CAT 2018 DILR Questions — Slot 2

Let the base rates (units of local currency L per unit) be $A=100k$, $B=120k$, $C=k$, matching the ratio $100:120:1$. The outlet buys at 5% below base ($0.95\times$base) and sells at 10% above base ($1.10\times$base).

Pin down $k$:

1. Selling A brought in 88000 L at a selling rate of $1.10\times100k=110k$, so units of A sold $=\dfrac{88000}{110k}=\dfrac{800}{k}$.
2. A's stock rose from 2500 to 3300, a gain of 800, so (A bought) $-$ (A sold) $=800$, giving A bought $=800+\dfrac{800}{k}$. The L spent buying A is $\left(800+\tfrac{800}{k}\right)(0.95)(100k)=76000k+76000$.
3. Sales of A and B are in ratio $5:9$, so sale of B $=88000\times\tfrac95=158400$ L. At B's selling rate $1.10\times120k=132k$, units of B sold $=\dfrac{1200}{k}$.
4. B's stock is unchanged, so B bought $=$ B sold $=\dfrac{1200}{k}$, and L spent buying B $=\dfrac{1200}{k}(0.95)(120k)=136800$.
5. L used to buy A and B are in ratio $5:3$: $\dfrac{76000k+76000}{136800}=\dfrac53\Rightarrow k=2$.

So the base rates are A = 200, B = 240, C = 2; buying rates 190, 228, 1.90; selling rates 220, 264, 2.20.

This question — units of A bought. With $k=2$, A sold $=\dfrac{800}{2}=400$; since A bought $-$ A sold $=800$, A bought $=800+400=1200$.

💡 Teacher tip: Carry the whole rate scale as one unknown $k$; the two buy/sell ratios pin it down without ever needing the absolute base rates first.

Answer: 1200 units

Let the base rates (units of local currency L per unit) be $A=100k$, $B=120k$, $C=k$, matching the ratio $100:120:1$. The outlet buys at 5% below base ($0.95\times$base) and sells at 10% above base ($1.10\times$base).

Pin down $k$:

1. Selling A brought in 88000 L at a selling rate of $1.10\times100k=110k$, so units of A sold $=\dfrac{88000}{110k}=\dfrac{800}{k}$.
2. A's stock rose from 2500 to 3300, a gain of 800, so (A bought) $-$ (A sold) $=800$, giving A bought $=800+\dfrac{800}{k}$. The L spent buying A is $\left(800+\tfrac{800}{k}\right)(0.95)(100k)=76000k+76000$.
3. Sales of A and B are in ratio $5:9$, so sale of B $=88000\times\tfrac95=158400$ L. At B's selling rate $1.10\times120k=132k$, units of B sold $=\dfrac{1200}{k}$.
4. B's stock is unchanged, so B bought $=$ B sold $=\dfrac{1200}{k}$, and L spent buying B $=\dfrac{1200}{k}(0.95)(120k)=136800$.
5. L used to buy A and B are in ratio $5:3$: $\dfrac{76000k+76000}{136800}=\dfrac53\Rightarrow k=2$.

So the base rates are A = 200, B = 240, C = 2; buying rates 190, 228, 1.90; selling rates 220, 264, 2.20.

This question — units of C sold. The L spent buying C equals the L earned selling C: $1.90\times(\text{C bought})=2.20\times(\text{C sold})$, so C bought: C sold $=22:19$. C's stock rose from 48000 to 51000, a gain of 3000, so C bought $-$ C sold $=3000$. With the ratio $22:19$ the difference is 3 parts $=3000$, so one part $=1000$ and C sold $=19\times1000=19000$.

Answer: 19000 (Choice B)

Let the base rates (units of local currency L per unit) be $A=100k$, $B=120k$, $C=k$, matching the ratio $100:120:1$. The outlet buys at 5% below base ($0.95\times$base) and sells at 10% above base ($1.10\times$base).

