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The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs). These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.
The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.
There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.
Ordering the PMs. If an NUR had a PM greater than some city other than the lowest city, it would also exceed every lower city, giving more than one such pair. So the single NUR-city "inversion" must involve the lowest city, Blusterburg, and the Humbleset NUR must lie between Blusterburg and Noodleton. All other NURs are below Blusterburg. Hence:
Integer PI. PI = \(0.5N + 0.25(c_1+c_2)\). Since \(0.5N\) is an integer, \(c_1+c_2\) must be divisible by 4, so the two city PMs (in tens) must add to an even number.
Humbleset has NUR 40 and Blusterburg 30, so its other city \(x\) makes \(30+x\) divisible by 4: \(x = 50, 70\) or \(90\), giving PI = 40, 45 or 50.
\(x=50\) (PI 40): remaining cities 60, 70, 80, 90 pair as {60,80} and {70,90}; the state with NUR 20 gets PI \(10+35=45\) or \(10+40=50\), above Humbleset. Rejected.
\(x=70\) (PI 45): remaining 50, 60, 80, 90 pair as {50,90} and {60,80}, both summing to 140; PIs \(5+35=40\) and \(10+35=45\) tie with Humbleset. Rejected.
\(x=90\) (PI 50): remaining 50, 60, 70, 80 pair as {50,70} and {60,80}. NUR 10 with {50,70} gives 35 and NUR 20 with {60,80} gives 45 (distinct, below 50). The other pairing gives 40 and 40, not distinct. Accepted.
Fogglia has the lowest PI, so:
State
NUR
Cities
PI
Fogglia
10
Noodleton (50), Quackford (70)
35
Whimshire
20
Splutterville (60), Mumpypore (80)
45
Humbleset
40
Blusterburg (30), Zingaloo (90)
50
Whimshire has NUR 20 and cities 60, 80: PI = \(10 + 140/4 = 45\).
Ordering the PMs. If an NUR had a PM greater than some city other than the lowest city, it would also exceed every lower city, giving more than one such pair. So the single NUR-city "inversion" must involve the lowest city, Blusterburg, and the Humbleset NUR must lie between Blusterburg and Noodleton. All other NURs are below Blusterburg. Hence:
Integer PI. PI = \(0.5N + 0.25(c_1+c_2)\). Since \(0.5N\) is an integer, \(c_1+c_2\) must be divisible by 4, so the two city PMs (in tens) must add to an even number.
Humbleset has NUR 40 and Blusterburg 30, so its other city \(x\) makes \(30+x\) divisible by 4: \(x = 50, 70\) or \(90\), giving PI = 40, 45 or 50.
\(x=50\) (PI 40): remaining cities 60, 70, 80, 90 pair as {60,80} and {70,90}; the state with NUR 20 gets PI \(10+35=45\) or \(10+40=50\), above Humbleset. Rejected.
\(x=70\) (PI 45): remaining 50, 60, 80, 90 pair as {50,90} and {60,80}, both summing to 140; PIs \(5+35=40\) and \(10+35=45\) tie with Humbleset. Rejected.
\(x=90\) (PI 50): remaining 50, 60, 70, 80 pair as {50,70} and {60,80}. NUR 10 with {50,70} gives 35 and NUR 20 with {60,80} gives 45 (distinct, below 50). The other pairing gives 40 and 40, not distinct. Accepted.
Fogglia has the lowest PI, so:
State
NUR
Cities
PI
Fogglia
10
Noodleton (50), Quackford (70)
35
Whimshire
20
Splutterville (60), Mumpypore (80)
45
Humbleset
40
Blusterburg (30), Zingaloo (90)
50
Fogglia has NUR 10 and cities 50, 70: PI = \(5 + 120/4 = 35\).
Ordering the PMs. If an NUR had a PM greater than some city other than the lowest city, it would also exceed every lower city, giving more than one such pair. So the single NUR-city "inversion" must involve the lowest city, Blusterburg, and the Humbleset NUR must lie between Blusterburg and Noodleton. All other NURs are below Blusterburg. Hence:
Integer PI. PI = \(0.5N + 0.25(c_1+c_2)\). Since \(0.5N\) is an integer, \(c_1+c_2\) must be divisible by 4, so the two city PMs (in tens) must add to an even number.
Humbleset has NUR 40 and Blusterburg 30, so its other city \(x\) makes \(30+x\) divisible by 4: \(x = 50, 70\) or \(90\), giving PI = 40, 45 or 50.
