $B_{13}=21+8+2=31$
• $B_{14}=15+y_{14}+12=30 \Rightarrow y_{14}=3$
• Day 16: $x_{15}+y_{14}=11 \Rightarrow x_{15}=8$; then $y_{15}=28-8-12=8$
• Day 17: $x_{16}+y_{15}=21 \Rightarrow x_{16}=13$; then $y_{16}=25-13-2=10$
• Day 18: $x_{17}+y_{16}=13 \Rightarrow x_{17}=3$; then $y_{17}=25-3-9=13$
• Day 19: $x_{18}+y_{17}=14 \Rightarrow x_{18}=1$

Fully decoded operations:

Solving Question 4

The average delivery time for orders booked on day $d$ is:

$\text{Average time} = \frac{x_d + 2y_d}{x_d + y_d}$

where $x_d$ orders take 1 day and $y_d$ orders take 2 days (lost orders are excluded since they are never delivered).

💡 Teacher tip: We can rewrite this as $1 + \frac{y_d}{x_d+y_d}$. So the average is minimised when the fraction of second-day deliveries is smallest — i.e., when the largest share of delivered orders went out the very next day.

Computing for each listed day:

The 14th has the smallest share of second-day deliveries ($3$ out of $18$ delivered), giving the lowest average time of $\approx 1.17$ days.

Answer: 14th (Choice C)

Setting up the scoring rules

First, let us tabulate what each player earns depending on how many of the four bid Hi:

A crucial observation: every round's total score is either −4, 0, or +4 — it is −4 when all four bid Hi, +4 when all four bid Lo, and 0 in all other cases.

Deducing rounds 1–3

Total after three rounds: $6 + 2 + (-2) + (-2) = 4$.

Since each round contributes −4, 0, or +4, the only way three rounds sum to +4 is: one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun's three-round total is 6. One of those rounds is the all-Lo round where Arun gets +1. So his two zero-sum-round scores must add up to $6 - 1 = 5$. The maximum a player can gain in a zero-sum round is +3 (being the sole Hi-bidder). The only way to make 5 from two values each ≤ 3 is +3 and +2. So Arun's round scores are:

• +3 → Arun alone bid Hi (1-Hi round)
• +2 → 2-Hi round with Arun bidding Hi
• +1 → all-Lo round

Using Clue 3 (Dipak: $D_1 > D_3 > D_2$, three distinct values summing to 2):

• In the Arun-alone-Hi round, Dipak bids Lo and gets −1.
• In the all-Lo round, Dipak gets +1.
• In the 2-Hi round, Dipak gets either +2 (if Hi) or −2 (if Lo).

Dipak's sum = $-1 + 1 + (\text{+2 or −2}) = +2$ or $-2$. We need the sum to be 2, so Dipak scored +2 in the 2-Hi round (he bid Hi).

Dipak's three scores are {+2, −1, +1}. With $D_1 > D_3 > D_2$: $D_1 = +2$, $D_3 = +1$, $D_2 = -1$.

This pins the round types:

• Round 1: Dipak +2 → the 2-Hi round (Arun & Dipak bid Hi)
• Round 2: Dipak −1 → the Arun-alone-Hi round
• Round 3: Dipak +1 → the all-Lo round

Verifying rounds 1–3

Totals: Arun $2+3+1=6$ ✓, Dipak $2-1+1=2$ ✓, Bankim $-2-1+1=-2$ ✓, Charu $-2-1+1=-2$ ✓.

Answering this question

From the solved table, Round 1 has Arun and Dipak bidding Hi, Bankim and Charu bidding Lo.

In the order Arun, Bankim, Charu, Dipak: Hi, Lo, Lo, Hi.

Answer: Hi, Lo, Lo, Hi (Choice A)

Setting up the scoring rules

Key fact: each round's total is either −4 (all Hi), 0 (1/2/3 Hi), or +4 (all Lo).

Deducing rounds 1–3

Total after 3 rounds: $6 + 2 + (-2) + (-2) = 4$ → one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun scored 6 in three rounds. One round is all-Lo (Arun gets +1), so his two zero-sum-round scores sum to 5. Since the max per zero-sum round is +3, the only split is +3 and +2. So Arun's scores are {+3, +2, +1}:

• +3 → Arun alone bid Hi
• +2 → 2-Hi round, Arun bid Hi
• +1 → all-Lo round

Using Clue 3 ($D_1 > D_3 > D_2$, sum = 2): Dipak gets −1 in the Arun-alone-Hi round, +1 in the all-Lo round, and +2 in the 2-Hi round (must be +2 to reach sum 2). So $D_1=+2$, $D_3=+1$, $D_2=-1$, giving:

Deducing rounds 4–6

Score changes from round 3 to round 6: Arun $+1$, Bankim $+1$, Charu $-3$, Dipak $-3$. Total $= -4$, so one all-Hi round (−4) and two zero-sum rounds (0 each).

