CAT 2022 DILR Questions — Slot 2

The fully solved table (households met / items sold / success rate):

How the table is built. Let $T$ be the two-day total of households met — the same for all three (given). Success rate $=$ items $\div$ households.

1. Lahur meets the same number and sells the same number on both days, so each day he sells $50$ items to $T/2$ households; his rate is $\dfrac{50}{T/2}=\dfrac{100}{T}$.
2. Hokli was stopped by a complaint from the very first household on day 2, so day 2 = $1$ household, $0$ items. That leaves $T-1$ households and $100$ items on day 1.
3. Tohri met $30$ more on day 2 than day 1, and the two add to $T$, so day 1 $=\dfrac{T-30}{2}$, day 2 $=\dfrac{T+30}{2}$.
4. Tohri's day-1 rate is twice Lahur's: $\dfrac{200}{T}$, giving day-1 items $=\dfrac{T-30}{2}\cdot\dfrac{200}{T}=\dfrac{100(T-30)}{T}$. His day-2 rate is $75\%$ of Lahur's: $\dfrac{75}{T}$, giving day-2 items $=\dfrac{T+30}{2}\cdot\dfrac{75}{T}=\dfrac{37.5(T+30)}{T}$.
5. Tohri sold $100$ in all: $\frac{100(T-30)+37.5(T+30)}{T}=100 \;\Rightarrow\; 137.5T-1875=100T \;\Rightarrow\; T=50.$

Putting $T=50$ back gives every number in the table above.

This question asks for the households met on day 1 by all three together. Read the day-1 column: $\underbrace{10}_{\text{Tohri}}+\underbrace{49}_{\text{Hokli}}+\underbrace{25}_{\text{Lahur}}=84.$

Answer: 84

The fully solved table (households met / items sold / success rate):

How the table is built. Let $T$ be the two-day total of households met — the same for all three (given). Success rate $=$ items $\div$ households.

1. Lahur meets the same number and sells the same number on both days, so each day he sells $50$ items to $T/2$ households; his rate is $\dfrac{50}{T/2}=\dfrac{100}{T}$.
2. Hokli was stopped by a complaint from the very first household on day 2, so day 2 = $1$ household, $0$ items. That leaves $T-1$ households and $100$ items on day 1.
3. Tohri met $30$ more on day 2 than day 1, and the two add to $T$, so day 1 $=\dfrac{T-30}{2}$, day 2 $=\dfrac{T+30}{2}$.
4. Tohri's day-1 rate is twice Lahur's: $\dfrac{200}{T}$, giving day-1 items $=\dfrac{T-30}{2}\cdot\dfrac{200}{T}=\dfrac{100(T-30)}{T}$. His day-2 rate is $75\%$ of Lahur's: $\dfrac{75}{T}$, giving day-2 items $=\dfrac{T+30}{2}\cdot\dfrac{75}{T}=\dfrac{37.5(T+30)}{T}$.
5. Tohri sold $100$ in all: $\frac{100(T-30)+37.5(T+30)}{T}=100 \;\Rightarrow\; 137.5T-1875=100T \;\Rightarrow\; T=50.$

Putting $T=50$ back gives every number in the table above.

This question asks how many items Tohri sold on day 1. Straight from the table, that is $\mathbf{40}$ items.

💡 Teacher tip: A quick sanity check — Tohri's day-1 rate is $40/10 = 4.0$, which is exactly twice Lahur's $2.0$, matching the clue.

Answer: 40

The fully solved table (households met / items sold / success rate):

How the table is built. Let $T$ be the two-day total of households met — the same for all three (given). Success rate $=$ items $\div$ households.