\(x=50\) (PI 40): remaining cities 60, 70, 80, 90 pair as {60,80} and {70,90}; the state with NUR 20 gets PI \(10+35=45\) or \(10+40=50\), above Humbleset. Rejected.
\(x=70\) (PI 45): remaining 50, 60, 80, 90 pair as {50,90} and {60,80}, both summing to 140; PIs \(5+35=40\) and \(10+35=45\) tie with Humbleset. Rejected.
\(x=90\) (PI 50): remaining 50, 60, 70, 80 pair as {50,70} and {60,80}. NUR 10 with {50,70} gives 35 and NUR 20 with {60,80} gives 45 (distinct, below 50). The other pairing gives 40 and 40, not distinct. Accepted.
Fogglia has the lowest PI, so:
State
NUR
Cities
PI
Fogglia
10
Noodleton (50), Quackford (70)
35
Whimshire
20
Splutterville (60), Mumpypore (80)
45
Humbleset
40
Blusterburg (30), Zingaloo (90)
50
Humbleset has NUR 40 and cities 30, 90: PI = \(20 + 120/4 = 50\).
Hence, the answer is 50.
Q4
LR - Arrangements MULTIPLE CHOICE
Which pair of cities definitely belong to the same state?
Ordering the PMs. If an NUR had a PM greater than some city other than the lowest city, it would also exceed every lower city, giving more than one such pair. So the single NUR-city "inversion" must involve the lowest city, Blusterburg, and the Humbleset NUR must lie between Blusterburg and Noodleton. All other NURs are below Blusterburg. Hence:
Integer PI. PI = \(0.5N + 0.25(c_1+c_2)\). Since \(0.5N\) is an integer, \(c_1+c_2\) must be divisible by 4, so the two city PMs (in tens) must add to an even number.
Humbleset has NUR 40 and Blusterburg 30, so its other city \(x\) makes \(30+x\) divisible by 4: \(x = 50, 70\) or \(90\), giving PI = 40, 45 or 50.
\(x=50\) (PI 40): remaining cities 60, 70, 80, 90 pair as {60,80} and {70,90}; the state with NUR 20 gets PI \(10+35=45\) or \(10+40=50\), above Humbleset. Rejected.
\(x=70\) (PI 45): remaining 50, 60, 80, 90 pair as {50,90} and {60,80}, both summing to 140; PIs \(5+35=40\) and \(10+35=45\) tie with Humbleset. Rejected.
\(x=90\) (PI 50): remaining 50, 60, 70, 80 pair as {50,70} and {60,80}. NUR 10 with {50,70} gives 35 and NUR 20 with {60,80} gives 45 (distinct, below 50). The other pairing gives 40 and 40, not distinct. Accepted.
Fogglia has the lowest PI, so:
State
NUR
Cities
PI
Fogglia
10
Noodleton (50), Quackford (70)
35
Whimshire
20
Splutterville (60), Mumpypore (80)
45
Humbleset
40
Blusterburg (30), Zingaloo (90)
50
Noodleton and Quackford are both in Fogglia. Every other option pairs cities from different states (Blusterburg-Mumpypore: H/W, Splutterville-Quackford: W/F, Mumpypore-Zingaloo: W/H).
Hence, the answer is (D) Noodleton, Quackford.
Q5
LR - Arrangements TYPE IN THE ANSWER
For how many of the cities and NURs is it possible to identify their PM and the state they belong to?
Ordering the PMs. If an NUR had a PM greater than some city other than the lowest city, it would also exceed every lower city, giving more than one such pair. So the single NUR-city "inversion" must involve the lowest city, Blusterburg, and the Humbleset NUR must lie between Blusterburg and Noodleton. All other NURs are below Blusterburg. Hence:
Integer PI. PI = \(0.5N + 0.25(c_1+c_2)\). Since \(0.5N\) is an integer, \(c_1+c_2\) must be divisible by 4, so the two city PMs (in tens) must add to an even number.
Humbleset has NUR 40 and Blusterburg 30, so its other city \(x\) makes \(30+x\) divisible by 4: \(x = 50, 70\) or \(90\), giving PI = 40, 45 or 50.
\(x=50\) (PI 40): remaining cities 60, 70, 80, 90 pair as {60,80} and {70,90}; the state with NUR 20 gets PI \(10+35=45\) or \(10+40=50\), above Humbleset. Rejected.
\(x=70\) (PI 45): remaining 50, 60, 80, 90 pair as {50,90} and {60,80}, both summing to 140; PIs \(5+35=40\) and \(10+35=45\) tie with Humbleset. Rejected.
