UPSC Insta–DART (Daily Aptitude and Reasoning Test) 7 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 Let x be a real number between 0 and 1. Which of the following statements is/are correct? x² > x³ x > √x Select the correct answer using the code given below: (a) I only (b) II only (c) Both I and II (d) Neither I nor II Correct Answer:(a) Given that: 0 < x < 1 Statement I: x2 > x3 This statement is correct. For example, let value of x be 0.5. x2 = 0.52 = 0.25 x3 = 0.53 = 0.125 Statement II: x > √x Let’s check with the help of an example. If x = 0.25 Then, √x = √0.25 = 0.5 So, it’s evident that this statement is not correct. Hence, option (a) is the correct answer. Incorrect Answer:(a) Given that: 0 < x < 1 Statement I: x2 > x3 This statement is correct. For example, let value of x be 0.5. x2 = 0.52 = 0.25 x3 = 0.53 = 0.125 Statement II: x > √x Let’s check with the help of an example. If x = 0.25 Then, √x = √0.25 = 0.5 So, it’s evident that this statement is not correct. Hence, option (a) is the correct answer.
#### 1. Question
Let x be a real number between 0 and 1. Which of the following statements is/are correct?
Select the correct answer using the code given below:
• (a) I only
• (b) II only
• (c) Both I and II
• (d) Neither I nor II
Answer:(a)
Given that: 0 < x < 1
Statement I: x2 > x3
This statement is correct.
For example, let value of x be 0.5. x2 = 0.52 = 0.25
x3 = 0.53 = 0.125
Statement II: x > √x
Let’s check with the help of an example. If x = 0.25
Then, √x = √0.25 = 0.5
So, it’s evident that this statement is not correct. Hence, option (a) is the correct answer.
Answer:(a)
Given that: 0 < x < 1
Statement I: x2 > x3
This statement is correct.
For example, let value of x be 0.5. x2 = 0.52 = 0.25
x3 = 0.53 = 0.125
Statement II: x > √x
Let’s check with the help of an example. If x = 0.25
Then, √x = √0.25 = 0.5
So, it’s evident that this statement is not correct. Hence, option (a) is the correct answer.
• Question 2 of 5 2. Question How many numbers are there less than 200 that cannot be written as a multiple of a perfect cube greater than 1? (a) 166 (b) 167 (c) 168 (d) 169 Correct Answer: (b) Solution: Given that, Perfect cubes greater than 1 and less than 200 are: 8, 27, 64, 125 Now, Total numbers less than 200 = 1 to 199 = 199 Multiples of 8 = 199/8 = 24 Multiples of 27 = 199/27 = 7 Multiples of 64 = 199/64 = 3 Multiples of 125 = 199/125 = 1 Overlap (64 is also counted in multiples of 8): Overlap = multiples of 64 = 199/64 = 3 Total numbers that CAN be written as multiple of a perfect cube > 1 = (24 + 7 + 3 + 1) − 3 = 32 Therefore, numbers that CANNOT be written = 199 − 32 = 167 Hence, option (b) is correct. Incorrect Answer: (b) Solution: Given that, Perfect cubes greater than 1 and less than 200 are: 8, 27, 64, 125 Now, Total numbers less than 200 = 1 to 199 = 199 Multiples of 8 = 199/8 = 24 Multiples of 27 = 199/27 = 7 Multiples of 64 = 199/64 = 3 Multiples of 125 = 199/125 = 1 Overlap (64 is also counted in multiples of 8): Overlap = multiples of 64 = 199/64 = 3 Total numbers that CAN be written as multiple of a perfect cube > 1 = (24 + 7 + 3 + 1) − 3 = 32 Therefore, numbers that CANNOT be written = 199 − 32 = 167 Hence, option (b) is correct.
#### 2. Question
How many numbers are there less than 200 that cannot be written as a multiple of a perfect cube greater than 1?