1. Lahur meets the same number and sells the same number on both days, so each day he sells $50$ items to $T/2$ households; his rate is $\dfrac{50}{T/2}=\dfrac{100}{T}$.
2. Hokli was stopped by a complaint from the very first household on day 2, so day 2 = $1$ household, $0$ items. That leaves $T-1$ households and $100$ items on day 1.
3. Tohri met $30$ more on day 2 than day 1, and the two add to $T$, so day 1 $=\dfrac{T-30}{2}$, day 2 $=\dfrac{T+30}{2}$.
4. Tohri's day-1 rate is twice Lahur's: $\dfrac{200}{T}$, giving day-1 items $=\dfrac{T-30}{2}\cdot\dfrac{200}{T}=\dfrac{100(T-30)}{T}$. His day-2 rate is $75\%$ of Lahur's: $\dfrac{75}{T}$, giving day-2 items $=\dfrac{T+30}{2}\cdot\dfrac{75}{T}=\dfrac{37.5(T+30)}{T}$.
5. Tohri sold $100$ in all: $\frac{100(T-30)+37.5(T+30)}{T}=100 \;\Rightarrow\; 137.5T-1875=100T \;\Rightarrow\; T=50.$

Putting $T=50$ back gives every number in the table above.

This question asks how many households Lahur met on day 2. Lahur met the same number both days, and that number is $25$. Since $25$ falls in the band $21$–$29$:

Answer: between 21 and 29

The fully solved table (households met / items sold / success rate):

How the table is built. Let $T$ be the two-day total of households met — the same for all three (given). Success rate $=$ items $\div$ households.

1. Lahur meets the same number and sells the same number on both days, so each day he sells $50$ items to $T/2$ households; his rate is $\dfrac{50}{T/2}=\dfrac{100}{T}$.
2. Hokli was stopped by a complaint from the very first household on day 2, so day 2 = $1$ household, $0$ items. That leaves $T-1$ households and $100$ items on day 1.
3. Tohri met $30$ more on day 2 than day 1, and the two add to $T$, so day 1 $=\dfrac{T-30}{2}$, day 2 $=\dfrac{T+30}{2}$.
4. Tohri's day-1 rate is twice Lahur's: $\dfrac{200}{T}$, giving day-1 items $=\dfrac{T-30}{2}\cdot\dfrac{200}{T}=\dfrac{100(T-30)}{T}$. His day-2 rate is $75\%$ of Lahur's: $\dfrac{75}{T}$, giving day-2 items $=\dfrac{T+30}{2}\cdot\dfrac{75}{T}=\dfrac{37.5(T+30)}{T}$.
5. Tohri sold $100$ in all: $\frac{100(T-30)+37.5(T+30)}{T}=100 \;\Rightarrow\; 137.5T-1875=100T \;\Rightarrow\; T=50.$

Putting $T=50$ back gives every number in the table above.

This question asks how many households Tohri met on day 1. From the table that is $10$, which falls in the band "10 or less."

Answer: 10 or less

The fully solved table (households met / items sold / success rate):

How the table is built. Let $T$ be the two-day total of households met — the same for all three (given). Success rate $=$ items $\div$ households.

1. Lahur meets the same number and sells the same number on both days, so each day he sells $50$ items to $T/2$ households; his rate is $\dfrac{50}{T/2}=\dfrac{100}{T}$.
2. Hokli was stopped by a complaint from the very first household on day 2, so day 2 = $1$ household, $0$ items. That leaves $T-1$ households and $100$ items on day 1.
3. Tohri met $30$ more on day 2 than day 1, and the two add to $T$, so day 1 $=\dfrac{T-30}{2}$, day 2 $=\dfrac{T+30}{2}$.
4. Tohri's day-1 rate is twice Lahur's: $\dfrac{200}{T}$, giving day-1 items $=\dfrac{T-30}{2}\cdot\dfrac{200}{T}=\dfrac{100(T-30)}{T}$. His day-2 rate is $75\%$ of Lahur's: $\dfrac{75}{T}$, giving day-2 items $=\dfrac{T+30}{2}\cdot\dfrac{75}{T}=\dfrac{37.5(T+30)}{T}$.
5. Tohri sold $100$ in all: $\frac{100(T-30)+37.5(T+30)}{T}=100 \;\Rightarrow\; 137.5T-1875=100T \;\Rightarrow\; T=50.$

Putting $T=50$ back gives every number in the table above.