\(x=90\) (PI 50): remaining 50, 60, 70, 80 pair as {50,70} and {60,80}. NUR 10 with {50,70} gives 35 and NUR 20 with {60,80} gives 45 (distinct, below 50). The other pairing gives 40 and 40, not distinct. Accepted.
Fogglia has the lowest PI, so:
State
NUR
Cities
PI
Fogglia
10
Noodleton (50), Quackford (70)
35
Whimshire
20
Splutterville (60), Mumpypore (80)
45
Humbleset
40
Blusterburg (30), Zingaloo (90)
50
The arrangement is unique: all 6 cities and all 3 NURs have a fixed PM and state.
Hence, the answer is 9.
There are six spherical balls, B1, B2, B3, B4, B5, and B6, and four circular hoops H1, H2, H3, and H4.
Each ball was tested on each hoop once, by attempting to pass the ball through the hoop. If the diameter of a ball is not larger than the diameter of the hoop, the ball passes through the hoop and makes a "ping". Any ball having a diameter larger than that of the hoop gets stuck on that hoop and does not make a ping.
The following additional information is known: 1. B1 and B6 each made a ping on H4, but B5 did not. 2. B4 made a ping on H3, but B1 did not. 3. All balls, except B3, made pings on H1. 4. None of the balls, except B2, made a ping on H2.
Q6
LR - Arrangements TYPE IN THE ANSWER
What was the total number of pings made by B1, B2, and B3?
A ball pings on a hoop exactly when ball \(\le\) hoop.
Only B2 pings on H2, so B2 \(\le\) H2 < every other ball.
B4 pings on H3 and B4 > H2, so H2 < B4 \(\le\) H3.
B1 does not ping on H3 but pings on H4: H3 < B1 \(\le\) H4.
B5 does not ping on H4 but pings on H1: H4 < B5 \(\le\) H1.
B3 alone fails on H1: H1 < B3.
B6 pings on H4 and H1, not on H2: H2 < B6 \(\le\) H4 (its position relative to H3 is unknown).
Combined chain: B2 \(\le\) H2 < B4 \(\le\) H3 < B1 \(\le\) H4 < B5 \(\le\) H1 < B3, with B6 somewhere between H2 and H4.
Ball
Pings on
Number of pings
B2
H1, H2, H3, H4
4
B4
H1, H3, H4
3
B1
H1, H4
2
B6
H1, H4 and possibly H3
2 or 3
B5
H1
1
B3
none
0
B4 < B5 < B3, B1 < B5 < B3 and B2 < B1 < B5 all follow from the chain. But B6 only satisfies H2 < B6 \(\le\) H4; it may be smaller than B1 (for example if B6 \(\le\) H3), so B1 < B6 < B3 is not necessarily true.
Hence, the answer is (B) B1 < B6 < B3.
Q8
LR - Arrangements MULTIPLE CHOICE
Which of the following statements about the relative sizes of the hoops is true?
A ball pings on a hoop exactly when ball \(\le\) hoop.
Only B2 pings on H2, so B2 \(\le\) H2 < every other ball.
B4 pings on H3 and B4 > H2, so H2 < B4 \(\le\) H3.
B1 does not ping on H3 but pings on H4: H3 < B1 \(\le\) H4.
B5 does not ping on H4 but pings on H1: H4 < B5 \(\le\) H1.
B3 alone fails on H1: H1 < B3.
B6 pings on H4 and H1, not on H2: H2 < B6 \(\le\) H4 (its position relative to H3 is unknown).
Combined chain: B2 \(\le\) H2 < B4 \(\le\) H3 < B1 \(\le\) H4 < B5 \(\le\) H1 < B3, with B6 somewhere between H2 and H4.
Ball
Pings on
Number of pings
B2
H1, H2, H3, H4
4
B4
H1, H3, H4
3
B1
H1, H4
2
B6
H1, H4 and possibly H3
2 or 3
B5
H1
1
B3
none
0
Total pings = \(4+3+2+1+0\) + (2 or 3 for B6) = 12 or 13, and both cases are possible depending on whether B6 \(\le\) H3.
Hence, the answer is (B) 12 or 13.