Pin down $k$:

1. Selling A brought in 88000 L at a selling rate of $1.10\times100k=110k$, so units of A sold $=\dfrac{88000}{110k}=\dfrac{800}{k}$.
2. A's stock rose from 2500 to 3300, a gain of 800, so (A bought) $-$ (A sold) $=800$, giving A bought $=800+\dfrac{800}{k}$. The L spent buying A is $\left(800+\tfrac{800}{k}\right)(0.95)(100k)=76000k+76000$.
3. Sales of A and B are in ratio $5:9$, so sale of B $=88000\times\tfrac95=158400$ L. At B's selling rate $1.10\times120k=132k$, units of B sold $=\dfrac{1200}{k}$.
4. B's stock is unchanged, so B bought $=$ B sold $=\dfrac{1200}{k}$, and L spent buying B $=\dfrac{1200}{k}(0.95)(120k)=136800$.
5. L used to buy A and B are in ratio $5:3$: $\dfrac{76000k+76000}{136800}=\dfrac53\Rightarrow k=2$.

So the base rates are A = 200, B = 240, C = 2; buying rates 190, 228, 1.90; selling rates 220, 264, 2.20.

This question — the base rate of B. The three ratio clues force $k=2$ (shown above), and $B=120k=120\times2=240$.

Answer: 240

Let the base rates (units of local currency L per unit) be $A=100k$, $B=120k$, $C=k$, matching the ratio $100:120:1$. The outlet buys at 5% below base ($0.95\times$base) and sells at 10% above base ($1.10\times$base).

Pin down $k$:

1. Selling A brought in 88000 L at a selling rate of $1.10\times100k=110k$, so units of A sold $=\dfrac{88000}{110k}=\dfrac{800}{k}$.
2. A's stock rose from 2500 to 3300, a gain of 800, so (A bought) $-$ (A sold) $=800$, giving A bought $=800+\dfrac{800}{k}$. The L spent buying A is $\left(800+\tfrac{800}{k}\right)(0.95)(100k)=76000k+76000$.
3. Sales of A and B are in ratio $5:9$, so sale of B $=88000\times\tfrac95=158400$ L. At B's selling rate $1.10\times120k=132k$, units of B sold $=\dfrac{1200}{k}$.
4. B's stock is unchanged, so B bought $=$ B sold $=\dfrac{1200}{k}$, and L spent buying B $=\dfrac{1200}{k}(0.95)(120k)=136800$.
5. L used to buy A and B are in ratio $5:3$: $\dfrac{76000k+76000}{136800}=\dfrac53\Rightarrow k=2$.

So the base rates are A = 200, B = 240, C = 2; buying rates 190, 228, 1.90; selling rates 220, 264, 2.20.

This question — the buying rate of C. The base rate of C is $k=2$, and buying is 5% below base: $0.95\times2=1.90$.

Answer: 1.90 (Choice D)

Let the total number of phones sold in 2016 be $V$. For each brand, revenue $=$ (its share of $V$) $\times$ unit price. Working in units of $V$:

The largest is Azra at $6{,}000V$. (The options are listed in a different order — Azra is the third choice.)

Answer: Azra (Choice C)

Profit $=$ profitability $\times$ revenue. Using $V$ for total 2016 volume:

Cxqi wins with $1{,}800V$ — its smaller revenue is more than made up for by the highest profit margin.

Answer: Cxqi (Choice C)

First update everything to 2017:

• Total volume grows 40%: now $1.4V$.
• Cxqi's share rises 15 percentage points ($15\% \to 30\%$); the other three each lose 5 points: Azra $40\% \to 35\%$, Bysi $25\% \to 20\%$, Dipq $20\% \to 15\%$ (these still add to 100%).
• Cxqi's price is cut 40%: $30{,}000 \to 18{,}000$. Other prices unchanged.
• Cxqi's profitability halves: $40\% \to 20\%$. Others unchanged.

Now compute 2017 profit $=$ profitability $\times$ share $\times 1.4V \times$ price:

Bysi is highest at $1{,}680V$.

Answer: Bysi (Choice A)

Compare each brand's profit in 2016 with 2017. (2017 uses total volume $1.4V$, with the new shares 35/20/15% for Azra/Bysi/Dipq at unchanged prices and margins, and Cxqi at 30% share, price 18,000, margin 20%.)

Only Cxqi falls — its price cut and halved margin outweigh its bigger volume. Profits rose for Azra, Bysi and Dipq.

Answer: Azra, Bysi, Dipq (Choice A)

Head counts. With Middle $=2\times$Old and Young $=2\times$Middle, and a total of 140: $O+2O+4O=140\Rightarrow O=20,\;M=40,\;Y=80$.

Ticket facts.

