CAT 2020 DILR Questions — Slot 1

Setting up the rules

A driver is eligible for bonus only if both conditions hold:

• No parameter rating is 1, and
• At least one parameter rating is 5.

So a driver gets no bonus if either he has a rating of 1 somewhere, or he has no rating of 5 at all.

Also recall Fact 2: no driver may have the same rating in three (or more) parameters — i.e., no value may appear 3 times.

For Damodaran, P2 = 3 is fixed, and the other four parameters are free (integers 1–5). We want the maximum final rating subject to “no bonus”.

Two routes to “no bonus”

Route 1 — No rating of 5 at all. Then the four free ratings come from {2, 3, 4} (we avoid 1 to keep the sum high, and avoid 5 to stay on this route). With no value allowed three times, the best set is {4, 4, 3, 2}:

Final rating = 16 / 5 = 3.2.

Route 2 — Include a 5, but also include a 1 (the 1 disqualifies the bonus). We can now use a 5 to push the sum up, while a single 1 keeps him bonus-ineligible. The best set for the four free slots is {5, 5, 4, 1} (three 5’s would violate Fact 2):

Check Fact 2: 5 appears twice, 3 appears once, 4 appears once, 1 appears once — no value three times. ✔

Final rating = 18 / 5 = 3.6.

Comparing

Route 2 (3.6) beats Route 1 (3.2), so the maximum possible final rating is 3.6.

💡 Teacher tip: When a condition can be failed in two different ways, check both ways — sometimes breaking the rule deliberately (here, inserting a 1) lets you reach a higher total than playing safe.

Answer: 3.6 (Choice C)

Setting up the rules

For Eman to get a bonus, both conditions must hold:

• No rating of 1 anywhere, and
• At least one rating of 5.

Also, Fact 2: no value may appear in three parameters.

For Eman, P5 = 2 is fixed. We want the minimum final rating while staying bonus-eligible.

Minimising the sum

All five ratings must be ≥ 2 (no 1’s allowed). To keep the sum low we would like to use 2’s, but P5 already gives us one 2, and Fact 2 lets us use at most two 2’s in total. So we may add only one more 2.

We are forced to include at least one 5 (bonus requirement). That leaves two slots, which we fill with the smallest available values, 3 and 3:

Check Fact 2: 2 appears twice, 3 appears twice, 5 appears once — no value three times. ✔

Final rating = 15 / 5 = 3.0.

Why we cannot go lower

• We cannot drop the 5 (bonus needs at least one 5).
• We cannot use three 2’s (Fact 2 forbids it), so the two non-5, non-2 slots must be at least 3 each.
• Hence 15 is the minimum possible sum.

💡 Teacher tip: For “minimum with constraints” problems, list the forced pieces first (here: one 5, at most two 2’s), then fill the remaining slots with the smallest legal values.

Answer: 3.0 (Choice D)

Setting up the rules

With all five drivers bonus-eligible, each driver must have:

• No rating of 1, and
• At least one rating of 5.

Plus the two facts:

• Fact 1: Arun and Barun have exactly one 5; Chandan has exactly two 5’s.
• Fact 2: No rating value appears in three parameters.

Monthly payment = Fixed Payment + (Bonus rate × Final Rating), where Final Rating = (sum of five ratings) / 5.

To find the minimum monthly payment any driver can receive, we minimise each driver’s final rating (subject to bonus eligibility), then pick the smallest payment.

Minimising each driver’s ratings

For each driver we keep the fixed parameter, add the required 5’s, and fill the rest with the smallest legal values (avoiding 1, and never repeating a value three times).

Arun — P4 = 4, exactly one 5. Fill the other three slots with 2, 2, 3:

• Ratings: 5, 2, 2, 4, 3 → sum = 16 → final = 3.2
• Payment = 1000 + 250 × 3.2 = 1800

Barun — P1 = 3, exactly one 5. Fill with 2, 2, 3:

• Ratings: 5, 2, 2, 3, 3 → sum = 15 → final = 3.0
• Payment = 1200 + 200 × 3.0 = 1800

Chandan — P3 = 2, exactly two 5’s. Fill the other two slots with 2, 3:

• Ratings: 5, 5, 2, 2, 3 → sum = 17 → final = 3.4
• Payment = 1400 + 100 × 3.4 = 1740

Damodaran — P2 = 3, at least one 5. Fill with 5, 2, 2, 3:

• Ratings: 5, 3, 2, 2, 3 → sum = 15 → final = 3.0
• Payment = 1300 + 150 × 3.0 = 1750

Eman — P5 = 2, at least one 5. Fill with 5, 2, 3, 3:

• Ratings: 5, 2, 3, 3, 2 → sum = 15 → final = 3.0
• Payment = 1100 + 200 × 3.0 = 1700

Summary table

The smallest achievable monthly payment is 1700, earned by Eman.