Ananya Raga, Bhaskar Tala, Charu Veena, and Devendra Sur are four musicians. Each of them started and completed their training as students under each of three Gurus — Pandit Meghnath, Ustad Samiran, and Acharya Raghunath between 2013 and 2024, including both the years. Each Guru trains any student for consecutive years only, for a span of 2, 3, or 4 years, with each Guru having a different span. During some of these years, a student may not have trained under these Gurus; however, they never trained under multiple Gurus in the same year. In none of these years, any of these Gurus trained more than two of these students at the same time. When two students train under the same Guru at the same time, they are referred to as Gurubhai, irrespective of their gender. The following additional facts are known. 1. Ustad Samiran never trained more than one of these students in the same year. 2. Acharya Raghunath did not train any of these students during 2015-2018, as well as during 2021-24. 3. Ananya and Devendra were never Gurubhai; neither were Bhaskar and Charu. All other pairs of musicians were Gurubhai for exactly 2 years. 4. In 2013, Ananya and Bhaskar started their trainings under Pandit Meghnath and under Ustad Samiran, respectively.
Q10
LR - Puzzles MULTIPLE CHOICE
In which of the following years were Ananya and Bhaskar Gurubhai?
Spans. Raghunath trains only in 2013-14 and 2019-20 (two separate 2-year windows), so his span is 2 years. Samiran trains at most one student per year; four students with a 4-year span would need 16 years, more than the 12 available, so Samiran's span is 3 and Meghnath's is 4. Samiran's four 3-year blocks then fill all 12 years exactly: 2013-15, 2016-18, 2019-21, 2022-24.
Raghunath. Each student spends 2013-14 or 2019-20 with him and at most two students at a time, so two students take each window. Bhaskar (Samiran 2013-15) and Ananya (Meghnath 2013-16) cannot use 2013-14, so they take 2019-20 and are Gurubhai then. Charu and Devendra take 2013-14 and are Gurubhai then. These pairs have now used their 2 years.
Meghnath. Samiran creates no Gurubhai, so the remaining pairs Ananya-Charu and Bhaskar-Devendra must each overlap for exactly 2 years under Meghnath. Charu's 4 years must overlap Ananya's 2013-16 by 2 years, so Charu: 2015-18. Then Ananya's Samiran block can only be 2022-24, Charu's is 2019-21, Devendra's is 2016-18. Bhaskar's Meghnath years must avoid 2013-15 and 2019-20, so 2021-24; Devendra must overlap him by exactly 2 years within 2019-24, so Devendra: 2019-22.
Spans. Raghunath trains only in 2013-14 and 2019-20 (two separate 2-year windows), so his span is 2 years. Samiran trains at most one student per year; four students with a 4-year span would need 16 years, more than the 12 available, so Samiran's span is 3 and Meghnath's is 4. Samiran's four 3-year blocks then fill all 12 years exactly: 2013-15, 2016-18, 2019-21, 2022-24.
Raghunath. Each student spends 2013-14 or 2019-20 with him and at most two students at a time, so two students take each window. Bhaskar (Samiran 2013-15) and Ananya (Meghnath 2013-16) cannot use 2013-14, so they take 2019-20 and are Gurubhai then. Charu and Devendra take 2013-14 and are Gurubhai then. These pairs have now used their 2 years.
Meghnath. Samiran creates no Gurubhai, so the remaining pairs Ananya-Charu and Bhaskar-Devendra must each overlap for exactly 2 years under Meghnath. Charu's 4 years must overlap Ananya's 2013-16 by 2 years, so Charu: 2015-18. Then Ananya's Samiran block can only be 2022-24, Charu's is 2019-21, Devendra's is 2016-18. Bhaskar's Meghnath years must avoid 2013-15 and 2019-20, so 2021-24; Devendra must overlap him by exactly 2 years within 2019-24, so Devendra: 2019-22.
Spans. Raghunath trains only in 2013-14 and 2019-20 (two separate 2-year windows), so his span is 2 years. Samiran trains at most one student per year; four students with a 4-year span would need 16 years, more than the 12 available, so Samiran's span is 3 and Meghnath's is 4. Samiran's four 3-year blocks then fill all 12 years exactly: 2013-15, 2016-18, 2019-21, 2022-24.
Raghunath. Each student spends 2013-14 or 2019-20 with him and at most two students at a time, so two students take each window. Bhaskar (Samiran 2013-15) and Ananya (Meghnath 2013-16) cannot use 2013-14, so they take 2019-20 and are Gurubhai then. Charu and Devendra take 2013-14 and are Gurubhai then. These pairs have now used their 2 years.
