CAT 2017 DILR Questions — Slot 1

Finding the stable route assignment

We need to figure out how the police department splits 4 cars between the two routes A–M–B and A–N–B so that no single car can shorten its own travel time by switching routes while the others stay put.

Step 1: Write each road's travel-time formula

If $n$ cars use a road, the per-car time is:

Step 2: Write each route's total time

Let $m$ cars take A–M–B and $n$ cars take A–N–B, with $m + n = 4$.

• A–M–B time $= (3m+3) + (0.9m+19.1) = 3.9m + 22.1$
• A–N–B time $= (n+19) + (3n+3) = 4n + 22$

Since $n = 4 - m$, we can rewrite the A–N–B time as: $4(4-m) + 22 = 38 - 4m$

Step 3: Find the balance point

The stable assignment sits near where the two route times are equal — if one route were much faster, cars on the slower route would switch.

$3.9m + 22.1 = 38 - 4m$ $7.9m = 15.9$ $m \approx 2.01$

This points to $m = 2$, so $n = 2$.

Step 4: Verify stability

With $m = 2,\; n = 2$:

• A car on A–N–B (30 min) switching to A–M–B would make it 3 cars there: $3.9(3)+22.1 = 33.8$ min — worse.
• A car on A–M–B (29.9 min) switching to A–N–B would make it 3 cars there: $4(3)+22 = 34$ min — worse.

No car can improve by switching, so this is the stable order.

Answer: 2

Difference in travel time between the two routes

From the stable assignment, 2 cars take A–M–B and 2 cars take A–N–B. We compute each route's time directly.

Route times at $m = 2,\; n = 2$

A–M–B (2 cars on A–M, 2 cars on M–B):

• A–M: $3(2)+3 = 9$ min
• M–B: $0.9(2)+19.1 = 20.9$ min
• Total: $9 + 20.9 = 29.9$ min

A–N–B (2 cars on A–N, 2 cars on N–B):

• A–N: $2+19 = 21$ min
• N–B: $3(2)+3 = 9$ min
• Total: $21 + 9 = 30$ min

Difference

$30 - 29.9 = 0.1 \text{ minutes}$

Answer: 0.1 (Choice B)

Finding the stable assignment with three routes

A new one-way road M→N is added, giving three routes: A–M–B, A–N–B, and A–M–N–B. The M–N road takes 7 min for one car, plus 1 min per additional car, so its formula is $7 + 1(z-1) = z + 6$.

💡 Teacher tip: The key insight is that roads are shared — A–M is used by both A–M–B and A–M–N–B cars, and N–B is used by both A–N–B and A–M–N–B cars. The congestion on a road depends on the total cars on it, not just the cars on one route.

Step 1: Set up variables and shared-road loads

Let $x$ cars take A–M–B, $y$ take A–N–B, $z$ take A–M–N–B, with $x + y + z = 4$.

Step 2: Write each route's total time

• A–M–B $= [3(x+z)+3] + [0.9x+19.1] = 3.9x + 3z + 22.1$
• A–N–B $= [y+19] + [3(y+z)+3] = 4y + 3z + 22$
• A–M–N–B $= [3(x+z)+3] + [z+6] + [3(y+z)+3] = 3x + 3y + 7z + 12$

Step 3: Try $z = 2$ (so $x + y = 2$)

The A–M–N–B time becomes: $3(x+y) + 7(2) + 12 = 3(2) + 14 + 12 = 32$

This is independent of the split between $x$ and $y$ — a promising sign.

Now set A–M–B = A–N–B to find the split: $3.9x + 3(2) + 22.1 = 4y + 3(2) + 22$ $3.9x + 28.1 = 4y + 28$

Substitute $y = 2 - x$: $3.9x + 28.1 = 4(2-x) + 28 = 36 - 4x$ $7.9x = 7.9$ $x = 1, \quad y = 1$

Step 4: Verify all three route times and stability

All three routes take exactly 32 minutes. Any car that switches to another route increases the load on that route's roads, pushing its time above 32. So no car can improve — this is the stable order.

Answer: 2

Minimum travel time under the police order (three routes)

From the stable assignment with the new M→N road, the police department orders:

Computing each route's time

The shared-road loads are:

• A–M carries $x + z = 1 + 2 = 3$ cars → time $3(3)+3 = 12$ min
• M–B carries $x = 1$ car → time $0.9(1)+19.1 = 20$ min
• A–N carries $y = 1$ car → time $1+19 = 20$ min
• N–B carries $y + z = 1 + 2 = 3$ cars → time $3(3)+3 = 12$ min
• M–N carries $z = 2$ cars → time $2+6 = 8$ min

Every route takes exactly 32 minutes. Since the police order is a stable assignment (no car can improve by switching), and all routes are equal at 32, the minimum travel time any car experiences is 32 minutes.

Answer: 32 (Choice B)

Let us first pin down the rule precisely, because almost every mistake in this set comes from misreading it.

A can reach B when all three hold:

1. A and B share a row or a column.
2. A is shorter than B (A's platform height < B's platform height).
3. Every person strictly between A and B is shorter than A.

