CAT 2018 DILR Questions — Slot 1

Two cells are "adjacent" if they touch horizontally, vertically or diagonally — the way a king moves in chess. So any cell not on an edge touches all 8 cells around it.

A lower bound that never goes away: look at any $2\times2$ block of cells. All four cells touch one another, so they must carry 4 different numerals. This means at least 4 numerals are needed for any grid of size $2$ or more.

A pattern that always works with 4 numerals is to repeat the tile $\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}$ across the whole grid. Any two neighbouring cells differ in their row or column position, so they always land on different numerals of the tile.

For a $3\times3$ grid: the lower bound above already forces at least 4 numerals (take any corner $2\times2$ block).

And 4 is enough — the repeating tile fills the grid cleanly: $\begin{matrix} 1 & 2 & 1 \\ 3 & 4 & 3 \\ 1 & 2 & 1 \end{matrix}$ Check the centre cell (4): all eight neighbours are 1s, 2s and 3s — all different from 4. Every other adjacency is fine too.

Answer: 4

Two cells are "adjacent" if they touch horizontally, vertically or diagonally — the way a king moves in chess. So any cell not on an edge touches all 8 cells around it.

A lower bound that never goes away: look at any $2\times2$ block of cells. All four cells touch one another, so they must carry 4 different numerals. This means at least 4 numerals are needed for any grid of size $2$ or more.

A pattern that always works with 4 numerals is to repeat the tile $\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}$ across the whole grid. Any two neighbouring cells differ in their row or column position, so they always land on different numerals of the tile.

For a $5\times5$ grid: the same lower bound applies — any $2\times2$ block needs 4 numerals, so at least 4.

And the repeating tile covers a $5\times5$ grid with exactly 4 numerals: $\begin{matrix} 1 & 2 & 1 & 2 & 1 \\ 3 & 4 & 3 & 4 & 3 \\ 1 & 2 & 1 & 2 & 1 \\ 3 & 4 & 3 & 4 & 3 \\ 1 & 2 & 1 & 2 & 1 \end{matrix}$ Every cell's neighbours all carry a different numeral, so 4 works.

Answer: 4

Two cells are "adjacent" if they touch horizontally, vertically or diagonally — the way a king moves in chess. So any cell not on an edge touches all 8 cells around it.

A lower bound that never goes away: look at any $2\times2$ block of cells. All four cells touch one another, so they must carry 4 different numerals. This means at least 4 numerals are needed for any grid of size $2$ or more.

A pattern that always works with 4 numerals is to repeat the tile $\begin{matrix} 1 & 2 \\ 3 & 4 \end{matrix}$ across the whole grid. Any two neighbouring cells differ in their row or column position, so they always land on different numerals of the tile.

Now one mistake is allowed — exactly one pair of adjacent cells may share a numeral — and we fill a $5\times5$ grid. What is the minimum number of numerals?

The question is whether that one free "mistake" lets us drop below 4 numerals. It does not. A $5\times5$ grid contains 16 overlapping $2\times2$ blocks, and each block is a set of four mutually-touching cells. If we used only 3 numerals, then every one of those blocks would be forced to repeat a numeral (four cells, three labels) — creating many clashing pairs, far more than the single mistake we're allowed.

So we still need at least 4 numerals — and 4 already works with zero mistakes (from the previous part).

💡 Teacher tip: One exception can only fix one bad pair. Three numerals create clashes all over the grid, so a single allowance can't rescue them.

Answer: 4 (Choice A)

Two cells are "adjacent" if they touch horizontally, vertically or diagonally (a king's move). This question adds a new, stronger condition: for any cell, all of its neighbours must carry different numerals from one another (not just different from the centre).

Lower bound: an interior cell of the grid has 8 neighbours, and now those 8 must all be different. That's 8 numerals just for the neighbours, plus the centre cell must differ from all of them — so at least 9 numerals are needed.

9 is enough using a repeating $3\times3$ block of the numerals 1–9: $\begin{matrix} 1 & 2 & 3 & 1 & 2 \\ 4 & 5 & 6 & 4 & 5 \\ 7 & 8 & 9 & 7 & 8 \\ 1 & 2 & 3 & 1 & 2 \\ 4 & 5 & 6 & 4 & 5 \end{matrix}$ Any two cells sharing a numeral are at least 3 apart in some direction, so they never touch; and the 8 neighbours of any interior cell sweep out the 8 other numerals of the block — all distinct. Edge and corner cells have fewer neighbours, still all different.

Answer: 9 (Choice D)

Let $a$, $b$, $c$ be the number of ₹100, ₹200 and ₹500 notes a customer receives. Since every withdrawal is exactly ₹5000:

$100a + 200b + 500c = 5000 \quad\Rightarrow\quad a + 2b + 5c = 50.$

The preference rule: if a customer prefers a denomination X, then the count of X-notes must be more than the total of the other two denominations combined.

The customer prefers ₹500, so we need $c > a + b$ together with $a + 2b + 5c = 50$.

Since $5c \le 50$, we have $c \le 10$. Work through the largest values of $c$ (small $c$ can never beat $a+b$):

Adding up: $3 + 3 + 1 = \mathbf{7}$ ways.

💡 Teacher tip: Because the ₹500 note carries the most value, preferring it lets the customer use few notes — so only the top end of the $c$ range works. Always start from the largest useful $c$.