Meghnath. Samiran creates no Gurubhai, so the remaining pairs Ananya-Charu and Bhaskar-Devendra must each overlap for exactly 2 years under Meghnath. Charu's 4 years must overlap Ananya's 2013-16 by 2 years, so Charu: 2015-18. Then Ananya's Samiran block can only be 2022-24, Charu's is 2019-21, Devendra's is 2016-18. Bhaskar's Meghnath years must avoid 2013-15 and 2019-20, so 2021-24; Devendra must overlap him by exactly 2 years within 2019-24, so Devendra: 2019-22.
Spans. Raghunath trains only in 2013-14 and 2019-20 (two separate 2-year windows), so his span is 2 years. Samiran trains at most one student per year; four students with a 4-year span would need 16 years, more than the 12 available, so Samiran's span is 3 and Meghnath's is 4. Samiran's four 3-year blocks then fill all 12 years exactly: 2013-15, 2016-18, 2019-21, 2022-24.
Raghunath. Each student spends 2013-14 or 2019-20 with him and at most two students at a time, so two students take each window. Bhaskar (Samiran 2013-15) and Ananya (Meghnath 2013-16) cannot use 2013-14, so they take 2019-20 and are Gurubhai then. Charu and Devendra take 2013-14 and are Gurubhai then. These pairs have now used their 2 years.
Meghnath. Samiran creates no Gurubhai, so the remaining pairs Ananya-Charu and Bhaskar-Devendra must each overlap for exactly 2 years under Meghnath. Charu's 4 years must overlap Ananya's 2013-16 by 2 years, so Charu: 2015-18. Then Ananya's Samiran block can only be 2022-24, Charu's is 2019-21, Devendra's is 2016-18. Bhaskar's Meghnath years must avoid 2013-15 and 2019-20, so 2021-24; Devendra must overlap him by exactly 2 years within 2019-24, so Devendra: 2019-22.
Ananya was with Samiran only in 2022-24, so (A) and (B) are false. Charu was with Samiran in 2019-21 (in 2018 she was with Meghnath), so (C) is true and (D) is false.
Hence, the answer is (C) Charu was training under Ustad Samiran in 2019.
Q14
LR - Puzzles TYPE IN THE ANSWER
In how many of the years between 2013-24, were only two of these four musicians training under these three Gurus?
Spans. Raghunath trains only in 2013-14 and 2019-20 (two separate 2-year windows), so his span is 2 years. Samiran trains at most one student per year; four students with a 4-year span would need 16 years, more than the 12 available, so Samiran's span is 3 and Meghnath's is 4. Samiran's four 3-year blocks then fill all 12 years exactly: 2013-15, 2016-18, 2019-21, 2022-24.
Raghunath. Each student spends 2013-14 or 2019-20 with him and at most two students at a time, so two students take each window. Bhaskar (Samiran 2013-15) and Ananya (Meghnath 2013-16) cannot use 2013-14, so they take 2019-20 and are Gurubhai then. Charu and Devendra take 2013-14 and are Gurubhai then. These pairs have now used their 2 years.
Meghnath. Samiran creates no Gurubhai, so the remaining pairs Ananya-Charu and Bhaskar-Devendra must each overlap for exactly 2 years under Meghnath. Charu's 4 years must overlap Ananya's 2013-16 by 2 years, so Charu: 2015-18. Then Ananya's Samiran block can only be 2022-24, Charu's is 2019-21, Devendra's is 2016-18. Bhaskar's Meghnath years must avoid 2013-15 and 2019-20, so 2021-24; Devendra must overlap him by exactly 2 years within 2019-24, so Devendra: 2019-22.
Exactly two musicians trained in 2017, 2018, 2023 and 2024.
Hence, the answer is 4.
The Sustainability Index (SI) of a country at a point in time is an integerbetween 1 and 100. This question is related to SI of six countries – A, B, C, D, E, and F – at three different points in time – 2016, 2020, and 2024. The plot represents the exact changes in their SI, with X-coordinate representing % increase in 2020 from 2016, i.e., (SI in 2020 minus SI in 2016) / (SI in 2016), and Y-coordinate representing % increase in 2024 from 2020. At any point in time, the country with highest SI is ranked 1, while the country with the lowest SI is ranked 6. The following additional facts are known. 1. In 2016, B, C, E, and A had ranks 1, 2, 3, and 4 respectively. 2. F had lower SI than any other country in 2016, 2020, and 2024. 3. In 2024, E was the only country with SI of 90. 4. The range of SI of the six countries was 60 in 2016 as well as in 2024.
Reading the plot: A (25%, 50%), B (-25%, -25%), C (-20%, 40%), D (100%, 20%), E (25%, 20%), F (100%, -25%). So SI2024 = SI2016 \(\times\) (1+X)(1+Y), and every SI is an integer.