💡 Teacher tip: When minimising payment, don’t just minimise the final rating — also weigh the fixed payment and bonus rate. Eman has the lowest fixed payment (1100) and a moderate bonus rate (200), which together beat Chandan’s lower bonus rate (100) but higher fixed base (1400).

Answer: 1700 (Choice D)

Setting up the rules

With all five drivers bonus-eligible, each driver must have:

• No rating of 1, and
• At least one rating of 5.

Plus the two facts:

• Fact 1: Arun and Barun have exactly one 5; Chandan has exactly two 5’s.
• Fact 2: No rating value appears in three parameters (so 5 can appear at most twice for any driver).

Monthly payment = Fixed Payment + (Bonus rate × Final Rating), where Final Rating = (sum of five ratings) / 5.

To find the maximum monthly payment any driver can receive, we maximise each driver’s final rating (subject to bonus eligibility), then pick the largest payment.

Maximising each driver’s ratings

For each driver we keep the fixed parameter, use the maximum allowed number of 5’s, and fill the remaining slots with the largest legal values (4’s and 3’s), never repeating a value three times.

Arun — P4 = 4, exactly one 5. The other three slots can be at most 4, 3, 3 (a second 4 is fine; a third 4 is not):

• Ratings: 5, 4, 3, 4, 3 → sum = 19 → final = 3.8
• Payment = 1000 + 250 × 3.8 = 1950

Barun — P1 = 3, exactly one 5. The other three slots can be 4, 4, 3:

• Ratings: 5, 4, 4, 3, 3 → sum = 19 → final = 3.8
• Payment = 1200 + 200 × 3.8 = 1960

Chandan — P3 = 2, exactly two 5’s. The other two slots can be 4, 4:

• Ratings: 5, 5, 4, 4, 2 → sum = 20 → final = 4.0
• Payment = 1400 + 100 × 4.0 = 1800

Damodaran — P2 = 3, at least one 5 (so up to two 5’s by Fact 2). Use two 5’s; the other two slots can be 4, 4:

• Ratings: 5, 3, 5, 4, 4 → sum = 21 → final = 4.2
• Payment = 1300 + 150 × 4.2 = 1930

Eman — P5 = 2, at least one 5 (up to two 5’s). Use two 5’s; the other two slots can be 4, 4:

• Ratings: 5, 5, 4, 4, 2 → sum = 20 → final = 4.0
• Payment = 1100 + 200 × 4.0 = 1900

Summary table

Picking the maximum

Although Damodaran reaches the highest final rating (4.2), his low bonus rate (150) caps his payment at 1930. Barun, with the same final rating as Arun (3.8) but a higher bonus rate than Damodaran’s effective payoff and a favourable fixed base, reaches 1960 — the largest of all.

💡 Teacher tip: The maximum payment is not always earned by the driver with the maximum final rating. Always multiply final rating by the bonus rate and add the fixed payment before comparing.

Answer: 1960 (Choice A)

Finding who missed the Mathematics examination

Key rule: If a student missed exactly one examination, the mark awarded for that missed subject equals the average of the best three among the four subjects they actually appeared for. If a student missed two examinations, both missing marks equal the average of the best two of the three real scores — so the two invented marks are equal.

Fact II tells us the student who missed Mathematics missed no other subject. So we are looking for someone whose Mathematics mark equals the average of the best three of their other four scores.

Let us test each student:

Only Carl has a Mathematics mark (90) that equals the average of the best three of his other four scores. This is consistent with Fact II: Carl missed Mathematics and no other subject.

Answer: Carl (Choice B)

Finding who missed the English examination

Recap of the marking rules:

• Missed one subject → that mark = average of the best 3 of the 4 real scores.
• Missed two subjects → both missing marks = average of the best 2 of the 3 real scores, so the two invented marks are equal.

We test whose English mark could be an invented (derived) mark.