After the all-Hi round (everyone −1), the remaining two zero-sum rounds must give: Arun $+2$, Bankim $+2$, Charu $-2$, Dipak $-2$.

Clue 4 says Arun was the only Hi-bidder in exactly 2 of 6 rounds. One is round 2, so exactly one more occurs in rounds 4–6. That Arun-alone-Hi round gives Arun +3, others −1 each. The remaining zero-sum round then gives: Arun $-1$, Bankim $+3$, Charu $-1$, Dipak $-1$ — which is a Bankim-alone-Hi round.

So rounds 4–6 (in some undetermined order) are:

Counting Arun's Hi bids

Across all six rounds, Arun bids Hi in:

1. Round 1 (2-Hi round) ✓
2. Round 2 (Arun alone Hi) ✓
3. Round 3 — No (all Lo)
4. Arun-alone-Hi round (among 4–6) ✓
5. All-Hi round (among 4–6) ✓
6. Bankim-alone-Hi round (among 4–6) — No

Total: 4 rounds.

Answer: 4

Setting up the scoring rules

Each round's total is either −4, 0, or +4.

Deducing rounds 1–3

Total after 3 rounds: $6 + 2 + (-2) + (-2) = 4$ → one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun scored 6 in three rounds. One round is all-Lo (+1), so his two zero-sum scores sum to 5. Max per zero-sum round is +3, so the only split is +3 and +2. Arun's scores: {+3, +2, +1}.

Using Clue 3 ($D_1 > D_3 > D_2$, sum = 2): Dipak gets −1 in the Arun-alone-Hi round, +1 in the all-Lo round, and +2 in the 2-Hi round. So $D_1=+2$, $D_3=+1$, $D_2=-1$.

Deducing rounds 4–6

Changes from round 3 to 6: Arun $+1$, Bankim $+1$, Charu $-3$, Dipak $-3$. Total $= -4$ → one all-Hi round and two zero-sum rounds.

After the all-Hi round (everyone −1), the two zero-sum rounds must give: Arun $+2$, Bankim $+2$, Charu $-2$, Dipak $-2$.

By Clue 4, one of these is an Arun-alone-Hi round (Arun +3, others −1). The other must then be a Bankim-alone-Hi round (Bankim +3, others −1).

So rounds 4–6 (in some order) are:

Counting Bankim's Lo bids

Bankim bids Lo in:

1. Round 1 (2-Hi round) ✓
2. Round 2 (Arun alone Hi) ✓
3. Round 3 (all Lo) ✓
4. Arun-alone-Hi round (among 4–6) ✓
5. Bankim-alone-Hi round — No (he bids Hi)
6. All-Hi round — No (he bids Hi)

Total: 4 rounds.

Answer: 4

Setting up the scoring rules

Each round's total is either −4, 0, or +4.

Deducing rounds 1–3

Total after 3 rounds: $6 + 2 + (-2) + (-2) = 4$ → one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun scored 6 in three rounds. One round is all-Lo (+1), so his two zero-sum scores sum to 5. Max per zero-sum round is +3, so the only split is +3 and +2. Arun's scores: {+3, +2, +1}.

Using Clue 3 ($D_1 > D_3 > D_2$, sum = 2): Dipak gets −1 in the Arun-alone-Hi round, +1 in the all-Lo round, and +2 in the 2-Hi round. So $D_1=+2$, $D_3=+1$, $D_2=-1$.

Deducing rounds 4–6

Changes from round 3 to 6: Arun $+1$, Bankim $+1$, Charu $-3$, Dipak $-3$. Total $= -4$ → one all-Hi round and two zero-sum rounds.

After the all-Hi round (everyone −1), the two zero-sum rounds must give: Arun $+2$, Bankim $+2$, Charu $-2$, Dipak $-2$.

By Clue 4, one of these is an Arun-alone-Hi round (Arun +3, others −1). The other must then be a Bankim-alone-Hi round (Bankim +3, others −1).

So rounds 4–6 (in some order) are:

Identifying rounds with identical bids

"Identical bids" means all four players bid the same thing — either all Hi or all Lo.