A clean way to apply this: stand at B and scan outwards in each of the four directions. As you move away from B, keep a running maximum of the heights you have passed. A cell reaches B exactly when its height is less than B's height and greater than that running maximum (i.e. it is a new “record” as you walk outward, yet still below B). The moment you meet a cell at least as tall as B, scanning stops in that direction.

Applying this to all 25 cells gives the following reacher-count table (the number of individuals who can reach the person on that platform):

We want the cells whose count is exactly 1. Reading the table, those are:

• (1,3) — height 2
• (3,1) — height 7
• (3,3) — height 4
• (3,5) — height 5
• (4,1) — height 3
• (4,5) — height 2
• (5,4) — height 3

That is 7 cells.

Let me sanity-check two of them so the method is convincing.

• (3,3), height 4: Up the column, (2,3)=3 is below 4 with nothing between, so it reaches; (1,3)=2 is then blocked by the 3. Down, (4,3)=5 is already taller than 4, so nothing. Left, (3,2)=8 is taller. Right, (3,4)=6 is taller. So exactly 1 reacher.
• (1,3), height 2: Only (1,2)=1 to its left qualifies (1 < 2, nothing between). To the right (1,4)=4 is taller; below (2,3)=3 is taller. So exactly 1 reacher.

💡 Teacher tip: The “running maximum while scanning outward” trick is what turns a messy between-cell check into a quick scan — use it whenever a rule says “everyone in between must be shorter than me.”

Answer: 7 (Choice C)

Look directly at condition (ii) of the reach rule: for A to reach B, A must be at a lower height than B.

Now consider any individual on a platform of height 1 m. Since 1 m is the lowest height appearing in the entire layout, there is simply nobody shorter available to act as a reacher. The other two conditions become irrelevant — condition (ii) already fails for every possible candidate.

We can confirm this against the full reacher-count table:

The three height-1 platforms are (1,2), (4,4) and (5,1), and each shows a reacher-count of 0.

💡 Teacher tip: Whenever a question singles out the minimum (or maximum) value in a comparison-based rule, ask first what the inequality itself forbids — here the minimum can never be “reached” because nobody is below it.

Answer: They cannot be reached by anyone (Choice D)

An individual who “cannot be reached by anyone” is simply a cell whose reacher-count is 0.

From the full reacher-count table:

the zero-count cells are:

• (1,2)
• (1,5)
• (2,4)
• (4,4)
• (5,1)

Now test each option to see whether it contains two of these:

• Last row (R5): only (5,1) → one.
• Fourth row (R4): only (4,4) → one.
• Fourth column (C4): (2,4) and (4,4) → two. ✓
• Middle column (C3): none.

So the fourth column is the only listed region holding two unreachable individuals.

💡 Teacher tip: For “find two in the same row/column” questions, list the qualifying cells first, then just read off which row/column they cluster in — don't re-derive counts for each option.

Answer: the fourth column (Choice C)

We need the one statement that is true for the whole layout. The reacher-count table is our working tool:

Check each option.

A. Each row has someone reached by 5 or more. Row maximums are R1 = 3, R2 = 5, R3 = 6, R4 = 4, R5 = 6. Rows 1 and 4 have nobody with 5 or more. False.

B. Each row has someone who cannot be reached by anyone (count 0). Row 3's counts are 1, 4, 1, 6, 1 — no zero. False.

C. Each row has at least two individuals reached by an equal number. Look for a repeated value in each row:

• R1: 3, 0, 1, 3, 0 → 3 repeats, 0 repeats. ✓
• R2: 4, 2, 2, 0, 5 → 2 repeats. ✓
• R3: 1, 4, 1, 6, 1 → 1 repeats (three times). ✓
• R4: 1, 4, 3, 0, 1 → 1 repeats. ✓
• R5: 0, 2, 2, 1, 6 → 2 repeats. ✓

Every row qualifies. True.

D. All height-9 individuals are reached by at least 5. The height-9 cells are (2,1), (4,2) and (5,5), with counts 4, 4 and 6. The first two are below 5. False.

💡 Teacher tip: For “which statement is true” questions with four independent claims, test the easiest-to-falsify claims first (here A, B, D each fail on a single counter-example); only C survives.

Answer: Each row has at least two individuals who can be reached by an equal number of individuals (Choice C)

Setting up the full region-by-school-type table

Before answering any question, we need to build the complete table of kids by region and school type. Let's work through it step by step.

Step 1 — Region sizes (villages × 50 kids each):

Step 2 — Overall school-type split (given as percentages of 30,000):

• Government (G): $55\% \times 30000 = 16{,}500$
• Private (P): $37\% \times 30000 = 11{,}100$
• Out of school (O): $8\% \times 30000 = 2{,}400$

Step 3 — Use the three clues to pin down each region's G and O:

1. Clue 1 (S): 60% of S's kids were in G → $G_S = 60\% \times 10000 = 6{,}000$. Also, in S every kid whose mother completed primary education was in school, so all out-of-school kids in S must come from the mothers-who-dropped-out group. The given table shows O in S (dropped-out group) = 300, so $O_S = 300$.

2. Clue 2 (NE): Among NE's out-of-school kids, 50% had mothers who dropped out. The table gives O in NE (dropped-out group) = 300, so $O_{NE} = 300 \times 2 = 600$.