Esha: English = 80. She has no equal pair among her marks, so she cannot have missed two subjects — she must have missed exactly one. Check: best 3 of {Hindi 85, Maths 95, Science 60, SS 55} = (95 + 85 + 60)/3 = 240/3 = 80 ✓. So Esha missed English only.

Foni: English = 83 and Social Science = 83 — an equal pair, suggesting two missed subjects. The three real scores would be {Hindi 72, Maths 78, Science 88}; best 2 = (88 + 78)/2 = 166/2 = 83 ✓. So Foni missed English + Social Science.

Checking the remaining students confirms their English marks are real scores, not invented ones.

Solved arrangement (examinations missed):

The two students who did not appear for English are Esha and Foni.

Answer: Esha and Foni (Choice D)

Finding who missed the Hindi examination

Setting up the counts:

Fact I says four students appeared in each of English, Hindi, Science, and Social Science. Since there are six students, each of these four subjects was missed by $6 - 4 = 2$ students. Fact II says Mathematics was missed by exactly 1 student (who missed no other subject).

Total misses across all students = $2 + 2 + 2 + 2 + 1 = 9$.

Let $x$ = students who missed one subject, $y$ = students who missed two. Then: $x + y = 6, \quad x + 2y = 9 \implies y = 3,\; x = 3$

Using Fact III: The two students who missed Hindi are: one who missed only Hindi, and one who missed Hindi + Science.

We test whose Hindi mark could be invented:

• Alva: Hindi = 75. If she missed only Hindi, the invented mark = best 3 of {English 80, Maths 70, Science 75, SS 60} = (80 + 75 + 70)/3 = 225/3 = 75 ✓.
• Deep: Hindi = 90, Science = 90 (an equal pair). If he missed Hindi + Science, the invented mark = best 2 of {English 70, Maths 100, SS 80} = (100 + 80)/2 = 90 ✓.

No other student's Hindi mark can be an invented mark. Therefore, the two Hindi-missers are exactly Alva and Deep — one missed only Hindi and the other missed Hindi + Science, but both definitely missed Hindi.

Solved arrangement:

Answer: Alva and Deep (Choice B)

Finding who missed the Science examination

Science was missed by exactly 2 students (from Fact I: four students appeared, so $6 - 4 = 2$ missed it).

From the Hindi analysis, we know one of Alva or Deep missed Hindi + Science (Fact III says one Hindi-misser also missed Science). So one Science-misser is in {Alva, Deep}.

The other Science-misser must be among the remaining students. We already know:

• Carl missed Mathematics only.
• Esha missed English only.
• Foni missed English + Social Science.

That leaves Bithi. Let us verify:

Bithi: Science = 85, Social Science = 85 (an equal pair). The three real scores would be {English 90, Hindi 80, Maths 55}; best 2 = (90 + 80)/2 = 170/2 = 85 ✓. So Bithi missed Science + Social Science.

Solved arrangement:

So Science was missed by Bithi and by exactly one of Alva or Deep. The data does not allow us to determine whether that second person is Alva or Deep.

Answer: Bithi and one out of Alva and Deep (Choice A)

Counting students who missed exactly one examination

Counting total misses from the subject side:

From Fact I, each of English, Hindi, Science, and Social Science was missed by 2 students. From Fact II, Mathematics was missed by 1 student.

$\text{Total misses} = 2 + 2 + 2 + 2 + 1 = 9$

Setting up equations:

Let $x$ = number of students who missed exactly one examination, and $y$ = number who missed exactly two.

• Every student missed at least one: $x + y = 6$
• Total misses counted per student: $x + 2y = 9$

Subtracting the first from the second: $y = 3, \quad x = 3$

Verification from the solved arrangement:

Bithi and Foni definitely missed 2 each. Carl and Esha definitely missed 1 each. One of Alva/Deep missed 1 and the other missed 2. So exactly 3 students missed one examination (Carl, Esha, and one of Alva/Deep).

Answer: 3

Counting students whose missed examinations we can pin down exactly

Let us walk through each student against the solved arrangement:

💡 Teacher tip: For Alva, Hindi = 75 works as a single-miss mark (best 3 of 80, 70, 75, 60 = 75), and Hindi = Science = 75 works as a double-miss mark (best 2 of 80, 70, 60 = 75). The same ambiguity exists for Deep (Hindi = Science = 90). Fact III tells us one of them missed only Hindi and the other missed Hindi + Science, but nothing in the data tells us which is which.