Answer: 7

Let $a$, $b$, $c$ be the number of ₹100, ₹200 and ₹500 notes a customer receives. Since every withdrawal is exactly ₹5000:

$100a + 200b + 500c = 5000 \quad\Rightarrow\quad a + 2b + 5c = 50.$

The preference rule: if a customer prefers a denomination X, then the count of X-notes must be more than the total of the other two denominations combined.

Here we have fifty ₹500 notes in stock, exactly 10 customers are served, and we want the maximum number of them who could have preferred ₹500.

How many ₹500 notes does one ₹500-preferring customer need at minimum? Such a customer needs $c > a + b$ with $a + 2b + 5c = 50$. Checking the values of $c$, no solution exists for $c \le 7$; the smallest workable value is $c = 8$ (for example $a=0,\ b=5,\ c=8$: here $8 > 5$ ✓ and $0 + 10 + 40 = 50$ ✓).

So each ₹500-preferring customer uses at least 8 of the ₹500 notes.

With only 50 such notes available:

$8k \le 50 \quad\Rightarrow\quad k \le 6 \quad(\text{since } 7 \times 8 = 56 > 50).$

Is 6 actually achievable? Yes: 6 customers each take $c = 8$, using $6 \times 8 = 48 \le 50$ notes; the remaining 4 customers prefer ₹100 or ₹200 and use no ₹500 notes at all.

Answer: 6

Let $a$, $b$, $c$ be the number of ₹100, ₹200 and ₹500 notes a customer receives. Since every withdrawal is exactly ₹5000:

$100a + 200b + 500c = 5000 \quad\Rightarrow\quad a + 2b + 5c = 50.$

The preference rule: if a customer prefers a denomination X, then the count of X-notes must be more than the total of the other two denominations combined.

Now the ATM has fifty ₹500 notes, unlimited ₹100 and ₹200 notes, and every withdrawal may use at most 20 notes. We want the maximum number of customers served.

Key observation: a withdrawal made from ₹100/₹200 notes alone needs far more than 20 notes. (For instance, preferring ₹200 with $c = 0$ forces $b \ge 17$ — already too many once you add the ₹100 notes.) So every customer must use some ₹500 notes, and to serve the most people we want each customer to use as few ₹500 notes as possible.

Try a customer preferring ₹200 with a small $c$:

• $c = 4$: then $a + 2b = 30$. Take $(a,b) = (0,15)$: preference holds since $15 > 0 + 4$ ✓, and total notes $= 0 + 15 + 4 = 19 \le 20$ ✓. This uses only 4 ₹500 notes.
• $c = 3$: the smallest note-count already exceeds 20, so this fails.

So the minimum is 4 ₹500 notes per customer. With 50 in stock:

$\left\lfloor \frac{50}{4} \right\rfloor = 12 \text{ customers} \quad(\text{using } 48 \text{ notes, 2 left over}).$

Answer: 12 (Choice A)

Let $a$, $b$, $c$ be the number of ₹100, ₹200 and ₹500 notes a customer receives. Since every withdrawal is exactly ₹5000:

$100a + 200b + 500c = 5000 \quad\Rightarrow\quad a + 2b + 5c = 50.$

The preference rule: if a customer prefers a denomination X, then the count of X-notes must be more than the total of the other two denominations combined.

We must serve 50 customers preferring ₹500 and 50 customers preferring ₹100, using the smallest possible total number of notes, and then count the ₹500 notes used.

To minimise the total notes, each customer should be served with the fewest notes their preference allows.

Customer preferring ₹500 — fewest notes: need $c > a + b$. The smallest total note-count comes from $(a,b,c) = (0,0,10)$: that's 10 notes, all ₹500, and $10 > 0$ ✓. So each such customer uses 10 ₹500 notes. $50 \times 10 = 500 \text{ ₹500 notes.}$

Customer preferring ₹100 — fewest notes: need $a > b + c$ with $a + 2b + 5c = 50$. The lightest option is $(a,b,c) = (10,0,8)$: total 18 notes, and $10 > 0 + 8$ ✓. Each such customer uses 8 ₹500 notes. $50 \times 8 = 400 \text{ ₹500 notes.}$

Total ₹500 notes $= 500 + 400 = \mathbf{900}$.

Answer: 900 (Choice A)

First fill in the blank figures using the two clues.

• 2016, Q2 (Apr, May, Jun) is an AP with April = 40 and the quarter total 150. So May + Jun = 110, and since April, May, Jun are in AP, May = 50, Jun = 60.
• 2016, Q4 (Oct, Nov, Dec) is an AP with October = 100 and the quarter total 360. So Nov + Dec = 260, and in AP, Nov = 120, Dec = 140.
• 2017, Q3 = 220 with Jul = 60, Sep = 70, so Aug = 220 − 60 − 70 = 90.
• 2017, Q4 = 500 with Oct = 150, Nov = 170, so Dec = 500 − 150 − 170 = 180.

Completed monthly sales:

Quarter and year totals:

December 2017 vs December 2016: Dec 2017 = 180, Dec 2016 = 140.

$\%\text{ increase} = \frac{180 - 140}{140} \times 100 = \frac{40}{140} \times 100 \approx 28.57\%.$

Answer: 28.57 (Choice C)

First fill in the blank figures using the two clues.

