UPSC Insta–DART (Daily Aptitude and Reasoning Test) 3 Nov 2025

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 Three Statements S1, S2 and S3 are given below followed by a Question: S1: P is younger than Q, but older than R and S. S2: Q is the oldest. S3: R is older than S. Question: Who among P, Q, R and S is the youngest? Which one of the following is correct in respect of the above Statements and the Question? (a) S1 alone is sufficient to answer the Question. (b) S1 and S2 together are sufficient to answer the Question. (c) S2 and S3 together are sufficient to answer the Question. (d) S1 and S3 together are sufficient to answer the Question. Correct Answer: D Solution: Given that, S1: P is younger than Q, but older than R and S → Q > P > R and Q > P > S; order between R and S unknown. S2: Q is the oldest → places Q at top. S3: R is older than S → R > S. Combining S1 and S3: Q > P > R > S, so S is the youngest. S2 is not needed once S1 fixes Q>P and S3 fixes R>S. Hence option (d) S1 and S3 together are sufficient. Incorrect Answer: D Solution: Given that, S1: P is younger than Q, but older than R and S → Q > P > R and Q > P > S; order between R and S unknown. S2: Q is the oldest → places Q at top. S3: R is older than S → R > S. Combining S1 and S3: Q > P > R > S, so S is the youngest. S2 is not needed once S1 fixes Q>P and S3 fixes R>S. Hence option (d) S1 and S3 together are sufficient.

#### 1. Question

Three Statements S1, S2 and S3 are given below followed by a Question:

S1: P is younger than Q, but older than R and S. S2: Q is the oldest. S3: R is older than S.

Question: Who among P, Q, R and S is the youngest?

Which one of the following is correct in respect of the above Statements and the Question?

• (a) S1 alone is sufficient to answer the Question.

• (b) S1 and S2 together are sufficient to answer the Question.

• (c) S2 and S3 together are sufficient to answer the Question.

• (d) S1 and S3 together are sufficient to answer the Question.

Answer: D

Solution:

Given that,

S1: P is younger than Q, but older than R and S → Q > P > R and Q > P > S; order between R and S unknown. S2: Q is the oldest → places Q at top. S3: R is older than S → R > S.

Combining S1 and S3: Q > P > R > S, so S is the youngest. S2 is not needed once S1 fixes Q>P and S3 fixes R>S.

Hence option (d) S1 and S3 together are sufficient.

Answer: D

Solution:

Given that,

S1: P is younger than Q, but older than R and S → Q > P > R and Q > P > S; order between R and S unknown. S2: Q is the oldest → places Q at top. S3: R is older than S → R > S.

Combining S1 and S3: Q > P > R > S, so S is the youngest. S2 is not needed once S1 fixes Q>P and S3 fixes R>S.

Hence option (d) S1 and S3 together are sufficient.

• Question 2 of 5 2. Question There are 30 equally spaced points on the circumference of a circle. What is the maximum number of equilateral triangles that can be formed? (a) 10 (b) 12 (c) 15 (d) 20 Correct Solution: Total central angle = 360°. Each gap between consecutive points = 360° ÷ 30 = 12°. For equilateral triangle, vertex gap = 120°. Number of steps = 120° ÷ 12° = 10. So from each point, an equilateral triangle can be formed by jumping 10 points each time. Hence, maximum triangles = 30 ÷ 3 = 10. Correct answer: (a) 10 Incorrect Solution: Total central angle = 360°. Each gap between consecutive points = 360° ÷ 30 = 12°. For equilateral triangle, vertex gap = 120°. Number of steps = 120° ÷ 12° = 10. So from each point, an equilateral triangle can be formed by jumping 10 points each time. Hence, maximum triangles = 30 ÷ 3 = 10. Correct answer: (a) 10

#### 2. Question

There are 30 equally spaced points on the circumference of a circle. What is the maximum number of equilateral triangles that can be formed?

Solution: Total central angle = 360°. Each gap between consecutive points = 360° ÷ 30 = 12°. For equilateral triangle, vertex gap = 120°. Number of steps = 120° ÷ 12° = 10. So from each point, an equilateral triangle can be formed by jumping 10 points each time. Hence, maximum triangles = 30 ÷ 3 = 10.

