UPSC Insta–DART (Daily Aptitude and Reasoning Test) 20 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 number N be a two-digit number. Statement I: N is the smallest two-digit number divisible by both 4 and 5. Statement II: When N is multiplied by a one-digit prime number, the result is a three-digit multiple of 11. Question: What is the value of N? Which one of the following is correct in respect of the Statements and the Question? (a) Statement I alone is sufficient to answer the question. (b) Statement II alone is sufficient to answer the question. (c) Both Statement I and Statement II together are sufficient to answer the question. (d) Either statement I or statement II alone is sufficient to answer the question. Correct Answer: (a) Solution: Using Statement I alone: N is the smallest two-digit number divisible by both 4 and 5. LCM(4, 5) = 20 ⇒ smallest two-digit multiple = 20. So, N is uniquely determined. Statement I is sufficient. Using Statement II alone: N × (one-digit prime) = three-digit multiple of 11. This allows many possible N. For example: 22 × 5 = 110 (valid) 44 × 3 = 132 (valid) 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient. Therefore, only Statement I alone is sufficient → (a). Incorrect Answer: (a) Solution: Using Statement I alone: N is the smallest two-digit number divisible by both 4 and 5. LCM(4, 5) = 20 ⇒ smallest two-digit multiple = 20. So, N is uniquely determined. Statement I is sufficient. Using Statement II alone: N × (one-digit prime) = three-digit multiple of 11. This allows many possible N. For example: 22 × 5 = 110 (valid) 44 × 3 = 132 (valid) 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient. Therefore, only Statement I alone is sufficient → (a).
#### 1. Question
Let number N be a two-digit number.
Statement I: N is the smallest two-digit number divisible by both 4 and 5. Statement II: When N is multiplied by a one-digit prime number, the result is a three-digit multiple of 11.
Question: What is the value of N?
Which one of the following is correct in respect of the Statements and the Question?
• (a) Statement I alone is sufficient to answer the question.
• (b) Statement II alone is sufficient to answer the question.
• (c) Both Statement I and Statement II together are sufficient to answer the question.
• (d) Either statement I or statement II alone is sufficient to answer the question.
Answer: (a)
Solution:
• Using Statement I alone: N is the smallest two-digit number divisible by both 4 and 5. LCM(4, 5) = 20 ⇒ smallest two-digit multiple = 20. So, N is uniquely determined. Statement I is sufficient.
• Using Statement II alone: N × (one-digit prime) = three-digit multiple of 11. This allows many possible N. For example: 22 × 5 = 110 (valid) 44 × 3 = 132 (valid) 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient.
• 22 × 5 = 110 (valid)
• 44 × 3 = 132 (valid)
• 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient.
Therefore, only Statement I alone is sufficient → (a).
Answer: (a)
Solution:
• Using Statement I alone: N is the smallest two-digit number divisible by both 4 and 5. LCM(4, 5) = 20 ⇒ smallest two-digit multiple = 20. So, N is uniquely determined. Statement I is sufficient.
• Using Statement II alone: N × (one-digit prime) = three-digit multiple of 11. This allows many possible N. For example: 22 × 5 = 110 (valid) 44 × 3 = 132 (valid) 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient.
• 22 × 5 = 110 (valid)
• 44 × 3 = 132 (valid)
• 55 × 2 = 110 (valid) So N is not unique. Statement II is not sufficient.
Therefore, only Statement I alone is sufficient → (a).