• 2016, Q2 (Apr, May, Jun) is an AP with April = 40 and the quarter total 150. So May + Jun = 110, and since April, May, Jun are in AP, May = 50, Jun = 60.
• 2016, Q4 (Oct, Nov, Dec) is an AP with October = 100 and the quarter total 360. So Nov + Dec = 260, and in AP, Nov = 120, Dec = 140.
• 2017, Q3 = 220 with Jul = 60, Sep = 70, so Aug = 220 − 60 − 70 = 90.
• 2017, Q4 = 500 with Oct = 150, Nov = 170, so Dec = 500 − 150 − 170 = 180.

Completed monthly sales:

Quarter and year totals:

Compare each 2017 quarter with the same quarter of 2016:

The largest rise is in Q1, about 58.3%.

Answer: Q1 (Choice B)

First fill in the blank figures using the two clues.

• 2016, Q2 (Apr, May, Jun) is an AP with April = 40 and the quarter total 150. So May + Jun = 110, and since April, May, Jun are in AP, May = 50, Jun = 60.
• 2016, Q4 (Oct, Nov, Dec) is an AP with October = 100 and the quarter total 360. So Nov + Dec = 260, and in AP, Nov = 120, Dec = 140.
• 2017, Q3 = 220 with Jul = 60, Sep = 70, so Aug = 220 − 60 − 70 = 90.
• 2017, Q4 = 500 with Oct = 150, Nov = 170, so Dec = 500 − 150 − 170 = 180.

Completed monthly sales:

Quarter and year totals:

We want the biggest percentage drop from one quarter to the immediately preceding quarter.

Going quarter by quarter, sales mostly rise; the drops are:

• 2016 Q1 → Q2: $240 \to 150$, a fall of $90/240 = 37.5\%$.
• 2017 Q1 → Q2: $380 \to 200$, a fall of $180/380 \approx 47.4\%$.
• (2016 Q4 → 2017 Q1 actually rises, $360 \to 380$.)

The steepest fall is Q2 of 2017, about 47.4%.

Answer: Q2 of 2017 (Choice A)

First fill in the blank figures using the two clues.

• 2016, Q2 (Apr, May, Jun) is an AP with April = 40 and the quarter total 150. So May + Jun = 110, and since April, May, Jun are in AP, May = 50, Jun = 60.
• 2016, Q4 (Oct, Nov, Dec) is an AP with October = 100 and the quarter total 360. So Nov + Dec = 260, and in AP, Nov = 120, Dec = 140.
• 2017, Q3 = 220 with Jul = 60, Sep = 70, so Aug = 220 − 60 − 70 = 90.
• 2017, Q4 = 500 with Oct = 150, Nov = 170, so Dec = 500 − 150 − 170 = 180.

Completed monthly sales:

Quarter and year totals:

We want the biggest month-on-month percentage increase. Check the candidate jumps:

October 2017 more than doubles from September — the largest jump.

Answer: October of 2017 (Choice B)

Every satellite lands in one region of a three-circle Venn diagram for broadcasting (B), communication (C) and surveillance (S) — plus a separate "only-others" group, because anything serving B, C or S is barred from serving O.

Name the regions:

• only B $=a$, only C $=b$, only S $=c$
• B and C only $=d$, B and S only $=e$, C and S only $=f$
• all three $=g$, and the separate only-O group $=O$.

Turn each fact into an equation:

1. All three: $g=100$.
2. Exclusive C equals exclusive S, and each is 30% of exclusive B: $b=c$ and $b=0.3a$. Write $a=10m$, so $b=c=3m$.
3. "Others" equals the C-and-S-but-not-B group: $O=f$.
4. Counts of B, C, S are in ratio $2:1:1$, i.e. $\#C=\#S$ and $\#B=2\#C$.

From $\#C=\#S$ (and $b=c$) we get $d=e$; call it $n$. From $\#B=2\#C$ the algebra collapses to $f=2m-50$. The grand total gives

$a+b+c+d+e+f+g+O=20m+2n=1600\;\Rightarrow\;n=800-10m.$

No region can be negative, so $f=2m-50\ge0$ and $n=800-10m\ge0$, i.e. $25\le m\le80$. Everything is now a single expression in $m$:

This question — what can be said about $\#C$? From the table, $\#C=850-5m$. At $m=25$ it is $725$; at $m=80$ it is $450$. As $m$ sweeps its whole feasible range, $\#C$ takes every value in between.

💡 Teacher tip: Squeeze all seven regions down to one free variable $m$ — then every "what can be said" question is just reading off the range of one straight-line expression.

Answer: Must be between 450 and 725 (Choice C)

Every satellite lands in one region of a three-circle Venn diagram for broadcasting (B), communication (C) and surveillance (S) — plus a separate "only-others" group, because anything serving B, C or S is barred from serving O.

Name the regions:

• only B $=a$, only C $=b$, only S $=c$
• B and C only $=d$, B and S only $=e$, C and S only $=f$
• all three $=g$, and the separate only-O group $=O$.

Turn each fact into an equation:

1. All three: $g=100$.
2. Exclusive C equals exclusive S, and each is 30% of exclusive B: $b=c$ and $b=0.3a$. Write $a=10m$, so $b=c=3m$.
3. "Others" equals the C-and-S-but-not-B group: $O=f$.
4. Counts of B, C, S are in ratio $2:1:1$, i.e. $\#C=\#S$ and $\#B=2\#C$.