This question asks which statement is FALSE. Line up the day-wise success rates:

Check each option:

• "Tohri had the highest success rate on day 2." On day 2 the rates are Tohri $1.5$, Hokli $0$, Lahur $2.0$ — Lahur is highest, not Tohri. This statement is FALSE.
• "Tohri's day-1 rate ($4.0$) beats his day-2 rate ($1.5$)" — true.
• "Tohri had the highest rate on day 1 ($4.0$)" — true.
• "Lahur had the lowest rate on day 1 ($2.0$)" — true.

Answer: Among the three, Tohri had the highest success rate on the second day.

Build one master grid. There are $320$ products, split two ways: cosmetic vs nutrition, and domestic vs foreign. Every product has FDA approval, EU approval, or both.

Step 1 — the four base groups. Domestic $=$ foreign $=160$ each. Nutrition $=140$, half of them ($70$) foreign, so $70$ nutrition are domestic; that leaves cosmetic $=180$, of which $160-70=90$ are foreign and $90$ domestic.

Step 2 — FDA approvals (total $200$; $70$ foreign, $120$ cosmetic; domestic FDA $=200-70=130$):

• Cosmetic & Domestic & FDA $=80$ (half of the $160$ domestic products are FDA-approved cosmetics).
• Cosmetic & Foreign & FDA $=120-80=40$.
• Nutrition & Foreign & FDA $=70-40=30$.
• Nutrition & Domestic & FDA $=130-80=50$.

Step 3 — both approvals. Only domestic products can have both; $60$ domestic do, no foreign do. Call the domestic cosmetic ones with both $=x$; then domestic nutrition with both $=60-x$. Since every product has at least one approval, in each cell: EU-only $=$ total $-$ FDA. This completes the grid:

Because $x-10\ge0$ and $60-x\ge0$, we must have $10\le x\le60$.

This question: foreign products that are FDA-approved cosmetics $=$ the Cosmetic & Foreign & FDA cell. Foreign products never have both approvals, so this is simply $\mathbf{40}$.

Answer: 40

Build one master grid. There are $320$ products, split two ways: cosmetic vs nutrition, and domestic vs foreign. Every product has FDA approval, EU approval, or both.

Step 1 — the four base groups. Domestic $=$ foreign $=160$ each. Nutrition $=140$, half of them ($70$) foreign, so $70$ nutrition are domestic; that leaves cosmetic $=180$, of which $160-70=90$ are foreign and $90$ domestic.

Step 2 — FDA approvals (total $200$; $70$ foreign, $120$ cosmetic; domestic FDA $=200-70=130$):

• Cosmetic & Domestic & FDA $=80$ (half of the $160$ domestic products are FDA-approved cosmetics).
• Cosmetic & Foreign & FDA $=120-80=40$.
• Nutrition & Foreign & FDA $=70-40=30$.
• Nutrition & Domestic & FDA $=130-80=50$.

Step 3 — both approvals. Only domestic products can have both; $60$ domestic do, no foreign do. Call the domestic cosmetic ones with both $=x$; then domestic nutrition with both $=60-x$. Since every product has at least one approval, in each cell: EU-only $=$ total $-$ FDA. This completes the grid:

Because $x-10\ge0$ and $60-x\ge0$, we must have $10\le x\le60$.

This question: cosmetics without FDA approval. Total cosmetics $=180$; cosmetics with FDA $=120$. So without FDA $=180-120=\mathbf{60}$ (these are the EU-only cosmetics: $50$ foreign $+\,10$ domestic).

Answer: 60

Build one master grid. There are $320$ products, split two ways: cosmetic vs nutrition, and domestic vs foreign. Every product has FDA approval, EU approval, or both.