3. Clue 3 (NE = W for G): We know total G = 16,500 and $G_S = 6{,}000$, so the remaining G for NE and W together = $16500 - 6000 = 10{,}500$. Since $G_{NE} = G_W$, each gets $10{,}500 \div 2 = 5{,}250$.

Step 4 — Fill P by subtraction (Total − G − O = P):

Answering the question

The question asks: What percentage of kids from S were studying in P?

From the S row: P in S = 3,700 and total kids in S = 10,000.

$\frac{3700}{10000} \times 100 = 37\%$

Answer: 37% (Choice A)

What we need

We need the number of kids in W whose mothers completed primary education and who were not in school (O).

Building the full table

Region sizes: NE = 7,500; W = 12,500; S = 10,000 (total 30,000).

Overall school split: G = 16,500; P = 11,100; O = 2,400.

Applying the three clues:

1. In S, 60% were in G → $G_S = 6{,}000$. Also, all mother-completed kids in S were in school, so every out-of-school kid in S had a mother who dropped out. The given table shows O in S (dropped-out group) = 300, so $O_S = 300$.
2. In NE, 50% of O kids had mothers who dropped out. The table gives O in NE (dropped-out group) = 300, so $O_{NE} = 600$.
3. G in NE = G in W. Remaining G after S = $16500 - 6000 = 10{,}500$, split equally → $G_{NE} = G_W = 5{,}250$.

Filling P by subtraction:

Answering the question

From the full table, W has 1,500 kids not in school (O).

The given table (mothers who dropped out) shows that of W's O kids, 1,200 had mothers who dropped out.

Therefore, the O kids in W whose mothers completed primary education:

$1500 - 1200 = 300$

💡 Teacher tip: The key distinction here is between the total O kids in W (1,500, from the full table) and the dropped-out-mother O kids in W (1,200, from the given table). The difference is exactly the completed-mother O kids we need.

Answer: 300 (Choice A)

Setting up the follow-up survey

Two years later, all formerly out-of-school (O) kids are now in school. We need to figure out how many of W's formerly-out-of-school kids joined G, then add that to W's original G count.

Step 1 — Recall the O kids by region (from the full table built using the clues):

Step 2 — Determine which redistribution rule applies to which region.

The three rules are:

• Rule A: 25% of former-O kids join G
• Rule B: 100% of former-O kids join G
• Rule C: 50% of former-O kids join G

Overall, 50% of all 2,400 former-O kids joined G, i.e., $0.50 \times 2400 = 1200$ kids joined G.

We need to assign {25%, 100%, 50%} to {NE, W, S} so that the total joining G equals 1,200:

$600 \times f_1 + 1500 \times f_2 + 300 \times f_3 = 1200$

where $\{f_1, f_2, f_3\} = \{0.25, 1.0, 0.5\}$ in some order.

Testing the arrangements:

The only arrangement that gives exactly 1,200 is: NE → 25%, W → 50%, S → 100%.

Step 3 — Compute G in W now.

• W's formerly-out-of-school kids who joined G: $1500 \times 50\% = 750$
• W's original G kids (who did not change schools): 5,250

$G \text{ in W now} = 5250 + 750 = 6000$

💡 Teacher tip: "No kid changed schools" means kids already in G or P stayed put — only the former-O kids moved into the school system.

Answer: 6,000 (Choice A)

What we need

We need the percentage of S's mothers-who-dropped-out kids who are now in G, after the follow-up survey.

Step 1 — Recall the redistribution assignment

From the follow-up survey conditions (worked out in the previous question), the three regions were assigned:

• NE → 25% of former-O kids join G
• W → 50% of former-O kids join G
• S → 100% of former-O kids join G

So all of S's formerly-out-of-school kids joined G.

Step 2 — Identify S's dropped-out-mother (D-group) kids

From Clue 1: in S, all kids whose mothers completed primary education were in school. This means every out-of-school kid in S had a mother who dropped out — there are no completed-mother O kids in S.

The given table shows S's D-group breakdown:

Step 3 — Apply the follow-up changes to S's D-group

• The 300 O kids all join G (since S → 100% to G).
• No kid changes school, so the 5,100 already in G stay in G, and the 300 in P stay in P.

Step 4 — Compute the percentage

$\frac{5400}{5700} \times 100 = 94.736\ldots \approx 94.7\%$

💡 Teacher tip: The key insight is that in S, all out-of-school kids belonged to the dropped-out-mother group (Clue 1), so all 300 former-O kids who joined G are part of the D-group we're measuring.

Answer: 94.7% (Choice A)

Setting up the Venn diagram for the 200 candidates

We focus only on the 200 candidates who are at or above the 90th percentile overall in CET, and sort them by which sections they cleared the 80th percentile in.

Let us label the seven regions of the three-circle Venn diagram (P, C, M):

• Clue 1 tells us the "none" region is 0.
• Clue 3 tells us the three "exactly one section" regions are equal; call each $x$.
• Clue 2 tells us $a + b + c = 150$.