From $\#C=\#S$ (and $b=c$) we get $d=e$; call it $n$. From $\#B=2\#C$ the algebra collapses to $f=2m-50$. The grand total gives

$a+b+c+d+e+f+g+O=20m+2n=1600\;\Rightarrow\;n=800-10m.$

No region can be negative, so $f=2m-50\ge0$ and $n=800-10m\ge0$, i.e. $25\le m\le80$. Everything is now a single expression in $m$:

This question — the minimum number serving B exclusively. The "only B" region is $10m$. It is smallest at the smallest feasible value $m=25$:

$10\times25=250.$

Answer: 250 (Choice A)

Every satellite lands in one region of a three-circle Venn diagram for broadcasting (B), communication (C) and surveillance (S) — plus a separate "only-others" group, because anything serving B, C or S is barred from serving O.

Name the regions:

• only B $=a$, only C $=b$, only S $=c$
• B and C only $=d$, B and S only $=e$, C and S only $=f$
• all three $=g$, and the separate only-O group $=O$.

Turn each fact into an equation:

1. All three: $g=100$.
2. Exclusive C equals exclusive S, and each is 30% of exclusive B: $b=c$ and $b=0.3a$. Write $a=10m$, so $b=c=3m$.
3. "Others" equals the C-and-S-but-not-B group: $O=f$.
4. Counts of B, C, S are in ratio $2:1:1$, i.e. $\#C=\#S$ and $\#B=2\#C$.

From $\#C=\#S$ (and $b=c$) we get $d=e$; call it $n$. From $\#B=2\#C$ the algebra collapses to $f=2m-50$. The grand total gives

$a+b+c+d+e+f+g+O=20m+2n=1600\;\Rightarrow\;n=800-10m.$

No region can be negative, so $f=2m-50\ge0$ and $n=800-10m\ge0$, i.e. $25\le m\le80$. Everything is now a single expression in $m$:

This question — if at least 100 satellites serve O. From the table $\#O=2m-50$. Requiring $2m-50\ge100$ gives $m\ge75$. Then

$\#S=850-5m\le850-5\times75=475,$

with equality at $m=75$. So the number serving S can be as high as 475 but no higher.

Answer: At most 475 (Choice A)

Every satellite lands in one region of a three-circle Venn diagram for broadcasting (B), communication (C) and surveillance (S) — plus a separate "only-others" group, because anything serving B, C or S is barred from serving O.

Name the regions:

• only B $=a$, only C $=b$, only S $=c$
• B and C only $=d$, B and S only $=e$, C and S only $=f$
• all three $=g$, and the separate only-O group $=O$.

Turn each fact into an equation:

1. All three: $g=100$.
2. Exclusive C equals exclusive S, and each is 30% of exclusive B: $b=c$ and $b=0.3a$. Write $a=10m$, so $b=c=3m$.
3. "Others" equals the C-and-S-but-not-B group: $O=f$.
4. Counts of B, C, S are in ratio $2:1:1$, i.e. $\#C=\#S$ and $\#B=2\#C$.

From $\#C=\#S$ (and $b=c$) we get $d=e$; call it $n$. From $\#B=2\#C$ the algebra collapses to $f=2m-50$. The grand total gives

$a+b+c+d+e+f+g+O=20m+2n=1600\;\Rightarrow\;n=800-10m.$

No region can be negative, so $f=2m-50\ge0$ and $n=800-10m\ge0$, i.e. $25\le m\le80$. Everything is now a single expression in $m$:

This question — if exactly 1200 satellites serve at least two of B, C, S. "At least two" is $d+e+f+g$. Using $d=e=n=800-10m$, $f=2m-50$ and $g=100$:

$d+e+f+g=2(800-10m)+(2m-50)+100=1650-18m.$

Set this to 1200: $1650-18m=1200\Rightarrow m=25$. At $m=25$ every region is pinned down, and in particular $\#C=850-5\times25=725$ — a single fixed value.

Now test the options: "only B is 250" (true, $10\times25$), "$\#B>1000$" (true, $1450$), "all 1600 serve B/C/S" (true, since $O=f=0$). The one that fails is "$\#C$ cannot be uniquely determined" — it can, it is exactly 725. That statement must be false.

Answer: "The number of satellites serving C cannot be uniquely determined" (Choice C)

Write H for high, M for medium, L for low. We fix pumps P1–P10 first.

1. P1–P5: three of these are H. None can be L, because clue 2 says P6 is the only L among P1–P10 — so the two non-H pumps here are M. To keep neighbouring pumps different, the three H's must sit apart, at P1, P3, P5. So P1–P5 = H M H M H (forced).
2. P6 = L (clue 2). By clue 3, P7 and P8 are the only equal consecutive pair, and neither is L.
3. P7–P10 use only H and M, with P7 = P8. Keeping neighbours different forces the shape $X\,X\,Y\,X$. If $X = M$ (M M H M), then all four remaining H's would have to squeeze into P11–P15 (P16–P20 have no H by clue 4) — but four H's in five slots must put two together, which isn't allowed. So $X = H$: P7–P10 = H H M H, giving P10 = H.
4. Counting: clue 5 says #H = 2 × #L. Testing #L = 3 leaves too few L's to keep the M's apart in P11–P20, so it fails. The only consistent totals are #H = 8, #L = 4, #M = 8.

