UPSC Insta–DART (Daily Aptitude and Reasoning Test) 11 Feb 2026
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 If the sum of the factors of 500 which are perfect squares is S, consider the following statements with reference to S: Statement I: S is divisible by 5. Statement II: The value of (S − 25) is a perfect square. Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Correct Answer: (a) Explanation: Prime factorisation of 500 is: 500 = 2² × 5³ Perfect square factors must have even powers only. Possible powers: 2 → 0, 2 5 → 0, 2 Perfect square factors are: 1, 4, 25, 100 Sum S = 1 + 4 + 25 + 100 = 130 Statement I: 130 is divisible by 5. So, Statement I is correct. Statement II: S − 25 = 130 − 25 = 105, which is not a perfect square. So Statement II is incorrect. Incorrect Answer: (a) Explanation: Prime factorisation of 500 is: 500 = 2² × 5³ Perfect square factors must have even powers only. Possible powers: 2 → 0, 2 5 → 0, 2 Perfect square factors are: 1, 4, 25, 100 Sum S = 1 + 4 + 25 + 100 = 130 Statement I: 130 is divisible by 5. So, Statement I is correct. Statement II: S − 25 = 130 − 25 = 105, which is not a perfect square. So Statement II is incorrect.
#### 1. Question
If the sum of the factors of 500 which are perfect squares is S, consider the following statements with reference to S:
Statement I: S is divisible by 5. Statement II: The value of (S − 25) is a perfect square.
Which of the above statements is/are correct?
• (a) 1 only
• (b) 2 only
• (c) Both 1 and 2
• (d) Neither 1 nor 2
Answer: (a)
Explanation: Prime factorisation of 500 is:
500 = 2² × 5³
Perfect square factors must have even powers only.
Possible powers: 2 → 0, 2 5 → 0, 2
Perfect square factors are: 1, 4, 25, 100
Sum S = 1 + 4 + 25 + 100 = 130
Statement I: 130 is divisible by 5. So, Statement I is correct.
Statement II: S − 25 = 130 − 25 = 105, which is not a perfect square.
So Statement II is incorrect.
Answer: (a)
Explanation: Prime factorisation of 500 is:
500 = 2² × 5³
Perfect square factors must have even powers only.
Possible powers: 2 → 0, 2 5 → 0, 2
Perfect square factors are: 1, 4, 25, 100
Sum S = 1 + 4 + 25 + 100 = 130
Statement I: 130 is divisible by 5. So, Statement I is correct.
Statement II: S − 25 = 130 − 25 = 105, which is not a perfect square.
So Statement II is incorrect.
• Question 2 of 5 2. Question In the expression 90 30 12 3 1, each * is to be replaced by one of the symbols +, −, ÷, with each symbol used at most two times. What is the smallest non-negative integer value of the expression? (a) 0 (b) 1 (c) 2 (d) 3 Correct Answer: (a) Explanation: Aim for 0, because 0 is the smallest possible non-negative integer. Choose: 90 ÷ 30 − 12 ÷ 3 + 1 Now evaluate: 90 ÷ 30 = 3 12 ÷ 3 = 4 So expression becomes: 3 − 4 + 1 = 0 This uses ÷ exactly two times, − once, + once (all within limits). Since 0 is achievable, it is the smallest non-negative integer possible. Incorrect Answer: (a) Explanation: Aim for 0, because 0 is the smallest possible non-negative integer. Choose: 90 ÷ 30 − 12 ÷ 3 + 1 Now evaluate: 90 ÷ 30 = 3 12 ÷ 3 = 4 So expression becomes: 3 − 4 + 1 = 0 This uses ÷ exactly two times, − once, + once (all within limits). Since 0 is achievable, it is the smallest non-negative integer possible.
#### 2. Question
In the expression 90 30 12 3 1, each * is to be replaced by one of the symbols +, −, ÷, with each symbol used at most two times. What is the smallest non-negative integer value of the expression?
Answer: (a)
Explanation: Aim for 0, because 0 is the smallest possible non-negative integer.
Choose: 90 ÷ 30 − 12 ÷ 3 + 1 Now evaluate: 90 ÷ 30 = 3 12 ÷ 3 = 4 So expression becomes: 3 − 4 + 1 = 0
This uses ÷ exactly two times, − once, + once (all within limits).