Step 1 — the four base groups. Domestic $=$ foreign $=160$ each. Nutrition $=140$, half of them ($70$) foreign, so $70$ nutrition are domestic; that leaves cosmetic $=180$, of which $160-70=90$ are foreign and $90$ domestic.

Step 2 — FDA approvals (total $200$; $70$ foreign, $120$ cosmetic; domestic FDA $=200-70=130$):

• Cosmetic & Domestic & FDA $=80$ (half of the $160$ domestic products are FDA-approved cosmetics).
• Cosmetic & Foreign & FDA $=120-80=40$.
• Nutrition & Foreign & FDA $=70-40=30$.
• Nutrition & Domestic & FDA $=130-80=50$.

Step 3 — both approvals. Only domestic products can have both; $60$ domestic do, no foreign do. Call the domestic cosmetic ones with both $=x$; then domestic nutrition with both $=60-x$. Since every product has at least one approval, in each cell: EU-only $=$ total $-$ FDA. This completes the grid:

Because $x-10\ge0$ and $60-x\ge0$, we must have $10\le x\le60$.

This question: domestic cosmetics with both approvals $=x$. The only limits come from keeping neighbouring cells non-negative: $x-10\ge0$ gives $x\ge10$, and $60-x\ge0$ gives $x\le60$. So $x$ is at least 10 and at most 60.

Answer: At least 10 and at most 60

Build one master grid. There are $320$ products, split two ways: cosmetic vs nutrition, and domestic vs foreign. Every product has FDA approval, EU approval, or both.

Step 1 — the four base groups. Domestic $=$ foreign $=160$ each. Nutrition $=140$, half of them ($70$) foreign, so $70$ nutrition are domestic; that leaves cosmetic $=180$, of which $160-70=90$ are foreign and $90$ domestic.

Step 2 — FDA approvals (total $200$; $70$ foreign, $120$ cosmetic; domestic FDA $=200-70=130$):

• Cosmetic & Domestic & FDA $=80$ (half of the $160$ domestic products are FDA-approved cosmetics).
• Cosmetic & Foreign & FDA $=120-80=40$.
• Nutrition & Foreign & FDA $=70-40=30$.
• Nutrition & Domestic & FDA $=130-80=50$.

Step 3 — both approvals. Only domestic products can have both; $60$ domestic do, no foreign do. Call the domestic cosmetic ones with both $=x$; then domestic nutrition with both $=60-x$. Since every product has at least one approval, in each cell: EU-only $=$ total $-$ FDA. This completes the grid:

Because $x-10\ge0$ and $60-x\ge0$, we must have $10\le x\le60$.

This question pins down $x$ using a new fact: $70$ cosmetics had no EU approval. Cosmetics with EU $=$ EU-only $(50+10)$ $+$ both $(x)=60+x$, so cosmetics without EU $=180-(60+x)=120-x$. Setting $120-x=70$ gives $x=50$.

Then nutrition products with both approvals $=60-x=60-50=\mathbf{10}$.

Answer: 10

Build one master grid. There are $320$ products, split two ways: cosmetic vs nutrition, and domestic vs foreign. Every product has FDA approval, EU approval, or both.

Step 1 — the four base groups. Domestic $=$ foreign $=160$ each. Nutrition $=140$, half of them ($70$) foreign, so $70$ nutrition are domestic; that leaves cosmetic $=180$, of which $160-70=90$ are foreign and $90$ domestic.

Step 2 — FDA approvals (total $200$; $70$ foreign, $120$ cosmetic; domestic FDA $=200-70=130$):

• Cosmetic & Domestic & FDA $=80$ (half of the $160$ domestic products are FDA-approved cosmetics).
• Cosmetic & Foreign & FDA $=120-80=40$.
• Nutrition & Foreign & FDA $=70-40=30$.
• Nutrition & Domestic & FDA $=130-80=50$.

