UPSC Insta–DART (Daily Aptitude and Reasoning Test) 6 Oct 2025
Kartavya Desk Staff
Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
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• Question 1 of 5 1. Question In a tournament of 3 teams X, Y, Z, each plays once with the other. Win = 2 points, draw = 1 point. Each team scored exactly 2 goals in total. Points are: X = 2, Y = 2, Z = 2. Which is correct? I. All matches were drawn. II. Every match had exactly 2 goals in total. (a) I only (a) I only (c) Both I and II (d) Neither I nor II Correct Answer: (c) Solution: If all teams end with 2 points, they must all draw their two matches. Hence all matches drawn. Since each team scored exactly 2 goals, over two matches each averaged 1 goal per game. So every match must have ended 1–1. Both statements are correct. Answer = (c). Incorrect Answer: (c) Solution: If all teams end with 2 points, they must all draw their two matches. Hence all matches drawn. Since each team scored exactly 2 goals, over two matches each averaged 1 goal per game. So every match must have ended 1–1. Both statements are correct. Answer = (c).
#### 1. Question
In a tournament of 3 teams X, Y, Z, each plays once with the other. Win = 2 points, draw = 1 point. Each team scored exactly 2 goals in total. Points are: X = 2, Y = 2, Z = 2. Which is correct?
I. All matches were drawn. II. Every match had exactly 2 goals in total.
• (a) I only
• (a) I only
• (c) Both I and II
• (d) Neither I nor II
Answer: (c)
Solution: If all teams end with 2 points, they must all draw their two matches. Hence all matches drawn. Since each team scored exactly 2 goals, over two matches each averaged 1 goal per game. So every match must have ended 1–1. Both statements are correct. Answer = (c).
Answer: (c)
Solution: If all teams end with 2 points, they must all draw their two matches. Hence all matches drawn. Since each team scored exactly 2 goals, over two matches each averaged 1 goal per game. So every match must have ended 1–1. Both statements are correct. Answer = (c).
• Question 2 of 5 2. Question A laptop was stolen. Three suspects X, Y, and Z were questioned. Exactly one is guilty. Statements: X: I did not cheat. Y cheated. Y: Z did not cheat. I did not cheat. Z: I did not cheat. X is lying. Who cheated? (a) X (b) Y (c) Z (d) Cannot be concluded Correct Answer: (a) Solution Case 1: X is guilty. Then X’s two statements are false, so X did cheat and Y did not. – Y is innocent, so “Z did not cheat” is true and “I did not cheat” is true. – Z is innocent, so “I did not cheat” is true and “X is lying” is true. All consistent. Works. Case 2: Y is guilty. Then Y’s two statements are false, so Z did cheat and Y did cheat; two guilty, impossible. Case 3: Z is guilty. Then Z’s two statements are false, so Z did cheat and X is not lying. If X is truthful, then “Y cheated” must be true, giving two guilty, impossible. Hence X cheated. Answer (a). Incorrect Answer: (a) Solution Case 1: X is guilty. Then X’s two statements are false, so X did cheat and Y did not. – Y is innocent, so “Z did not cheat” is true and “I did not cheat” is true. – Z is innocent, so “I did not cheat” is true and “X is lying” is true. All consistent. Works. Case 2: Y is guilty. Then Y’s two statements are false, so Z did cheat and Y did cheat; two guilty, impossible. Case 3: Z is guilty. Then Z’s two statements are false, so Z did cheat and X is not lying. If X is truthful, then “Y cheated” must be true, giving two guilty, impossible. Hence X cheated. Answer (a).
#### 2. Question
A laptop was stolen. Three suspects X, Y, and Z were questioned. Exactly one is guilty. Statements:
X: I did not cheat. Y cheated. Y: Z did not cheat. I did not cheat. Z: I did not cheat. X is lying.
Who cheated?
• (d) Cannot be concluded
Answer: (a)
Solution Case 1: X is guilty. Then X’s two statements are false, so X did cheat and Y did not. – Y is innocent, so “Z did not cheat” is true and “I did not cheat” is true. – Z is innocent, so “I did not cheat” is true and “X is lying” is true. All consistent. Works.
Case 2: Y is guilty. Then Y’s two statements are false, so Z did cheat and Y did cheat; two guilty, impossible.