So the total gives us: $3x + a + b + c + d = 200 \implies 3x + 150 + d = 200 \implies 3x + d = 50$

Using the ratio clue (Clue 4)

"Below 80th percentile in P" means the candidate is not in P, i.e. they are in Only C, Only M, or C&M only: $\text{Below 80th in P} = x + x + c = 2x + c$

Similarly: $\text{Below 80th in C} = 2x + b, \qquad \text{Below 80th in M} = 2x + a$

Clue 4 says these are in the ratio $4: 2: 1$, so we can write: $2x + c = 4k,\quad 2x + b = 2k,\quad 2x + a = k$

Adding all three: $6x + (a+b+c) = 7k \implies 6x + 150 = 7k$

For $k$ to be a positive integer, $6x + 150$ must be divisible by 7. Combined with $3x + d = 50$ (so $x \leq 16$), the valid solutions are:

Answering the question

A candidate sits the BIE separate test if they are at or above the 80th percentile in P but are not eligible for AIE's AET.

Among the 200 candidates (all at or above 90th overall), AET eligibility requires being at or above the 80th percentile in at least two sections. So the only candidates among the 200 who have P but are not AET-eligible are the "Only P" candidates — they have P but not two sections.

That count is simply $x$, which is either 3 or 10.

Answer: 3 or 10 (Choice A)

Recalling the solved Venn diagram

From the set-up, the 200 candidates at or above the 90th percentile overall split into Venn regions with two possible cases:

These satisfy $3x + d = 50$ and $a + b + c = 150$, with the below-80th ratios $4:2:1$.

Applying the extra condition

The question states that the number of candidates at or above the 80th percentile in all three sections (i.e. $d$) is a multiple of 5.

• $d = 41$ → not a multiple of 5.
• $d = 20$ → is a multiple of 5. ✓

So we are in Case 2: $x = 10$, $d = 20$, $a = 10$, $b = 40$, $c = 100$.

Finding candidates at or above 80th in both P and M

Candidates at or above the 80th percentile in both P and M come from two regions:

• P & M only (not C): $b = 40$
• All three (P, C, and M): $d = 20$

$b + d = 40 + 20 = 60$

Answer: 60

Recalling the solved Venn diagram

From the set-up, the 200 candidates at or above the 90th percentile overall split into Venn regions with two possible cases:

Applying the extra condition

The number of candidates at or above the 80th percentile in all three sections is $d$, which must be a multiple of 5.

• $d = 41$ → not a multiple of 5.
• $d = 20$ → multiple of 5. ✓

So we use Case 2: $d = 20$.

Counting AET shortlists

AIE shortlists for the AET every candidate (among the 200) who is at or above the 80th percentile in at least two sections. These are:

• Candidates at or above 80th in exactly two sections: $a + b + c = 150$ (given by Clue 2).
• Candidates at or above 80th in all three sections: $d = 20$.

$\text{AET shortlists} = 150 + d = 150 + 20 = 170$

Answer: 170

Recalling the solved Venn diagram

From the set-up, the 200 candidates at or above the 90th percentile overall split into Venn regions with two possible cases:

Applying the extra condition

The number of candidates at or above the 90th percentile overall and at or above the 80th percentile in P is: $x + a + b + d$

(These are all regions that include P: Only P, P&C only, P&M only, and All three.)

• Case 1 ($x=3$): $3 + 18 + 42 + 41 = 104$
• Case 2 ($x=10$): $10 + 10 + 40 + 20 = 80$

The condition "more than 100" selects Case 1: $x = 3$, $a = 18$, $b = 42$, $c = 90$, $d = 41$.

Counting BIE separate-test candidates

We are told there are 400 candidates in total at or above the 80th percentile in P. These split into two streams:

1. Candidates going through AIE's AET who are also at or above 80th in P: These are candidates among the 200 who are at or above 80th in P and in at least two sections (AET requires at least two sections at or above 80th). This is the P&C, P&M, and All-three regions: $a + b + d = 18 + 42 + 41 = 101$

2. Everyone else at or above 80th in P who is not going through AET must sit BIE's separate test: $400 - 101 = 299$

💡 Teacher tip: The key insight is that BIE's separate test captures all P-qualified candidates who don't go through AIE's AET — not just those among the 200, but all 400. We subtract the AET-eligible-with-P count from 400.

Answer: 299 (Choice A)

Setting up the team compositions

Team sizes. There are 21 employees across 5 teams. The challenging team has one more employee than each standard team, so the only split that works is:

$4 + 4 + 4 + 4 + 5 = 21$

Standard teams have 4 employees; the challenging team has 5.

Month 1 SE counts. We are told the SE counts step down by 1 from T1 to T5, and there are 10 SE in total. If T5 has $x$ SE, then:

$(x+4) + (x+3) + (x+2) + (x+1) + x = 10 \implies 5x + 10 = 10 \implies x = 0$

So the Month 1 SE counts are T1=4, T2=3, T3=2, T4=1, T5=0, and the RE counts (team size minus SE) are T1=1, T2=1, T3=2, T4=3, T5=4.

Transition rules (applied each month in order):

1. Rule (a): The new challenging team receives 2 SE from the previous month's challenging team, and sends 1 RE back to that team.
2. Rule (b): After Rule (a), check two conditional swaps:

• If T1 has ≥1 SE and T5 has ≥1 RE: move 1 SE from T1→T5 and 1 RE from T5→T1.
• If T2 has ≥1 SE and T4 has ≥1 RE: move 1 SE from T2→T4 and 1 RE from T4→T2.

