UPSC Insta–DART (Daily Aptitude and Reasoning Test) 9 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 360 which are perfect squares is S, consider the following statements with reference to S: Statement I: The value of (S + 1) is odd. Statement II: The value of (S − 16) 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: First, find the prime factorisation of 360. 360 = 2³ × 3² × 5¹ For a factor to be a perfect square, all prime exponents must be even. Possible even powers: 2 → 0, 2 3 → 0, 2 5 → 0 So perfect square factors are: 1, 4, 9, 36 Sum S = 1 + 4 + 9 + 36 = 50 Now check the statements. Statement I: S + 1 = 50 + 1 = 51, which is odd. So, Statement I is correct. Statement II: S − 16 = 50 − 16 = 34, which is not a perfect square. So, Statement II is incorrect. Hence, only Statement I is correct. Incorrect Answer: (a) Explanation: First, find the prime factorisation of 360. 360 = 2³ × 3² × 5¹ For a factor to be a perfect square, all prime exponents must be even. Possible even powers: 2 → 0, 2 3 → 0, 2 5 → 0 So perfect square factors are: 1, 4, 9, 36 Sum S = 1 + 4 + 9 + 36 = 50 Now check the statements. Statement I: S + 1 = 50 + 1 = 51, which is odd. So, Statement I is correct. Statement II: S − 16 = 50 − 16 = 34, which is not a perfect square. So, Statement II is incorrect. Hence, only Statement I is correct.
#### 1. Question
If the sum of the factors of 360 which are perfect squares is S, consider the following statements with reference to S:
Statement I: The value of (S + 1) is odd. Statement II: The value of (S − 16) 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: First, find the prime factorisation of 360.
360 = 2³ × 3² × 5¹
For a factor to be a perfect square, all prime exponents must be even.
Possible even powers: 2 → 0, 2 3 → 0, 2 5 → 0
So perfect square factors are: 1, 4, 9, 36
Sum S = 1 + 4 + 9 + 36 = 50
Now check the statements.
Statement I: S + 1 = 50 + 1 = 51, which is odd. So, Statement I is correct.
Statement II: S − 16 = 50 − 16 = 34, which is not a perfect square. So, Statement II is incorrect.
Hence, only Statement I is correct.
Answer: (a)
Explanation: First, find the prime factorisation of 360.
360 = 2³ × 3² × 5¹
For a factor to be a perfect square, all prime exponents must be even.
Possible even powers: 2 → 0, 2 3 → 0, 2 5 → 0
So perfect square factors are: 1, 4, 9, 36
Sum S = 1 + 4 + 9 + 36 = 50
Now check the statements.
Statement I: S + 1 = 50 + 1 = 51, which is odd. So, Statement I is correct.
Statement II: S − 16 = 50 − 16 = 34, which is not a perfect square. So, Statement II is incorrect.
Hence, only Statement I is correct.
• Question 2 of 5 2. Question A certain number gives a remainder of 47 when divided by 96. What will be the remainder when the number is divided by 12? (a) 3 (b) 5 (c) 7 (d) 11 Correct Answer: (d) Solution: Let the number be N. N = 96 × k + 47 Now, 96 = 12 × 8 So, N = (12 × 8 × k) + 47 Write 47 in terms of 12: 47 = 12 × 3 + 11 So, N = 12(8k + 3) + 11 Hence, the remainder when N is divided by 12 is 11. Incorrect Answer: (d) Solution: Let the number be N. N = 96 × k + 47 Now, 96 = 12 × 8 So, N = (12 × 8 × k) + 47 Write 47 in terms of 12: 47 = 12 × 3 + 11 So, N = 12(8k + 3) + 11 Hence, the remainder when N is divided by 12 is 11.
#### 2. Question
A certain number gives a remainder of 47 when divided by 96. What will be the remainder when the number is divided by 12?
Answer: (d)
Solution: Let the number be N.
N = 96 × k + 47
Now, 96 = 12 × 8
So, N = (12 × 8 × k) + 47
Write 47 in terms of 12: 47 = 12 × 3 + 11
So, N = 12(8k + 3) + 11
Hence, the remainder when N is divided by 12 is 11.
Answer: (d)
Solution: Let the number be N.
N = 96 × k + 47
Now, 96 = 12 × 8
So, N = (12 × 8 × k) + 47
Write 47 in terms of 12: 47 = 12 × 3 + 11
So, N = 12(8k + 3) + 11
Hence, the remainder when N is divided by 12 is 11.
• Question 3 of 5 3. Question From the first 100 natural numbers, how many are not divisible by any of 2, 3, or 5? (a) 21 (b) 22 (c) 23 (d) 26 Correct Answer: (d) Explanation Total numbers: 100 Remove multiples of 2: (50 numbers). Remaining: 50 odd numbers. Remove multiples of 3 from the remaining: We need odd multiples of 3 up to 100 (3, 9, 15, …, 99). Count = 17 numbers. Remaining: $50 – 17 = 33$. Remove multiples of 5 from the remaining: We need odd multiples of 5 that are not multiples of 3. (5, 25, 35, 55, 65, 85, 95). Note: 15, 45, 75 were already removed in step 3. Count = 7 numbers. Remaining: 33 – 7 = 26. Incorrect Answer: (d) Explanation Total numbers: 100 Remove multiples of 2: (50 numbers). Remaining: 50 odd numbers. Remove multiples of 3 from the remaining: We need odd multiples of 3 up to 100 (3, 9, 15, …, 99). Count = 17 numbers. Remaining: $50 – 17 = 33$. Remove multiples of 5 from the remaining: We need odd multiples of 5 that are not multiples of 3. (5, 25, 35, 55, 65, 85, 95). Note: 15, 45, 75 were already removed in step 3. Count = 7 numbers. Remaining: 33 – 7 = 26.