So we can be definite about Carl, Esha, Bithi, and Foni — that is 4 students.

Answer: 4

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012 (A is active in 2012 via its multi-year, and Fact IV says 5 contracts in 2012).
• Y's two single-year contracts cover 2010 and 2019. Fact VII says B and D each had exactly one contract in 2012 — their multi-year contracts already supply that, so their single-year contracts must avoid 2012. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Contracts active each year:

Answering the question

We need the year (among the options) with two or more contracts:

In 2015, two contracts are active: B–Z (2010–2016) and D–X (2012–2015). The other three options each have only 1 contract.

Answer: 2015 (Choice C)

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012.
• Y's two single-year contracts cover 2010 and 2019. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Answering the question

We check each option against the solved table:

• A: B had a contract with Z in 2017 — B's Z-contract runs 2010–2016; it ended in 2016. False.
• B: B had a contract with Y in 2019 — B's Y-contract is the single year 2010, not 2019. False.
• C: D had a contract with X in 2011 — D's X-contract runs 2012–2015, not 2011. False.
• D: D had a contract with Y in 2019 — D's single-year Y-contract is exactly 2019. True.

Answer: D had a contract with Y in 2019 (Choice D)

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012.
• Y's two single-year contracts cover 2010 and 2019. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Contracts active each year:

Answering the question

We need the number of years with exactly one active contract. Reading from the table:

• 2016: only B–Z (2010–2016) → 1 contract
• 2017: only C–Z (2017–2019) → 1 contract
• 2018: only C–Z (2017–2019) → 1 contract

Every other year has 2 or more. So there are 3 years with exactly one contract.

Answer: 3 (Choice A)

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012.
• Y's two single-year contracts cover 2010 and 2019. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Contracts active each year:

Answering the question

In 2010, the active contracts are:

1. A–X (multi-year, 2010–2012) — covers 2010 ✓
2. B–Z (multi-year, 2010–2016) — covers 2010 ✓
3. B–Y (single-year, 2010) — covers 2010 ✓

No other contract touches 2010: C's multi-year starts in 2017, D's multi-year starts in 2012, and the W contracts are in 2012. So 2010 has exactly 3 contracts.

Answer: exactly 3 (Choice B)

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012.
• Y's two single-year contracts cover 2010 and 2019. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Answering the question

An institute has multiple contracts in the same year only if its two contracts overlap in time. We check each institute:

• A: multi-year (2010–2012) and single-year (2012) both cover 2012 → two contracts in 2012. ✓
• B: multi-year (2010–2016) and single-year (2010) both cover 2010 → two contracts in 2010. ✓
• C: multi-year (2017–2019) and single-year (2012) do not overlap. ✗
• D: multi-year (2012–2015) and single-year (2019) do not overlap. ✗

So only A and B had multiple contracts during the same year.

Answer: A and B only (Choice C)

Setting up the full arrangement

There are 8 contracts in total: each of the 4 institutes signs exactly 2, and each of the 4 vendors signs exactly 2. Fact V tells us 4 are multi-year (B = 7 years, D = 4 years, A = 3 years, C = 3 years) and 4 are single-year — so each institute has one multi-year and one single-year contract.

Step 1 — Assign vendors to contract types.

• W was active only in 2012 (Fact III), so both W contracts must be single-year, both in 2012.
• Z must be active in every year 2010–2019 (Fact I) using only 2 contracts. The 7-year contract (B) covers 7 of the 10 years; Z's second contract must cover the remaining gap. So Z takes B's 7-year contract and one 3-year contract.
• X must cover all of 2010–2015 with no activity after (Fact II). Two multi-year contracts can do this, so X takes the remaining two multi-year contracts (A's 3-year and D's 4-year).
• Y only needs 2010 and 2019 (Fact III), so Y takes the remaining two single-year contracts.

Step 2 — Place the multi-year contracts.

• B's 7-year (with Z) must cover 7 consecutive years. Since Z must cover all of 2010–2019, and C's 3-year (with Z) must cover the remaining 3 years, the only split is B–Z: 2010–2016 and C–Z: 2017–2019.
• X must cover 2010–2015 with A's 3-year and D's 4-year. Fact VII says D had no contract in 2010, so D's contract cannot start in 2010. The only way to cover 2010–2015 with a 3-year and a 4-year (D not starting at 2010) is A–X: 2010–2012 and D–X: 2012–2015.

