CAT 2024 DILR Questions — Slot 3

Finding the smallest population in 2024

The key relationship given in the set-up is:

$\text{GDP per capita} = \frac{\text{GDP}}{\text{Population}}$

Rearranging this:

$\text{Population} = \frac{\text{GDP}}{\text{GDP per capita}}$

The table gives each country's GDP as a fraction of Country 9's GDP, and its GDP per capita as a fraction of Country 10's GDP per capita. Since both columns use the same reference country across all rows, dividing Column 2 by Column 3 gives each country's population relative to a common baseline. We don't need absolute numbers — we only need to compare these ratios.

💡 Teacher tip: You never need the actual GDP of Country 9 or the actual GDP per capita of Country 10. Because every row is expressed against the same reference, the ratio (Column 2 ÷ Column 3) preserves the correct ordering of populations.

Scanning the last column, the smallest value is 0.171, which belongs to Country 8.

Answer: A. Country 8

Ratio of Country 4's GDP to Country 5's GDP in 2026

We need the GDP ratio two years after 2024 (i.e., in 2026). Each country's GDP grows at its own annual GDP growth rate, compounded yearly.

From the table:

After 2 years of compounding:

$\text{GDP}_{C4}(2026) = 0.12 \times (1.005)^2$

$\text{GDP}_{C5}(2026) = 0.10 \times (1.007)^2$

The required ratio is:

$\frac{\text{GDP}_{C4}(2026)}{\text{GDP}_{C5}(2026)} = \frac{0.12 \times (1.005)^2}{0.10 \times (1.007)^2} = \frac{0.12}{0.10} \times \left(\frac{1.005}{1.007}\right)^2$

Computing each piece:

• $\frac{0.12}{0.10} = 1.2$
• $(1.005)^2 = 1.010025$
• $(1.007)^2 = 1.014049$

$\text{Ratio} = 1.2 \times \frac{1.010025}{1.014049} = 1.2 \times 0.99603 \approx 1.195$

💡 Teacher tip: The population growth rate is irrelevant here — the question asks only about GDP, which depends solely on the GDP growth rate. Don't let the extra column distract you.

Answer: A. 1.195

Largest population in 2027 among Countries 1, 4, 5, and 7

Step 1 — Find each country's relative population in 2024.

Since $\text{Population} = \text{GDP} \div \text{GDP per capita}$, we divide Column 2 by Column 3:

Step 2 — Project each population forward 3 years (2024 → 2027).

Each population is multiplied by $(1 + r)^3$ where $r$ is the population growth rate.

• Country 1: $0.366 \times (0.9988)^3 \approx 0.366 \times 0.9964 \approx 0.365$
• Country 4: $0.316 \times (1.0049)^3 \approx 0.316 \times 1.0148 \approx 0.321$
• Country 5: $0.278 \times (1.0031)^3 \approx 0.278 \times 1.0093 \approx 0.281$
• Country 7: $0.267 \times (0.9989)^3 \approx 0.267 \times 0.9967 \approx 0.266$

💡 Teacher tip: Country 4 has the highest population growth rate among the four, so it closes the gap slightly — but the starting gap between Country 1 (0.366) and Country 4 (0.316) is far too large to close in just three years at these small rates. Country 1 stays on top.

Even after three years of growth, Country 1 remains the most populous.

Answer: A. Country 1

Counting countries whose GDP per capita falls by 2027

Key idea: GDP per capita $= \text{GDP} \div \text{Population}$. Over time, GDP grows at the GDP growth rate and population grows at the population growth rate. So the annual growth rate of GDP per capita is approximately:

$\text{GDP per capita growth rate} \approx (\text{GDP growth rate}) - (\text{Population growth rate})$

If this difference is positive, GDP per capita rises over the three years. If it is negative, GDP per capita falls.

Step 1 — Compute the difference for every country.

💡 Teacher tip: Watch the signs carefully. Countries 1, 2, and 7 have negative population growth rates (their populations are shrinking). Subtracting a negative number adds, which makes the GDP-per-capita growth even larger. Country 4 is the closest call — GDP grows only 0.01% faster than population — but it is still positive, so GDP per capita still rises (barely).

Step 2 — Count the countries where GDP per capita falls.