Step 3 — both approvals. Only domestic products can have both; $60$ domestic do, no foreign do. Call the domestic cosmetic ones with both $=x$; then domestic nutrition with both $=60-x$. Since every product has at least one approval, in each cell: EU-only $=$ total $-$ FDA. This completes the grid:

Because $x-10\ge0$ and $60-x\ge0$, we must have $10\le x\le60$.

This question fixes $x$ from a different fact: $50$ nutrition products had no EU approval, so nutrition with EU $=140-50=90$. Nutrition with EU $=$ EU-only $(40+20)$ $+$ both $(60-x)=120-x$. Setting $120-x=90$ gives $x=30$.

Domestic cosmetics without EU $=$ (domestic cosmetics) $-$ (those with EU) $=90-(\underbrace{10}_{\text{EU-only}}+\underbrace{30}_{\text{both}})=90-40=\mathbf{50}$.

Answer: 50

First turn the two plots into two working tables.

Profit $=$ Revenue $-$ Cost (Rs. crores):

Employees lost in a year $=$ (strength at start) $+$ (new hires) $-$ (strength at start of next year):

This question: the single highest annual profit. Each company's best year is A $=30$, B $=50$, C $=70$, D $=22$... reading the profit table, the largest number anywhere is Company C in 2021 at $70$ crores.

Answer: Company C

First turn the two plots into two working tables.

Profit $=$ Revenue $-$ Cost (Rs. crores):

Employees lost in a year $=$ (strength at start) $+$ (new hires) $-$ (strength at start of next year):

This question: the highest annual loss. Only one cell in the profit table is negative — Company D in 2020, at $-35$ crores. Every other company stayed non-negative in every year, so D holds the biggest loss.

Answer: Company D

First turn the two plots into two working tables.

Profit $=$ Revenue $-$ Cost (Rs. crores):

Employees lost in a year $=$ (strength at start) $+$ (new hires) $-$ (strength at start of next year):

This question: lowest profit-to-cost ratio in 2019.

The smallest is Company A at $0.125$.

Answer: Company A

First turn the two plots into two working tables.

Profit $=$ Revenue $-$ Cost (Rs. crores):

Employees lost in a year $=$ (strength at start) $+$ (new hires) $-$ (strength at start of next year):

This question: who lost the fewest employees across 2019 and 2020 combined? Add the two "employees lost" columns:

The least is Company B at $40$.

Answer: Company B

First turn the two plots into two working tables.

Profit $=$ Revenue $-$ Cost (Rs. crores):

Employees lost in a year $=$ (strength at start) $+$ (new hires) $-$ (strength at start of next year):

This question: highest profit per employee in 2020, where "employees" for 2020 is the average of the strength at the start of 2020 and the start of 2021.

The highest is Company B at about $0.204$.

Answer: Company B

Note: This question was found to be incorrect and was withdrawn by CAT. It is included here only because it appeared in the original paper; it does not have a valid, unambiguous answer as printed.

For the record, CAT's official answer key marked the answer to this question as 35.

Answer (per CAT's official key): 35

Shortest distance between any two places (use a direct road, or go via the warehouse $W$, whichever is shorter):

There is no direct road $A$–$C$ or $B$–$D$, so those legs go through $W$: $A$–$C=5+12=17$, $B$–$D=10+2=12$. Where both a direct road and a via-$W$ path exist, take the smaller — e.g. $A$–$D$ is $8$ direct but $7$ via $W$, so we use $7$.

The routing rule: start at $W$, visit the four places in decreasing order of demand; if two places are tied on demand, go to the one closest to where you are now; each leg is travelled by the shortest path above.

Demands (with probabilities): A $=50\,(40\%)$ or $70\,(60\%)$; B $=40\,(30\%)$ or $60\,(70\%)$; C $=70\,(30\%)$ or $100\,(70\%)$; D $=30\,(40\%)$ or $50\,(60\%)$.