Step 3 — Place the single-year contracts.

• W's two contracts are both in 2012. Fact VI says C had a contract in 2012, and C's multi-year (2017–2019) doesn't cover 2012, so C–W: 2012. The other W contract goes to A–W: 2012.
• Y's two single-year contracts cover 2010 and 2019. Since D had no contract in 2010 (Fact VII), D's single-year goes to 2019: D–Y: 2019. B's single-year goes to 2010: B–Y: 2010.

Solved arrangement — the eight contracts

Vendor view:

Answering the question

We need to find all entities (both institutes and vendors) that had more than one contract in some year — i.e., whose two contracts overlap in time.

Institutes:

Vendors:

💡 Teacher tip: The key insight is that two contracts "overlap" if they share at least one calendar year. For vendors, even though W's two contracts are with different institutes, they both fall in 2012 — so W had two contracts active simultaneously that year.

Combining both lists: the entities with more than one contract in some year are institutes A, B and vendors W, X.

Answer: A, B, W, and X (Choice B)

Step 1 — Count the category sizes

The set-up tells us how many musicians fall into each expertise category:

These add to $3+3+1+2+1 = 10$, so every musician fits one of these five categories. There is no Mridangam+Ghatam-only person and no all-three expert.

Step 2 — Place the fixed musicians

• Fact 2: D is an expert in both tabla and ghatam ⇒ D is the unique Tabla+Ghatam person.
• Fact 4: I is not an expert in tabla.
• Fact 5: H and I are not experts in mridangam, and exactly one of them is an expert in ghatam.

Since I is neither in tabla nor in mridangam, the only category I can occupy is Ghatam only (the unique Ghatam-only slot). Then, because exactly one of H and I is in ghatam, and that one is I, H is not in ghatam. H is also not in mridangam (Fact 5), so H must be Tabla only.

Step 3 — Check the options

We need someone definitely in tabla only (not mridangam, not ghatam):

• F — Fact 3 says F is in tabla, but F could be Tabla+Mridangam or Tabla only. Not definite.
• C — C's category is not yet pinned; could be Tabla only or Mridangam only. Not definite.
• A — Fact 1 says A is in mridangam; A could be Tabla+Mridangam or Mridangam only. Not definite.
• H — fixed as Tabla only. ✅

💡 Teacher tip: When a clue says "exactly one of X, Y has property P," first check whether one of them is forced into (or out of) P by other clues — the other follows automatically.

Answer: H (Option D)

Step 1 — Recall the fixed placements

Working through the clues (as in the previous question):

• D is the unique Tabla + Ghatam expert (Fact 2).
• I is Ghatam only (not tabla by Fact 4, not mridangam by Fact 5).
• H is Tabla only (not mridangam by Fact 5, not ghatam since I is the ghatam one).

Step 2 — Pin down J

Fact 4 says neither I nor J is an expert in tabla.

• I is already placed as Ghatam only.
• J is not in tabla, and the single Ghatam-only slot is already taken by I, so J cannot be Ghatam only.
• The only remaining category for a non-tabla musician is Mridangam only.

So J is definitely a mridangam-only expert.

Step 3 — Check the options

We need someone definitely in mridangam only (not tabla, not ghatam):

• B — Fact 1: B is in mridangam, but could be Tabla+Mridangam or Mridangam only. Not definite.
• J — fixed as Mridangam only. ✅
• G — Fact 3: G is in tabla, so G cannot be mridangam only. Eliminated.
• E — E's category is not pinned; could be Tabla only or Mridangam only. Not definite.

Solved arrangement so far

Answer: J (Option B)

Step 1 — Identify who can be a Tabla+Mridangam expert

The two Tabla+Mridangam slots are filled by:

• One of {A, B} — Fact 1 says both A and B are in mridangam, but only one is also in tabla.
• One of {F, G} — Fact 3 says both F and G are in tabla, but only one is also in mridangam.

So every Tabla+Mridangam expert must come from the set {A, B, F, G}, with exactly one from each pair.

Step 2 — Examine each option

We need the pair that cannot contain any Tabla+Mridangam musician.