Every single difference is positive, meaning GDP per capita increases for all eight countries from 2024 to 2027. Therefore, the number of countries whose GDP per capita in 2027 is lower than in 2024 is:

$\boxed{0}$

Answer: 0

Finding the total number of subscribers

Let $T$ be the total number of subscribers in 2023. Since the total grew by 10% (clue 1), the 2024 total is $1.1T$.

From the bar chart:

We need $T$ to get actual numbers. Use clues 3 and 4.

Clue 4 tells us that in 2023:

• Kid one-app = 10,000
• Elder multi-app = 15,000

Clue 3 says: Kid-multi (2023) = Elder-one-app (2023). Let this common value be $x$.

So in 2023:

• Kid total = $10{,}000 + x = 0.15T$
• Elder total = $x + 15{,}000 = 0.20T$

From the Elder equation: $x = 0.20T - 15{,}000$.

Substitute into the Kid equation: $10{,}000 + 0.20T - 15{,}000 = 0.15T$ $0.05T = 5{,}000$ $\boxed{T = 100{,}000}$

So the 2024 total = $1.1 \times 100{,}000 = 110{,}000$.

Answering the question

In 2024, Others = 50% of the total: $0.50 \times 110{,}000 = 55{,}000$

Answer: C. 55,000

Setting up the numbers

Let $T$ = total subscribers in 2023; then 2024 total = $1.1T$.

From the bar chart, in 2023: Kid = $0.15T$, Elder = $0.20T$.

Clue 4 gives us two fixed 2023 values:

• Kid one-app = 10,000
• Elder multi-app = 15,000

Clue 3 says Kid-multi (2023) = Elder-one-app (2023). Call this common value $x$.

So:

• Kid total (2023) = $10{,}000 + x = 0.15T$
• Elder total (2023) = $x + 15{,}000 = 0.20T$

From the Elder equation: $x = 0.20T - 15{,}000$.

Substituting into the Kid equation: $10{,}000 + 0.20T - 15{,}000 = 0.15T$ $0.05T = 5{,}000 \implies T = 100{,}000$

Then $x = 0.20 \times 100{,}000 - 15{,}000 = 5{,}000$.

Solved 2023 Kid breakdown

Answering the question

We need the percentage of Kid subscribers who used multiple apps in 2023: $\frac{5{,}000}{15{,}000} = \frac{1}{3} = 33.33\%$

Answer: A. 33.33%

Finding the total and category counts

Let $T$ = total subscribers in 2023; then 2024 total = $1.1T$.

From the bar chart:

• 2023 Elder = $0.20T$
• 2024 Elder = $0.30 \times 1.1T = 0.33T$

We need $T$. Using clues 3 and 4:

In 2023, let $x$ = Kid-multi = Elder-one-app (clue 3). Clue 4 gives Kid-one-app = 10,000 and Elder-multi = 15,000.

• Kid total: $10{,}000 + x = 0.15T$
• Elder total: $x + 15{,}000 = 0.20T$

From the Elder equation: $x = 0.20T - 15{,}000$.

Substitute into the Kid equation: $10{,}000 + 0.20T - 15{,}000 = 0.15T$ $0.05T = 5{,}000 \implies T = 100{,}000$

Elder counts

Percentage increase

$\text{Increase} = 33{,}000 - 20{,}000 = 13{,}000$ $\text{Percentage increase} = \frac{13{,}000}{20{,}000} \times 100 = 65\%$

Answer: C. 65%

Building the full 2024 picture

First, find $T$ (2023 total). Let $x$ = Kid-multi (2023) = Elder-one-app (2023) by clue 3.

• Kid (2023): $10{,}000 + x = 0.15T$
• Elder (2023): $x + 15{,}000 = 0.20T$

Solving: $x = 0.20T - 15{,}000$, so $10{,}000 + 0.20T - 15{,}000 = 0.15T \implies T = 100{,}000$.

Thus 2024 total = $110{,}000$.

Now apply clue 2 for 2024:

• Kid total = $0.20 \times 110{,}000 = 22{,}000$; one-app = $\frac{1}{2} \times 22{,}000 = 11{,}000$, so multi = 11,000.
• Elder total = $0.30 \times 110{,}000 = 33{,}000$; one-app = $\frac{2}{3} \times 33{,}000 = 22{,}000$, so multi = 11,000.
• Others total = $0.50 \times 110{,}000 = 55{,}000$ — but no clue tells us how Others split between one-app and multi-app.