• Economy total is 55; Young bought 38 of them, so Old + Mid Economy $=17$ (an odd number — keep this in mind).
• Young bought half of all Platinum: $YP=\tfrac{P}{2}$, so Old + Mid Platinum $=\tfrac{P}{2}$.
• Old bought equal Gold and Economy: $OG=OE$.
• Since Economy $=55$, Platinum + Gold $=85$, i.e. $G=85-P$; and Young Gold $=80-\tfrac{P}{2}-38=42-\tfrac{P}{2}$.

This question — if Old-Platinum equals Mid-Platinum, what could total Platinum be? Then $OP=MP=\tfrac{P}{4}$, so $P$ must be a multiple of 4 — only 32 or 36 among the options. The Old row gives $OP+OG+OE=20$ with $OG=OE$, so $OG=\dfrac{20-OP}{2}$.

• $P=36\Rightarrow OP=9\Rightarrow OG=5.5$ — not a whole number, impossible.
• $P=32\Rightarrow OP=8\Rightarrow OG=6$ — valid.

💡 Teacher tip: The odd total $OE+ME=17$ combined with $OG=OE$ quietly does most of the work across this whole set.

Answer: 32 (Choice D)

Head counts. With Middle $=2\times$Old and Young $=2\times$Middle, and a total of 140: $O+2O+4O=140\Rightarrow O=20,\;M=40,\;Y=80$.

Ticket facts.

• Economy total is 55; Young bought 38 of them, so Old + Mid Economy $=17$ (an odd number — keep this in mind).
• Young bought half of all Platinum: $YP=\tfrac{P}{2}$, so Old + Mid Platinum $=\tfrac{P}{2}$.
• Old bought equal Gold and Economy: $OG=OE$.
• Since Economy $=55$, Platinum + Gold $=85$, i.e. $G=85-P$; and Young Gold $=80-\tfrac{P}{2}-38=42-\tfrac{P}{2}$.

This question — if Old-Platinum equals Mid-Economy, find Old-Gold. Let $OP=ME=a$. Since $OE+ME=17$, we get $OE=17-a$, and $OG=OE=17-a$. The Old row totals 20:

$OP+OG+OE=a+(17-a)+(17-a)=34-a=20\Rightarrow a=14.$

So $OG=17-14=3$.

Answer: 3

Head counts. With Middle $=2\times$Old and Young $=2\times$Middle, and a total of 140: $O+2O+4O=140\Rightarrow O=20,\;M=40,\;Y=80$.

Ticket facts.

• Economy total is 55; Young bought 38 of them, so Old + Mid Economy $=17$ (an odd number — keep this in mind).
• Young bought half of all Platinum: $YP=\tfrac{P}{2}$, so Old + Mid Platinum $=\tfrac{P}{2}$.
• Old bought equal Gold and Economy: $OG=OE$.
• Since Economy $=55$, Platinum + Gold $=85$, i.e. $G=85-P$; and Young Gold $=80-\tfrac{P}{2}-38=42-\tfrac{P}{2}$.

This question — if Old-Gold is strictly more than Young-Gold, find Mid-Gold. From the Gold row, $MG=(85-P)-OG-\left(42-\tfrac{P}{2}\right)=\left(43-\tfrac{P}{2}\right)-OG$. Since Young-Gold is $42-\tfrac{P}{2}$, this is $MG=(YG+1)-OG$.

Now $OG>YG$ means $OG\ge YG+1$, which forces $MG\le0$. A ticket count can't be negative, so $MG=0$.

Answer: 0

Head counts. With Middle $=2\times$Old and Young $=2\times$Middle, and a total of 140: $O+2O+4O=140\Rightarrow O=20,\;M=40,\;Y=80$.

Ticket facts.

• Economy total is 55; Young bought 38 of them, so Old + Mid Economy $=17$ (an odd number — keep this in mind).
• Young bought half of all Platinum: $YP=\tfrac{P}{2}$, so Old + Mid Platinum $=\tfrac{P}{2}$.
• Old bought equal Gold and Economy: $OG=OE$.
• Since Economy $=55$, Platinum + Gold $=85$, i.e. $G=85-P$; and Young Gold $=80-\tfrac{P}{2}-38=42-\tfrac{P}{2}$.

This question — which must be false? Check "Old and Mid bought equal numbers of Economy tickets," i.e. $OE=ME$. But $OE+ME=17$, an odd number, so two equal whole numbers can't add to it. Impossible — this statement must be false. (The other three can each be arranged.)