• Question 2 of 5 2. Question If N2 = 12345678987654321, then how many digits does the number N have? (a) 8 (b) 9 (c) 10 (d) 11 Correct Answer:(b) Let’s look at the pattern of squares of numbers consisting only of ones: 12 = 1 (1 digit in N, 1 digit in N2, highest digit in N2 is 1) 112 = 121 (2 digits in N, 3 digits in N2, highest digit in N2 is 2) 1112 = 12321 (3 digits in N, 5 digits in N2, highest digit in N2 is 3) 11112 = 1234321 (4 digits in N, 7 digits in N2, highest digit in N2 is 4) 111112 = 123454321 (5 digits in N, 9 digits in N2, highest digit in N2 is 5) …….and so on. In our case, N2=12345678987654321. The highest digit in the sequence is 9, and the number of digits in it is 17. This means that N must be a number consisting of nine ‘1’s. So, N = 111,111,111. Thus, the number N has 9 digits. Incorrect Answer:(b) Let’s look at the pattern of squares of numbers consisting only of ones: 12 = 1 (1 digit in N, 1 digit in N2, highest digit in N2 is 1) 112 = 121 (2 digits in N, 3 digits in N2, highest digit in N2 is 2) 1112 = 12321 (3 digits in N, 5 digits in N2, highest digit in N2 is 3) 11112 = 1234321 (4 digits in N, 7 digits in N2, highest digit in N2 is 4) 111112 = 123454321 (5 digits in N, 9 digits in N2, highest digit in N2 is 5) …….and so on. In our case, N2=12345678987654321. The highest digit in the sequence is 9, and the number of digits in it is 17. This means that N must be a number consisting of nine ‘1’s. So, N = 111,111,111. Thus, the number N has 9 digits.
#### 2. Question
If N2 = 12345678987654321, then how many digits does the number N have?
Answer:(b)
Let’s look at the pattern of squares of numbers consisting only of ones: 12 = 1 (1 digit in N, 1 digit in N2, highest digit in N2 is 1)
112 = 121 (2 digits in N, 3 digits in N2, highest digit in N2 is 2) 1112 = 12321 (3 digits in N, 5 digits in N2, highest digit in N2 is 3)
11112 = 1234321 (4 digits in N, 7 digits in N2, highest digit in N2 is 4) 111112 = 123454321 (5 digits in N, 9 digits in N2, highest digit in N2 is 5)
…….and so on.
In our case, N2=12345678987654321. The highest digit in the sequence is 9, and the number of digits in it is 17. This means that N must be a number consisting of nine ‘1’s. So, N = 111,111,111.
Thus, the number N has 9 digits.
Answer:(b)
Let’s look at the pattern of squares of numbers consisting only of ones: 12 = 1 (1 digit in N, 1 digit in N2, highest digit in N2 is 1)
112 = 121 (2 digits in N, 3 digits in N2, highest digit in N2 is 2) 1112 = 12321 (3 digits in N, 5 digits in N2, highest digit in N2 is 3)
11112 = 1234321 (4 digits in N, 7 digits in N2, highest digit in N2 is 4) 111112 = 123454321 (5 digits in N, 9 digits in N2, highest digit in N2 is 5)
…….and so on.
In our case, N2=12345678987654321. The highest digit in the sequence is 9, and the number of digits in it is 17. This means that N must be a number consisting of nine ‘1’s. So, N = 111,111,111.
Thus, the number N has 9 digits.
• Question 3 of 5 3. Question If X+Y=10 , then which of the following must be true? Both and must be positive for any value of and . If is negative, then must be positive for any value of and . Both and cannot be zero simultaneously. 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 nor 3 Correct Incorrect
#### 3. Question
If X+Y=10 , then which of the following must be true?
• Both and must be positive for any value of and .
• If is negative, then must be positive for any value of and .
• Both and cannot be zero simultaneously.
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 nor 3
• Question 4 of 5 4. Question A person X wants to distribute some pencils among 4 children A, B, C and D. Suppose A gets three times the number of pencils received by B, four times that of C and six times that of D. What is the minimum number of pencils X should buy so that the number of pencils each one gets is an even number? (a) 168 (b) 192 (c) 216 (d) 240 Correct Answer: (a) Solution: Given that, X distributes some pencils among 4 children A, B, C and D. A gets three times the number of pencils received by B, four times that of C and six times that of D. Now, Let A get P pencils. B = P/3, C = P/4 and D = P/6 Thus, LCM of 3, 4 and 6 = 12 P = 12, 24, 36, 48, 60, 72, 84, 96 … (possible values) The pencils each should get must be even in number. Taking P = 24, C = 6 (even), D = 4 (even), B = 8 (even) Total = 24 + 8 + 6 + 4 = 42 (not in options) Taking P = 96, A = 96, B = 32, C = 24, D = 16 Total = 96 + 32 + 24 + 16 = 168 Hence option (a) is correct. Incorrect Answer: (a) Solution: Given that, X distributes some pencils among 4 children A, B, C and D. A gets three times the number of pencils received by B, four times that of C and six times that of D. Now, Let A get P pencils. B = P/3, C = P/4 and D = P/6 Thus, LCM of 3, 4 and 6 = 12 P = 12, 24, 36, 48, 60, 72, 84, 96 … (possible values) The pencils each should get must be even in number. Taking P = 24, C = 6 (even), D = 4 (even), B = 8 (even) Total = 24 + 8 + 6 + 4 = 42 (not in options) Taking P = 96, A = 96, B = 32, C = 24, D = 16 Total = 96 + 32 + 24 + 16 = 168 Hence option (a) is correct.
