CAT 2023 DILR Questions — Slot 2

How each firm behaves. A firm raises Rs. 1 crore in its first and last year, climbs to a single peak, then falls back to 1. Every year-to-year change is $+1$, $+2$, $-1$ or $-2$ crore, and no two neighbouring years are equal. So from a firm's span (first to last year) and its known total we can reconstruct its year-by-year path.

• Drjbna (2011–2015, total 10): the only mountain that fits is $1 \to 2 \to 4 \to 2 \to 1$. (Fully fixed.)
• Czechy (2013–2017, total 9): only $1 \to 2 \to 3 \to 2 \to 1$. (Fully fixed.)
• Bzygoo (2012–2015, four years): total forced to 7, but the middle can be $1,2,3,1$ or $1,3,2,1$. (Not unique.)
• Alfloo (2009–2016, total 21): two paths that differ only at the peak — 2012/2013 is $4,5$ or $5,4$; both give 2014:3, 2015:2, 2016:1. (Not unique.)
• Elavalaki (2010–?, total 13): three possible paths — $1,3,5,3,1$ (ends 2014); $1,2,3,4,2,1$; or $1,2,4,3,2,1$ (both end 2015). (Not unique.)

Year-by-year table (bold = fixed; a slash lists the alternatives):

This question — which firms are fully pinned down every year?

A firm is certain only if exactly one valid schedule exists. Drjbna ($1,2,4,2,1$) and Czechy ($1,2,3,2,1$) each have a single path. Bzygoo, Alfloo and Elavalaki each allow two or more. So only Czechy and Drjbna are fully determined.

Answer: Only Czechy and Drjbna

Recall the reconstructed paths (each firm rises to one peak then falls, steps of 1 or 2, starting and ending at 1):

This question — total raised in 2015

Add the 2015 column. Fixed values: Alfloo 2, Bzygoo 1 (its last year), Czechy 3, Drjbna 1 (its last year) — that is 7. Elavalaki is either closed in 2014 (contributes 0) or still running in 2015 (contributes 1). So the 2015 total is $7$ or $8$.

Answer: It is either Rs. 7 crores or Rs. 8 crores.

Recall the reconstructed paths:

This question — largest possible 2013 total

Take each firm's biggest feasible 2013 value on its own: Alfloo 5, Bzygoo 3, Czechy 1, Drjbna 4, Elavalaki 4. Sum $=5+3+1+4+4=17$.

Answer: 17

Recall the reconstructed paths:

This question — smallest 2012 total, given Elavalaki raised 3 in 2013

Elavalaki raising 3 in 2013 keeps the paths $1,3,5,3,1$ or $1,2,4,3,2,1$, whose 2012 values are 5 or 4 — take the smaller, 4. Now make every other 2012 value as small as possible: Alfloo 4 (its lower peak-year option), Bzygoo 1 (first year), Czechy 0 (not founded until 2013), Drjbna 2. Sum $=4+1+0+2+4=11$.

Answer: 11

Recall the reconstructed paths:

This question — if the 2014 total is 12, what is impossible?

The 2014 column has Alfloo 3, Czechy 2, Drjbna 2 (that is 7), plus Bzygoo (2 or 3) and Elavalaki (1 or 2). To reach 12 we need Bzygoo + Elavalaki $=5$, so Bzygoo $=3$ and Elavalaki $=2$. Bzygoo $=3$ in 2014 fixes its path to $1,2,3,1$, so its 2013 value is 2. Elavalaki $=2$ in 2014 means its 2013 value is 3 or 4.

Now test the options: Alfloo can equal Drjbna in 2013 (both 4) — possible; Alfloo can equal Bzygoo in 2014 (both 3) — possible; Bzygoo (3) can exceed Elavalaki (2) in 2014 — possible. But Bzygoo's 2013 is 2 while Elavalaki's is 3 or 4 — they can never be equal.

Answer: Bzygoo raised the same amount of money as Elavalaki in 2013.

The big idea first. Each box holds three sacks of 1–9 coins, and each box's average is a whole number, so each box total is a multiple of 3. The nine averages are distinct integers, and the only nine distinct integers available (1–9) that also let all totals work out are exactly $1,2,\dots,9$. So the grand total is $3(1+2+\cdots+9)=135$, every row and every column totals 45, and the nine averages themselves form a grid where each row and column of averages adds to 15.