Answer: (b)
Given that,
Perfect cubes greater than 1 and less than 200 are: 8, 27, 64, 125
Total numbers less than 200 = 1 to 199 = 199
Multiples of 8 = 199/8 = 24 Multiples of 27 = 199/27 = 7 Multiples of 64 = 199/64 = 3 Multiples of 125 = 199/125 = 1
Overlap (64 is also counted in multiples of 8):
Overlap = multiples of 64 = 199/64 = 3
Total numbers that CAN be written as multiple of a perfect cube > 1 = (24 + 7 + 3 + 1) − 3 = 32
Therefore, numbers that CANNOT be written = 199 − 32 = 167
Hence, option (b) is correct.
Answer: (b)
Given that,
Perfect cubes greater than 1 and less than 200 are: 8, 27, 64, 125
Total numbers less than 200 = 1 to 199 = 199
Multiples of 8 = 199/8 = 24 Multiples of 27 = 199/27 = 7 Multiples of 64 = 199/64 = 3 Multiples of 125 = 199/125 = 1
Overlap (64 is also counted in multiples of 8):
Overlap = multiples of 64 = 199/64 = 3
Total numbers that CAN be written as multiple of a perfect cube > 1 = (24 + 7 + 3 + 1) − 3 = 32
Therefore, numbers that CANNOT be written = 199 − 32 = 167
Hence, option (b) is correct.
• Question 3 of 5 3. Question If the sum of n terms of the series 1 + 2 + 4 + 8 + 16 + 32 + … is more than 2000, then what is the minimum value of n? (a) 10 (b) 11 (c) 12 (d) 13 Correct Answer: (b) Solution: Given that, The series 1 + 2 + 4 + 8 + … is a GP with first term a = 1 and common ratio r = 2. Sum of n terms: S = 2^n − 1 Now, 2^n − 1 > 2000 2^n > 2001 Now check powers of 2: 2^10 = 1024 2^11 = 2048 2048 > 2001 So minimum n = 11 Hence, option (b) is correct. Incorrect Answer: (b) Solution: Given that, The series 1 + 2 + 4 + 8 + … is a GP with first term a = 1 and common ratio r = 2. Sum of n terms: S = 2^n − 1 Now, 2^n − 1 > 2000 2^n > 2001 Now check powers of 2: 2^10 = 1024 2^11 = 2048 2048 > 2001 So minimum n = 11 Hence, option (b) is correct.
#### 3. Question
If the sum of n terms of the series 1 + 2 + 4 + 8 + 16 + 32 + … is more than 2000, then what is the minimum value of n?
Answer: (b)
Given that,
The series 1 + 2 + 4 + 8 + … is a GP with first term a = 1 and common ratio r = 2.
Sum of n terms:
S = 2^n − 1
2^n − 1 > 2000 2^n > 2001
Now check powers of 2:
2^10 = 1024 2^11 = 2048
2048 > 2001
So minimum n = 11
Hence, option (b) is correct.
Answer: (b)
Given that,
The series 1 + 2 + 4 + 8 + … is a GP with first term a = 1 and common ratio r = 2.
Sum of n terms:
S = 2^n − 1
2^n − 1 > 2000 2^n > 2001
Now check powers of 2:
2^10 = 1024 2^11 = 2048
2048 > 2001
So minimum n = 11
Hence, option (b) is correct.