Forced backbone (P1–P10):

Within P11–P20: exactly two H's (both inside P11–P15, non-adjacent), three L's, five M's; P16–P20 hold no H.

Which statement MUST be true?

The one fact forced by the deductions above is P10 = H (step 3: the only alternative shape for P7–P10 would jam four H's into P11–P15, which is impossible). The other options name P20, P13 and P12 — pumps that are not pinned down by the clues and can vary.

Answer: The contamination level at P10 was recorded as high (Choice D)

Write H for high, M for medium, L for low. We fix pumps P1–P10 first.

1. P1–P5: three of these are H. None can be L, because clue 2 says P6 is the only L among P1–P10 — so the two non-H pumps here are M. To keep neighbouring pumps different, the three H's must sit apart, at P1, P3, P5. So P1–P5 = H M H M H (forced).
2. P6 = L (clue 2). By clue 3, P7 and P8 are the only equal consecutive pair, and neither is L.
3. P7–P10 use only H and M, with P7 = P8. Keeping neighbours different forces the shape $X\,X\,Y\,X$. If $X = M$ (M M H M), then all four remaining H's would have to squeeze into P11–P15 (P16–P20 have no H by clue 4) — but four H's in five slots must put two together, which isn't allowed. So $X = H$: P7–P10 = H H M H, giving P10 = H.
4. Counting: clue 5 says #H = 2 × #L. Testing #L = 3 leaves too few L's to keep the M's apart in P11–P20, so it fails. The only consistent totals are #H = 8, #L = 4, #M = 8.

Forced backbone (P1–P10):

Within P11–P20: exactly two H's (both inside P11–P15, non-adjacent), three L's, five M's; P16–P20 hold no H.

What can we say about the number of medium (M) pumps?

From the counting step, the alternatives #L = 2 or #L = 3 both fail — there aren't enough L's to separate the M's across P11–P20 — and the layout M M H M for P7–P10 is impossible. That leaves a single consistent split of the 20 pumps:

$\#H = 8, \quad \#L = 4, \quad \#M = 8.$

So the number of medium pumps is fixed at exactly 8.

Answer: Exactly 8 (Choice C)

Write H for high, M for medium, L for low. We fix pumps P1–P10 first.

1. P1–P5: three of these are H. None can be L, because clue 2 says P6 is the only L among P1–P10 — so the two non-H pumps here are M. To keep neighbouring pumps different, the three H's must sit apart, at P1, P3, P5. So P1–P5 = H M H M H (forced).
2. P6 = L (clue 2). By clue 3, P7 and P8 are the only equal consecutive pair, and neither is L.
3. P7–P10 use only H and M, with P7 = P8. Keeping neighbours different forces the shape $X\,X\,Y\,X$. If $X = M$ (M M H M), then all four remaining H's would have to squeeze into P11–P15 (P16–P20 have no H by clue 4) — but four H's in five slots must put two together, which isn't allowed. So $X = H$: P7–P10 = H H M H, giving P10 = H.
4. Counting: clue 5 says #H = 2 × #L. Testing #L = 3 leaves too few L's to keep the M's apart in P11–P20, so it fails. The only consistent totals are #H = 8, #L = 4, #M = 8.

Forced backbone (P1–P10):

Within P11–P20: exactly two H's (both inside P11–P15, non-adjacent), three L's, five M's; P16–P20 hold no H.

Extra condition for this question: P11 = L. Which statement MUST then be true?

We know P10 = H, so P11 being L is fine. The two H's of the P11–P15 stretch must now lie in P12–P15 (since P11 is L), must be non-adjacent, and P12 cannot be L (or it would tie with P11).

Testing the possible H-placements together with the fixed counts (three L's and five M's still to be placed across P11–P20, P16–P20 having no H):

• Putting H's at {P12, P14} forces the rest of P16–P20 to hold too many M's with too few L's to separate them — no valid completion.
• The placements that do complete consistently (for example H's at P13 and P15, with P12 = M) all end up with P14 = M — e.g. P16–P20 = M, L, M, L, M works.

So whenever P11 = L, P14 must be M.

Answer: The contamination level at P14 was recorded as medium (Choice D)

Write H for high, M for medium, L for low. We fix pumps P1–P10 first.

1. P1–P5: three of these are H. None can be L, because clue 2 says P6 is the only L among P1–P10 — so the two non-H pumps here are M. To keep neighbouring pumps different, the three H's must sit apart, at P1, P3, P5. So P1–P5 = H M H M H (forced).
2. P6 = L (clue 2). By clue 3, P7 and P8 are the only equal consecutive pair, and neither is L.
3. P7–P10 use only H and M, with P7 = P8. Keeping neighbours different forces the shape $X\,X\,Y\,X$. If $X = M$ (M M H M), then all four remaining H's would have to squeeze into P11–P15 (P16–P20 have no H by clue 4) — but four H's in five slots must put two together, which isn't allowed. So $X = H$: P7–P10 = H H M H, giving P10 = H.
4. Counting: clue 5 says #H = 2 × #L. Testing #L = 3 leaves too few L's to keep the M's apart in P11–P20, so it fails. The only consistent totals are #H = 8, #L = 4, #M = 8.