Answer: The numbers of Old and Middle-aged visitors buying Economy tickets were equal (Choice D)

Each of the 23 boxes is one product: its area is the product's revenue, and its centre gives its popularity and potential scores, which drop it into one of the four categories (using the score cut-offs in the table). Sort the boxes by category, noting which are already shaded (Alfa/Bravo/Charlie) and which are erased, along with each box's area.

For this question we don't even need to recover the erased shadings — a category's total revenue is just the sum of its box areas, shaded or not:

• Promising: $2 + 9 + 3 = 14$
• Blockbuster: $6 + 2 + 3 + 6 + 6 + 9 + 4 = 36$
• Doubtful: $9 + 4 + 1 + 6 + 1 + 6 + 2 = 29$
• No-hope: $2 + 4 + 4 + 3 + 1 + 1 = 15$

The biggest is Blockbuster at 36.

💡 Teacher tip: The question asks about category revenue, so add the areas within each category directly and skip the harder work of un-erasing every box.

Answer: Blockbuster (Choice B)

This one needs the erased shadings recovered. Set up each category with its boxes; the shaded ones are known and the eight facts pin down the erased ones.

Reconstruction, fact by fact:

1. Doubtful (erased boxes worth 9, 6, 1): Fact 4 says Charlie has none here and Alfa has exactly one more than Bravo. So the three erased boxes split as Alfa two, Bravo one, Charlie zero. Fact 5 says Bravo's Doubtful revenue beats Alfa's, so Bravo takes the big box worth 9, and Alfa takes the 6 and the 1.
2. Promising (erased 9, 3, 2): Fact 3 says all three companies have equal counts here — one box each.
3. No-hope (erased 3, 1, 1): Fact 2 says Charlie has more than Bravo but fewer than Alfa. The only split that works is Alfa 1, Charlie 2, Bravo 0 among these erased boxes. Fact 7 (Bravo's No-hope revenue equals Charlie's, which is 4 from Bravo's shaded box) forces Charlie's two erased boxes to total 4 — the 3 and one of the 1's.
4. Blockbuster (erased 9, 4): Fact 1 says Alfa and Bravo have equal counts. Alfa already has two shaded here and Bravo one, so the two erased boxes go one to Bravo and one to Charlie. Fact 6 (Charlie's Blockbuster revenue beats Bravo's) makes Bravo take the smaller box (4) and Charlie the bigger (9).
5. Promising split: Fact 8 (Alfa's grand total equals Charlie's) is the tie-breaker; the totals work out with Alfa's Promising box the 2, Charlie's the 9, and Bravo's the 3.

Now count Bravo's boxes in each category:

Sequence (No-hope, Doubtful, Promising, Blockbuster) = 1, 3, 1, 2.

Answer: 1, 3, 1, 2 (Choice A)

Using the recovered assignment (erased boxes pinned down by the eight facts), here are each company's revenues by category, in area units:

Test each statement:

• A. Alfa's Blockbuster (9) equals Charlie's Promising (9). True.
• B. Bravo's Blockbuster (10) greater than Alfa's Doubtful (12)? No — $10 < 12$. False.
• C. Bravo's No-hope (4) equals Charlie's No-hope (4). True.
• D. Total No-hope (15) less than total Doubtful (29). True.

The statement that is NOT correct is B.

Answer: Choice B

Once the erased boxes are assigned (Bravo takes the big Doubtful box worth 9 so its Doubtful revenue tops Alfa's, the smaller Blockbuster box worth 4 by Fact 6, and the Promising box worth 3 by Fact 8), just add up Bravo's areas:

Total $= 3 + 10 + 17 + 4 = 34$ area units. With the smallest box (1 area unit) worth Rs. 1 crore, Bravo's revenue is about Rs. 34 crore.

Answer: 34 (Choice C)

Seven candidates go into three rooms (101, 102, 103). First work out how many people each room holds, using the statements:

• Fatima found three people (including Akil) already inside when she entered — so her room has $3 + 1 = 4$ people.
• Erina was alone — her room has 1 person.
• That leaves $7 - 4 - 1 = 2$ for the third room, and Ganeshan says his room (102) had exactly two candidates. So Room 102 holds 2, and the 4-person and 1-person rooms are 101 and 103 in some order.

Which is which? Balaram says he was the third person into Room 101, so Room 101 has at least 3 people — it must be the 4-person room. Then Room 103 is the single-person room (Erina).

