CAT 2024 Slot 1 · DILR · Set 2: Amiya and Ramya Campaign Election
The full set and all 5 questions (Q5–Q9) from CAT 2024 Slot 1, with answers and solutions.
Two students, Amiya and Ramya are the only candidates in an election for the position of class representative. Students will vote based on the intensity level of Amiya’s and Ramya’s campaigns and the type of campaigns they run. Each campaign is said to have a level of 1 if it is a staid campaign and a level of 2 if it is a vigorous campaign. Campaigns can be of two types, they can either focus on issues, or on attacking the other candidate.
If Amiya and Ramya both run campaigns focusing on issues, then
• The percentage of students voting in the election will be 20 times the sum of the levels of campaigning of the two students. For example, if Amiya and Ramya both run vigorous campaigns, then 20 × (2+2)%, that is, 80% of the students will vote in the election.
• Among voting students, the percentage of votes for each candidate will be proportional to the levels of their campaigns. For example, if Amiya runs a staid (i.e., level 1) campaign while Ramya runs a vigorous (i.e., level 2) campaign, then Amiya will receive 1/3 of the votes cast, and Ramya will receive the other 2/3.
The above-mentioned percentages change as follows if at least one of them runs a campaign attacking their opponent.
• If Amiya runs a campaign attacking Ramya and Ramya runs a campaign focusing on issues, then 10% of the students who would have otherwise voted for Amiya will vote for Ramya, and another 10% who would have otherwise voted for Amiya, will not vote at all.
• If Ramya runs a campaign attacking Amiya and Amiya runs a campaign focusing on issues, then 20% of the students who would have otherwise voted for Ramya will vote for Amiya, and another 5% who would have otherwise voted for Ramya, will not vote at all.
• If both run campaigns attacking each other, then 10% of the students who would have otherwise voted for them had they run campaigns focusing on issues, will not vote at all.
Question 5 · Quant Based Puzzles and Optimisation
If both of them run staid campaigns attacking the other, then what percentage of students will vote in the election?
- A
60%
- B
64%
- C
40%
- D
36%
Answer & solution
Answer: Option D
Solving the set
Let Amiya's campaign level be a and Ramya's be r (1 = staid, 2 = vigorous). All percentages below are percentages of all students.
Step 1 – Both on issues. Turnout = 20(a + r)%. Amiya gets a/(a + r) of these votes, i.e., 20(a + r) × a/(a + r) = 20a%. Similarly, Ramya gets 20r%. So on issues each candidate simply gets 20 × her own level.
Step 2 – Amiya attacks, Ramya on issues. Of Amiya's 20a, 10% (= 2a) move to Ramya and another 10% (= 2a) stop voting. Amiya = 20a − 4a = 16a; Ramya = 20r + 2a.
Step 3 – Ramya attacks, Amiya on issues. Of Ramya's 20r, 20% (= 4r) move to Amiya and 5% (= r) stop voting. Ramya = 20r − 5r = 15r; Amiya = 20a + 4r.
Step 4 – Both attack. Each loses 10% of her issue-based votes: Amiya = 18a, Ramya = 18r.
|
Campaign types |
Amiya's votes |
Ramya's votes |
Turnout |
|
Both on issues |
20a |
20r |
20a + 20r |
|
Amiya attacks, Ramya on issues |
0.8 × 20a = 16a |
20r + 0.1 × 20a = 20r + 2a |
18a + 20r |
|
Ramya attacks, Amiya on issues |
20a + 0.2 × 20r = 20a + 4r |
0.75 × 20r = 15r |
20a + 19r |
|
Both attack |
0.9 × 20a = 18a |
0.9 × 20r = 18r |
18a + 18r |
Step 5 – Turnout for every combination of levels (in %):
|
Campaign types |
a = 1, r = 1 |
a = 1, r = 2 |
a = 2, r = 1 |
a = 2, r = 2 |
|
Both on issues |
40 |
60 |
60 |
80 |
|
Amiya attacks |
38 |
58 |
56 |
76 |
|
Ramya attacks |
39 |
58 |
59 |
78 |
|
Both attack |
36 |
54 |
54 |
72 |
Step 6 – Winning margins. For each campaign type, the margin is largest when the winner is vigorous (level 2) and the loser is staid (level 1).
|
Campaign types |
Margin formula |
Largest margin |
At |
|
Both on issues |
20(a − r) or 20(r − a) |
20 |
levels 2 and 1 |
|
Amiya attacks |
Ramya − Amiya = 20r − 14a |
40 − 14 = 26 |
a = 1, r = 2 |
|
Ramya attacks |
Amiya − Ramya = 20a − 11r |
40 − 11 = 29 |
a = 2, r = 1 |
|
Both attack |
18(a − r) or 18(r − a) |
18 |
levels 2 and 1 |
This question
Both staid means a = r = 1. Both attacking gives turnout = 18a + 18r = 18 + 18 = 36% (see Step 4 / Step 5).