Since 0 is achievable, it is the smallest non-negative integer possible.
Answer: (a)
Explanation: Aim for 0, because 0 is the smallest possible non-negative integer.
Choose: 90 ÷ 30 − 12 ÷ 3 + 1 Now evaluate: 90 ÷ 30 = 3 12 ÷ 3 = 4 So expression becomes: 3 − 4 + 1 = 0
This uses ÷ exactly two times, − once, + once (all within limits).
Since 0 is achievable, it is the smallest non-negative integer possible.
• Question 3 of 5 3. Question A certain number leaves a remainder of 125 when divided by 168. What will be the remainder when the number is divided by 24? (a) 5 (b) 11 (c) 13 (d) 17 Correct Answer: (a) Solution: Let the number be N. N = 168 × m + 125 Now, 168 = 24 × 7 So, N = (24 × 7 × m) + 125 Write 125 in terms of 24: 125 = 24 × 5 + 5 So, N = 24(7m + 5) + 5 Hence, the remainder when N is divided by 24 is 5. Incorrect Answer: (a) Solution: Let the number be N. N = 168 × m + 125 Now, 168 = 24 × 7 So, N = (24 × 7 × m) + 125 Write 125 in terms of 24: 125 = 24 × 5 + 5 So, N = 24(7m + 5) + 5 Hence, the remainder when N is divided by 24 is 5.
#### 3. Question
A certain number leaves a remainder of 125 when divided by 168. What will be the remainder when the number is divided by 24?
Answer: (a)
Solution: Let the number be N.
N = 168 × m + 125
Now, 168 = 24 × 7
So, N = (24 × 7 × m) + 125
Write 125 in terms of 24: 125 = 24 × 5 + 5
So, N = 24(7m + 5) + 5
Hence, the remainder when N is divided by 24 is 5.
Answer: (a)
Solution: Let the number be N.
N = 168 × m + 125
Now, 168 = 24 × 7
So, N = (24 × 7 × m) + 125
Write 125 in terms of 24: 125 = 24 × 5 + 5
So, N = 24(7m + 5) + 5
Hence, the remainder when N is divided by 24 is 5.
• Question 4 of 5 4. Question From the first 300 natural numbers, how many are not divisible by any of 2, 3, 5, or 7? (a) 50 (b) 61 (c) 69 (d) 63 Correct Answer: (c) Explanation: we need numbers up to 300 that have no factors of 2, 3, 5, or 7. These fall into three groups: The Number 1: Count = 1 Prime Numbers (11 to 300): There are 62 primes total. Exclude 2, 3, 5, and 7. Remaining = 62 – 4 = 58 Composite Numbers (Products of Primes ≥ 11): Multiples of 11: 121, 143, 187, 209, 253 (5 numbers) Multiples of 13: 169, 221, 247, 299 (4 numbers) Multiples of 17: 289 (1 number) Total = 10. Total Count: 1 + 58 + 10 = 69 Incorrect Answer: (c) Explanation: we need numbers up to 300 that have no factors of 2, 3, 5, or 7. These fall into three groups: The Number 1: Count = 1 Prime Numbers (11 to 300): There are 62 primes total. Exclude 2, 3, 5, and 7. Remaining = 62 – 4 = 58 Composite Numbers (Products of Primes ≥ 11): Multiples of 11: 121, 143, 187, 209, 253 (5 numbers) Multiples of 13: 169, 221, 247, 299 (4 numbers) Multiples of 17: 289 (1 number) Total = 10. Total Count: 1 + 58 + 10 = 69
#### 4. Question
From the first 300 natural numbers, how many are not divisible by any of 2, 3, 5, or 7?
Answer: (c)
Explanation:
we need numbers up to 300 that have no factors of 2, 3, 5, or 7. These fall into three groups:
• The Number 1: Count = 1
• Count = 1
• Prime Numbers (11 to 300): There are 62 primes total. Exclude 2, 3, 5, and 7. Remaining = 62 – 4 = 58
• There are 62 primes total. Exclude 2, 3, 5, and 7.
• Remaining = 62 – 4 = 58
• Composite Numbers (Products of Primes ≥ 11): Multiples of 11: 121, 143, 187, 209, 253 (5 numbers) Multiples of 13: 169, 221, 247, 299 (4 numbers) Multiples of 17: 289 (1 number) Total = 10.