Applying these rules month by month gives the complete table (shown as SE, RE):

💡 Teacher tip: Always verify your table — each row should sum to 10 SE and 11 RE, and the challenging team should be the only 5-person team.

Answering the question

A team's composition remained unchanged across two successive months if both its SE count and RE count stayed the same.

T2 across all five months:

T2 remained unchanged 1 time (M3→M4).

T4 across all five months:

T4 remained unchanged 0 times.

The pair is (1, 0).

Answer: (1, 0) — Choice B

Setting up the team compositions

Team sizes. There are 21 employees across 5 teams. The challenging team has one more employee than each standard team:

$4 + 4 + 4 + 4 + 5 = 21$

So standard teams have 4 employees and the challenging team has 5.

Month 1 SE counts. The SE counts decrease by 1 from T1 to T5, totaling 10 SE. If T5 has $x$ SE:

$(x+4)+(x+3)+(x+2)+(x+1)+x = 10 \implies x = 0$

Month 1 SE: T1=4, T2=3, T3=2, T4=1, T5=0; RE: T1=1, T2=1, T3=2, T4=3, T5=4.

Transition rules:

1. Rule (a): The new challenging team gets 2 SE from the previous challenging team and sends 1 RE back.
2. Rule (b): If T1 has SE and T5 has RE, swap 1 SE (T1→T5) and 1 RE (T5→T1). If T2 has SE and T4 has RE, swap 1 SE (T2→T4) and 1 RE (T4→T2).

Working through all five months (shown as SE, RE):

Answering the question

We need the SE counts of T1 and T5 in Month 3. Reading directly from the Month 3 row:

So the number of SE in T1 and T5 for the third month are 0 and 2, respectively.

Answer: (0, 2) — Choice A

Credit points per employee per project

Each project's credit points are equally shared among all employees on that team.

• Standard project: 100 points, team of 4 → each member earns $\frac{100}{4} = 25$ points.
• Challenging project: 200 points, team of 5 → each member earns $\frac{200}{5} = 40$ points.

💡 Teacher tip: The key insight is that team sizes are fixed (4 for standard, 5 for challenging), so the per-person credit is always 25 or 40 — no other value is possible in any single month.

Total credit over five months

Each employee works on one project each month for 5 months. In any given month, an employee earns:

• 40 points if they are on that month's challenging team, or
• 25 points if they are on a standard team.

If an employee is on the challenging team in $q$ of the 5 months (where $q$ can be 0, 1, 2, 3, 4, or 5), their total is:

$\text{Total} = 40q + 25(5 - q) = 125 + 15q$

The possible totals are:

The value 150 does not appear in this list, so it cannot be earned by any employee.

Answer: 150 — Choice B

What does 185 points tell us?

From the credit formula (standard project = 25 per member, challenging project = 40 per member), an employee's 5-month total is:

$\text{Total} = 125 + 15q$

where $q$ is the number of months they were on the challenging team. Setting this equal to 185:

$125 + 15q = 185 \implies 15q = 60 \implies q = 4$

So Aneek was on the challenging team in exactly 4 of the 5 months, and on a standard team in exactly 1 month.

Since the challenging team in month $m$ is $T_m$ (Month 1→T1, Month 2→T2, etc.), Aneek's "off" month — the one month he was not on the challenging team — determines which four teams he worked in:

We need to find which option is impossible.

The full team composition table

For reference, here is the complete monthly table (SE, RE):

Testing Option D (off in Month 2)

Option D requires Aneek to be on the challenging team in Months 1, 3, 4, 5 — meaning he is in T1 in Month 1, T3 in Month 3, T4 in Month 4, and T5 in Month 5 — and is not on the challenging team (T2) in Month 2.

Step 1: Where can Aneek go from T1 in Month 1 to Month 2?

During the Month 1→2 transition:

• Rule (a): T2 (new challenging team) takes 2 SE from T1 (old challenging team), and T1 receives 1 RE from T2. So SE can move from T1→T2.
• Rule (b): T1↔T5 swap (1 SE each way) and T2↔T4 swap (1 SE each way).

Notice that T3 is not involved in any transfer during this transition. So an employee starting in T1 can only end up in T1, T2, T4, or T5 in Month 2 — never T3.

Since Month 2 is Aneek's off month, he is not in T2. So in Month 2, Aneek is in T1, T4, or T5.

Step 2: Can Aneek reach T3 in Month 3?

During the Month 2→3 transition:

• Rule (a): T3 (new challenging team) takes 2 SE from T2 (old challenging team), and T3 sends 1 RE to T2. So the only way to enter T3 is from T2.
• Rule (b): Only T1↔T5 and T2↔T4 swaps occur — T3 is not involved.

So to be in T3 in Month 3, Aneek must have been in T2 or T3 in Month 2. But:

• He is not in T2 (that's his off month — T2 is the challenging team in Month 2).
• He is not in T3 (we showed in Step 1 that he cannot reach T3 from T1 in one transition).

💡 Teacher tip: The critical observation is that T3 is a "bystander" in both the M1→M2 and M2→M3 transitions — it neither sends nor receives anyone. The only way into T3 during M2→M3 is via Rule (a) from T2, but Aneek cannot be in T2 during Month 2.