Minimising multi-app users

💡 Teacher tip: The Kid and Elder multi-app counts are fixed by clue 2. Only the Others category is flexible — so to minimise total multi-app users, push as many Others as possible into one-app (i.e., set Others-multi = 0).

Minimum multi-app total = $11{,}000 + 11{,}000 + 0 = 22{,}000$.

$\text{Minimum percentage} = \frac{22{,}000}{110{,}000} \times 100 = 20\%$

Answer: B. 20.0%

We need to determine how many times the AC was switched off between 11:01 pm and 1:59 am.

Step 1: Read the inside-temperature graph.

Step 2: Identify on/off periods.

When the AC is on (cooling), the inside temperature falls. When the AC is off, the inside temperature rises (the room warms toward the outside temperature). Checking each 30-minute segment:

Step 3: Count the switch-offs.

The AC was turned off at 0:00 and at 1:00 — both times fall within 11:01 pm to 1:59 am. That gives 2 switch-offs.

Answer: B. 2

We need the outside temperature at 1:00 am. The key is the AC off-period that begins at 1:00.

Step 1: Identify the off-period starting at 1:00.

From the graph, the inside temperature rises from 26 °C to 30 °C between 1:00 and 1:30, which means the AC was switched off at 1:00.

Step 2: Apply the off-period rule.

When the AC is switched off, the inside temperature rises at a constant rate so as to reach the outside temperature at the moment of switching off in exactly 1 hour.

• At 1:00, inside temperature = 26 °C.
• In 30 minutes, it rose to 30 °C — a rise of 4 °C.
• Since the rate is constant, in a full hour the rise would be $4 \times 2 = 8$ °C.
• Therefore, the outside temperature at 1:00 = $26 + 8 = 34$ °C.

Answer: 34

We need the outside temperature at 9 pm. To get there, we first determine the outside temperature at known times and then use the constant rate of change.

Step 1: Find outside temperatures using the off-periods.

When the AC is off, the inside temperature rises at a constant rate to reach the outside temperature (at the moment of switching off) in 1 hour.

So: outside(0:00) = 36 °C, outside(1:00) = 34 °C.

Step 2: Find the rate of outside temperature change.

The outside temperature falls at a constant rate from 7 pm to 3 am. Between 0:00 and 1:00 it dropped from 36 to 34:

$\text{Rate} = \frac{36 - 34}{1} = 2 \text{ °C per hour (falling)}$

Step 3: Extrapolate back to 9 pm.

9 pm (21:00) is 3 hours before midnight (0:00). Since the outside temperature falls by 2 °C every hour, it was 2 °C higher for each hour we go back:

$\text{outside}(21{:}00) = 36 + 2 \times 3 = 36 + 6 = 42 \text{ °C}$

Answer: 42

We need to count how many times the AC was turned on or had its temperature setting altered between 11:01 pm and 1:59 am. Switch-offs do not count.

Step 1: Identify all on/off periods from the graph.

Step 2: Determine the mode for the first hour (23:00–0:00).

The inside temperature falls 38 → 32 → 26 in two 30-minute steps. There are two possibilities:

• REGULAR mode set to 26 °C: takes 60 minutes to reach 26. At the 30-minute mark it would be at the midpoint $\frac{38+26}{2} = 32$, matching the graph. But this involves no setting change at 23:30.
• POWER mode: set to 32 for 23:00–23:30 (reaches 32 in 30 min), then change the setting to 26 at 23:30 (reaches 26 in 30 min). This involves a setting change at 23:30.

The problem states POWER mode was used for longer than REGULAR mode. If the first hour were REGULAR (60 min of REGULAR), the remaining cooling time is only 60 minutes (0:30–1:00 and 1:30–2:00), so POWER could be at most 60 min — not strictly greater than REGULAR. This violates the constraint.

Therefore, the first hour must be POWER mode, which requires a setting change at 23:30 (from 32 to 26).

💡 Teacher tip: The constraint "POWER longer than REGULAR" is the decisive clue. It forces the first hour into POWER mode, which in turn forces a setting change at 23:30.

Step 3: List all events between 11:01 pm and 1:59 am.

Settings change only at:00 or:30. Here is the complete timeline:

Step 4: Count.

Three events qualify: the setting change at 23:30, the turn-on at 0:30, and the turn-on at 1:30.

Answer: D. Exactly 3

We need the maximum value of (outside temperature − inside temperature) between 11:01 pm and 1:59 am.