Reading the two tables box by box (Table 1 gives some medians; Table 2 gives how many sacks exceed 5, with $*$ = exactly one of {min $=1$, median $=1$, max $=9$} and $**$ = two or more):

1. (1,2) median 9 → $\{a,9,9\}$; two sacks $>5$ means $a\le5$; whole-number average forces $a=3$, so $\{3,9,9\}$, avg 7.
2. (2,1) median 2 with $**$ → needs two of the conditions, only min $=1$ and max $=9$ fit → $\{1,2,9\}$, avg 4.
3. (3,1) median 8, all three $>5$, $*$ → max $=9$ → $\{7,8,9\}$, avg 8.
4. Column-1 averages add to 15 → (1,1) avg 3; Row-1 averages → (1,3) avg 5.
5. Filling the rest by the same median/condition logic gives every box.

The completed grid (each cell shows the three sacks and their average):

Every row and column totals 45, the nine averages are exactly 1–9, and all medians and $*$/$**$ markers check out — so the grid is unique.

This question — total coins in the 3rd row

All row totals are equal and the grand total is 135, so each row is $135/3=45$. Reading Row 3 directly: $24+18+3=45$.

Answer: 45

Recall the fully solved grid (forced by every box average being a whole number, so the nine averages are exactly 1–9 and every row/column totals 45):

This question — boxes with at least one sack of 9 coins

Scan for a 9: (1,2) {3,9,9}, (2,1) {1,2,9}, (2,3) {9,9,9}, (3,1) {7,8,9}, (3,2) {1,8,9}. That is 5 boxes; the other four have no 9.

Answer: 5

Recall the fully solved grid:

This question — boxes where the average equals the median

Compare each box's middle (sorted) value with its average:

The other five boxes differ. So 4 boxes have average equal to median.

Answer: 4

Recall the fully solved grid:

This question — how many sacks hold exactly one coin

Count the 1's: (1,1) {1,1,7} → 2, (1,3) {1,6,8} → 1, (2,1) {1,2,9} → 1, (2,2) {1,2,3} → 1, (3,2) {1,8,9} → 1, (3,3) {1,1,1} → 3. Total $=2+1+1+1+1+3=9$.

Answer: 9

Recall the fully solved grid:

This question — boxes whose three sacks are all different

Look for boxes with no repeated value: (1,3) {1,6,8}, (2,1) {1,2,9}, (2,2) {1,2,3}, (3,1) {7,8,9}, (3,2) {1,8,9}. That is 5 boxes; the other four each repeat a value.

Answer: 5

Getting the daily totals. Let $T_d$ be a day's combined score. The 2-day averages give $T_1+T_2=30$, $T_2+T_3=31$, $T_3+T_4=32$, $T_4+T_5=34$. Since Day 3 and Day 4 totals are equal, $2T_3=32$, so $T_3=T_4=16$; then $T_1=15$, $T_2=15$, $T_5=18$.

Filling the scores using the ranks (rank 1 = highest that day, ties share the best rank), the clue that Chatur only scores multiples of 3, and the special clues:

1. Day 3 (Chatur highest, Akhil = Bimal, total 16): $2A+C=16$ with $C$ a multiple of 3 → $C=6$, $A=B=5$.
2. Bimal's Day 1 equals his Day 3, so Bimal Day 1 = 5. Day 1 (Akhil > Bimal > Chatur, total 15) → Chatur 3, Akhil 7.
3. Chatur's minimum (3) happens only on Day 1 and equals Akhil's Day 4, so Akhil Day 4 = 3. Day 4 (Bimal > Chatur > Akhil, total 16) → Chatur 6, Bimal 7.
4. Day 2 (Chatur > Akhil > Bimal, total 15): Chatur's Day-2 score is the unique highest in the whole event, so Chatur 9; then Akhil + Bimal $=6$.
5. Day 5 (Bimal > Chatur > Akhil, total 18): Chatur is a multiple of 3, not 3 and not 9, so 6; then Akhil + Bimal $=12$.

The solved score grid (only the Akhil/Bimal splits on Days 2 and 5 stay open):

On Day 2, $(A,B)=(5,1)$ or $(4,2)$; on Day 5, $(A,B)=(5,7)$ or $(4,8)$.

This question — Akhil's Day 1 score

On Day 1 the order is Akhil > Bimal > Chatur, total 15. Bimal's Day 1 equals his Day 3 score (5), and Chatur is the lowest multiple of 3 below 5, i.e. 3. So Akhil $=15-5-3=7$.