B: \(\times \tfrac{3}{4} \times \tfrac{3}{4}\), so B2016 is a multiple of 16. C: \(\times \tfrac{4}{5}\times\tfrac{7}{5}\), so C2016 is a multiple of 25.
2016 ranks B > C > E (60) > A: C2016 = 75 and B2016 = 80 or 96.
2024: A2016 < 60 and A grows by \(\times\tfrac{15}{8}\), so A2016 is a multiple of 8 and at most 48; 48 gives 90 (not allowed, only E has 90). D2016 < A2016 \(\le\) 40 gives D2024 = 2.4D2016 \(\le\) 84. C2024 = 84. So the 2024 maximum is E = 90 and, with range 60, F2024 = 30.
F: \(\times 2 \times 0.75 = \times 1.5\), so F2016 = 20, F2020 = 40. 2016 range 60 gives B2016 = 80, so B2020 = 60, B2024 = 45.
F is lowest in 2020, so A2020 = 1.25A2016 > 40, i.e. A2016 > 32, so A2016 = 40. D2016 is a multiple of 5 between 20 and 40: 25, 30 or 35 (not fixed).
Reading the plot: A (25%, 50%), B (-25%, -25%), C (-20%, 40%), D (100%, 20%), E (25%, 20%), F (100%, -25%). So SI2024 = SI2016 \(\times\) (1+X)(1+Y), and every SI is an integer.
B: \(\times \tfrac{3}{4} \times \tfrac{3}{4}\), so B2016 is a multiple of 16. C: \(\times \tfrac{4}{5}\times\tfrac{7}{5}\), so C2016 is a multiple of 25.
2016 ranks B > C > E (60) > A: C2016 = 75 and B2016 = 80 or 96.
2024: A2016 < 60 and A grows by \(\times\tfrac{15}{8}\), so A2016 is a multiple of 8 and at most 48; 48 gives 90 (not allowed, only E has 90). D2016 < A2016 \(\le\) 40 gives D2024 = 2.4D2016 \(\le\) 84. C2024 = 84. So the 2024 maximum is E = 90 and, with range 60, F2024 = 30.
F: \(\times 2 \times 0.75 = \times 1.5\), so F2016 = 20, F2020 = 40. 2016 range 60 gives B2016 = 80, so B2020 = 60, B2024 = 45.
F is lowest in 2020, so A2020 = 1.25A2016 > 40, i.e. A2016 > 32, so A2016 = 40. D2016 is a multiple of 5 between 20 and 40: 25, 30 or 35 (not fixed).
Reading the plot: A (25%, 50%), B (-25%, -25%), C (-20%, 40%), D (100%, 20%), E (25%, 20%), F (100%, -25%). So SI2024 = SI2016 \(\times\) (1+X)(1+Y), and every SI is an integer.
B: \(\times \tfrac{3}{4} \times \tfrac{3}{4}\), so B2016 is a multiple of 16. C: \(\times \tfrac{4}{5}\times\tfrac{7}{5}\), so C2016 is a multiple of 25.
2016 ranks B > C > E (60) > A: C2016 = 75 and B2016 = 80 or 96.
2024: A2016 < 60 and A grows by \(\times\tfrac{15}{8}\), so A2016 is a multiple of 8 and at most 48; 48 gives 90 (not allowed, only E has 90). D2016 < A2016 \(\le\) 40 gives D2024 = 2.4D2016 \(\le\) 84. C2024 = 84. So the 2024 maximum is E = 90 and, with range 60, F2024 = 30.
F: \(\times 2 \times 0.75 = \times 1.5\), so F2016 = 20, F2020 = 40. 2016 range 60 gives B2016 = 80, so B2020 = 60, B2024 = 45.
F is lowest in 2020, so A2020 = 1.25A2016 > 40, i.e. A2016 > 32, so A2016 = 40. D2016 is a multiple of 5 between 20 and 40: 25, 30 or 35 (not fixed).
Reading the plot: A (25%, 50%), B (-25%, -25%), C (-20%, 40%), D (100%, 20%), E (25%, 20%), F (100%, -25%). So SI2024 = SI2016 \(\times\) (1+X)(1+Y), and every SI is an integer.
B: \(\times \tfrac{3}{4} \times \tfrac{3}{4}\), so B2016 is a multiple of 16. C: \(\times \tfrac{4}{5}\times\tfrac{7}{5}\), so C2016 is a multiple of 25.
2016 ranks B > C > E (60) > A: C2016 = 75 and B2016 = 80 or 96.
