CAT 2021 DILR Questions — Slot 1

Let us build the complete arrangement from the clues, then read off the answer.

Step 1 — Split the 18 papers into two groups

The total papers by A, D, G, H is double the total by B, C, E, F. If the smaller group (B+C+E+F) totals $x$, then the larger group (A+D+G+H) totals $2x$, and $x + 2x = 18$, so $x = 6$.

• A + D + G + H = 12
• B + C + E + F = 6

Since every author writes at most 3 papers, four authors summing to 12 must each write exactly 3: A = D = G = H = 3.

Step 2 — Assign countries

Four authors are Indian, two Japanese, two Chinese.

• A is Automation (clue 2). Since no Japanese/Chinese author is in Automation (clue 3), A is Indian.
• C and E are the two Chinese authors (clue 5).
• D is Japanese (clue 8).
• A and H are from different countries (clue 4). A is Indian, and H cannot be Chinese (C and E already are), so H is Japanese.

Thus: India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.

Step 3 — Assign areas

Automation has 2 authors, Logistics has 4, so Manufacturing has $8 - 4 - 2 = 2$.

• Indians are never in Manufacturing; Japanese/Chinese are never in Automation (clue 3). So both Automation authors are Indian, and both Manufacturing authors are Japanese/Chinese.
• A is Automation. Among Indians, F and B are Logistics (clues 1 and 6), so the second Automation author is G.
• Manufacturing (2 authors, from Japan/China) contains D. C and E are in different areas (clue 5), so exactly one of them is Manufacturing. Clue 9 says C and H are in different areas; H is Logistics, so C is Manufacturing and E is Logistics.

Step 4 — Determine the smaller-group counts

F = 1 (clue 1). So B + C + E = 5. Clue 5 says C and E wrote the same number of papers, and E's papers are in consecutive issues, so E must have at least 2. The only split of 5 with C = E and E ≥ 2 is C = E = 2, B = 1.

This question

Reading the paper counts directly from the table:

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

Let us build the complete arrangement from the clues, then sum the Indian authors' papers.

Step 1 — Split the 18 papers into two groups

The total papers by A, D, G, H is double the total by B, C, E, F. If the smaller group totals $x$, then $x + 2x = 18$, so $x = 6$.

• A + D + G + H = 12
• B + C + E + F = 6

Since each author writes at most 3 papers, four authors summing to 12 forces A = D = G = H = 3.

Step 2 — Assign countries

Four authors are Indian, two Japanese, two Chinese.

• A is Automation (clue 2). No Japanese/Chinese author is in Automation (clue 3), so A is Indian.
• C and E are the two Chinese authors (clue 5).
• D is Japanese (clue 8).
• A and H are from different countries (clue 4). A is Indian and H cannot be Chinese, so H is Japanese.

Thus: India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.

Step 3 — Assign areas and counts

• Automation has 2 authors (both Indian): A and G.
• Manufacturing has 2 authors (from Japan/China): C and D.
• Logistics has 4 authors: B, E, F, H.

For the smaller group: F = 1 (clue 1), and B + C + E = 5 with C = E (clue 5) and E ≥ 2 (consecutive issues), giving C = E = 2, B = 1.

Complete solved table

This question

The Indian authors are A, B, F, and G. Adding their papers:

$3 + 1 + 1 + 3 = 8$

Answer: 8

Let us build the complete arrangement, then test each statement against the issue-by-issue schedule.

Step 1 — Split the 18 papers into two groups

A+D+G+H is double B+C+E+F. If the smaller group totals $x$, then $3x = 18$, so $x = 6$. Thus A+D+G+H = 12 and B+C+E+F = 6. Since each author writes at most 3, we get A = D = G = H = 3.

Step 2 — Countries and areas

• A is Automation → Indian (no Japanese/Chinese in Automation). C and E are Chinese (clue 5). D is Japanese (clue 8). A and H are different countries (clue 4) → H is Japanese.
• India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.
• Automation (2, both Indian): A, G. Manufacturing (2, from Japan/China): C, D. Logistics (4): B, E, F, H.
• F = 1; B + C + E = 5 with C = E and E ≥ 2 → C = E = 2, B = 1.

Step 3 — Assign issues

• A skips October (clue 2), has 3 papers → Jan, Apr, Jul. H copies A (clue 4).
• D skips July (clue 8) → Jan, Apr, Oct.
• B is in April only (clue 6, since B = 1).
• G (3 papers) shares no issue with B (clue 7) → Jan, Jul, Oct.
• C and E split the remaining slots: E must be consecutive, C must not → E = Apr, Jul and C = Jan, Oct.