• {F, G}: exactly one of them is Tabla+Mridangam ⇒ can contain one. ❌ (not our answer)
• {A, B}: exactly one of them is Tabla+Mridangam ⇒ can contain one. ❌
• {C, F}: F could be the Tabla+Mridangam one ⇒ can contain one. ❌
• {C, E}: C and E fill the remaining single-instrument slots (one Tabla only, one Mridangam only) — neither can be Tabla+Mridangam. ✅

Solved arrangement — instrument expertise

💡 Teacher tip: The key move is to notice that the Tabla+Mridangam slots are already fully accounted for by {A,B} and {F,G}. Anyone outside that set — like C and E — is structurally barred from being Tabla+Mridangam.

Answer: C and E (Option B)

Step 1 — Start from the solved arrangement

From the clues we already have these fixed placements:

The remaining musicians {A, B, C, E, F, G} fill:

• 2 Tabla-only slots (one already taken by H, so 2 left)
• 2 Mridangam-only slots (one already taken by J, so 2 left)
• 2 Tabla+Mridangam slots

with the pairings:

• {A, B}: one Tabla+Mridangam, one Mridangam only.
• {F, G}: one Tabla+Mridangam, one Tabla only.
• {C, E}: one Tabla only, one Mridangam only.

Step 2 — Apply the extra conditions

Condition 1: C is an expert in mridangam.

• C is not in tabla (since {C, E} split into one Tabla only and one Mridangam only).
• So C is Mridangam only, which forces E to be Tabla only.

Condition 2: F is not an expert in mridangam.

• Fact 3 says F is in tabla. Since F is not in mridangam, F is Tabla only, which forces G to be Tabla+Mridangam.

Step 3 — Collect the three Tabla-only musicians

We need exactly three Tabla-only experts. Reading off:

• H — fixed Tabla only.
• E — now Tabla only (from Condition 1).
• F — now Tabla only (from Condition 2).

That gives the three: H, E, F.

Final arrangement under the extra conditions

(A and B remain interchangeable, but they do not affect the Tabla-only count.)

💡 Teacher tip: Each "pair" clue ({A,B}, {F,G}, {C,E}) acts like a seesaw: fixing one member's category automatically fixes the other's. Use that to cascade deductions quickly.

Answer: E, F and H (Option A)

Setting up the Venn diagram

The 1000 patients are split equally: 500 in the treatment group (who receive medicines A, B, C, D) and 500 in the control group (placebo only). All the medicine clues apply to the 500 treatment-group patients, so our Venn diagram has four overlapping sets — A, B, C, D — summing to 500.

Filling in the regions step by step

Step 1 — Exactly-one-type regions.

We are told 75 patients were treated with exactly one type, and three of the four single-medicine regions are given:

• A only = 25, C only = 20, D only = 10

So B only = 75 − (25 + 20 + 10) = 20.

Step 2 — Exactly-three-type regions.

We are told 100 patients took exactly three types. Two of the four three-type regions are given:

• ABC only = 40, ACD only = 20

So ABD only + BCD only = 100 − 40 − 20 = 40.

Step 3 — Use the A-total (250) to pin ABD only.

Every region that includes A must sum to 250:

$25 + 20 + 30 + 35 + 40 + (\text{ABD only}) + 20 + 50 = 250$

$220 + \text{ABD only} = 250 \implies \text{ABD only} = 30$

Then BCD only = 40 − 30 = 10.

Step 4 — Use the C-total (210) to pin BC only.

Every region that includes C must sum to 210:

$20 + 30 + (\text{BC only}) + 20 + 40 + 20 + 10 + 50 = 210$

$190 + \text{BC only} = 210 \implies \text{BC only} = 20$

Step 5 — Use the grand total (500) to pin BD only.

Adding up all 14 known regions:

$25+20+20+10+20+30+35+20+20+40+30+20+10+50 = 350$

$\text{BD only} = 500 - 350 = 150$

Complete solved Venn diagram

Answering: How many were treated with medicine type B?

We sum every region that includes B:

$B = 20 + 20 + 20 + 150 + 40 + 30 + 10 + 50 = 340$

Answer: 340

Setting up the Venn diagram

The 1000 patients are split equally: 500 in the treatment group (who receive medicines A, B, C, D) and 500 in the control group (placebo only). All medicine clues apply to the 500 treatment-group patients.

Finding the BCD-only region

The question asks for patients treated with B, C, and D but not A — this is the "BCD only" region of the Venn diagram.