Check: on issues the turnout would be 20 × (1 + 1) = 40%, and 10% of those voters drop out: 40 × 0.9 = 36%.
Hence, the answer is option 4.
Question 6 · Quant Based Puzzles and Optimisation
What is the minimum percentage of students who will vote in the election?
- A
36%
- B
38%
- C
32%
- D
40%
Answer & solution
Answer: Option A
Solving the set
Let Amiya's campaign level be a and Ramya's be r (1 = staid, 2 = vigorous). All percentages below are percentages of all students.
Step 1 – Both on issues. Turnout = 20(a + r)%. Amiya gets a/(a + r) of these votes, i.e., 20(a + r) × a/(a + r) = 20a%. Similarly, Ramya gets 20r%. So on issues each candidate simply gets 20 × her own level.
Step 2 – Amiya attacks, Ramya on issues. Of Amiya's 20a, 10% (= 2a) move to Ramya and another 10% (= 2a) stop voting. Amiya = 20a − 4a = 16a; Ramya = 20r + 2a.
Step 3 – Ramya attacks, Amiya on issues. Of Ramya's 20r, 20% (= 4r) move to Amiya and 5% (= r) stop voting. Ramya = 20r − 5r = 15r; Amiya = 20a + 4r.
Step 4 – Both attack. Each loses 10% of her issue-based votes: Amiya = 18a, Ramya = 18r.
|
Campaign types |
Amiya's votes |
Ramya's votes |
Turnout |
|
Both on issues |
20a |
20r |
20a + 20r |
|
Amiya attacks, Ramya on issues |
0.8 × 20a = 16a |
20r + 0.1 × 20a = 20r + 2a |
18a + 20r |
|
Ramya attacks, Amiya on issues |
20a + 0.2 × 20r = 20a + 4r |
0.75 × 20r = 15r |
20a + 19r |
|
Both attack |
0.9 × 20a = 18a |
0.9 × 20r = 18r |
18a + 18r |
Step 5 – Turnout for every combination of levels (in %):
|
Campaign types |
a = 1, r = 1 |
a = 1, r = 2 |
a = 2, r = 1 |
a = 2, r = 2 |
|
Both on issues |
40 |
60 |
60 |
80 |
|
Amiya attacks |
38 |
58 |
56 |
76 |
|
Ramya attacks |
39 |
58 |
59 |
78 |
|
Both attack |
36 |
54 |
54 |
72 |
Step 6 – Winning margins. For each campaign type, the margin is largest when the winner is vigorous (level 2) and the loser is staid (level 1).
|
Campaign types |
Margin formula |
Largest margin |
At |
|
Both on issues |
20(a − r) or 20(r − a) |
20 |
levels 2 and 1 |
|
Amiya attacks |
Ramya − Amiya = 20r − 14a |
40 − 14 = 26 |
a = 1, r = 2 |
|
Ramya attacks |
Amiya − Ramya = 20a − 11r |
40 − 11 = 29 |
a = 2, r = 1 |
|
Both attack |
18(a − r) or 18(r − a) |
18 |
levels 2 and 1 |
This question
From the turnout table in Step 5, every formula increases with a and r, so the minimum of each type is at a = r = 1: 40, 38, 39 and 36.
The minimum turnout is 36%, when both run staid campaigns attacking each other.
Hence, the answer is option 1.
Question 7 · Quant Based Puzzles and Optimisation
If Amiya runs a campaign focusing on issues, then what is the maximum percentage of votes that she can get?
- A
44%
- B
40%
- C
48%
- D
36%
Answer & solution
Answer: Option C
Solving the set
Let Amiya's campaign level be a and Ramya's be r (1 = staid, 2 = vigorous). All percentages below are percentages of all students.
Step 1 – Both on issues. Turnout = 20(a + r)%. Amiya gets a/(a + r) of these votes, i.e., 20(a + r) × a/(a + r) = 20a%. Similarly, Ramya gets 20r%. So on issues each candidate simply gets 20 × her own level.
Step 2 – Amiya attacks, Ramya on issues. Of Amiya's 20a, 10% (= 2a) move to Ramya and another 10% (= 2a) stop voting. Amiya = 20a − 4a = 16a; Ramya = 20r + 2a.