Complete solved tables

This question

Statement A — Every issue had at least one paper from each country (India, Japan, China).

Every issue has at least one Indian, one Japanese, and one Chinese paper. Statement A is true.

Statement B — Every issue had at most two papers from each area.

Look at April: the authors are A (Automation), B (Logistics), D (Manufacturing), E (Logistics), H (Logistics). That is three Logistics papers — B, E, and H. This exceeds two, so Statement B is false.

Answer: Only Statement A (Option C)

Let us build the complete arrangement, then test each option against the issue-by-issue schedule.

Step 1 — Split the 18 papers into two groups

A+D+G+H is double B+C+E+F. If the smaller group totals $x$, then $3x = 18$, so $x = 6$. Thus A+D+G+H = 12 and B+C+E+F = 6. Since each author writes at most 3, we get A = D = G = H = 3.

Step 2 — Countries and areas

• A is Automation → Indian. C and E are Chinese (clue 5). D is Japanese (clue 8). A and H are different countries (clue 4) → H is Japanese.
• India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.
• Automation (2): A, G. Manufacturing (2): C, D. Logistics (4): B, E, F, H.
• F = 1; B + C + E = 5 with C = E and E ≥ 2 → C = E = 2, B = 1.

Step 3 — Assign issues

• A skips October → Jan, Apr, Jul. H copies A.
• D skips July → Jan, Apr, Oct.
• B is in April only.
• G shares no issue with B → Jan, Jul, Oct.
• C and E split remaining: E consecutive, C not → E = Apr, Jul, C = Jan, Oct.

Complete solved tables

This question — test each option

Option A: Every issue had at least one Automation paper. Automation authors are A (Jan, Apr, Jul) and G (Jan, Jul, Oct). Together they cover all four issues. True.

Option B: Every issue had exactly one Chinese paper. Chinese authors: C (Jan, Oct) and E (Apr, Jul) — exactly one per issue. True.

Option C: Every issue had exactly two Logistics papers. Logistics authors are B, E, F, H. Counting per issue:

• January: only H → 1 Logistics paper.
• April: B, E, H → 3 Logistics papers.
• July: E, H → 2.
• October: F → 1.

January and October each have only one Logistics paper, and April has three. So this statement is false.

Option D: Every issue had exactly two Indian papers. Indian authors: A, B, F, G.

• January: A, G → 2.
• April: A, B → 2.
• July: A, G → 2.
• October: F, G → 2.

True.

Answer: Every issue had exactly two papers by authors from Logistics area. (Option C)

Let us build the complete arrangement, then test each option against the issue-by-issue schedule.

Step 1 — Split the 18 papers into two groups

A+D+G+H is double B+C+E+F. If the smaller group totals $x$, then $3x = 18$, so $x = 6$. Thus A+D+G+H = 12 and B+C+E+F = 6. Since each author writes at most 3, we get A = D = G = H = 3.

Step 2 — Countries and areas

• A is Automation → Indian. C and E are Chinese (clue 5). D is Japanese (clue 8). A and H are different countries (clue 4) → H is Japanese.
• India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.
• Automation (2): A, G. Manufacturing (2): C, D. Logistics (4): B, E, F, H.
• F = 1; B + C + E = 5 with C = E and E ≥ 2 → C = E = 2, B = 1.

Step 3 — Assign issues

• A skips October → Jan, Apr, Jul. H copies A.
• D skips July → Jan, Apr, Oct.
• B is in April only.
• G shares no issue with B → Jan, Jul, Oct.
• C and E split remaining: E consecutive, C not → E = Apr, Jul, C = Jan, Oct.

Complete solved tables

This question — test each option

The Manufacturing authors are C (Jan, Oct) and D (Jan, Apr, Oct). Counting Manufacturing papers per issue:

Now check each statement:

• Option A: Exactly two Manufacturing papers in January → C and D both appear. True.
• Option B: Exactly one Manufacturing paper in April → only D. True.
• Option C: Exactly one Logistics paper in October → Logistics authors in October: only F. True.
• Option D: Exactly two Manufacturing papers in July → July has zero Manufacturing papers (neither C nor D appears). False.

Answer: There were exactly two papers by authors from Manufacturing area in the July issue. (Option D)

Let us build the complete arrangement, then sum the paper counts by area.