Step 1 — Narrow down using the exactly-three-types total.

We are told 100 patients took exactly three types of medicine. Two of the four three-type regions are given directly:

• ABC only = 40 (given in clue e)
• ACD only = 20 (given in clue e)

So the remaining two three-type regions must satisfy:

$\text{ABD only} + \text{BCD only} = 100 - 40 - 20 = 40$

Step 2 — Use the A-total (250) to pin ABD only.

Every region that includes A sums to 250. Listing them all:

$25 + 20 + 30 + 35 + 40 + (\text{ABD only}) + 20 + 50 = 250$

$220 + \text{ABD only} = 250 \implies \text{ABD only} = 30$

Step 3 — Solve for BCD only.

$\text{BCD only} = 40 - 30 = 10$

💡 Teacher tip: The region "treated with B, C and D but not A" is exactly the BCD-only region — it is the part of the triple-overlap of B, C, D that excludes A entirely.

Complete solved Venn diagram (for reference)

Answer: 10

Setting up the Venn diagram

The 1000 patients are split equally: 500 in the treatment group (who receive medicines A, B, C, D) and 500 in the control group (placebo only). All medicine clues apply to the 500 treatment-group patients.

Building up to the BD-only region

The question asks for patients treated with B and D only (no A, no C). This is the last unknown region we solve for.

Step 1 — Fill the exactly-one-type regions.

Given: A only = 25, C only = 20, D only = 10, and exactly-one total = 75.

$\text{B only} = 75 - (25 + 20 + 10) = 20$

Step 2 — Fill the exactly-two-type regions (partially given).

From the clues: AB only = 20, AC only = 30, AD only = 35, CD only = 20.

BC only and BD only are unknown for now.

Step 3 — Fill the exactly-three-type regions.

Given: ABC only = 40, ACD only = 20, and exactly-three total = 100.

$\text{ABD only} + \text{BCD only} = 100 - 40 - 20 = 40$

Using the A-total = 250:

$25 + 20 + 30 + 35 + 40 + (\text{ABD only}) + 20 + 50 = 250 \implies \text{ABD only} = 30$

$\text{BCD only} = 40 - 30 = 10$

Step 4 — Use the C-total (210) to find BC only.

$20 + 30 + (\text{BC only}) + 20 + 40 + 20 + 10 + 50 = 210 \implies \text{BC only} = 20$

Step 5 — Use the grand total (500) to find BD only.

Now every region except BD only is known. Summing all 14 known regions:

$25+20+20+10+20+30+35+20+20+40+30+20+10+50 = 350$

$\text{BD only} = 500 - 350 = 150$

💡 Teacher tip: When every other region is pinned down by the given totals (A = 250, C = 210, exactly-one = 75, exactly-three = 100, all-four = 50), the grand total of 500 acts as the final equation that closes the system.

Complete solved Venn diagram

Answer: 150

Setting up the Venn diagram

The 1000 patients are split equally: 500 in the treatment group (who receive medicines A, B, C, D) and 500 in the control group (placebo only). All medicine clues apply to the 500 treatment-group patients.

Solving the full Venn diagram

Step 1 — Exactly-one-type regions.

Given A only = 25, C only = 20, D only = 10, and exactly-one total = 75:

$\text{B only} = 75 - (25 + 20 + 10) = 20$

Step 2 — Exactly-three-type regions.

Given ABC only = 40, ACD only = 20, and exactly-three total = 100:

$\text{ABD only} + \text{BCD only} = 40$

Using A-total = 250:

$25 + 20 + 30 + 35 + 40 + (\text{ABD only}) + 20 + 50 = 250 \implies \text{ABD only} = 30$

$\text{BCD only} = 40 - 30 = 10$

Step 3 — Find BC only using C-total = 210.

$20 + 30 + (\text{BC only}) + 20 + 40 + 20 + 10 + 50 = 210 \implies \text{BC only} = 20$

Step 4 — Find BD only using grand total = 500.

$\text{BD only} = 500 - 350 = 150$

Complete solved Venn diagram

Answering: How many were treated with medicine type D?

We sum every region that includes D — these are the rows where D appears:

$D = 10 + 35 + 150 + 20 + 30 + 20 + 10 + 50 = 325$

💡 Teacher tip: To find the total for any single medicine, add up every Venn region that contains that medicine — from the single-only region through the all-four overlap.

Answer: 325