2024: A2016 < 60 and A grows by \(\times\tfrac{15}{8}\), so A2016 is a multiple of 8 and at most 48; 48 gives 90 (not allowed, only E has 90). D2016 < A2016 \(\le\) 40 gives D2024 = 2.4D2016 \(\le\) 84. C2024 = 84. So the 2024 maximum is E = 90 and, with range 60, F2024 = 30.
F: \(\times 2 \times 0.75 = \times 1.5\), so F2016 = 20, F2020 = 40. 2016 range 60 gives B2016 = 80, so B2020 = 60, B2024 = 45.
F is lowest in 2020, so A2020 = 1.25A2016 > 40, i.e. A2016 > 32, so A2016 = 40. D2016 is a multiple of 5 between 20 and 40: 25, 30 or 35 (not fixed).
Country
2016
2020
2024
A
40
50
75
B
80
60
45
C
75
60
84
D
25/30/35
50/60/70
60/72/84
E
60
75
90
F
20
40
30
B2024 = 80 \(\times\) 0.75 \(\times\) 0.75 = 45.
Hence, the answer is (D) 45.
The following charts depict details of research papers written by four authors, Arman, Brajen, Chintan, and Devon. The papers were of four types, single-author, two-author, three-author, and four-author, that is, written by one, two, three, or all four of these authors, respectively. No other authors were involved in writing these papers.
The following additional facts are known. 1. Each of the authors wrote at least one of each of the four types of papers. 2. The four authors wrote different numbers of single-author papers. 3. Both Chintan and Devon wrote more three-author papers than Brajen. 4. The number of single-author and two-author papers written by Brajen were the same.
Q19
LR - Selection & Distribution TYPE IN THE ANSWER
What was the total number of two-author and three-author papers written by Brajen?
Totals: 10 single, 4 two-author, 3 three-author, 2 four-author papers; author totals A 5, B 8, C 12, D 10. Every author is on both four-author papers, so for each author: singles + two-author + three-author = total - 2.
Arman: 5 - 2 = 3 with at least one of each type, so 1 single, 1 two-author, 1 three-author.
Three-author papers: 3 papers give 9 slots, each author in at most 3. Arman has 1, so B + C + D = 8. Since C and D each exceed B, B = 2 and C = D = 3. So C and D are on all three: the papers are A-C-D, B-C-D, B-C-D.
Singles: four distinct numbers, each at least 1, adding to 10, so they are 1, 2, 3, 4. Arman has 1.
Brajen: 8 - 2 - 2 = 4 = singles + two-author, and these are equal, so 2 and 2.
Chintan and Devon have singles 3 and 4 in some order. Chintan: singles + two-author = 12 - 2 - 3 = 7; Devon: 10 - 2 - 3 = 5.
Author
Single
Two-author
Three-author
Four-author
Total
Arman
1
1
1
2
5
Brajen
2
2
2
2
8
Chintan
3 or 4
4 or 3
3
2
12
Devon
4 or 3
1 or 2
3
2
10
Both cases are realisable: (C single 3, two-author 4; D single 4, two-author 1) with two-author papers A-C, B-C, B-C, C-D; or (C single 4, two-author 3; D single 3, two-author 2) with A-C, B-C, B-D, C-D.
Brajen wrote 2 two-author and 2 three-author papers: \(2+2 = 4\).
Hence, the answer is 4.
Q20
LR - Selection & Distribution MULTIPLE CHOICE
Which of the following statements is/are NECESSARILY true? i. Chintan wrote exactly three two-author papers. ii. Chintan wrote more single-author papers than Devon.
Totals: 10 single, 4 two-author, 3 three-author, 2 four-author papers; author totals A 5, B 8, C 12, D 10. Every author is on both four-author papers, so for each author: singles + two-author + three-author = total - 2.
Arman: 5 - 2 = 3 with at least one of each type, so 1 single, 1 two-author, 1 three-author.
Three-author papers: 3 papers give 9 slots, each author in at most 3. Arman has 1, so B + C + D = 8. Since C and D each exceed B, B = 2 and C = D = 3. So C and D are on all three: the papers are A-C-D, B-C-D, B-C-D.
Singles: four distinct numbers, each at least 1, adding to 10, so they are 1, 2, 3, 4. Arman has 1.
Brajen: 8 - 2 - 2 = 4 = singles + two-author, and these are equal, so 2 and 2.
Chintan and Devon have singles 3 and 4 in some order. Chintan: singles + two-author = 12 - 2 - 3 = 7; Devon: 10 - 2 - 3 = 5.