Step 1 — Split the 18 papers into two groups

A+D+G+H is double B+C+E+F. If the smaller group totals $x$, then $3x = 18$, so $x = 6$. Thus A+D+G+H = 12 and B+C+E+F = 6. Since each author writes at most 3, we get A = D = G = H = 3.

Step 2 — Countries and areas

• A is Automation → Indian. C and E are Chinese (clue 5). D is Japanese (clue 8). A and H are different countries (clue 4) → H is Japanese.
• India = {A, B, F, G}, Japan = {D, H}, China = {C, E}.
• Automation (2): A, G. Manufacturing (2): C, D. Logistics (4): B, E, F, H.
• F = 1; B + C + E = 5 with C = E and E ≥ 2 → C = E = 2, B = 1.

Complete solved table

This question — sum papers by area

Verification: $6 + 5 + 7 = 18$ ✓

Answer: 6, 5, 7 (Option A)

What we need

We must find the total 2019 sales across every subcategory that belongs to the office supplies category, then pick the closest option.

The office supplies category consists of the nine middle subcategories: Binders, Art, Appliances, Envelopes, Fasteners, Labels, Paper, Storage, and Supplies. For each, the 2019 value is read from the vertical tick line on the bar chart.

Step 1 — Read each 2019 office-supply value

Step 2 — Add them up

$3.60 + 0.45 + 1.85 + 0.35 + 0.05 + 0.20 + 1.50 + 4.30 + 1.05 = 13.35$

Step 3 — Match to the options

The options are 18.0, 16.5, 13.5, and 12.5. Our computed total of 13.35 is closest to 13.5.

Answer: 13.5 (Option C)

What we need

We need the percentage increase in total Furniture sales from 2019 to 2020, to the closest option.

Furniture consists of four subcategories: Bookcases, Chairs, Furnishings, and Tables.

Step 1 — Read the Furniture values

Step 2 — Compute the increase

$\text{Increase} = 15.80 - 14.55 = 1.25 \text{ ₹ million}$

Step 3 — Percentage increase

$\frac{1.25}{14.55} \times 100 \approx 8.6\%$

💡 Teacher tip: When the base is large and the increase is small, the percentage stays small — here roughly $1.25/15 \approx 1/12 \approx 8\%$, a handy sanity check before computing precisely.

Step 4 — Match to the options

The options are 20%, 8%, 25%, and 1%. Our value of about 8.6% is closest to 8%.

Answer: 8% (Option B)

What we need

We must count subcategories that satisfy both conditions:

1. 2019 sales of ₹4 million or more, and
2. a 2020 increase exceeding 25% over 2019.

Step 1 — Identify subcategories with 2019 sales ≥ ₹4 million

Scanning the 2019 tick values across all 17 subcategories:

These five clear the ₹4 million bar.

Step 2 — Check the 2020 growth for each

💡 Teacher tip: Machines is a trap — its 2019 value clears ₹4 million, but its 2020 bar is shorter than the 2019 tick, so it cannot satisfy the growth condition.

Step 3 — Count

Only Phones satisfies both conditions.

Answer: 1

What we need

The improvement index of a category is defined as the maximum percentage increase in sales (2019 → 2020) among any of its subcategories. We must order the three categories by this index from smallest to largest.

Step 1 — Find the best-growing subcategory in each category

Furniture subcategories:

Best = Bookcases ≈ 15.8%.

Technology subcategories:

Best = Accessories ≈ 44.3%.

Office supply subcategories:

Best = Fasteners = 100%.

Step 2 — Compare the three indices

💡 Teacher tip: A tiny base value (like Fasteners at 0.05) can produce a huge percentage jump from a small absolute increase — so the "improvement index" rewards categories that contain at least one small-base, fast-growing subcategory.

Step 3 — Write in increasing order

Furniture (16%) < Technology (44%) < Office supply (100%).

Answer: furniture, technology, office supply (Option A)

Finding Gowtham's acquaintances

We need to build the full relationship grid first, then read off Gowtham's acquaintances.

Step 1 — Fill the blanks using the row-sum rule.

Each person has exactly 6 relationships (one with each of the other six people), so Friends + Acquaintances + Strangers = 6.

• Amudha: Friends $= 6 - 1 - 4 = 1$
• Gowtham: Friends $= 6 - 3 - 2 = 1$

Step 2 — Identify each person's single friend.

Clue 2 tells us Amudha, Dhinesh, and Fani are Ezhil's friends. Since Amudha has exactly 1 friend, that friend must be Ezhil. Since Fani has exactly 1 friend, that friend must also be Ezhil. Clue 3 tells us Chandran and Gowtham are friends; since Gowtham has exactly 1 friend, it is Chandran.