Scanning all six rounds:

• Round 3: all Lo ✓
• All-Hi round (one of rounds 4–6): all Hi ✓
• Every other round has a mix of Hi and Lo bids.

Total: 2 rounds.

Answer: 2

Setting up the scoring rules

Each round's total is either −4, 0, or +4.

Deducing rounds 1–3

Total after 3 rounds: $6 + 2 + (-2) + (-2) = 4$ → one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun scored 6 in three rounds. One round is all-Lo (+1), so his two zero-sum scores sum to 5. Max per zero-sum round is +3, so the only split is +3 and +2. Arun's scores: {+3, +2, +1}.

Using Clue 3 ($D_1 > D_3 > D_2$, sum = 2): Dipak gets −1 in the Arun-alone-Hi round, +1 in the all-Lo round, and +2 in the 2-Hi round. So $D_1=+2$, $D_3=+1$, $D_2=-1$.

Deducing rounds 4–6

Changes from round 3 to 6: Arun $+1$, Bankim $+1$, Charu $-3$, Dipak $-3$. Total $= -4$ → one all-Hi round and two zero-sum rounds.

After the all-Hi round (everyone −1), the two zero-sum rounds must give: Arun $+2$, Bankim $+2$, Charu $-2$, Dipak $-2$.

By Clue 4, one of these is an Arun-alone-Hi round (Arun +3, others −1). The other must then be a Bankim-alone-Hi round (Bankim +3, others −1).

So Dipak's scores in rounds 4–6 are:

Counting rounds where Dipak gained exactly 1 point

Dipak's per-round scores across all six rounds:

💡 Teacher tip: A player gains exactly +1 only when everyone bids Lo (the 0-Hi round). In every other round type, a player's score is either +2, +3, −1, −2, or −3 — never +1.

Dipak scores +1 only in Round 3 (the all-Lo round). That is 1 round.

Answer: 1

Setting up the scoring rules

Each round's total is either −4, 0, or +4.

Deducing rounds 1–3

Total after 3 rounds: $6 + 2 + (-2) + (-2) = 4$ → one all-Lo round (+4) and two zero-sum rounds (0 each).

Arun scored 6 in three rounds. One round is all-Lo (+1), so his two zero-sum scores sum to 5. Max per zero-sum round is +3, so the only split is +3 and +2. Arun's scores: {+3, +2, +1}.

Using Clue 3 ($D_1 > D_3 > D_2$, sum = 2): Dipak gets −1 in the Arun-alone-Hi round, +1 in the all-Lo round, and +2 in the 2-Hi round. So $D_1=+2$, $D_3=+1$, $D_2=-1$.

Deducing rounds 4–6

Changes from round 3 to 6: Arun $+1$, Bankim $+1$, Charu $-3$, Dipak $-3$. Total $= -4$ → one all-Hi round and two zero-sum rounds.

After the all-Hi round (everyone −1), the two zero-sum rounds must give: Arun $+2$, Bankim $+2$, Charu $-2$, Dipak $-2$.

By Clue 4, one of these is an Arun-alone-Hi round (Arun +3, others −1). The other must then be a Bankim-alone-Hi round (Bankim +3, others −1).

So rounds 4–6 (in some order) are:

Identifying where Arun was DEFINITELY the only Hi-bidder

Arun-alone-Hi occurs in:

• Round 2 — fixed and certain ✓
• Exactly one of rounds 4, 5, or 6 — but we cannot determine which one, since the order of the last three rounds is not fixed.

The question asks for the round where Arun was definitely the only Hi-bidder. Only Round 2 is guaranteed.

Answer: Second round (Choice A)

Finding the diseased patient from partial test results

Here is the patient-to-vial assignment table:

Key principle: Since exactly one patient has the disease, the set of positive vials is exactly that patient's four vials. So:

• A positive vial must be among the diseased patient's vials.
• A negative vial must not be among the diseased patient's vials.

Given clues: Vial C is positive; vials A, E, H are negative.

We filter step by step:

1. C is positive → the patient's vial-set contains C → patients {5, 6, 7, 8, 13, 14, 15, 16}.
2. A is negative → the patient's vial-set does not contain A → restrict to patients {1–8} → intersecting with step 1 gives {5, 6, 7, 8}.
3. E is negative → no E in the patient's set. Among {5, 6, 7, 8}: patient 7 (B,C,E,H) and patient 8 (B,C,E,G) both contain E, so remove them → {5, 6}.
4. H is negative → no H. Patient 5 (B,C,F,H) contains H; patient 6 (B,C,F,G) does not → patient 6.