Now place people. Room 101 (4 people) is Fatima's room, Akil was already inside, and Balaram was third in. Chitra says she was the last into her room; she cannot be the fourth person of Room 101 because that fourth person is Fatima (Fatima entered when three were already there). So Chitra is not in 101 — she goes to Room 102 with Ganeshan. That leaves Divya as the fourth member of Room 101.

So Divya was in Room 101.

Answer: Definitely Room 101 (Choice A)

From the room reconstruction, Room 102 holds exactly two people: Ganeshan and Chitra.

Chitra says she was the last to enter her room, so in Room 102 she came second — after Ganeshan. That means when Ganeshan walked in, Chitra wasn't there yet, and she is the only other person in the room. So no one was already inside.

Answer: No one (Choice D)

Use the log together with the solved rooms:

Fatima is the fourth into Room 101, and she arrives at 7:40. So Balaram (3rd) and Divya (2nd) must both arrive before 7:40. Ganeshan enters Room 102 before Chitra, and Chitra is at 7:30 — so Ganeshan is also before 7:40. That's three people (Divya, Balaram, Ganeshan) filling the three unknown slots before 7:40: the second 7:10 person, 7:15, and 7:25.

That uses up every early unknown. The only name left unplaced is Erina, and the only empty slot is 7:45. So Erina arrived at 7:45.

Answer: 7:45 am (Choice A)

From the arrangement, Room 101 fills in the order Akil, Divya, Balaram, Fatima — so in arrival time Divya must come before Balaram. We already know Erina is the 7:45 arrival, which leaves the three slots 7:10 (the second person), 7:15 and 7:25 for Divya, Balaram and Ganeshan, with Divya before Balaram.

Now add the new condition: Ganeshan arrived before Divya. That forces the full order Ganeshan < Divya < Balaram across those three early slots: $\text{Ganeshan} = 7{:}10,\quad \text{Divya} = 7{:}15,\quad \text{Balaram} = 7{:}25.$

Any other arrangement either puts Divya after Balaram or Ganeshan after Divya, both not allowed.

So Balaram entered at 7:25.

Answer: 7:25 am (Choice A)

Grades carry points A 50, B 40, C 30, D 20, F 0. The four weights are 0.1, 0.2, 0.3, 0.4 in some order, one each for faculty (F), reputation (R), placement (P), infrastructure (I).

Turn the three ranking clues into inequalities:

1. High Q beats Best Ed. Comparing grade-points, this reduces to $2w_I>w_F+w_R$.
2. Best Ed beats Cosmopolitan, which reduces to $w_R>w_I$.
3. Education Aid beats A-one, which reduces to $w_I>w_P$.

So $w_R>w_I>w_P$, and $2w_I>w_F+w_R$. The only way to fit $\{0.1,0.2,0.3,0.4\}$ to that is

$w_F=0.1,\quad w_R=0.4,\quad w_P=0.2,\quad w_I=0.3$

(check: $0.4>0.3>0.2$, and $2\times0.3=0.6>0.1+0.4=0.5$). So Score $=0.1F+0.4R+0.2P+0.3I$.

This question — the weight of faculty quality. From the forced solution, $w_F=0.1$.

💡 Teacher tip: Convert each ranking clue into an inequality first — together they leave exactly one legal weight order, so you never test all 24 permutations.

Answer: 0.1 (Choice D)

Grades carry points A 50, B 40, C 30, D 20, F 0. The four weights are 0.1, 0.2, 0.3, 0.4 in some order, one each for faculty (F), reputation (R), placement (P), infrastructure (I).

Turn the three ranking clues into inequalities:

1. High Q beats Best Ed. Comparing grade-points, this reduces to $2w_I>w_F+w_R$.
2. Best Ed beats Cosmopolitan, which reduces to $w_R>w_I$.
3. Education Aid beats A-one, which reduces to $w_I>w_P$.

So $w_R>w_I>w_P$, and $2w_I>w_F+w_R$. The only way to fit $\{0.1,0.2,0.3,0.4\}$ to that is

$w_F=0.1,\quad w_R=0.4,\quad w_P=0.2,\quad w_I=0.3$

(check: $0.4>0.3>0.2$, and $2\times0.3=0.6>0.1+0.4=0.5$). So Score $=0.1F+0.4R+0.2P+0.3I$.

This question — how many colleges get AAA? AAA needs a score of at least 45. From the table those are A-one (47), Education Aid (48) and Fancy (45) — three colleges.