Forced backbone (P1–P10):

Within P11–P20: exactly two H's (both inside P11–P15, non-adjacent), three L's, five M's; P16–P20 hold no H.

Extra condition for this question: P15 = M. Which statement MUST be FALSE?

Since P15 = M, its neighbour P16 cannot be M. And P16–P20 contain no H (clue 4), so P16 = L is forced.

Now test option B's claim that P11 and P16 are the same, i.e. P11 = L. If P11 = L, the two H's of P11–P15 must sit in P12–P14 (P11 = L, P15 = M); the only non-adjacent choice is {P12, P14}, leaving P13 as M or L. Either way, completing P17–P20 forces two M's next to each other — which isn't allowed. So P11 = L is impossible; P11 must be M while P16 = L, and the two can never match.

Hence "P11 and P16 were recorded as the same" cannot happen.

Answer: Contamination levels at P11 and P16 were recorded as the same — MUST be FALSE (Choice B)

Label the committees R (research), T (teaching), A (administration), and the three types B (bureaucrats), E (educationalists), P (politicians). Total people = 24.

• Bureaucrats (clue 1): $B_R = B_T$ and $B_R = \tfrac34 B_A$. So write $B_R = B_T = 3k$ and $B_A = 4k$ — total bureaucrats $= 10k$.
• Politicians (clue 3): 60% in A, 20% in T, 20% in R, so the total is a multiple of 5: $P_A = 3p,\ P_T = P_R = p$ — total politicians $= 5p$.
• Educationalists (clue 2): $E_T < E_R$ and $E_R = \tfrac12(E_T + E_A)$. That makes the three educationalist counts sum to $3E_R$.

Putting it together: $10k + 3E_R + 5p = 24$. With $k \ge 1$, $p \ge 1$ and $E_R \ge 2$, the only whole-number fit is $k = 1,\ p = 1,\ E_R = 3$ — which leaves $E_T \in \{1, 2\}$.

Solved table (two allowed variants):

Fixed totals: bureaucrats = 10, educationalists = 9, politicians = 5. Only the split of educationalists between T and A (and so the sizes of T and A) stays flexible.

Which statement MUST be FALSE?

• "In T, #E = #P": possible — take $E_T = 1 = P_T$.
• "In A, #B = #E": possible — take $E_A = 4 = B_A$.
• "Size of research < size of teaching": research is always 7, while teaching is 5 or 6. So 7 < (size of T) is never true.
• "Size of research < size of administration": 7 < 11 or 12 — always true.

The one that can never hold is the third.

Answer: The size of the research committee is less than the size of the teaching committee (Choice C)

Label the committees R (research), T (teaching), A (administration), and the three types B (bureaucrats), E (educationalists), P (politicians). Total people = 24.

• Bureaucrats (clue 1): $B_R = B_T$ and $B_R = \tfrac34 B_A$. So write $B_R = B_T = 3k$ and $B_A = 4k$ — total bureaucrats $= 10k$.
• Politicians (clue 3): 60% in A, 20% in T, 20% in R, so the total is a multiple of 5: $P_A = 3p,\ P_T = P_R = p$ — total politicians $= 5p$.
• Educationalists (clue 2): $E_T < E_R$ and $E_R = \tfrac12(E_T + E_A)$. That makes the three educationalist counts sum to $3E_R$.

Putting it together: $10k + 3E_R + 5p = 24$. With $k \ge 1$, $p \ge 1$ and $E_R \ge 2$, the only whole-number fit is $k = 1,\ p = 1,\ E_R = 3$ — which leaves $E_T \in \{1, 2\}$.

Solved table (two allowed variants):

Fixed totals: bureaucrats = 10, educationalists = 9, politicians = 5. Only the split of educationalists between T and A (and so the sizes of T and A) stays flexible.

Number of bureaucrats in the administration committee.

From clue 1, $B_A = 4k$. The overall total of 24 forces $k = 1$, so $B_A = 4$.

Answer: 4

Label the committees R (research), T (teaching), A (administration), and the three types B (bureaucrats), E (educationalists), P (politicians). Total people = 24.

• Bureaucrats (clue 1): $B_R = B_T$ and $B_R = \tfrac34 B_A$. So write $B_R = B_T = 3k$ and $B_A = 4k$ — total bureaucrats $= 10k$.
• Politicians (clue 3): 60% in A, 20% in T, 20% in R, so the total is a multiple of 5: $P_A = 3p,\ P_T = P_R = p$ — total politicians $= 5p$.
• Educationalists (clue 2): $E_T < E_R$ and $E_R = \tfrac12(E_T + E_A)$. That makes the three educationalist counts sum to $3E_R$.

Putting it together: $10k + 3E_R + 5p = 24$. With $k \ge 1$, $p \ge 1$ and $E_R \ge 2$, the only whole-number fit is $k = 1,\ p = 1,\ E_R = 3$ — which leaves $E_T \in \{1, 2\}$.

Solved table (two allowed variants):

Fixed totals: bureaucrats = 10, educationalists = 9, politicians = 5. Only the split of educationalists between T and A (and so the sizes of T and A) stays flexible.

Number of educationalists in the research committee.

$E_R$ is the average of $E_T$ and $E_A$, and the total-people equation $10k + 3E_R + 5p = 24$ (with $k, p \ge 1$) has the single solution giving $E_R = 3$.