Conclusion: It is impossible for Aneek to be in T1 (Month 1), skip T2 (Month 2), and then appear in T3 (Month 3). Therefore Option D cannot be true.

The other three options (off in Months 5, 3, and 1 respectively) are each achievable, as the transition chains do not require passing through an unreachable team.

Answer: Aneek worked only in teams T1, T3, T4, and T5 — Choice D

Setting up the key idea

The airline's principle says: any person in any of the 10 cities must be able to travel to any other city in the morning and return by the evening of the same day.

In this question, we are told that every journey must be possible using only direct (non-stop) flights. That means every pair of cities must be connected by a direct flight.

How many flights does a single pair need?

Take any two cities, say X and Y. To satisfy the principle for both directions of travel:

1. A morning flight X → Y (so someone in X can leave in the morning).
2. A morning flight Y → X (so someone in Y can leave in the morning).
3. An evening flight X → Y (so someone in Y can return in the evening).
4. An evening flight Y → X (so someone in X can return in the evening).

So each pair of cities needs 4 direct flights.

Counting the pairs

There are 10 cities, so the number of unordered pairs is:

$\binom{10}{2} = \frac{10 \times 9}{2} = 45$

Total flights

$45 \text{ pairs} \times 4 \text{ flights per pair} = 180$

💡 Teacher tip: The trap here is forgetting that "morning and return by evening" applies to both cities in a pair — so each pair needs flights in both directions at both times, giving 4 flights, not 2.

Answer: 180 (Choice C)

Understanding the hub structure

We choose 3 of the 10 cities as hubs. The rules are:

• Each hub is connected to every other city (including the other two hubs) by direct flights both ways, morning and evening.
• The only direct flights scheduled are those with at least one end at a hub.
• No trip may pass through more than one hub.

Because every non-hub city connects to a hub, and hubs connect to each other, any city can reach any other city via at most one hub — satisfying the "at most one hub per trip" rule.

Counting connected pairs

Let the 3 hubs be H1, H2, H3 and the 7 non-hubs be the remaining cities.

Note: there are no direct flights between two non-hub cities, since those flights don't involve a hub.

Flights per pair

As established in the first question, each connected pair needs 4 flights (morning and evening, each direction).

Total flights

$24 \times 4 = 96$

💡 Teacher tip: The key insight is that non-hub-to-non-hub pairs don't need direct flights — they connect through a hub. This is what saves flights compared to the fully-connected network in Question 1.

Answer: 96 (Choice C)

Understanding the group structure

The 10 cities are divided into 4 groups:

Direct flights are allowed only between:

1. Two cities both in G1 (i.e., among A, B, C).
2. A and any city in G2.
3. B and any city in G3.
4. C and any city in G4.

Why the principle is satisfied

Cities A, B, C form a fully connected core (every pair among them is linked). Then:

• A feeds G2,
• B feeds G3,
• C feeds G4.

So any city can reach the core through its designated link, and from the core reach any other group. Every city can travel to any other city (morning) and return (evening).

Counting connected pairs

Total flights

Each pair needs 4 flights (morning and evening, each direction):

$10 \times 4 = 40$

💡 Teacher tip: The structure here is a star-of-stars: A, B, C are the central core, and each of them is the sole gateway for one group. This minimises flights while keeping the network connected.

Answer: 40

Recalling the original arrangement (from Question 3)

Before the operational difficulties, the network had 10 connected pairs (and thus 40 flights):

What changes due to operational difficulties at A?

The new rule: the only flights operating at A are those to and from B.

This means the following links are removed:

1. A–C link (1 pair) — A can no longer connect to C.
2. A to each city in G2 (3 pairs) — A can no longer serve G2.

That is 4 pairs removed.

What replaces the lost G2 connections?

The 3 cities formerly in G2 must be reassigned to G3 or G4:

• If a city goes to G3, it links to B.
• If a city goes to G4, it links to C.

Either way, each of the 3 former G2 cities gets one new link (to B or to C). That is 3 new pairs added.

Net change in connected pairs

This net change of −1 pair holds regardless of how the 3 cities are split between G3 and G4 — each city contributes exactly one new pair either way.

Does the network stay connected?

• A–B still exists (A connects to the core via B).
• B–C still exists (the core remains connected).
• All former G2 cities now connect through B or C.

So the underlying principle is still satisfied.

Maximum reduction in flights

Each pair corresponds to 4 flights. A reduction of 1 pair means:

$1 \times 4 = 4 \text{ flights}$

💡 Teacher tip: The reduction is the same no matter how you split the G2 cities between G3 and G4, because each reassigned city creates exactly one replacement link. The only net loss is the A–C link, which costs 4 flights.

Answer: 4

Setting up the problem

Healthy Bites has two employees (Anish and Bani) and one automatic fryer. The key preparation times are:

The fryer runs on its own, so employees are free to do other tasks while it operates. However, fries can never be prepared before an order is placed.

Applying the rule for this question

The rule states that only one client's order can be processed at a time. Client 1 placed the order at 10:00, so we handle Client 1's order first — no work on Client 2 or Client 3 may begin until Client 1 is fully served.

Client 1's order

Client 1 ordered: 1 burger, 3 portions of fries, 1 order of ice cream.