Step 1: Establish outside temperatures at each half-hour mark.

From the off-period analysis (inside rises at a constant rate to reach the outside temperature at switch-off in 1 hour):

• Switched off at 0:00 (inside = 26): rose 5 °C in 30 min → 10 °C in 1 hour → outside(0:00) = $26 + 10 = 36$
• Switched off at 1:00 (inside = 26): rose 4 °C in 30 min → 8 °C in 1 hour → outside(1:00) = $26 + 8 = 34$

Outside falls at a constant rate: $\frac{36 - 34}{1} = 2$ °C/hour. Also, since the AC was first turned on at 23:00 (and was off before that), the inside had risen to match the outside, so outside(23:00) = inside(23:00) = 38.

Step 2: Verify the maximum doesn't occur inside a segment.

Within each 30-minute segment, both inside and outside temperatures change linearly, so their difference also changes monotonically:

• During cooling (AC on): inside falls faster than outside, so the difference increases throughout the segment.
• During warming (AC off): inside rises while outside falls, so the difference decreases throughout the segment.

Since the difference is monotonic within each segment, the overall maximum must occur at one of the endpoints listed above.

Step 3: Identify the peak.

The largest difference is 10 °C at 0:00 — the moment the inside has just been cooled to 26 °C while the outside is still 36 °C. This time falls within 11:01 pm to 1:59 am.

Answer: 10

We need to determine the cash requirement at the (R-C, V2) intersection. Let's build the full arrangement from the clues.

Step 1 — Verify the grand total

Summing the horizontal road totals: $22 + 20 + 20 = 62$.

Summing the vertical road totals: $15 + 21 + 26 = 62$. ✓

So the six ATMs together hold Rs. 62 Lakhs.

Step 2 — Locate the minimum (7) and maximum (15)

Clue 1 says the ATMs with Rs. 7 Lakhs (minimum) and Rs. 15 Lakhs (maximum) are on the same road. Their sum is $7 + 15 = 22$.

We check which road could hold exactly these two (or more) ATMs:

R-A = {7, 15}, exactly two ATMs.

Step 3 — Determine the remaining four cash values

The other four ATMs sum to $62 - 22 = 40$ and must split into R-B (20) and R-C (20).

Four distinct integers between 8 and 14 (since 7 is the min and 15 is the max) summing to 40:

• {8, 9, 10, 13}: sum = 40, but no two pairs each total 20 (8+13 = 21, 9+10 = 19). ✗
• {8, 9, 11, 12}: sum = 40, and 8+12 = 20, 9+11 = 20. ✓

So R-B = {8, 12}, R-C = {9, 11}, and the six values are {7, 8, 9, 11, 12, 15}.

Step 4 — Place the ATMs using Clue 2

The second-highest value is 12. Clue 2 says it is 12 km from the ATM at (R-C, V3), so there is an ATM at (R-C, V3).

Since 12 belongs to R-B, we test each possible column on R-B. Recall the road distances: V1–V2 = 4 km, V2–V3 = 7 km (along any horizontal road); R-A–R-B = 3 km, R-B–R-C = 5 km (along any vertical road).

So (R-B, V2) = 12.

Step 5 — Place R-C's two values

R-C = {9, 11} with an ATM at (R-C, V3). The V3 column totals 26.

• If (R-C, V3) = 11: remaining V3 = $26 - 11 = 15$ for the R-A and R-B cells in column V3. This is achievable (see grid below).
• If (R-C, V3) = 9: remaining V3 = $26 - 9 = 17$. R-A can contribute 7 or 15; R-B's remaining ATM is 8. The maximum fill is $15 + 8 = 23$, but we need exactly 17. If R-B contributes 8, R-A must contribute 9 — but 9 is already used by R-C. If R-B contributes nothing, R-A must contribute 17 — impossible (max is 15). ✗

So (R-C, V3) = 11 and therefore (R-C, V2) = 9.

Step 6 — Complete the grid

Column V2 = 21 = 12 (at R-B, V2) + 9 (at R-C, V2) + (R-A, V2). So (R-A, V2) = 0 → no ATM there.

R-A's 7 and 15 must occupy V1 and V3; R-B's 8 sits at V1 or V3. Two equally valid layouts remain (cells shown a / b differ between layouts; bold = fixed in both):

💡 Teacher tip: Even though two layouts exist, the (R-C, V2) cell is 9 in both — the V2 column must total 21, it already holds 12 and 9, and R-A contributes nothing at V2.