Answer: 3

Grades carry points A 50, B 40, C 30, D 20, F 0. The four weights are 0.1, 0.2, 0.3, 0.4 in some order, one each for faculty (F), reputation (R), placement (P), infrastructure (I).

Turn the three ranking clues into inequalities:

1. High Q beats Best Ed. Comparing grade-points, this reduces to $2w_I>w_F+w_R$.
2. Best Ed beats Cosmopolitan, which reduces to $w_R>w_I$.
3. Education Aid beats A-one, which reduces to $w_I>w_P$.

So $w_R>w_I>w_P$, and $2w_I>w_F+w_R$. The only way to fit $\{0.1,0.2,0.3,0.4\}$ to that is

$w_F=0.1,\quad w_R=0.4,\quad w_P=0.2,\quad w_I=0.3$

(check: $0.4>0.3>0.2$, and $2\times0.3=0.6>0.1+0.4=0.5$). So Score $=0.1F+0.4R+0.2P+0.3I$.

This question — the highest overall score. Reading the score column, the maximum is 48 (Education Aid).

Answer: 48

Grades carry points A 50, B 40, C 30, D 20, F 0. The four weights are 0.1, 0.2, 0.3, 0.4 in some order, one each for faculty (F), reputation (R), placement (P), infrastructure (I).

Turn the three ranking clues into inequalities:

1. High Q beats Best Ed. Comparing grade-points, this reduces to $2w_I>w_F+w_R$.
2. Best Ed beats Cosmopolitan, which reduces to $w_R>w_I$.
3. Education Aid beats A-one, which reduces to $w_I>w_P$.

So $w_R>w_I>w_P$, and $2w_I>w_F+w_R$. The only way to fit $\{0.1,0.2,0.3,0.4\}$ to that is

$w_F=0.1,\quad w_R=0.4,\quad w_P=0.2,\quad w_I=0.3$

(check: $0.4>0.3>0.2$, and $2\times0.3=0.6>0.1+0.4=0.5$). So Score $=0.1F+0.4R+0.2P+0.3I$.

This question — how many colleges score between 31 and 40 inclusive? The scores are 47, 26, 25, 27, 48, 45, 13, 27. None of them lands in the 31–40 band, so the count is 0.

Answer: 0 (Choice A)

This is a classic "crack the substitution" set. Each word's letters become digits, and then those digits are sorted into non-decreasing order before being written. So for each coded word we know the multiset of digits its letters use — just not the order.

First line up every word with its given code:

Now peel the letters off one at a time, always starting from the shortest words:

1. "is" = {3, 5}, so $\{i, s\} = \{3, 5\}$.
2. "as" = {5, 6}. Since $s$ is 3 or 5, and 5 is the shared value, $s = 5$; then $a = 6$, and back in "is", $i = 3$.
3. "India" = {1, 3, 3, 6, 6}. With $i = 3$ (twice) and $a = 6$, the two letters $n, d$ must supply the leftover $\{1, 6\}$.
4. "bird" = {1, 3, 3, 4}. With $i = 3$, the letters $b, r, d$ supply $\{1, 3, 4\}$. Since $d$ has to be 1 or 6, it must be $d = 1$. That forces $n = 6$, and leaves $\{b, r\} = \{3, 4\}$.
5. "designated" = {1,1,3,5,5,5,6,6,7,8}. Remove the known letters $d, d, s, i, n, a$ (that is $1, 1, 5, 3, 6, 6$) and $\{5, 5, 7, 8\}$ remains for $e, e, g, t$. So $e = 5$, and $\{g, t\} = \{7, 8\}$.
6. "the" = {4, 5, 8}. With $e = 5$ we get $\{t, h\} = \{4, 8\}$; since $t$ is 7 or 8, $t = 8$, $h = 4$, and therefore $g = 7$.
7. "national" = {1,3,6,6,6,6,8,9}. Remove $n, n, a, a, t, i$ and the letters $o, l$ supply $\{1, 9\}$.
8. "of" = {7, 9}, so $o$ is 7 or 9; combined with $\{1, 9\}$ it must be $o = 9$, giving $l = 1$ and $f = 7$.
9. "Peacock" = {5,6,8,8,9,9,9}. With $e = 5, a = 6, o = 9$ removed, $\{8, 8, 9, 9\}$ is left for $P, c, c, k$. If $c$ were 8 then $P$ and $k$ would both be 9 — three letters on digit 9, but 9 codes only two. So $c = 9$, and $P = k = 8$.