All three items can be started simultaneously at 10:00:

Finding the completion time

An order is completely served only when its last item is ready.

• Ice cream is ready at 10:02.
• Fries are ready at 10:05.
• The burger — which takes the longest at 10 minutes — is ready at 10:10.

The burger is the binding item, so Client 1's order is completely served at 10:10.

Answer: 10:10 (Choice B)

Understanding the rule

The rule is the same as before: only one client's order can be processed at a time. Orders are handled strictly one after another, in the order they were placed: Client 1 → Client 2 → Client 3.

Step 1 — Client 1 (placed 10:00)

Client 1 ordered: 1 burger, 3 portions of fries, 1 ice cream.

All three items start at 10:00 in parallel:

The burger is the last item ready, so Client 1 is completely served at 10:10.

Step 2 — Client 2 (placed 10:05, but can only start at 10:10)

Client 2 ordered: 2 portions of fries, 1 ice cream (no burger).

Anish has nothing to do (no burger in this order) and cannot start Client 3 yet. The fries take longer, so Client 2 is completely served at 10:15.

Step 3 — Client 3 (placed 10:07, but can only start at 10:15)

Client 3 ordered: 1 burger, 1 portion of fries.

The burger takes 10 minutes and is the last item ready, so Client 3 is completely served at 10:25.

Answer: 10:25 (Choice C)

Understanding the new rule

Now employees may work on multiple orders at the same time. The only constraint is that we give priority to earlier orders — we never delay an earlier client's order to favour a later one.

Building the schedule

At 10:00 — Client 1 arrives (1 burger, 3 portions fries, 1 ice cream):

At 10:05 — Client 2 arrives (2 portions fries, 1 ice cream):

• Bani has been free since 10:02, so Bani starts Client 2's ice cream at 10:05 → ready at 10:07.
• The fryer finished Client 1's fries at 10:05, so it starts Client 2's 2 portions of fries at 10:05 → ready at 10:10.
• Neither of these tasks interferes with Anish, who is still making Client 1's burger (10:00–10:10). So Client 1 is not delayed.

When is Client 2 completely served?

Client 2's two items:

The fries are the binding item: they cannot start before 10:05 (the order time — fries are never made in anticipation) and they take 5 minutes, so the earliest they can be ready is 10:10.

Answer: 10:10 (Choice A)

Understanding the task

We need to count, between 10:00 and 10:30, the number of minutes during which exactly one of the two employees (Anish or Bani) is idle. The automatic fryer is not an employee, so it never counts.

We use the same multi-order, priority-to-earlier-clients schedule from the previous question.

Full employee schedule

Let me trace what each employee does minute by minute.

Anish:

• 10:00–10:10: Making Client 1's burger.
• After 10:10: Nothing left to do — Client 2 has no burger, and Client 3's burger is already being handled by Bani (see below). So Anish is idle from 10:10 onward.

Bani:

• 10:00–10:02: Making Client 1's ice cream.
• 10:02–10:05: Idle (Client 2 hasn't arrived yet; Client 1's other items are handled by Anish and the fryer).
• 10:05–10:07: Making Client 2's ice cream.
• 10:07–10:17: Making Client 3's burger (Client 3 arrives at 10:07; Anish is still busy with Client 1's burger until 10:10, so Bani takes the burger to finish it as early as possible).
• After 10:17: Idle — all three orders are fully served, and Client 4 doesn't arrive until 10:35.

💡 Teacher tip: The key insight is that at 10:07, when Client 3's order arrives, Bani is free but Anish is not. Since the goal is to serve as early as possible, Bani starts the burger right away (finishing at 10:17) rather than waiting for Anish to be free at 10:10 (which would finish at 10:20).

Minute-by-minute idle check

Total

$3 + 7 = 10 \text{ minutes}$

Answer: 10 (Choice B)

We need Amda's score in F (freedom to life choices).

The frequency table (our anchor)

Each column sums to 10 (the ten countries). No country scores 0 or 8 in any parameter.

Using Clue 1

Clue 1 says Amda and Calla have identical scores in all three parameters, their total is 7, and they are jointly the lowest.

Because the two countries share the same triple $(S,F,C)$, each of those three score values is used by at least two countries — one for Amda and one for Calla. So every score Amda/Calla holds must have a frequency of 2 or more in its column.

Values with frequency $\ge 2$ in each column:

Finding the triple that sums to 7

We need $S + F + C = 7$ picking one value from each row above.

1. $S = 3$: then $F + C = 4$. The pair $F = 1,\ C = 3$ works $\Rightarrow (3,1,3) = 7$. ✓
2. $S = 4$: then $F + C = 3$. The pair $F = 1,\ C = 2$ works $\Rightarrow (4,1,2) = 7$. ✓
3. $S = 5$: then $F + C = 2$. No eligible pair gives 2 (C = 1 has frequency only 1). ✗

The two valid triples are $(3,1,3)$ and $(4,1,2)$ — and both have $F = 1$.

💡 Teacher tip: When two entities share identical attribute values, each shared value must appear at least twice in the data. That single observation shrinks the search space dramatically.

Answer: 1

We need Zooma's score in S. Zooma's total is 17 (Clue 2), and Clue 3 links the happy countries to the top scores.