Answer: Choice B — the ATM at (R-C, V2) has a cash requirement of Rs. 9 Lakhs.

We first determine the six cash values, then count how many are Rs. 10 Lakhs or more.

Step 1 — Grand total

Horizontal road totals: $22 + 20 + 20 = 62$.

Vertical road totals: $15 + 21 + 26 = 62$. ✓

Step 2 — Locate the minimum (7) and maximum (15)

Clue 1 places them on the same road; their sum is $7 + 15 = 22$, which matches R-A exactly. No other road can hold both (R-B and R-C total only 20; V3 = 26 would require a third ATM worth 4, below the minimum of 7). So R-A = {7, 15}.

Step 3 — Find the remaining four values

The other four ATMs sum to $62 - 22 = 40$ and split into R-B (20) and R-C (20). Four distinct integers between 8 and 14 summing to 40:

• {8, 9, 10, 13}: no pair of pairs each totals 20. ✗
• {8, 9, 11, 12}: $8 + 12 = 20$ and $9 + 11 = 20$. ✓

So R-B = {8, 12}, R-C = {9, 11}.

Step 4 — The six cash values

The values that are Rs. 10 Lakhs or more are 11, 12, and 15 — three ATMs.

Answer: 3

We need to check whether each statement holds in every valid layout.

Step 1 — Determine the row (horizontal road) assignments

From the clues (detailed in the full solution):

• R-A = 22 = 7 + 15 → two ATMs
• R-B = 20 = 8 + 12 → two ATMs
• R-C = 20 = 9 + 11 → two ATMs

Step 2 — Determine the column (vertical road) assignments

Using Clue 2 (the second-highest value, 12, is 12 km from the ATM at (R-C, V3)), we pin (R-B, V2) = 12, (R-C, V3) = 11, and (R-C, V2) = 9. Column V2 = 21 = 12 + 9, so R-A has no ATM at V2. R-A's 7 and 15 go to V1 and V3; R-B's 8 goes to V1 or V3. This yields two valid layouts:

Step 3 — Evaluate Statement A

Statement A: Each of R-A, R-B, and R-C has two ATMs.

From Step 1: R-A = {7, 15}, R-B = {8, 12}, R-C = {9, 11} — each has exactly two ATMs in both layouts. ✅

Step 4 — Evaluate Statement B

Statement B: Each of V1, V2, and V3 has two ATMs.

• Layout 1 (R-A,V1 = 7; R-B,V1 = 8; R-A,V3 = 15; R-B,V3 empty):

• V1: {7, 8} → 2 ATMs

• V2: {12, 9} → 2 ATMs

• V3: {15, 11} → 2 ATMs ✅

• Layout 2 (R-A,V1 = 15; R-B,V1 empty; R-A,V3 = 7; R-B,V3 = 8):

• V1: {15} → 1 ATM

• V2: {12, 9} → 2 ATMs

• V3: {7, 8, 11} → 3 ATMs ❌

Since Statement B fails in Layout 2, it is not definitely true.

💡 Teacher tip: "Definitely true" means true in every valid arrangement. Even one counterexample disqualifies a statement.

Answer: Choice A — Only Statement A.

We need the road distance between the ATM with the second-highest cash requirement and the ATM with the second-lowest cash requirement.

Step 1 — Identify the six values

From the full deduction (min = 7, max = 15 on R-A; remaining four split into R-B = {8, 12} and R-C = {9, 11}):

Sorted values: 7, 8, 9, 11, 12, 15

• Second-lowest = 8
• Second-highest = 12

Step 2 — Locate 12

Using Clue 2, the second-highest (12) is 12 km from the ATM at (R-C, V3). Testing positions on R-B:

• (R-B, V2) → (R-B, V3) → (R-C, V3) = $7 + 5 = 12$ km ✓

So (R-B, V2) = 12 in every valid layout.

Step 3 — Locate 8

The value 8 belongs to R-B = {8, 12}. Since 12 is at V2, the 8 must be at V1 or V3. Both layouts are valid:

Step 4 — Compute the distance in each layout

Both ATMs are on R-B, so the distance is purely horizontal along R-B.