Step 3 – Ramya attacks, Amiya on issues. Of Ramya's 20r, 20% (= 4r) move to Amiya and 5% (= r) stop voting. Ramya = 20r − 5r = 15r; Amiya = 20a + 4r.
Step 4 – Both attack. Each loses 10% of her issue-based votes: Amiya = 18a, Ramya = 18r.
|
Campaign types |
Amiya's votes |
Ramya's votes |
Turnout |
|
Both on issues |
20a |
20r |
20a + 20r |
|
Amiya attacks, Ramya on issues |
0.8 × 20a = 16a |
20r + 0.1 × 20a = 20r + 2a |
18a + 20r |
|
Ramya attacks, Amiya on issues |
20a + 0.2 × 20r = 20a + 4r |
0.75 × 20r = 15r |
20a + 19r |
|
Both attack |
0.9 × 20a = 18a |
0.9 × 20r = 18r |
18a + 18r |
Step 5 – Turnout for every combination of levels (in %):
|
Campaign types |
a = 1, r = 1 |
a = 1, r = 2 |
a = 2, r = 1 |
a = 2, r = 2 |
|
Both on issues |
40 |
60 |
60 |
80 |
|
Amiya attacks |
38 |
58 |
56 |
76 |
|
Ramya attacks |
39 |
58 |
59 |
78 |
|
Both attack |
36 |
54 |
54 |
72 |
Step 6 – Winning margins. For each campaign type, the margin is largest when the winner is vigorous (level 2) and the loser is staid (level 1).
|
Campaign types |
Margin formula |
Largest margin |
At |
|
Both on issues |
20(a − r) or 20(r − a) |
20 |
levels 2 and 1 |
|
Amiya attacks |
Ramya − Amiya = 20r − 14a |
40 − 14 = 26 |
a = 1, r = 2 |
|
Ramya attacks |
Amiya − Ramya = 20a − 11r |
40 − 11 = 29 |
a = 2, r = 1 |
|
Both attack |
18(a − r) or 18(r − a) |
18 |
levels 2 and 1 |
This question
Amiya focuses on issues, so only two rows apply.
• Ramya also on issues: Amiya gets 20a, at most 40%.
• Ramya attacks: Amiya gets 20a + 4r, at most 40 + 8 = 48% (a = 2, r = 2).
The maximum is 48%: both run vigorous campaigns, Amiya on issues and Ramya attacking Amiya.
Hence, the answer is option 3.
Question 8 · Quant Based Puzzles and Optimisation
If Ramya runs a campaign attacking Amiya, then what is the minimum percentage of votes that she is guaranteed to get?
- A
15%
- B
30%
- C
12%
- D
18%
Answer & solution
Answer: Option A
Solving the set
Let Amiya's campaign level be a and Ramya's be r (1 = staid, 2 = vigorous). All percentages below are percentages of all students.
Step 1 – Both on issues. Turnout = 20(a + r)%. Amiya gets a/(a + r) of these votes, i.e., 20(a + r) × a/(a + r) = 20a%. Similarly, Ramya gets 20r%. So on issues each candidate simply gets 20 × her own level.
Step 2 – Amiya attacks, Ramya on issues. Of Amiya's 20a, 10% (= 2a) move to Ramya and another 10% (= 2a) stop voting. Amiya = 20a − 4a = 16a; Ramya = 20r + 2a.
Step 3 – Ramya attacks, Amiya on issues. Of Ramya's 20r, 20% (= 4r) move to Amiya and 5% (= r) stop voting. Ramya = 20r − 5r = 15r; Amiya = 20a + 4r.
Step 4 – Both attack. Each loses 10% of her issue-based votes: Amiya = 18a, Ramya = 18r.
|
Campaign types |
Amiya's votes |
Ramya's votes |
Turnout |
|
Both on issues |
20a |
20r |
20a + 20r |
|
Amiya attacks, Ramya on issues |
0.8 × 20a = 16a |
20r + 0.1 × 20a = 20r + 2a |
18a + 20r |
|
Ramya attacks, Amiya on issues |
20a + 0.2 × 20r = 20a + 4r |
0.75 × 20r = 15r |
20a + 19r |
|
Both attack |
0.9 × 20a = 18a |
0.9 × 20r = 18r |
18a + 18r |
Step 5 – Turnout for every combination of levels (in %):
|
Campaign types |
a = 1, r = 1 |
a = 1, r = 2 |
a = 2, r = 1 |
a = 2, r = 2 |
|
Both on issues |
40 |
60 |
60 |
80 |
|
Amiya attacks |
38 |
58 |
56 |
76 |
|
Ramya attacks |
39 |
58 |
59 |
78 |
|
Both attack |
36 |
54 |
54 |
72 |
Step 6 – Winning margins. For each campaign type, the margin is largest when the winner is vigorous (level 2) and the loser is staid (level 1).