Verification: Patient 6's vials are B, C, F, G — contains C ✓, no A ✓, no E ✓, no H ✓.

Answer: Patient 6 (Choice C).

Choosing the next vial to test

Here is the patient-to-vial assignment table:

Key principle: The positive vials are exactly the diseased patient's four vials; every negative vial is absent from the patient's set.

Given clues: Vial A is positive; vials D, G are negative. We may test one more vial to identify the patient uniquely.

Filtering step by step:

1. A is positive → patient has A → patients {9, 10, 11, 12, 13, 14, 15, 16}.
2. D is negative → patient has no D. Among 9–16, patients 9, 10, 11, 12 all contain D, so remove them → {13, 14, 15, 16}.
3. G is negative → patient has no G. Patients 14 (A,C,F,G) and 16 (A,C,E,G) contain G, so remove them → {13, 15}.

The disease is either patient 13 (A, C, F, H) or patient 15 (A, C, E, H). Let us compare:

These two patients differ in exactly one vial-pair: E vs. F. Testing vial E settles it:

• If E tests positive → patient 15.
• If E tests negative → patient 13.

💡 Teacher tip: When two candidates differ in only one position, testing either of the two differing vials will identify the patient. Here vial F would also work, but it is not among the answer choices — vial E is.

Answer: Vial E (Choice B).

Identifying the impossible combination of test results

Here is the patient-to-vial assignment table:

Key principle: A combination of test results is possible only if at least one patient's vial-set is consistent with it — every vial reported positive must be in that patient's set, and every vial reported negative must not be. If no patient fits, the combination is impossible.

We check each option.

Option A: A and E positive; C and D negative.

• A and E both positive → the patient must have both A and E → candidates: {11 (A,D,E,H), 12 (A,D,E,G), 15 (A,C,E,H), 16 (A,C,E,G)}.
• C and D both negative → the patient must have neither C nor D. But:
• Patient 11 has D ✗
• Patient 12 has D ✗
• Patient 15 has C ✗
• Patient 16 has C ✗
• All four candidates are disqualified. No patient fits → this combination is NOT possible.

For completeness, we verify the other three options are each possible:

Option B: B positive; C, F, H negative.

• Patient 4 (B, D, E, G) has B, and none of C, F, H. ✓ Possible.

Option C: A and G positive; D and E negative.

• Patient 14 (A, C, F, G) has A and G, and neither D nor E. ✓ Possible.

Option D: B and D positive; F and H negative.

• Patient 4 (B, D, E, G) has B and D, and neither F nor H. ✓ Possible.

Only Option A has no consistent patient.

Answer: Vials A and E positive, vials C and D negative (Choice A).

Possible numbers of positive vials after the mixing accident

Here is the patient-to-vial assignment table:

Understanding the structure: Each patient's vial-set is a unique 4-vial code — one vial chosen from each of four pairs: (A/B), (C/D), (E/F), (G/H). Two distinct patients always differ in at least one of these four positions.

What the accident does: Two patients' blood is mixed into one combined sample before distribution. That combined sample is placed into every vial that either patient was assigned to — i.e., it reaches the union of the two patients' vial-sets. Exactly one patient has the disease, so we consider two cases.

Case 1 — the diseased patient is NOT one of the two mixed patients.

The disease travels only with the diseased patient's own, unmixed sample. Exactly that patient's 4 vials test positive.

→ 4 positive vials.

Case 2 — the diseased patient IS one of the two mixed patients.

The combined sample carries the disease, so every vial in the union of the two patients' 4-vial sets tests positive. The number of positive vials equals the size of this union.

Since each patient picks one vial from each of the four pairs, two distinct patients share a vial in every position where they make the same choice. If they differ in $k$ of the four positions, they share $4 - k$ vials, so:

$\text{Union size} = 4 + 4 - (4 - k) = 4 + k$

Every value from 5 to 8 is achievable:

• $k=1$: patients 1 (B,D,F,H) and 2 (B,D,F,G) differ only in position 4 → union size 5.
• $k=2$: patients 1 and 4 (B,D,E,G) differ in positions 3 and 4 → union size 6.
• $k=3$: patients 1 and 8 (B,C,E,G) differ in positions 2, 3, 4 → union size 7.
• $k=4$: patients 1 (B,D,F,H) and 16 (A,C,E,G) differ in all four positions → union size 8.

Since two distinct patients can never be identical, the union can never be as small as 4 in this case.