Answer: 7

Recall the solved score grid (daily totals come from the 2-day averages; the rest from the ranks and Chatur's multiples-of-3 rule):

This question — who has the highest total score?

Chatur's total is fixed at $3+9+6+6+6=30$. Akhil can reach at most $15+5+5=25$ and Bimal at most $17+2+8=27$ — both below 30 in every case. So Chatur always wins.

Answer: Chatur

Recall the solved score grid:

This question — Bimal's smallest possible total

Bimal's total is $5+B_2+5+7+B_5 = 17+B_2+B_5$, with $B_2\in\{1,2\}$ and $B_5\in\{7,8\}$. The minimum is $17+1+7=25$.

Answer: 25

Recall the solved score grid:

This question — Akhil's Day 2 score if Bimal's total is a multiple of 3

Bimal's total is $17+B_2+B_5$ with $B_2\in\{1,2\}$, $B_5\in\{7,8\}$, giving 25, 26, 26 or 27. Only 27 is a multiple of 3, which forces $B_2=2$ and $B_5=8$. With Bimal scoring 2 on Day 2 and the Day-2 pair adding to 6, Akhil $=6-2=4$.

Answer: 4

Recall the solved score grid:

This question — Bimal's total if Akhil scores 24

Akhil's total is $15+A_2+A_5$ with $A_2,A_5\in\{4,5\}$. To reach 24 we need $A_2+A_5=9$, i.e. $(5,4)$ or $(4,5)$. Each pairs with a Bimal split of $(1,8)$ or $(2,7)$, both adding to 9, so Bimal's total $=17+9=26$.

Answer: 26

We first rebuild the full ride timetable, then simply total Bipasha's tickets.

Each ride lasts one hour and holds one visitor at a time. Let us reconstruct everyone's day.

1. Chitra never waits and finishes by 11 am after spending ₹50. Since Anjali waits 30 minutes for Chitra to finish Ride-1, Chitra is on Ride-1 from 10–11. Her only other ride, taken with no wait, must be Ride-3 from 9–10 (₹30 + ₹20 = ₹50).
2. Anjali waits 30 minutes and takes Ride-1 at 11–12. To also have spent ₹50 by 12:15 she buys Ride-3 next, so Ride-3 12–1. She then takes Ride-2 (1–2) and Ride-4 (2–3).
3. Bipasha begins her first of three rides at 11:30 with Ride-2 (₹50) — that one ticket makes her total ₹50 by 12:15, matching the others. She then takes Ride-1 (12:30–1:30), a 1-hour coffee break, and finally Ride-4 (3–4), waiting 30 minutes for Anjali to finish it first.

Solved timetable (built from all four clues):

Ticket prices: Ride-1 = ₹20, Ride-2 = ₹50, Ride-3 = ₹30, Ride-4 = ₹40.

This question — Bipasha's total spend. Bipasha rides Ride-2, Ride-1 and Ride-4:

$50 + 20 + 40 = 110$

Answer: C — ₹110

The key is to fix the exact finishing time of each of Anjali's rides.

Each ride lasts one hour and holds one visitor at a time. Let us reconstruct everyone's day.

1. Chitra never waits and finishes by 11 am after spending ₹50. Since Anjali waits 30 minutes for Chitra to finish Ride-1, Chitra is on Ride-1 from 10–11. Her only other ride, taken with no wait, must be Ride-3 from 9–10 (₹30 + ₹20 = ₹50).
2. Anjali waits 30 minutes and takes Ride-1 at 11–12. To also have spent ₹50 by 12:15 she buys Ride-3 next, so Ride-3 12–1. She then takes Ride-2 (1–2) and Ride-4 (2–3).
3. Bipasha begins her first of three rides at 11:30 with Ride-2 (₹50) — that one ticket makes her total ₹50 by 12:15, matching the others. She then takes Ride-1 (12:30–1:30), a 1-hour coffee break, and finally Ride-4 (3–4), waiting 30 minutes for Anjali to finish it first.

Solved timetable (built from all four clues):

Ticket prices: Ride-1 = ₹20, Ride-2 = ₹50, Ride-3 = ₹30, Ride-4 = ₹40.

This question — what Anjali had completed by 2:00 pm. Anjali finishes Ride-1 at 12:00, Ride-3 at 1:00 and Ride-2 at 2:00. Her Ride-4 runs 2–3, so it is not yet complete at 2 pm.