• Multiples of 11: 121, 143, 187, 209, 253 (5 numbers)
• Multiples of 13: 169, 221, 247, 299 (4 numbers)
• Multiples of 17: 289 (1 number)
• Total = 10.
Total Count:
1 + 58 + 10 = 69
Answer: (c)
Explanation:
we need numbers up to 300 that have no factors of 2, 3, 5, or 7. These fall into three groups:
• The Number 1: Count = 1
• Count = 1
• Prime Numbers (11 to 300): There are 62 primes total. Exclude 2, 3, 5, and 7. Remaining = 62 – 4 = 58
• There are 62 primes total. Exclude 2, 3, 5, and 7.
• Remaining = 62 – 4 = 58
• Composite Numbers (Products of Primes ≥ 11): Multiples of 11: 121, 143, 187, 209, 253 (5 numbers) Multiples of 13: 169, 221, 247, 299 (4 numbers) Multiples of 17: 289 (1 number) Total = 10.
• Multiples of 11: 121, 143, 187, 209, 253 (5 numbers)
• Multiples of 13: 169, 221, 247, 299 (4 numbers)
• Multiples of 17: 289 (1 number)
• Total = 10.
Total Count:
1 + 58 + 10 = 69
• Question 5 of 5 5. Question What is the maximum value of n such that 27 × 81 × 625 × 125 × 45 is divisible by 45ⁿ? (a) 2 (b) 3 (c) 4 (d) 5 Correct Answer: (b) Explanation: 45ⁿ = 3²ⁿ × 5ⁿ Prime factorisation of the given expression: 27 = 3³ 81 = 3⁴ 625 = 5⁴ 125 = 5³ 45 = 3² × 5¹ Total powers: Power of 3 = 3 + 4 + 2 = 9 Power of 5 = 4 + 3 + 1 = 8 For divisibility by 45ⁿ: 3²ⁿ ≤ 9 ⇒ n ≤ 4 5ⁿ ≤ 8 ⇒ n ≤ 3 The limiting factor is the power of 5. Therefore, maximum value of n = 3. Hence, option (b) is correct. Incorrect Answer: (b) Explanation: 45ⁿ = 3²ⁿ × 5ⁿ Prime factorisation of the given expression: 27 = 3³ 81 = 3⁴ 625 = 5⁴ 125 = 5³ 45 = 3² × 5¹ Total powers: Power of 3 = 3 + 4 + 2 = 9 Power of 5 = 4 + 3 + 1 = 8 For divisibility by 45ⁿ: 3²ⁿ ≤ 9 ⇒ n ≤ 4 5ⁿ ≤ 8 ⇒ n ≤ 3 The limiting factor is the power of 5. Therefore, maximum value of n = 3. Hence, option (b) is correct.
#### 5. Question
What is the maximum value of n such that 27 × 81 × 625 × 125 × 45 is divisible by 45ⁿ?
Answer: (b)
Explanation: 45ⁿ = 3²ⁿ × 5ⁿ
Prime factorisation of the given expression: 27 = 3³ 81 = 3⁴ 625 = 5⁴ 125 = 5³ 45 = 3² × 5¹
Total powers: Power of 3 = 3 + 4 + 2 = 9 Power of 5 = 4 + 3 + 1 = 8
For divisibility by 45ⁿ: 3²ⁿ ≤ 9 ⇒ n ≤ 4 5ⁿ ≤ 8 ⇒ n ≤ 3
The limiting factor is the power of 5.
Therefore, maximum value of n = 3.
Hence, option (b) is correct.
Answer: (b)
Explanation: 45ⁿ = 3²ⁿ × 5ⁿ
Prime factorisation of the given expression: 27 = 3³ 81 = 3⁴ 625 = 5⁴ 125 = 5³ 45 = 3² × 5¹
Total powers: Power of 3 = 3 + 4 + 2 = 9 Power of 5 = 4 + 3 + 1 = 8
For divisibility by 45ⁿ: 3²ⁿ ≤ 9 ⇒ n ≤ 4 5ⁿ ≤ 8 ⇒ n ≤ 3
The limiting factor is the power of 5.
Therefore, maximum value of n = 3.
Hence, option (b) is correct.
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