Step 3 — Use the Amudha–Bharatan linkage (clues 4 and 5).

Clue 4 says Amudha's friends = Bharatan's acquaintances. Amudha's only friend is Ezhil, so Bharatan's only acquaintance is Ezhil.

Clue 5 says Bharatan's friends = Amudha's acquaintances. Amudha has exactly 1 acquaintance, so Bharatan has exactly 1 friend — the same person.

That person is an acquaintance of Amudha and a friend of Bharatan. It cannot be Ezhil (already Amudha's friend), Chandran (stranger to Amudha by clue 1), Fani (whose only friend is Ezhil), or Gowtham (whose only friend is Chandran). So it is Dhinesh.

Step 4 — Pin down Gowtham's relationships.

Amudha's 4 strangers are Bharatan, Chandran, Fani, and Gowtham (her only non-strangers are Ezhil and Dhinesh). So Amudha–Gowtham = stranger. Similarly, Bharatan's 4 strangers are Amudha, Chandran, Fani, and Gowtham, so Bharatan–Gowtham = stranger.

Gowtham has exactly 2 strangers (Amudha and Bharatan) and 1 friend (Chandran). His remaining 3 relationships — with Dhinesh, Ezhil, and Fani — must all be acquaintances.

The complete relationship grid (F = friend, A = acquaintance, S = stranger):

Reading Gowtham's row: friend = Chandran, strangers = Amudha & Bharatan, acquaintances = Dhinesh, Ezhil, and Fani.

Answer: A. Dhinesh, Ezhil and Fani

Which pair of pairs shares the same relationship type?

We use the fully solved relationship grid (F = friend, A = acquaintance, S = stranger):

Now check each option by reading the two relationships off the grid:

• Option A: (Amudha, Gowtham) = stranger; (Ezhil, Fani) = friend → different.
• Option B: (Bharatan, Chandran) = stranger; (Dhinesh, Ezhil) = friend → different.
• Option C: (Chandran, Ezhil) = stranger; (Dhinesh, Gowtham) = acquaintance → different.
• Option D: (Bharatan, Ezhil) = acquaintance; (Fani, Gowtham) = acquaintance → same!

Answer: D. (Bharatan, Ezhil) and (Fani, Gowtham)

Who is an acquaintance of Amudha?

From the given table, Amudha has exactly 1 acquaintance and 4 strangers, so she also has $6 - 1 - 4 = 1$ friend.

Identifying Amudha's friend: Clue 2 says Amudha, Dhinesh, and Fani are Ezhil's friends. Since Amudha has exactly one friend, that friend is Ezhil.

Using the Amudha–Bharatan linkage:

Clue 5 states: every acquaintance of Amudha is a friend of Bharatan, and every friend of Bharatan is an acquaintance of Amudha. This means Amudha's acquaintances and Bharatan's friends are the same set of people.

So we need to find Bharatan's single friend. Clue 4 tells us Bharatan's acquaintances = Amudha's friends = {Ezhil}, so Bharatan's only acquaintance is Ezhil. Since Amudha has exactly 1 acquaintance, Bharatan has exactly 1 friend.

Who is Bharatan's friend? It must be someone who is an acquaintance of Amudha. It cannot be:

• Ezhil (already Amudha's friend, not acquaintance)
• Chandran (stranger to Amudha by clue 1)
• Fani (whose only friend is Ezhil)
• Gowtham (whose only friend is Chandran)

That leaves Dhinesh — he is an acquaintance of Amudha and a friend of Bharatan.

💡 Teacher tip: Clues 4 and 5 create a two-way mirror between Amudha and Bharatan: her friends ↔ his acquaintances, and her acquaintances ↔ his friends. Once you know one person's counts, you immediately know the other's.

Answer: A. Dhinesh

Who is an acquaintance of Chandran?

From the given table, Chandran has exactly 1 acquaintance. Let us determine who it is by elimination.

What we already know about Chandran's relationships:

1. Clue 1: Amudha, Bharatan, and Chandran are mutual strangers → Chandran is a stranger to Amudha and Bharatan.
2. Clue 3: Chandran and Gowtham are friends → Gowtham is Chandran's friend.
3. Dhinesh has 2 strangers. From the grid, Dhinesh's friends are Bharatan and Ezhil, and his acquaintances are Amudha and Gowtham. That accounts for 4 of his 6 relationships; the remaining 2 (Chandran and Fani) must be his strangers. So Chandran–Dhinesh = stranger.