The frequency table

Identifying the three happy countries (Clue 3)

Clue 3 says the three happy countries each hold the highest score in exactly one parameter. Reading the top of each column:

• Highest S score present = 7 (held by exactly 1 country).
• Highest F score present = 7 (held by exactly 1 country).
• Highest C score present = 6 (held by exactly 1 country; score 7 never occurs in C).

So the three happy countries are: the S = 7 holder, the F = 7 holder, and the C = 6 holder. Zooma (total 17) is one of them.

Testing which leader Zooma is

Case 1 — Zooma holds S = 7. Then $F + C = 10$.

• $F = 4,\ C = 6$: but C = 6 is another leader's score, so Zooma would lead two parameters — forbidden.
• $F = 5,\ C = 5$: C = 5 has frequency 0. Impossible.
• $F = 6,\ C = 4$: F = 6 has frequency 0. Impossible.

No valid split, so Zooma is not the S = 7 holder.

Case 2 — Zooma holds F = 7. Then $S + C = 10$, with $S \le 6$ (S = 7 belongs to another leader) and $C \le 4$ (C = 6 belongs to another leader).

• $S = 6,\ C = 4$: both frequencies are $\ge 1$. ✓
• $S = 5,\ C = 5$: C = 5 has frequency 0. ✗

So here Zooma's $S = 6$.

Case 3 — Zooma holds C = 6. Then $S + F = 11$, with $S \le 6$ and $F \le 5$.

• $S = 6,\ F = 5$: both frequencies are $\ge 1$. ✓
• $S = 5,\ F = 6$: F = 6 has frequency 0. ✗

So here too Zooma's $S = 6$.

💡 Teacher tip: "Highest score in exactly one parameter" means each happy country is a column-leader in precisely one column — so the three leaders are forced by the data, and Zooma must be one of them.

In every valid case, Zooma's score in S is 6.

Answer: 6

Benga and Delma are two happy countries tied at the same total. We want the maximum value of that tie.

The frequency table

Who are the three happy countries?

From Clue 3, the three happy countries are the three column-leaders: the S = 7 holder, the F = 7 holder, and the C = 6 holder. Zooma (total 17) is one of them, so Benga and Delma must be the other two leaders.

Bounding the tie

Take a concrete valid placement: Zooma is the F = 7 holder with $(S,F,C) = (6,7,4)$, total 17. Then Benga and Delma are the S = 7 holder and the C = 6 holder.

Look at the S = 7 holder's maximum possible total:

• $F \le 5$ (F = 7 is already Zooma's).
• $C \le 3$ (C = 4 is already Zooma's; C = 6 is the other leader's).
• Maximum total $= 7 + 5 + 3 = \mathbf{15}$.

The C = 6 holder could in principle reach a higher total, but since Benga and Delma must be tied, their common value cannot exceed the smaller of the two ceilings — that is, 15.

Is a tie at 15 actually reachable?

Yes. For example:

• S = 7 holder $= (7, 5, 3) = 15$
• C = 6 holder $= (5, 4, 6) = 15$
• Zooma $= (6, 7, 4) = 17$

One can verify the remaining five countries fill the rest of the frequency table consistently.

Why not 16?

To reach 16, the S = 7 holder would need $F + C = 9$ with $F \le 5$ and $C \le 3$ (since C = 4 is Zooma's). The maximum is $5 + 3 = 8$, giving only 15. Pushing higher would require the C = 4 slot that Zooma already occupies.

💡 Teacher tip: When two entities must be tied, the tie value is capped by whichever entity has the lower ceiling. Find the binding constraint first.

Answer: 15 (Choice B)

Given Benga = 16 and Delma = 15, we want the maximum number of countries with total score exactly 13.

Step 1 — Total of all scores across the 10 countries

From the frequency table, sum each parameter column (score × frequency):

Grand total of all 10 countries' scores $= 44 + 36 + 29 = \mathbf{109}$.

Step 2 — Account for the known countries

• Three happy countries: Zooma (17) + Benga (16) + Delma (15) $= 48$.
• Amda and Calla: $7 + 7 = 14$ (Clue 1, jointly the lowest).
• Remaining five countries: $109 - 48 - 14 = \mathbf{47}$.

Step 3 — Bound the number of 13s

These five remaining countries are non-happy (the only happy ones are Zooma, Benga, Delma), so each totals at most 13. Also, since Amda and Calla are the unique lowest at 7, every other country totals at least 8.

Suppose $k$ of these five countries have total exactly 13. The remaining $(5 - k)$ countries each total at least 8, so:

$47 - 13k \ge 8(5 - k)$ $47 - 13k \ge 40 - 8k$ $7 \ge 5k$ $k \le 1.4$

Since $k$ is an integer, $\boxed{k \le 1}$.

Step 4 — Confirm $k = 1$ is achievable

One neutral country at 13 and the other four at, say, 8, 9, 9, 8 gives $13 + 8 + 9 + 9 + 8 = 47$. ✓ All values lie in the valid 8–13 range, and the full frequency table can be matched with a concrete assignment.

💡 Teacher tip: When a question asks for a maximum count under a sum constraint, set up an inequality: the "special" countries consume a fixed amount, and the rest must each stay above a floor. Solve for the count.

Answer: 1 (Choice B)