Putting it together:

The question asks for L, pinned down in step 8: $l = 1$.

Answer: 1 (Choice A)

Working through the nine coded words (matching each word to its sorted digit-code and peeling letters off from the shortest words first) fixes almost every letter: $s=5,\ a=6,\ i=3,\ d=1,\ n=6,\ e=5,\ t=8,\ h=4,\ g=7,\ o=9,\ l=1,\ f=7,\ c=9,\ P=k=8.$

The only pair we cannot separate is $b$ and $r$. They come from "bird" = {1, 3, 3, 4}: with $i = 3$ and $d = 1$ already known, $b$ and $r$ take the remaining $\{3, 4\}$ — but nothing in the data tells us which is which.

So $b$ is either 3 or 4.

Answer: 3 or 4 (Choice A)

After cracking the code, the digit → letter table looks like this. Remember each of digits 1–8 codes three letters and digit 9 codes two; a blank means that slot's letter never appears in the sentence, so we can't name it.

A digit's list is "completely identified" only when every slot is filled by a known letter. Only two digits reach that bar: digit 8 (t, P, k — all three) and digit 9 (o, c — both). Every other digit still has at least one unknown letter.

So the count is 2.

Answer: 2 (Choice B)

We use the cracked code. The capacity rule is the key: each of digits 1–8 holds exactly three letters, digit 9 holds exactly two, and here is how many slots are already taken by known letters:

For a set of letters to share one digit, that digit must have enough free room for the new (so-far-unknown) letters in the set. Test each option:

• S, E, Z: S and E are both 5; digit 5 has one open slot, so Z could also be 5. Possible.
• I, B, M: I is 3; B can be 3 (it is 3-or-4), and M could take digit 3's open slot. Possible.
• S, U, V: S is 5, so all three would have to be 5 — but digit 5 has only one free slot while U and V are two new letters. Two into one won't fit. Impossible.
• X, Y, Z: none appears in the sentence, and digit 2 has three empty slots, so all three can sit on 2. Possible.

The set that cannot share a digit is S, U, V.

Answer: Choice C

Fun Sports has 39 students spread across three sports, and every student does at least one. The clean way to handle a three-set problem is to name the seven regions of the Venn diagram and turn each clue into an equation.

Turning the clues into equations:

1. Only L is double all-three: $c = 2g$.
2. Total in G is 17: $a + d + e + g = 17$.
3. Only G is one less than only L: $a = c - 1 = 2g - 1$.
4. Only K equals (K and L): $b = f + g$.
5. L has the largest enrolment.
6. Ten of the G students do a second sport: $d + e + g = 10$.

From clue 6, $d+e+g=10$, and since $|G| = a + (d+e+g) = a + 10 = 17$, we get $a = 7$. Then clue 3 gives $2g - 1 = 7$, so $g = 4$ and $c = 2g = 8$. Also $d + e = 10 - g = 6$.

Now use the total of 39: $7 + b + 8 + 6 + f + 4 = 39$, so $b + f = 14$. With clue 4 ($b = f + 4$): $f = 5$, $b = 9$.

Every region is now fixed except how the 6 in $d+e$ split:

Finally use clue 5. Here $|K| = 18 + d$ and $|L| = 23 - d$; for L to stay the largest we need $23 - d \ge 18 + d$, i.e. $d \le 2$. So $d \in \{0, 1, 2\}$.

The question wants the minimum of "G and L but not K", which is region $e = 6 - d$. To make it smallest, take the largest allowed $d = 2$: $e_{\min} = 6 - 2 = 4.$

Answer: 4

Use the same solved diagram. From the clues, every region is fixed except the split of the 6 students across the two G-and-L overlaps, controlled by one number $d$:

This gives the two group totals $|K| = 9 + d + 5 + 4 = 18 + d$ and $|L| = 8 + (6-d) + 5 + 4 = 23 - d$.

We're told $|K|: |L| = 19: 22$: $\frac{18+d}{23-d} = \frac{19}{22} \Rightarrow 22(18+d) = 19(23-d).$ $396 + 22d = 437 - 19d \Rightarrow 41d = 41 \Rightarrow d = 1.$

So $|L| = 23 - 1 = 22$.

Answer: 22 (Choice C)