Every one of the 16 possible demand-days, worked out:

The one tie-break that matters most: when A and C are both $70$ (top demand), A is nearer the warehouse ($5<12$), so A is visited first.

This question: total widgets $=250$. Scanning the last column, only combination #16 gives $70+60+70+50=250$. Here A and C tie at $70$, and A is nearer $W$, so the order is $W\to A\to C\to B\to D$: $5+17+4+12=\mathbf{38}\text{ km}.$

Answer: 38 km

Shortest distance between any two places (use a direct road, or go via the warehouse $W$, whichever is shorter):

There is no direct road $A$–$C$ or $B$–$D$, so those legs go through $W$: $A$–$C=5+12=17$, $B$–$D=10+2=12$. Where both a direct road and a via-$W$ path exist, take the smaller — e.g. $A$–$D$ is $8$ direct but $7$ via $W$, so we use $7$.

The routing rule: start at $W$, visit the four places in decreasing order of demand; if two places are tied on demand, go to the one closest to where you are now; each leg is travelled by the shortest path above.

Demands (with probabilities): A $=50\,(40\%)$ or $70\,(60\%)$; B $=40\,(30\%)$ or $60\,(70\%)$; C $=70\,(30\%)$ or $100\,(70\%)$; D $=30\,(40\%)$ or $50\,(60\%)$.

Every one of the 16 possible demand-days, worked out:

The one tie-break that matters most: when A and C are both $70$ (top demand), A is nearer the warehouse ($5<12$), so A is visited first.

This question: we want the day to deliver exactly $260$ widgets and end at Bikrampore ($B$). Totalling $260$ happens in combinations #6, #10 and #13 — but of those only #10 ends at $B$. There A$=70$, B$=40$, C$=100$, D$=50$, so the order (by demand) is C, A, D, B and the route $W\to C\to A\to D\to B$ finishes at $B$.

Its probability is the product of the four independent demand chances: $0.60\times0.30\times0.70\times0.60=0.0756=\mathbf{7.56\%}.$

Answer: 7.56%

Shortest distance between any two places (use a direct road, or go via the warehouse $W$, whichever is shorter):

There is no direct road $A$–$C$ or $B$–$D$, so those legs go through $W$: $A$–$C=5+12=17$, $B$–$D=10+2=12$. Where both a direct road and a via-$W$ path exist, take the smaller — e.g. $A$–$D$ is $8$ direct but $7$ via $W$, so we use $7$.

The routing rule: start at $W$, visit the four places in decreasing order of demand; if two places are tied on demand, go to the one closest to where you are now; each leg is travelled by the shortest path above.

Demands (with probabilities): A $=50\,(40\%)$ or $70\,(60\%)$; B $=40\,(30\%)$ or $60\,(70\%)$; C $=70\,(30\%)$ or $100\,(70\%)$; D $=30\,(40\%)$ or $50\,(60\%)$.

Every one of the 16 possible demand-days, worked out:

The one tie-break that matters most: when A and C are both $70$ (top demand), A is nearer the warehouse ($5<12$), so A is visited first.

This question is conditional: given Ahmednagar is visited first, what is the chance the route is $40$ km?

Ahmednagar is first only when A ties C at the top demand — that is A$=70$ and C$=70$ (combinations #11, #12, #15, #16). So $P(\text{A first})=P(A=70)\times P(C=70)=0.60\times0.30=0.18.$

Among those four, the distance is $40$ km only in #12 (A$=70$, B$=40$, C$=70$, D$=50$), route $W\to A\to C\to D\to B=5+17+6+12=40$. Its chance is $0.60\times0.30\times0.30\times0.60=0.0324.$

The conditional probability is $\frac{0.0324}{0.18}=0.18=\mathbf{18\%}.$

💡 Teacher tip: "Given A is first" means we only compare inside that world — divide by $P(\text{A first})$, don't use the full $16$-day space.

Answer: 18%