Answer: 3

Label the committees R (research), T (teaching), A (administration), and the three types B (bureaucrats), E (educationalists), P (politicians). Total people = 24.

• Bureaucrats (clue 1): $B_R = B_T$ and $B_R = \tfrac34 B_A$. So write $B_R = B_T = 3k$ and $B_A = 4k$ — total bureaucrats $= 10k$.
• Politicians (clue 3): 60% in A, 20% in T, 20% in R, so the total is a multiple of 5: $P_A = 3p,\ P_T = P_R = p$ — total politicians $= 5p$.
• Educationalists (clue 2): $E_T < E_R$ and $E_R = \tfrac12(E_T + E_A)$. That makes the three educationalist counts sum to $3E_R$.

Putting it together: $10k + 3E_R + 5p = 24$. With $k \ge 1$, $p \ge 1$ and $E_R \ge 2$, the only whole-number fit is $k = 1,\ p = 1,\ E_R = 3$ — which leaves $E_T \in \{1, 2\}$.

Solved table (two allowed variants):

Fixed totals: bureaucrats = 10, educationalists = 9, politicians = 5. Only the split of educationalists between T and A (and so the sizes of T and A) stays flexible.

Which quantity CANNOT be found uniquely?

• Size of research = 7 — fixed.
• Total bureaucrats = 10 — fixed.
• Total educationalists = 9 — fixed.
• Size of teaching = 5 or 6 — not fixed, because the educationalists can split as $E_T = 1$ (size 5) or $E_T = 2$ (size 6).

Only the teaching committee's size stays undetermined.

Answer: The size of the teaching committee (Choice A)

The rule. Composite $=2\cdot DI+WE+GA$. A candidate is disqualified if she scores below 14 (that is 70% of 20) in two or more sections. Of those who survive, the four highest composites are recruited.

Solving the missing marks step by step.

1. Jatin and Indu each scored 100% (a 20) in exactly one section, and Jatin's composite is 10 more than Indu's.

• Jatin already has WE 16 and GA 14, so his 20 is in DI: Jatin = DI 20, WE 16, GA 14, composite 70.
• So Indu's composite is 60. Indu has WE 8. If her 20 were in DI, her GA would be $60-40-8=12$, leaving two sub-14 sections (WE and GA) — she'd be disqualified, but she was recruited. So her 20 is in GA: $2\cdot DI+8+20=60\Rightarrow DI=16$. Indu = DI 16, WE 8, GA 20, composite 60.

1. Danish, Harini and Indu share a GA mark, so both Danish and Harini have GA 20. Danish = DI 8, WE 15, GA 20, composite 51.
2. Ajay is the unique top scorer in WE. The known WE marks top out at Ester's 18, so Ajay needs 19 or 20. His composite is $2\cdot8+WE+16=32+WE$. WE 19 would tie Danish's composite of 51, which isn't allowed (no two composites are equal), so Ajay = WE 20, composite 52.
3. Geeta is the lowest of the four recruited. The clear top three composites are Jatin 70, Indu 60, Ester 58. Geeta's is $2\cdot14+WE+6=34+WE$; to edge out Ajay's 52 she needs $WE>18$, and since Ajay's 20 is unique she takes WE 19, composite 53.

Solved table (bold = deduced):

Recruited (top four composites): Jatin 70, Indu 60, Ester 58, Geeta 53.

This question — which must be true?

1. Jatin's composite beats Danish's: $70>51$ — true.
2. Indu scored less than Chetna in DI: $16<19$ — true.
3. Jatin scored more than Indu in GA: $14$ vs $20$ — false.

So statements 1 and 2 hold.

Answer: Both 1 and 2 (Choice D)

The rule. Composite $=2\cdot DI+WE+GA$. A candidate is disqualified if she scores below 14 (that is 70% of 20) in two or more sections. Of those who survive, the four highest composites are recruited.

Solving the missing marks step by step.

1. Jatin and Indu each scored 100% (a 20) in exactly one section, and Jatin's composite is 10 more than Indu's.

• Jatin already has WE 16 and GA 14, so his 20 is in DI: Jatin = DI 20, WE 16, GA 14, composite 70.
• So Indu's composite is 60. Indu has WE 8. If her 20 were in DI, her GA would be $60-40-8=12$, leaving two sub-14 sections (WE and GA) — she'd be disqualified, but she was recruited. So her 20 is in GA: $2\cdot DI+8+20=60\Rightarrow DI=16$. Indu = DI 16, WE 8, GA 20, composite 60.

1. Danish, Harini and Indu share a GA mark, so both Danish and Harini have GA 20. Danish = DI 8, WE 15, GA 20, composite 51.
2. Ajay is the unique top scorer in WE. The known WE marks top out at Ester's 18, so Ajay needs 19 or 20. His composite is $2\cdot8+WE+16=32+WE$. WE 19 would tie Danish's composite of 51, which isn't allowed (no two composites are equal), so Ajay = WE 20, composite 52.
3. Geeta is the lowest of the four recruited. The clear top three composites are Jatin 70, Indu 60, Ester 58. Geeta's is $2\cdot14+WE+6=34+WE$; to edge out Ajay's 52 she needs $WE>18$, and since Ajay's 20 is unique she takes WE 19, composite 53.