• Layout 1: 8 is at (R-B, V1), 12 is at (R-B, V2). Distance = V1→V2 = 4 km.
• Layout 2: 8 is at (R-B, V3), 12 is at (R-B, V2). Distance = V3→V2 = 7 km.

Both layouts satisfy all clues, so the distance could be either value.

💡 Teacher tip: When two valid arrangements exist and a question asks "what best can be said," the answer must cover both possibilities.

Answer: Choice D — Either 4 km or 7 km.

We need to count the ATMs whose location and cash requirement are both uniquely determined — meaning they are fixed in every valid layout.

Step 1 — Establish the two valid layouts

From the full deduction:

• R-A = {7, 15}, R-B = {8, 12}, R-C = {9, 11}
• (R-B, V2) = 12, (R-C, V2) = 9, (R-C, V3) = 11 (fixed by clues)
• (R-A, V2) is empty (V2 = 21 = 12 + 9)
• R-A's 7 and 15 go to V1 and V3 (in either order); R-B's 8 goes to V1 or V3

Step 2 — Identify which ATMs are fixed in both layouts

💡 Teacher tip: An ATM is "uniquely determined" only if both its position and its value are the same in every valid arrangement. If either can vary, it doesn't count.

Exactly 3 ATMs — those at (R-B, V2), (R-C, V2), and (R-C, V3) — are fully determined.

Answer: 3

Setting up the full table

Every row sums to 100 g. We combine this with the divisibility rules (missing carbs are non-zero multiples of 5; missing protein/fat/other are non-zero multiples of 4) and the three ordering constraints to fill every blank.

Key deductions (worked in dependency order):

1. M3 and P1 are linked by Clue 5.

• M3: $56 + \text{Protein} + 12 + \text{Other} = 100 \Rightarrow \text{Protein} + \text{Other} = 32$, both multiples of 4.
• P1: $66 + \text{Protein} + \text{Fat} + 10 = 100 \Rightarrow \text{Protein} + \text{Fat} = 24$, both multiples of 4.
• Clue 5 says P1's protein $= 2 \times$ M3's protein. Testing the multiples of 4: if M3's protein were 4, P1's protein would be 8 — but P1 must beat every millet on protein (Clue 1), and M1 already has 10, so 8 is too low. The only fit is M3 protein = 8, M3 other = 24 and P1 protein = 16, P1 fat = 8.

1. P2 (pseudo-cereal): $\text{Carb} + 14 + \text{Fat} + 8 = 100 \Rightarrow \text{Carb} + \text{Fat} = 78$. Carb must be a multiple of 5, fat a multiple of 4, and P2's carb must exceed every millet's carb (M1 = 62, M3 = 56). The only pair with carb $> 62$ is carb = 70, fat = 8.

2. Cereals must out-carb every pseudo-cereal (Clue 2): pseudo-cereals reach 66 (P1) and 70 (P2), so each cereal's carb $> 70$.

• C1: $\text{Carb} + \text{Protein} = 88$. The only multiple-of-5 / multiple-of-4 split with carb $> 70$ is 80, 8.
• C2: $\text{Carb} + \text{Protein} = 87$. The only split with carb $> 70$ is 75, 12.

1. M2 (millet): $\text{Carb} + \text{Protein} = 77$. Carb must be a multiple of 5, protein a multiple of 4, and M2 must stay below every pseudo-cereal on both carb ($< 66$) and protein ($< 14$). The only fit is carb = 65, protein = 12.

2. M1: $\text{Fat} + \text{Other} = 28$ (both multiples of 4), but the individual split is not pinned down.

Solved table:

Answering the question

M1's carbohydrate is 62 g. We count every other grain whose carbohydrate exceeds 62:

That is 5 grains (C1, C2, M2, P1, P2).

Answer: 5

Setting up the full table

Every row sums to 100 g. Missing carbs are non-zero multiples of 5; missing protein/fat/other are non-zero multiples of 4. The ordering constraints (Clues 1–2) and the protein link (Clue 5) pin down every value except M1's fat/other split.

Key deductions:

1. M3 & P1 (linked by Clue 5): M3: Protein + Other = 32 (both ×4). P1: Protein + Fat = 24 (both ×4), with P1-protein = 2 × M3-protein. If M3-protein = 4, then P1-protein = 8, but P1 must beat every millet on protein and M1 already has 10 — too low. So M3-protein = 8, M3-other = 24, P1-protein = 16, P1-fat = 8.