#### 4. Question
A person X wants to distribute some pencils among 4 children A, B, C and D. Suppose A gets three times the number of pencils received by B, four times that of C and six times that of D. What is the minimum number of pencils X should buy so that the number of pencils each one gets is an even number?
Answer: (a)
Solution:
Given that,
X distributes some pencils among 4 children A, B, C and D.
A gets three times the number of pencils received by B, four times that of C and six times that of D.
Let A get P pencils.
B = P/3, C = P/4 and D = P/6
Thus, LCM of 3, 4 and 6 = 12
P = 12, 24, 36, 48, 60, 72, 84, 96 … (possible values)
The pencils each should get must be even in number.
Taking P = 24,
C = 6 (even), D = 4 (even), B = 8 (even)
Total = 24 + 8 + 6 + 4 = 42 (not in options)
Taking P = 96,
A = 96, B = 32, C = 24, D = 16
Total = 96 + 32 + 24 + 16 = 168
Hence option (a) is correct.
Answer: (a)
Solution:
Given that,
X distributes some pencils among 4 children A, B, C and D.
A gets three times the number of pencils received by B, four times that of C and six times that of D.
Let A get P pencils.
B = P/3, C = P/4 and D = P/6
Thus, LCM of 3, 4 and 6 = 12
P = 12, 24, 36, 48, 60, 72, 84, 96 … (possible values)
The pencils each should get must be even in number.
Taking P = 24,
C = 6 (even), D = 4 (even), B = 8 (even)
Total = 24 + 8 + 6 + 4 = 42 (not in options)
Taking P = 96,
A = 96, B = 32, C = 24, D = 16
Total = 96 + 32 + 24 + 16 = 168
Hence option (a) is correct.
• Question 5 of 5 5. Question What is the sum of all 3-digit numbers less than 400 formed by the digits 1, 2 and 3, where none of the digits is repeated? (a) 1332 (b) 1368 (c) 1392 (d) 1428 Correct Answer: (a) Solution: Given that, 3-digit numbers less than 400 formed by the digits 1, 2 and 3 Digits must not be repeated Now, From 100 to 400 Possible numbers are: 123, 132, 213, 231, 312 and 321 Sum = 123 + 132 + 213 + 231 + 312 + 321 = 1332 Hence option (a) is correct. Incorrect Answer: (a) Solution: Given that, 3-digit numbers less than 400 formed by the digits 1, 2 and 3 Digits must not be repeated Now, From 100 to 400 Possible numbers are: 123, 132, 213, 231, 312 and 321 Sum = 123 + 132 + 213 + 231 + 312 + 321 = 1332 Hence option (a) is correct.
#### 5. Question
What is the sum of all 3-digit numbers less than 400 formed by the digits 1, 2 and 3, where none of the digits is repeated?
Answer: (a)
Solution:
Given that,
3-digit numbers less than 400 formed by the digits 1, 2 and 3
Digits must not be repeated
From 100 to 400
Possible numbers are:
123, 132, 213, 231, 312 and 321
Sum = 123 + 132 + 213 + 231 + 312 + 321 = 1332
Hence option (a) is correct.
Answer: (a)
Solution:
Given that,
3-digit numbers less than 400 formed by the digits 1, 2 and 3
Digits must not be repeated
From 100 to 400
Possible numbers are:
123, 132, 213, 231, 312 and 321
Sum = 123 + 132 + 213 + 231 + 312 + 321 = 1332
Hence option (a) is correct.
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