Solved table (bold = deduced):

Recruited (top four composites): Jatin 70, Indu 60, Ester 58, Geeta 53.

This question — which must be false?

Check "Bala scored the same as Jatin in DI." Jatin's DI is 20. If Bala also had DI 20, Bala's composite would be $2\cdot20+9+11=60$ — exactly Indu's composite, and no two candidates may share a composite. So Bala can never have DI 20; this statement is always false. (The other three options are each achievable.)

Answer: "Bala scored same as Jatin in DI" (Choice A)

The rule. Composite $=2\cdot DI+WE+GA$. A candidate is disqualified if she scores below 14 (that is 70% of 20) in two or more sections. Of those who survive, the four highest composites are recruited.

Solving the missing marks step by step.

1. Jatin and Indu each scored 100% (a 20) in exactly one section, and Jatin's composite is 10 more than Indu's.

• Jatin already has WE 16 and GA 14, so his 20 is in DI: Jatin = DI 20, WE 16, GA 14, composite 70.
• So Indu's composite is 60. Indu has WE 8. If her 20 were in DI, her GA would be $60-40-8=12$, leaving two sub-14 sections (WE and GA) — she'd be disqualified, but she was recruited. So her 20 is in GA: $2\cdot DI+8+20=60\Rightarrow DI=16$. Indu = DI 16, WE 8, GA 20, composite 60.

1. Danish, Harini and Indu share a GA mark, so both Danish and Harini have GA 20. Danish = DI 8, WE 15, GA 20, composite 51.
2. Ajay is the unique top scorer in WE. The known WE marks top out at Ester's 18, so Ajay needs 19 or 20. His composite is $2\cdot8+WE+16=32+WE$. WE 19 would tie Danish's composite of 51, which isn't allowed (no two composites are equal), so Ajay = WE 20, composite 52.
3. Geeta is the lowest of the four recruited. The clear top three composites are Jatin 70, Indu 60, Ester 58. Geeta's is $2\cdot14+WE+6=34+WE$; to edge out Ajay's 52 she needs $WE>18$, and since Ajay's 20 is unique she takes WE 19, composite 53.

Solved table (bold = deduced):

Recruited (top four composites): Jatin 70, Indu 60, Ester 58, Geeta 53.

This question — every candidate except Ajay and Danish has a different DI mark, and Bala's composite is below Chetna's (54). Find the largest DI Bala can have.

The other candidates whose DI must all differ from Bala's have these DI marks: Chetna 19, Ester 12, Falak 15, Geeta 14, Harini 5, Indu 16, Jatin 20. So Bala's DI must avoid $\{5,12,14,15,16,19,20\}$.

Bala's composite is $2\cdot DI+20<54$, so $DI<17$, i.e. $DI\le16$.

The largest value $\le16$ not on the forbidden list is 13 ($16,15,14$ are all taken).

Answer: 13

The rule. Composite $=2\cdot DI+WE+GA$. A candidate is disqualified if she scores below 14 (that is 70% of 20) in two or more sections. Of those who survive, the four highest composites are recruited.

Solving the missing marks step by step.

1. Jatin and Indu each scored 100% (a 20) in exactly one section, and Jatin's composite is 10 more than Indu's.

• Jatin already has WE 16 and GA 14, so his 20 is in DI: Jatin = DI 20, WE 16, GA 14, composite 70.
• So Indu's composite is 60. Indu has WE 8. If her 20 were in DI, her GA would be $60-40-8=12$, leaving two sub-14 sections (WE and GA) — she'd be disqualified, but she was recruited. So her 20 is in GA: $2\cdot DI+8+20=60\Rightarrow DI=16$. Indu = DI 16, WE 8, GA 20, composite 60.

1. Danish, Harini and Indu share a GA mark, so both Danish and Harini have GA 20. Danish = DI 8, WE 15, GA 20, composite 51.
2. Ajay is the unique top scorer in WE. The known WE marks top out at Ester's 18, so Ajay needs 19 or 20. His composite is $2\cdot8+WE+16=32+WE$. WE 19 would tie Danish's composite of 51, which isn't allowed (no two composites are equal), so Ajay = WE 20, composite 52.
3. Geeta is the lowest of the four recruited. The clear top three composites are Jatin 70, Indu 60, Ester 58. Geeta's is $2\cdot14+WE+6=34+WE$; to edge out Ajay's 52 she needs $WE>18$, and since Ajay's 20 is unique she takes WE 19, composite 53.

Solved table (bold = deduced):

Recruited (top four composites): Jatin 70, Indu 60, Ester 58, Geeta 53.

This question — every candidate has a different WE mark. Find the largest WE Harini can have.

The WE marks already fixed are Chetna 4, Falak 7, Indu 8, Bala 9, Danish 15, Jatin 16, Ester 18, Geeta 19, Ajay 20. Harini must avoid all of these, leaving $\{0,1,2,3,5,6,10,11,12,13,14,17\}$.

One more rule: no two composites are equal. Harini's composite is $2\cdot5+WE+20=30+WE$. Setting that equal to an existing composite, $30+WE=47$ (Falak) gives $WE=17$ — so 17 is blocked (every other existing composite would need $WE\ge21$, impossible).

Removing 17, the largest remaining choice is 14 (composite 44, distinct from everyone).

Answer: 14