2. P2: Carb + Fat = 78 (carb ×5, fat ×4). P2's carb must exceed every millet's (≥ 62). Only fit: carb = 70, fat = 8.

3. Cereals must out-carb every pseudo-cereal (max pseudo-cereal carb = 70): C1: Carb + Protein = 88 → only split with carb > 70 is 80, 8. C2: Carb + Protein = 87 → only split with carb > 70 is 75, 12.

4. M2: Carb + Protein = 77 (carb ×5, protein ×4). M2 must stay below every pseudo-cereal on carb (< 66) and protein (< 14). Only fit: carb = 65, protein = 12.

Solved table:

Answering the question

For M2 specifically: $\text{Carb} + \text{Protein} = 100 - 7 - 16 = 77$. The carbohydrate must be a non-zero multiple of 5 and the protein a non-zero multiple of 4. As a millet, M2 must have less carbohydrate than every pseudo-cereal (so carb $< 66$) and less protein than every pseudo-cereal (so protein $< 14$, since P2 has 14).

The possible (carb, protein) pairs summing to 77 with the divisibility rules are (5, 72), (25, 52), (45, 32), (65, 12). Only (65, 12) satisfies protein $< 14$.

Answer: 12

Setting up the full table

Every row sums to 100 g. Missing carbs are non-zero multiples of 5; missing protein/fat/other are non-zero multiples of 4. Clues 1–2 give ordering constraints; Clue 5 links P1's and M3's protein.

Key deductions:

1. M3 & P1 (linked by Clue 5): M3: $56 + \text{Protein} + 12 + \text{Other} = 100 \Rightarrow \text{Protein} + \text{Other} = 32$, both multiples of 4. P1: $66 + \text{Protein} + \text{Fat} + 10 = 100 \Rightarrow \text{Protein} + \text{Fat} = 24$, both multiples of 4. Clue 5: P1-protein = 2 × M3-protein.

2. Testing the multiples of 4 for M3's protein:

• If M3-protein = 4 → P1-protein = 8. But Clue 1 requires P1's protein to exceed every millet's, and M1 already has protein = 10. Since 8 < 10, this fails.
• If M3-protein = 8 → P1-protein = 16, P1-fat = 8, M3-other = 24. Check: P1-protein (16) beats every millet's protein (M1 = 10, M3 = 8, M2 will be 12) ✓.
• Larger values of M3-protein (12, 16, …) would make P1-protein = 24+, but P1-protein + P1-fat = 24 and fat must be ≥ 4, so P1-protein ≤ 20. Hence no larger value works.

So M3-protein = 8 and M3-other = 24.

1. Remaining grains (P2, C1, C2, M2) are filled similarly using the ordering and divisibility rules, giving:

Solved table:

Answering the question

For M3: $\text{Protein} + \text{Other} = 100 - 56 - 12 = 32$, both non-zero multiples of 4. With P1's protein $= 2 \times$ M3's protein and the requirement that P1's protein exceeds every millet's, the only valid split is protein = 8, other = 24.

Answer: 24

Setting up the full table

Every row sums to 100 g. Missing carbs are non-zero multiples of 5; missing protein/fat/other are non-zero multiples of 4. The ordering constraints (Clues 1–2) and the protein link (Clue 5) determine every protein value.

Key deductions:

1. M3 & P1 (Clue 5): M3-protein + M3-other = 32; P1-protein + P1-fat = 24; P1-protein = 2 × M3-protein. The only multiple-of-4 solution keeping P1's protein above every millet's is M3-protein = 8, P1-protein = 16, P1-fat = 8, M3-other = 24.

2. P2: Carb + Fat = 78; must out-carb every millet → carb = 70, fat = 8. Protein is already given as 14.

3. Cereals must out-carb every pseudo-cereal (max = 70): C1: Carb + Protein = 88 → carb = 80, protein = 8. C2: Carb + Protein = 87 → carb = 75, protein = 12.

4. M2: Carb + Protein = 77; must stay below pseudo-cereals on both → carb = 65, protein = 12.

5. M1: Protein is already given as 10 (fat + other = 28, not individually fixed).

Solved table:

Answering the question

Collect the seven protein values:

Sort them in ascending order:

$8,\ 8,\ 10,\ \mathbf{12},\ 12,\ 14,\ 16$

With 7 values, the median is the 4th value, which is 12.

Answer: 12