#### 3. Question
From the first 100 natural numbers, how many are not divisible by any of 2, 3, or 5?
Answer: (d)
Explanation
Total numbers: 100
Remove multiples of 2: (50 numbers). Remaining: 50 odd numbers.
Remove multiples of 3 from the remaining:
We need odd multiples of 3 up to 100 (3, 9, 15, …, 99).
Count = 17 numbers.
Remaining: $50 – 17 = 33$.
Remove multiples of 5 from the remaining:
We need odd multiples of 5 that are not multiples of 3.
(5, 25, 35, 55, 65, 85, 95). Note: 15, 45, 75 were already removed in step 3.
Count = 7 numbers.
Remaining: 33 – 7 = 26.
Answer: (d)
Explanation
Total numbers: 100
Remove multiples of 2: (50 numbers). Remaining: 50 odd numbers.
Remove multiples of 3 from the remaining:
We need odd multiples of 3 up to 100 (3, 9, 15, …, 99).
Count = 17 numbers.
Remaining: $50 – 17 = 33$.
Remove multiples of 5 from the remaining:
We need odd multiples of 5 that are not multiples of 3.
(5, 25, 35, 55, 65, 85, 95). Note: 15, 45, 75 were already removed in step 3.
Count = 7 numbers.
Remaining: 33 – 7 = 26.
• Question 4 of 5 4. Question What is the maximum value of n such that 16 × 64 × 125 × 625 × 20 is divisible by 20ⁿ? (a) 2 (b) 3 (c) 4 (d) 5 Correct Answer: (b) Explanation: 20ⁿ = 2²ⁿ × 5ⁿ Prime factorisation of the given expression: 16 = 2⁴ 64 = 2⁶ 125 = 5³ 625 = 5⁴ 20 = 2² × 5¹ Total powers: Power of 2 = 4 + 6 + 2 = 12 Power of 5 = 3 + 4 + 1 = 8 For divisibility by 20ⁿ, we need: 2²ⁿ ≤ 12 ⇒ n ≤ 6 5ⁿ ≤ 8 ⇒ n ≤ 3 The limiting factor is the lower value, i.e. n = 3. Hence, option (b) is correct. Incorrect Answer: (b) Explanation: 20ⁿ = 2²ⁿ × 5ⁿ Prime factorisation of the given expression: 16 = 2⁴ 64 = 2⁶ 125 = 5³ 625 = 5⁴ 20 = 2² × 5¹ Total powers: Power of 2 = 4 + 6 + 2 = 12 Power of 5 = 3 + 4 + 1 = 8 For divisibility by 20ⁿ, we need: 2²ⁿ ≤ 12 ⇒ n ≤ 6 5ⁿ ≤ 8 ⇒ n ≤ 3 The limiting factor is the lower value, i.e. n = 3. Hence, option (b) is correct.
#### 4. Question
What is the maximum value of n such that 16 × 64 × 125 × 625 × 20 is divisible by 20ⁿ?
Answer: (b)
Explanation: 20ⁿ = 2²ⁿ × 5ⁿ
Prime factorisation of the given expression: 16 = 2⁴ 64 = 2⁶ 125 = 5³ 625 = 5⁴ 20 = 2² × 5¹
Total powers: Power of 2 = 4 + 6 + 2 = 12 Power of 5 = 3 + 4 + 1 = 8
For divisibility by 20ⁿ, we need: 2²ⁿ ≤ 12 ⇒ n ≤ 6 5ⁿ ≤ 8 ⇒ n ≤ 3
The limiting factor is the lower value, i.e. n = 3.
Hence, option (b) is correct.
Answer: (b)
Explanation: 20ⁿ = 2²ⁿ × 5ⁿ
Prime factorisation of the given expression: 16 = 2⁴ 64 = 2⁶ 125 = 5³ 625 = 5⁴ 20 = 2² × 5¹
Total powers: Power of 2 = 4 + 6 + 2 = 12 Power of 5 = 3 + 4 + 1 = 8
For divisibility by 20ⁿ, we need: 2²ⁿ ≤ 12 ⇒ n ≤ 6 5ⁿ ≤ 8 ⇒ n ≤ 3
The limiting factor is the lower value, i.e. n = 3.
Hence, option (b) is correct.
• Question 5 of 5 5. Question If N² = 123454321, then how many digits does the number N have? (a) 4 (b) 5 (c) 6 (d) 7 Correct Answer: (b) Explanation: Observe the pattern of squares of numbers made only of 1’s: 1² = 1 11² = 121 111² = 12321 1111² = 1234321 11111² = 123454321 The given number 123454321 matches the square of 11111. So N = 11111, which has 5 digits. Hence, option (b) is correct. Incorrect Answer: (b) Explanation: Observe the pattern of squares of numbers made only of 1’s: 1² = 1 11² = 121 111² = 12321 1111² = 1234321 11111² = 123454321 The given number 123454321 matches the square of 11111. So N = 11111, which has 5 digits. Hence, option (b) is correct.
#### 5. Question
If N² = 123454321, then how many digits does the number N have?
Answer: (b)
Explanation: Observe the pattern of squares of numbers made only of 1’s: 1² = 1 11² = 121 111² = 12321 1111² = 1234321 11111² = 123454321
The given number 123454321 matches the square of 11111. So N = 11111, which has 5 digits.
Hence, option (b) is correct.
Answer: (b)
Explanation: Observe the pattern of squares of numbers made only of 1’s: 1² = 1 11² = 121 111² = 12321 1111² = 1234321 11111² = 123454321
The given number 123454321 matches the square of 11111. So N = 11111, which has 5 digits.
Hence, option (b) is correct.
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