• Question 4 of 5 4. Question A six-digit number N is formed using the digits 0, 3, 4 and 7 only. Each of the digits is used at least once. It was found that N is divisible by 4. What is the hundred’s digit of the smallest possible such six-digit number? (a) 0 (b) 3 (c) 4 (d) 7 Correct Answer: (d) Solution: The “Push-Right” Strategy: To get the smallest number, we want the largest digits as far to the right as possible. Start Small: The number must start with 3 0 0 … to be minimal. The “7” Problem: We must use the digit 7, but: It cannot be the Unit digit (it’s odd). It cannot be the Tens digit (neither 70 nor 74 is divisible by 4). Conclusion: Since 7 cannot be in the last two spots, the furthest right we can place it is the Hundreds place. (The number is 300,704) Hence, option (d) is correct. Incorrect Answer: (d) Solution: The “Push-Right” Strategy: To get the smallest number, we want the largest digits as far to the right as possible. Start Small: The number must start with 3 0 0 … to be minimal. The “7” Problem: We must use the digit 7, but: It cannot be the Unit digit (it’s odd). It cannot be the Tens digit (neither 70 nor 74 is divisible by 4). Conclusion: Since 7 cannot be in the last two spots, the furthest right we can place it is the Hundreds place. (The number is 300,704) Hence, option (d) is correct.
#### 4. Question
A six-digit number N is formed using the digits 0, 3, 4 and 7 only. Each of the digits is used at least once. It was found that N is divisible by 4. What is the hundred’s digit of the smallest possible such six-digit number?
Answer: (d)
Solution: The “Push-Right” Strategy: To get the smallest number, we want the largest digits as far to the right as possible.
• Start Small: The number must start with 3 0 0 … to be minimal.
• The “7” Problem: We must use the digit 7, but: It cannot be the Unit digit (it’s odd). It cannot be the Tens digit (neither 70 nor 74 is divisible by 4).
• It cannot be the Unit digit (it’s odd).
• It cannot be the Tens digit (neither 70 nor 74 is divisible by 4).
• Conclusion: Since 7 cannot be in the last two spots, the furthest right we can place it is the Hundreds place.
(The number is 300,704)
Hence, option (d) is correct.
Answer: (d)
Solution: The “Push-Right” Strategy: To get the smallest number, we want the largest digits as far to the right as possible.
• Start Small: The number must start with 3 0 0 … to be minimal.
• The “7” Problem: We must use the digit 7, but: It cannot be the Unit digit (it’s odd). It cannot be the Tens digit (neither 70 nor 74 is divisible by 4).
• It cannot be the Unit digit (it’s odd).
• It cannot be the Tens digit (neither 70 nor 74 is divisible by 4).
• Conclusion: Since 7 cannot be in the last two spots, the furthest right we can place it is the Hundreds place.
(The number is 300,704)
Hence, option (d) is correct.
• Question 5 of 5 5. Question The positive integers a and b leave remainders of 4 and 5, respectively, when divided by 7. What is the remainder when (a + b) is divided by 7? (a) 1 (b) 2 (c) 3 (d) 4 Correct Answer: (b) Explanation If a number leaves remainder r on division by 7, it can be written as 7k + r. So, a = 7m + 4 b = 7n + 5 Then, a + b = (7m + 4) + (7n + 5) = 7(m + n) + 9 = 7(m + n) + 2 Hence, the remainder when (a + b) is divided by 7 is 2. Incorrect Answer: (b) Explanation If a number leaves remainder r on division by 7, it can be written as 7k + r. So, a = 7m + 4 b = 7n + 5 Then, a + b = (7m + 4) + (7n + 5) = 7(m + n) + 9 = 7(m + n) + 2 Hence, the remainder when (a + b) is divided by 7 is 2.
#### 5. Question
The positive integers a and b leave remainders of 4 and 5, respectively, when divided by 7. What is the remainder when (a + b) is divided by 7?
Answer: (b)
Explanation
If a number leaves remainder r on division by 7, it can be written as 7k + r.
So, a = 7m + 4 b = 7n + 5
Then, a + b = (7m + 4) + (7n + 5) = 7(m + n) + 9 = 7(m + n) + 2
Hence, the remainder when (a + b) is divided by 7 is 2.
Answer: (b)
Explanation
If a number leaves remainder r on division by 7, it can be written as 7k + r.
So, a = 7m + 4 b = 7n + 5
Then, a + b = (7m + 4) + (7n + 5) = 7(m + n) + 9 = 7(m + n) + 2
Hence, the remainder when (a + b) is divided by 7 is 2.
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