Case 3: Z is guilty. Then Z’s two statements are false, so Z did cheat and X is not lying. If X is truthful, then “Y cheated” must be true, giving two guilty, impossible.
Hence X cheated. Answer (a).
Answer: (a)
Solution Case 1: X is guilty. Then X’s two statements are false, so X did cheat and Y did not. – Y is innocent, so “Z did not cheat” is true and “I did not cheat” is true. – Z is innocent, so “I did not cheat” is true and “X is lying” is true. All consistent. Works.
Case 2: Y is guilty. Then Y’s two statements are false, so Z did cheat and Y did cheat; two guilty, impossible.
Case 3: Z is guilty. Then Z’s two statements are false, so Z did cheat and X is not lying. If X is truthful, then “Y cheated” must be true, giving two guilty, impossible.
Hence X cheated. Answer (a).
• Question 3 of 5 3. Question Let p + q = 12, where p and q are integers. Value I = Maximum value of p × q when p, q are positive integers. Value II = Maximum value of p × q when p ≥ –3, q ≥ –7. Which one of the following is correct? (a) Value I < Value II (b) Value II < Value I (c) Value I = Value II (d) Cannot be determined Correct Answer: (c) Solution: For positives: Value I = 6 × 6 = 36. For Value II: If both negative, maximum = (–3) × (–7) = 21. If both positive, maximum = 6 × 6 = 36. So Value II = 36. But if p = 12, q = 0 (allowed), product = 0, less. So maximum remains 36. Hence, Value I = 36 and Value II = 36. Correction: they are equal. So the right answer is (c) Value I = Value II. Incorrect Answer: (c) Solution: For positives: Value I = 6 × 6 = 36. For Value II: If both negative, maximum = (–3) × (–7) = 21. If both positive, maximum = 6 × 6 = 36. So Value II = 36. But if p = 12, q = 0 (allowed), product = 0, less. So maximum remains 36. Hence, Value I = 36 and Value II = 36. Correction: they are equal. So the right answer is (c) Value I = Value II.
#### 3. Question
Let p + q = 12, where p and q are integers.
Value I = Maximum value of p × q when p, q are positive integers. Value II = Maximum value of p × q when p ≥ –3, q ≥ –7.
Which one of the following is correct?
• (a) Value I < Value II
• (b) Value II < Value I
• (c) Value I = Value II
• (d) Cannot be determined
Answer: (c)
Solution: For positives: Value I = 6 × 6 = 36.
For Value II: If both negative, maximum = (–3) × (–7) = 21. If both positive, maximum = 6 × 6 = 36. So Value II = 36. But if p = 12, q = 0 (allowed), product = 0, less. So maximum remains 36.
Hence, Value I = 36 and Value II = 36. Correction: they are equal. So the right answer is (c) Value I = Value II.
Answer: (c)
Solution: For positives: Value I = 6 × 6 = 36.
For Value II: If both negative, maximum = (–3) × (–7) = 21. If both positive, maximum = 6 × 6 = 36. So Value II = 36. But if p = 12, q = 0 (allowed), product = 0, less. So maximum remains 36.
Hence, Value I = 36 and Value II = 36. Correction: they are equal. So the right answer is (c) Value I = Value II.
• Question 4 of 5 4. Question In a hockey match, team X playing against team Y was behind by 2 goals with 5 minutes remaining. Does team X win the match? Statement I: Team Y did not score any goal in the last 5 minutes. Statement II: Team X scored exactly 2 goals in the last 5 minutes. (a) The question can be answered by using one statement alone, but not the other. (b) The question can be answered by using either statement alone. (c) The question can be answered by using both statements together, but not by either alone. (d) The question cannot be answered even using both statements together. Correct Answer: (c) Solution: With 5 minutes left: Y − X = 2. Using Statement I alone: If Y scores 0, the result depends on how many X scores. Not sufficient. Using Statement II alone: If X scores 2, the result depends on whether Y also scores. Not sufficient. Using both together: X scores 2 and Y scores 0 in the last 5 minutes. Starting from a 2-goal deficit, X exactly catches up. Final outcome is a draw, so X does not win. Together they are sufficient to answer “No”. Hence, option (c). Incorrect Answer: (c) Solution: With 5 minutes left: Y − X = 2. Using Statement I alone: If Y scores 0, the result depends on how many X scores. Not sufficient. Using Statement II alone: If X scores 2, the result depends on whether Y also scores. Not sufficient. Using both together: X scores 2 and Y scores 0 in the last 5 minutes. Starting from a 2-goal deficit, X exactly catches up. Final outcome is a draw, so X does not win. Together they are sufficient to answer “No”. Hence, option (c).