So Chandran is a stranger to Amudha, Bharatan, and Dhinesh, and a friend of Gowtham. That uses 4 of his 6 relationships. His remaining two are with Ezhil and Fani — one is his acquaintance, the other is a stranger.

Resolving Ezhil vs Fani:

Ezhil's friends are exactly Amudha, Dhinesh, and Fani (clue 2). So Ezhil is not a friend of Chandran. We also know Ezhil's acquaintances are Bharatan and Gowtham (from the full grid). That leaves Chandran as Ezhil's one stranger. So Chandran–Ezhil = stranger.

By elimination, Chandran–Fani = acquaintance.

The relevant portion of the solved grid:

Answer: B. Fani

How many friends does Ezhil have?

This one follows directly from the clues.

Clue 2 explicitly states: Amudha, Dhinesh, and Fani are Ezhil's friends.

Since all relationships are mutual, Ezhil is a friend of Amudha, Dhinesh, and Fani — giving Ezhil at least 3 friends.

Now we confirm Ezhil has exactly 3 friends by checking the full grid:

Reading Ezhil's row: friends = Amudha, Dhinesh, Fani (3); acquaintances = Bharatan, Gowtham (2); stranger = Chandran (1). The total is $3 + 2 + 1 = 6$ ✓.

No additional friends are possible — every other relationship is accounted for as an acquaintance or stranger.

Answer: 3

How many people are either a friend or a friend-of-a-friend of Ezhil?

We need to find everyone who is either a direct friend of Ezhil, or a friend of one of Ezhil's friends (excluding Ezhil herself).

Step 1 — Identify Ezhil's direct friends.

From the solved grid:

Ezhil's friends are: Amudha, Dhinesh, Fani (3 people).

Step 2 — Find the friends of each of Ezhil's friends.

Reading each row from the full grid:

• Amudha's only friend is Ezhil (already counted).
• Dhinesh's friends are Bharatan and Ezhil. Bharatan is new.
• Fani's only friend is Ezhil (already counted).

Step 3 — Combine the sets.

• Direct friends of Ezhil: {Amudha, Dhinesh, Fani}
• Friends-of-friends (new): {Bharatan}

Total = {Amudha, Dhinesh, Fani, Bharatan} = 4 people.

💡 Teacher tip: "Friend-of-a-friend" means a person reachable by following exactly two friend-edges in the friendship network. Draw the friendship links as a graph and count everyone within distance 2 of Ezhil (excluding Ezhil herself).

Answer: 4

Setting up the selling prices

First, we work out the selling price (revenue) of each finished product. The selling price equals the cost of raw materials + profit, where profit = 2 × (number of raw materials used).

Leftover raw materials are sold in the last hour at a ₹1 loss per unit: milk → ₹4, mango → ₹2, apple → ₹1 (banana leftovers are not allowed by Fact 4).

Decoding Hour 1 revenues

In Hours 1–3, revenue comes only from finished products (Fact 2: at most one unit of any product per person per hour). Also, only Ganga may sell a Mango smoothie (Fact 1).

Ganga — Hour 1 = ₹23: We need two product prices that sum to 23. The only pair is 12 + 11 = 23, i.e., Apple smoothie (₹11) plus a ₹12 item. The ₹12 item could be Fruit salad or Mango smoothie (only Ganga can sell the latter). So Ganga sells 0 or 1 Fruit salad in Hour 1.

Kaveri — Hour 1 = ₹19: No two-product combination sums to 19 (10+11=21, 10+12=22, 11+12=23, 19+anything > 19). But a single Mixed fruit smoothie = ₹19. So Kaveri sells 0 Fruit salad in Hour 1.

Narmada — Hour 1 = ₹31: We need two products summing to 31: 19 + 12 = 31. The ₹19 item is Mixed fruit smoothie; the ₹12 item must be Fruit salad (Narmada cannot sell Mango smoothie). So Narmada sells 1 Fruit salad in Hour 1.

Counting Fruit salads in Hour 1

Total = 1 + (0 or 1) = 1 or 2.

Answer: Either 1 or 2.

Selling prices (quick reference)

Leftover sale prices (₹1 loss per unit): milk → ₹4, mango → ₹2, apple → ₹1. No banana leftovers (Fact 4). Each woman has exactly two different leftover materials (Fact 3), so the possible leftover pairs and their values are:

Decoding Hour 4 (the last hour)

Hour 4 revenue = finished products sold + leftover sales. We subtract each possible leftover total from each woman's Hour 4 revenue and check if the remainder can be formed from product prices.