Author
Single
Two-author
Three-author
Four-author
Total
Arman
1
1
1
2
5
Brajen
2
2
2
2
8
Chintan
3 or 4
4 or 3
3
2
12
Devon
4 or 3
1 or 2
3
2
10
Both cases are realisable: (C single 3, two-author 4; D single 4, two-author 1) with two-author papers A-C, B-C, B-C, C-D; or (C single 4, two-author 3; D single 3, two-author 2) with A-C, B-C, B-D, C-D.
Chintan's two-author count can be 4 or 3, so (i) is not necessary. Chintan's singles can be 3 (fewer than Devon's 4) or 4 (more than Devon's 3), so (ii) is not necessary either.
Hence, the answer is (A) Neither i nor ii.
Q21
LR - Selection & Distribution MULTIPLE CHOICE
Which of the following statements is/are NECESSARILY true? i. Arman wrote three-author papers only with Chintan and Devon. ii. Brajen wrote three-author papers only with Chintan and Devon.
Totals: 10 single, 4 two-author, 3 three-author, 2 four-author papers; author totals A 5, B 8, C 12, D 10. Every author is on both four-author papers, so for each author: singles + two-author + three-author = total - 2.
Arman: 5 - 2 = 3 with at least one of each type, so 1 single, 1 two-author, 1 three-author.
Three-author papers: 3 papers give 9 slots, each author in at most 3. Arman has 1, so B + C + D = 8. Since C and D each exceed B, B = 2 and C = D = 3. So C and D are on all three: the papers are A-C-D, B-C-D, B-C-D.
Singles: four distinct numbers, each at least 1, adding to 10, so they are 1, 2, 3, 4. Arman has 1.
Brajen: 8 - 2 - 2 = 4 = singles + two-author, and these are equal, so 2 and 2.
Chintan and Devon have singles 3 and 4 in some order. Chintan: singles + two-author = 12 - 2 - 3 = 7; Devon: 10 - 2 - 3 = 5.
Author
Single
Two-author
Three-author
Four-author
Total
Arman
1
1
1
2
5
Brajen
2
2
2
2
8
Chintan
3 or 4
4 or 3
3
2
12
Devon
4 or 3
1 or 2
3
2
10
Both cases are realisable: (C single 3, two-author 4; D single 4, two-author 1) with two-author papers A-C, B-C, B-C, C-D; or (C single 4, two-author 3; D single 3, two-author 2) with A-C, B-C, B-D, C-D.
Since Chintan and Devon are on all three three-author papers, Arman's only three-author paper is with Chintan and Devon, and both of Brajen's are with Chintan and Devon. Both statements are necessarily true.
Hence, the answer is (D) Both i and ii.
Q22
LR - Selection & Distribution TYPE IN THE ANSWER
If Devon wrote more than one two-author papers, then how many two-author papers did Chintan write?
Totals: 10 single, 4 two-author, 3 three-author, 2 four-author papers; author totals A 5, B 8, C 12, D 10. Every author is on both four-author papers, so for each author: singles + two-author + three-author = total - 2.
Arman: 5 - 2 = 3 with at least one of each type, so 1 single, 1 two-author, 1 three-author.
Three-author papers: 3 papers give 9 slots, each author in at most 3. Arman has 1, so B + C + D = 8. Since C and D each exceed B, B = 2 and C = D = 3. So C and D are on all three: the papers are A-C-D, B-C-D, B-C-D.
Singles: four distinct numbers, each at least 1, adding to 10, so they are 1, 2, 3, 4. Arman has 1.
Brajen: 8 - 2 - 2 = 4 = singles + two-author, and these are equal, so 2 and 2.
Chintan and Devon have singles 3 and 4 in some order. Chintan: singles + two-author = 12 - 2 - 3 = 7; Devon: 10 - 2 - 3 = 5.
Author
Single
Two-author
Three-author
Four-author
Total
Arman
1
1
1
2
5
Brajen
2
2
2
2
8
Chintan
3 or 4
4 or 3
3
2
12
Devon
4 or 3
1 or 2
3
2
10
Both cases are realisable: (C single 3, two-author 4; D single 4, two-author 1) with two-author papers A-C, B-C, B-C, C-D; or (C single 4, two-author 3; D single 3, two-author 2) with A-C, B-C, B-D, C-D.
If Devon wrote more than one two-author paper, Devon has 2 two-author papers and 3 singles, so Chintan has 4 singles and \(7-4 = 3\) two-author papers.