|
Campaign types |
Margin formula |
Largest margin |
At |
|
Both on issues |
20(a − r) or 20(r − a) |
20 |
levels 2 and 1 |
|
Amiya attacks |
Ramya − Amiya = 20r − 14a |
40 − 14 = 26 |
a = 1, r = 2 |
|
Ramya attacks |
Amiya − Ramya = 20a − 11r |
40 − 11 = 29 |
a = 2, r = 1 |
|
Both attack |
18(a − r) or 18(r − a) |
18 |
levels 2 and 1 |
This question
Ramya attacks, so two rows apply.
• Amiya on issues: Ramya gets 15r, at least 15% (r = 1).
• Amiya also attacks: Ramya gets 18r, at least 18% (r = 1).
Whatever Amiya does, Ramya gets at least 15%, and 15% is actually possible (Ramya staid, Amiya on issues). So the guaranteed minimum is 15%.
Hence, the answer is option 1.
Question 9 · Quant Based Puzzles and Optimisation
What is the maximum possible voting margin with which one of the candidates can win?
- A
29%
- B
26%
- C
20%
- D
28%
Answer & solution
Answer: Option A
Solving the set
Let Amiya's campaign level be a and Ramya's be r (1 = staid, 2 = vigorous). All percentages below are percentages of all students.
Step 1 – Both on issues. Turnout = 20(a + r)%. Amiya gets a/(a + r) of these votes, i.e., 20(a + r) × a/(a + r) = 20a%. Similarly, Ramya gets 20r%. So on issues each candidate simply gets 20 × her own level.
Step 2 – Amiya attacks, Ramya on issues. Of Amiya's 20a, 10% (= 2a) move to Ramya and another 10% (= 2a) stop voting. Amiya = 20a − 4a = 16a; Ramya = 20r + 2a.
Step 3 – Ramya attacks, Amiya on issues. Of Ramya's 20r, 20% (= 4r) move to Amiya and 5% (= r) stop voting. Ramya = 20r − 5r = 15r; Amiya = 20a + 4r.
Step 4 – Both attack. Each loses 10% of her issue-based votes: Amiya = 18a, Ramya = 18r.
|
Campaign types |
Amiya's votes |
Ramya's votes |
Turnout |
|
Both on issues |
20a |
20r |
20a + 20r |
|
Amiya attacks, Ramya on issues |
0.8 × 20a = 16a |
20r + 0.1 × 20a = 20r + 2a |
18a + 20r |
|
Ramya attacks, Amiya on issues |
20a + 0.2 × 20r = 20a + 4r |
0.75 × 20r = 15r |
20a + 19r |
|
Both attack |
0.9 × 20a = 18a |
0.9 × 20r = 18r |
18a + 18r |
Step 5 – Turnout for every combination of levels (in %):
|
Campaign types |
a = 1, r = 1 |
a = 1, r = 2 |
a = 2, r = 1 |
a = 2, r = 2 |
|
Both on issues |
40 |
60 |
60 |
80 |
|
Amiya attacks |
38 |
58 |
56 |
76 |
|
Ramya attacks |
39 |
58 |
59 |
78 |
|
Both attack |
36 |
54 |
54 |
72 |
Step 6 – Winning margins. For each campaign type, the margin is largest when the winner is vigorous (level 2) and the loser is staid (level 1).
|
Campaign types |
Margin formula |
Largest margin |
At |
|
Both on issues |
20(a − r) or 20(r − a) |
20 |
levels 2 and 1 |
|
Amiya attacks |
Ramya − Amiya = 20r − 14a |
40 − 14 = 26 |
a = 1, r = 2 |
|
Ramya attacks |
Amiya − Ramya = 20a − 11r |
40 − 11 = 29 |
a = 2, r = 1 |
|
Both attack |
18(a − r) or 18(r − a) |
18 |
levels 2 and 1 |
This question
From the margins table in Step 6, the largest margin is 29%, when Amiya runs a vigorous campaign on issues (a = 2) and Ramya runs a staid campaign attacking Amiya (r = 1).
Check: Amiya = 20 × 2 + 4 × 1 = 44%, Ramya = 15 × 1 = 15%. Margin = 44 − 15 = 29%.
Shortcut: Attacking hurts the attacker most when she is weak and the opponent is strong, so test the 'Ramya attacks' case with a = 2, r = 1.
Hence, the answer is option 1.
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