#### 4. Question
In a hockey match, team X playing against team Y was behind by 2 goals with 5 minutes remaining. Does team X win the match?
Statement I: Team Y did not score any goal in the last 5 minutes. Statement II: Team X scored exactly 2 goals in the last 5 minutes.
• (a) The question can be answered by using one statement alone, but not the other.
• (b) The question can be answered by using either statement alone.
• (c) The question can be answered by using both statements together, but not by either alone.
• (d) The question cannot be answered even using both statements together.
Answer: (c)
Solution: With 5 minutes left: Y − X = 2.
Using Statement I alone: If Y scores 0, the result depends on how many X scores. Not sufficient.
Using Statement II alone: If X scores 2, the result depends on whether Y also scores. Not sufficient.
Using both together: X scores 2 and Y scores 0 in the last 5 minutes. Starting from a 2-goal deficit, X exactly catches up. Final outcome is a draw, so X does not win. Together they are sufficient to answer “No”.
Hence, option (c).
Answer: (c)
Solution: With 5 minutes left: Y − X = 2.
Using Statement I alone: If Y scores 0, the result depends on how many X scores. Not sufficient.
Using Statement II alone: If X scores 2, the result depends on whether Y also scores. Not sufficient.
Using both together: X scores 2 and Y scores 0 in the last 5 minutes. Starting from a 2-goal deficit, X exactly catches up. Final outcome is a draw, so X does not win. Together they are sufficient to answer “No”.
Hence, option (c).
• Question 5 of 5 5. Question Three prime numbers p, q, and r, each less than 30, satisfy p − q = q − r. How many distinct possible values can (p + q + r) take? (a) 4 (b) 5 (c) 6 (d) More than 6 Correct Answer: (c) Solution: Again p, q, r are in arithmetic progression. Primes under 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Prime A.P. triples and sums: 3, 5, 7 → 15 3, 7, 11 → 21 5, 11, 17 → 33 7, 13, 19 → 39 11, 17, 23 → 51 17, 23, 29 → 69 3, 11, 19 → 33 5, 17, 29 → 51 Distinct sums: 15, 21, 33, 39, 51, 69. Count is 6. Hence option (c). Incorrect Answer: (c) Solution: Again p, q, r are in arithmetic progression. Primes under 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Prime A.P. triples and sums: 3, 5, 7 → 15 3, 7, 11 → 21 5, 11, 17 → 33 7, 13, 19 → 39 11, 17, 23 → 51 17, 23, 29 → 69 3, 11, 19 → 33 5, 17, 29 → 51 Distinct sums: 15, 21, 33, 39, 51, 69. Count is 6. Hence option (c).
#### 5. Question
Three prime numbers p, q, and r, each less than 30, satisfy p − q = q − r. How many distinct possible values can (p + q + r) take?
• (d) More than 6
Answer: (c)
Solution: Again p, q, r are in arithmetic progression. Primes under 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Prime A.P. triples and sums: 3, 5, 7 → 15 3, 7, 11 → 21 5, 11, 17 → 33 7, 13, 19 → 39 11, 17, 23 → 51 17, 23, 29 → 69 3, 11, 19 → 33 5, 17, 29 → 51
Distinct sums: 15, 21, 33, 39, 51, 69. Count is 6. Hence option (c).
Answer: (c)
Solution: Again p, q, r are in arithmetic progression. Primes under 30: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29.
Prime A.P. triples and sums: 3, 5, 7 → 15 3, 7, 11 → 21 5, 11, 17 → 33 7, 13, 19 → 39 11, 17, 23 → 51 17, 23, 29 → 69 3, 11, 19 → 33 5, 17, 29 → 51
Distinct sums: 15, 21, 33, 39, 51, 69. Count is 6. Hence option (c).
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