Ganga — Hour 4 = ₹30:

• Leftover ₹6 → finished = 24 = 12 + 12 = Mango smoothie + Fruit salad ✓
• Leftover ₹5 → finished = 25 → no combination of product prices gives 25 ✗
• Leftover ₹3 → finished = 27 → no combination gives 27 ✗

So Ganga's leftovers are milk + mango (₹6).

Narmada — Hour 4 = ₹22:

• Leftover ₹6 → finished = 16 → impossible ✗
• Leftover ₹5 → finished = 17 → impossible ✗
• Leftover ₹3 → finished = 19 = Mixed fruit smoothie ✓

So Narmada's leftovers are mango + apple (₹3).

Kaveri — Hour 4 = ₹27:

• Leftover ₹6 → finished = 21 = 11 + 10 = Apple smoothie + Banana smoothie ✓
• Leftover ₹5 → finished = 22 = 12 + 10 = Fruit salad + Banana smoothie ✓
• Leftover ₹3 → finished = 24 = 12 + 12, but both ₹12 items would need to be Fruit salad (Kaveri can't sell Mango smoothie), and selling the same product twice in one hour is not allowed ✗

So Kaveri's leftovers are milk + mango or milk + apple — not uniquely determined.

Summary of leftovers

Checking each option

• A. Ganga did not sell any leftover mangoes → Ganga's leftovers include mango → False.
• B. Ganga did not sell any leftover apples → Ganga's leftovers are milk + mango, with no apple → Necessarily true ✓
• C. Narmada sold one unit of leftover milk → Narmada's leftovers are mango + apple, no milk → False.
• D. Kaveri sold one unit of leftover mangoes → Kaveri might have milk + apple instead → Not necessarily true.

Answer: Ganga did not sell any leftover apples.

Selling prices (quick reference)

Leftover sale prices: milk → ₹4, mango → ₹2, apple → ₹1. Possible leftover pairs: milk+mango (₹6), milk+apple (₹5), mango+apple (₹3).

Full decoded schedule

Using the same decoding process (matching each hour's revenue to product-price combinations, with only Ganga allowed to sell Mango smoothie), the full schedule is:

Counting milk units

Milk used = (units of every smoothie sold, since all smoothies use milk) + (milk leftover, if any). Fruit salad does not use milk.

Narmada:

• Smoothies: Mixed(H1) + Apple+Banana(H2) + Apple(H3) + Mixed(H4) = 1 + 2 + 1 + 1 = 5 milk used
• Leftover: mango + apple → 0 milk leftover
• Total milk = 5

Ganga:

• Smoothies: Apple(H1) + possibly Mango(H1) + Apple+Banana(H2) + Banana+Mixed(H3) + Mango(H4) = 1 + (0 or 1) + 2 + 2 + 1 = 6 or 7 milk used
• Leftover: milk + mango → 1 milk leftover
• Total milk = 7 or 8

Kaveri:

• Smoothies: Mixed(H1) + Banana(H2) + Apple+Mixed(H3) + H4(2 smoothies in Case 1, or 1 smoothie in Case 2) = 1 + 1 + 2 + (2 or 1) = 6 or 5 milk used
• Leftover: milk + mango or milk + apple → 1 milk leftover (both cases)
• Total milk = 7 or 6

Combining all three

$\text{Total milk} = 5 + (7 \text{ or } 8) + (6 \text{ or } 7)$

The possible values are:

• 5 + 7 + 6 = 18
• 5 + 8 + 6 = 19
• 5 + 7 + 7 = 19
• 5 + 8 + 7 = 20

So the total ranges over 18, 19, or 20.

Answer: Either 18 or 19 or 20 units.

Selling prices (quick reference)

Leftover sale prices: milk → ₹4, mango → ₹2, apple → ₹1. Possible leftover pairs: milk+mango (₹6), milk+apple (₹5), mango+apple (₹3).

Full decoded schedule

Using the new condition: 3 leftover mangoes

From the decoded schedule, the leftover mangoes come from:

Ganga and Narmada together contribute 2 leftover mangoes. For the total to be 3, Kaveri must also have a mango leftover — this forces Case 1 for Kaveri.

So Kaveri's Hour 4 = Apple smoothie + Banana smoothie (with leftovers milk + mango).

Counting apple smoothies sold during the day

With Kaveri in Case 1, let us count apple smoothies for each woman across all four hours:

$\text{Total apple smoothies} = 2 + 2 + 2 = 6$

Answer: 6