Combining both cases:

• Case 1 gives 4 positive vials.
• Case 2 gives 5, 6, 7, or 8 positive vials.

The full set of possible positive-vial counts is:

$\{4,\ 5,\ 6,\ 7,\ 8\}$

Answer: {4, 5, 6, 7, 8} (Choice C).

Setting up the 4-set Venn diagram

We have 600 schools and four facilities (F1–F4), so we build a Venn diagram with four overlapping circles. Each school falls into exactly one region based on which combination of facilities it has.

Directly from the clues:

Key variable — the "exactly three" regions.

Clue 4 says the number of schools with exactly three facilities is the same regardless of which three. There are four such triple regions (F1F2F3, F1F2F4, F1F3F4, F2F3F4). Let each one equal $t$.

Finding $t$

Step 1 — Use clue 8: $n(F_1 \cap F_2) = 162$.

The region $F_1 \cap F_2$ includes four parts: F1&F2 only, the two triple regions containing both F1 and F2 (each $= t$), and the all-four region: $n(F_1 \cap F_2) = b_{12} + t + t + 40 = b_{12} + 2t + 40 = 162$ $\Rightarrow b_{12} = 122 - 2t$

where $b_{12}$ is the number having only F1 and F2.

Step 2 — Use clue 5: $n(F_2) = 313$.

Every school with F2 is counted: only F2, the three pairwise regions touching F2, the three triple regions touching F2, and the all-four region: $n(F_2) = 30 + b_{12} + 26 + b_{24} + 3t + 40 = 313$ $\Rightarrow b_{12} + b_{24} + 3t = 217$

Step 3 — Use clue 7: Among schools having F4, 45 had only F2 (meaning F2 & F4 only) and 24 had only F3 (meaning F3 & F4 only).

So $b_{24} = 45$ and $b_{34} = 24$.

Substituting $b_{24} = 45$ into the equation from Step 2: $b_{12} + 45 + 3t = 217 \Rightarrow b_{12} + 3t = 172$

Now substitute $b_{12} = 122 - 2t$: $(122 - 2t) + 3t = 172$ $122 + t = 172$ $t = 50$

💡 Teacher tip: The breakthrough is spotting that clue 8 gives a relationship between $b_{12}$ and $t$, while clue 5 (combined with clue 7's $b_{24}=45$) gives a second equation in the same two unknowns. Two equations, two unknowns — solve.

Answering the question

Each of the four "exactly three" regions equals $t = 50$. The total number of schools with exactly three facilities is: $4t = 4 \times 50 = 200$

Answer: 200 (Choice C)

Setting up the 4-set Venn diagram

We have 600 schools and four facilities (F1–F4). Let each of the four "exactly three facilities" regions equal $t$ (clue 4 makes them all equal).

Directly known from clues:

Finding $t$ and the remaining regions

From clue 8 ($n(F_1 \cap F_2) = 162$): the overlap of F1 and F2 consists of F1&F2 only, two triple regions (each $t$), and the all-four region: $b_{12} + 2t + 40 = 162 \Rightarrow b_{12} = 122 - 2t$

From clue 5 ($n(F_2) = 313$): summing all regions inside F2: $30 + b_{12} + 26 + 45 + 3t + 40 = 313 \Rightarrow b_{12} + 3t = 172$

Substituting $b_{12} = 122 - 2t$: $122 - 2t + 3t = 172 \Rightarrow t = 50, \quad b_{12} = 22$

Finding $b_{13}$ and $b_{14}$ using clue 9 ($n(F_1) = n(F_4)$) and the grand total of 600:

• $n(F_1) = 25 + 22 + b_{13} + b_{14} + 3(50) + 40 = 237 + b_{13} + b_{14}$
• $n(F_4) = 20 + b_{14} + 45 + 24 + 3(50) + 40 = 279 + b_{14}$

Setting them equal: $237 + b_{13} + b_{14} = 279 + b_{14} \Rightarrow b_{13} = 42$.

Using the total: $80 + 101 + (22+42+b_{14}+26+45+24) + 200 + 40 = 600 \Rightarrow b_{14} = 20$.

Fully solved Venn diagram

Answering the question

We need $n(F_2 \cap F_4)$ — every school that has both F2 and F4, regardless of what else it has. This spans four Venn regions:

$n(F_2 \cap F_4) = 45 + 50 + 50 + 40 = 185$

💡 Teacher tip: "Schools having F2 and F4" means at least F2 and F4 — not only F2 and F4. Always sum every region inside the intersection of the two circles.

Answer: 185 (Choice A)