So by 2:00 pm she has completed Ride-1, Ride-2 and Ride-3.

Answer: B — Ride-1, Ride-2, and Ride-3

We look for the one ride that appears in all three visitors' schedules.

Each ride lasts one hour and holds one visitor at a time. Let us reconstruct everyone's day.

1. Chitra never waits and finishes by 11 am after spending ₹50. Since Anjali waits 30 minutes for Chitra to finish Ride-1, Chitra is on Ride-1 from 10–11. Her only other ride, taken with no wait, must be Ride-3 from 9–10 (₹30 + ₹20 = ₹50).
2. Anjali waits 30 minutes and takes Ride-1 at 11–12. To also have spent ₹50 by 12:15 she buys Ride-3 next, so Ride-3 12–1. She then takes Ride-2 (1–2) and Ride-4 (2–3).
3. Bipasha begins her first of three rides at 11:30 with Ride-2 (₹50) — that one ticket makes her total ₹50 by 12:15, matching the others. She then takes Ride-1 (12:30–1:30), a 1-hour coffee break, and finally Ride-4 (3–4), waiting 30 minutes for Anjali to finish it first.

Solved timetable (built from all four clues):

Ticket prices: Ride-1 = ₹20, Ride-2 = ₹50, Ride-3 = ₹30, Ride-4 = ₹40.

This question — the ride taken by all three. Ride-1 is taken by Chitra (10–11), Anjali (11–12) and Bipasha (12:30–1:30). No other ride is common to all three.

💡 Teacher tip: Because only one visitor can be on a ride at a time, a shared ride shows up as three different one-hour slots — exactly what we see for Ride-1.

Answer: A — Ride-1

We just count how many rides Anjali and Chitra each took.

Each ride lasts one hour and holds one visitor at a time. Let us reconstruct everyone's day.

1. Chitra never waits and finishes by 11 am after spending ₹50. Since Anjali waits 30 minutes for Chitra to finish Ride-1, Chitra is on Ride-1 from 10–11. Her only other ride, taken with no wait, must be Ride-3 from 9–10 (₹30 + ₹20 = ₹50).
2. Anjali waits 30 minutes and takes Ride-1 at 11–12. To also have spent ₹50 by 12:15 she buys Ride-3 next, so Ride-3 12–1. She then takes Ride-2 (1–2) and Ride-4 (2–3).
3. Bipasha begins her first of three rides at 11:30 with Ride-2 (₹50) — that one ticket makes her total ₹50 by 12:15, matching the others. She then takes Ride-1 (12:30–1:30), a 1-hour coffee break, and finally Ride-4 (3–4), waiting 30 minutes for Anjali to finish it first.

Solved timetable (built from all four clues):

Ticket prices: Ride-1 = ₹20, Ride-2 = ₹50, Ride-3 = ₹30, Ride-4 = ₹40.

This question — total rides by Anjali and Chitra. Anjali takes all four rides (Ride-1, Ride-2, Ride-3, Ride-4) and Chitra takes two (Ride-3, Ride-1):

$4 + 2 = 6$

Answer: 6

We total the ticket prices of the four rides Anjali took.

Each ride lasts one hour and holds one visitor at a time. Let us reconstruct everyone's day.

1. Chitra never waits and finishes by 11 am after spending ₹50. Since Anjali waits 30 minutes for Chitra to finish Ride-1, Chitra is on Ride-1 from 10–11. Her only other ride, taken with no wait, must be Ride-3 from 9–10 (₹30 + ₹20 = ₹50).
2. Anjali waits 30 minutes and takes Ride-1 at 11–12. To also have spent ₹50 by 12:15 she buys Ride-3 next, so Ride-3 12–1. She then takes Ride-2 (1–2) and Ride-4 (2–3).
3. Bipasha begins her first of three rides at 11:30 with Ride-2 (₹50) — that one ticket makes her total ₹50 by 12:15, matching the others. She then takes Ride-1 (12:30–1:30), a 1-hour coffee break, and finally Ride-4 (3–4), waiting 30 minutes for Anjali to finish it first.

Solved timetable (built from all four clues):

Ticket prices: Ride-1 = ₹20, Ride-2 = ₹50, Ride-3 = ₹30, Ride-4 = ₹40.

This question — Anjali's total spend. Anjali rides Ride-1, Ride-3, Ride-2 and Ride-4:

$20 + 30 + 50 + 40 = 140$

Answer: ₹140