Correct answer: (a) 10

• Question 3 of 5 3. Question All members of a committee visited Jaipur and stayed in a hotel. On the first day, 70% attended a workshop, 55% attended a keynote, and 15% stayed back in the hotel. Which of the following conclusion(s) can be drawn? 40% members attended both workshop and keynote. 30% members attended only the workshop. Select the correct answer using the code given below: (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Correct Answer: C Solution: Given that, Workshop = 70%, Keynote = 55%, Stayed back = 15% ⇒ At least one activity = 85%. Let total = 100. Both = 70 + 55 − 85 = 40. Only workshop = 70 − 40 = 30. Only keynote = 55 − 40 = 15. Both = 40% → Correct. Only workshop = 30% → Correct. Hence, option (c) is correct. Incorrect Answer: C Solution: Given that, Workshop = 70%, Keynote = 55%, Stayed back = 15% ⇒ At least one activity = 85%. Let total = 100. Both = 70 + 55 − 85 = 40. Only workshop = 70 − 40 = 30. Only keynote = 55 − 40 = 15. Both = 40% → Correct. Only workshop = 30% → Correct. Hence, option (c) is correct.

#### 3. Question

All members of a committee visited Jaipur and stayed in a hotel. On the first day, 70% attended a workshop, 55% attended a keynote, and 15% stayed back in the hotel. Which of the following conclusion(s) can be drawn?

• 40% members attended both workshop and keynote.

• 30% members attended only the workshop.

Select the correct answer using the code given below:

• (a) 1 only

• (b) 2 only

• (c) Both 1 and 2

• (d) Neither 1 nor 2

Given that, Workshop = 70%, Keynote = 55%, Stayed back = 15% ⇒ At least one activity = 85%.

Let total = 100. Both = 70 + 55 − 85 = 40. Only workshop = 70 − 40 = 30. Only keynote = 55 − 40 = 15.

• Both = 40% → Correct.

• Only workshop = 30% → Correct.

Hence, option (c) is correct.

Given that, Workshop = 70%, Keynote = 55%, Stayed back = 15% ⇒ At least one activity = 85%.

Let total = 100. Both = 70 + 55 − 85 = 40. Only workshop = 70 − 40 = 30. Only keynote = 55 − 40 = 15.

• Both = 40% → Correct.

• Only workshop = 30% → Correct.

Hence, option (c) is correct.

• Question 4 of 5 4. Question What is the maximum value of n such that 700 × 343 × 175 × 4900 × 1125 × 77 is divisible by 35^n? (a) 6 (b) 7 (c) 8 (d) 9 Correct Answer: (c) Solution: 35^n = 7^n × 5^n. Factorise: 700 = 2^2 × 5^2 × 7 343 = 7^3 175 = 5^2 × 7 4900 = 2^2 × 5^2 × 7^2 1125 = 3^2 × 5^3 77 = 7 × 11 Power of 5 = 2 + 0 + 2 + 2 + 3 + 0 = 9. Power of 7 = 1 + 3 + 1 + 2 + 0 + 1 = 8. Minimum = 8, so n = 8. Incorrect Answer: (c) Solution: 35^n = 7^n × 5^n. Factorise: 700 = 2^2 × 5^2 × 7 343 = 7^3 175 = 5^2 × 7 4900 = 2^2 × 5^2 × 7^2 1125 = 3^2 × 5^3 77 = 7 × 11 Power of 5 = 2 + 0 + 2 + 2 + 3 + 0 = 9. Power of 7 = 1 + 3 + 1 + 2 + 0 + 1 = 8. Minimum = 8, so n = 8.

#### 4. Question

What is the maximum value of n such that 700 × 343 × 175 × 4900 × 1125 × 77 is divisible by 35^n?

Answer: (c) Solution: 35^n = 7^n × 5^n. Factorise: 700 = 2^2 × 5^2 × 7 343 = 7^3 175 = 5^2 × 7 4900 = 2^2 × 5^2 × 7^2 1125 = 3^2 × 5^3 77 = 7 × 11 Power of 5 = 2 + 0 + 2 + 2 + 3 + 0 = 9. Power of 7 = 1 + 3 + 1 + 2 + 0 + 1 = 8. Minimum = 8, so n = 8.

• Question 5 of 5 5. Question What is X in the sequence 5, 10, 8, 16, 14, 28, X? (a) 24 (b) 25 (c) 26 (d) 30 Correct Answer: (c) Solution: Pattern alternates: ×2, −2, ×2, −2, ×2, −2 5×2=10, 10−2=8, 8×2=16, 16−2=14, 14×2=28, 28−2=26 So X = 26. Incorrect Answer: (c) Solution: Pattern alternates: ×2, −2, ×2, −2, ×2, −2 5×2=10, 10−2=8, 8×2=16, 16−2=14, 14×2=28, 28−2=26 So X = 26.

#### 5. Question

What is X in the sequence 5, 10, 8, 16, 14, 28, X?

Answer: (c) Solution: Pattern alternates: ×2, −2, ×2, −2, ×2, −2 5×2=10, 10−2=8, 8×2=16, 16−2=14, 14×2=28, 28−2=26 So X = 26.

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