UPSC Insta–DART (Daily Aptitude and Reasoning Test) 21 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 While writing all the numbers from 400 to 699, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place? (a) 45 (b) 30 (c) 31 (d) 66 Correct Answer: (c) Solution: Given that, From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place. Now, Considering 400 to 499 Hundreds digit = 4 So we need: 4 > tens > units Possible pairs (tens, units): tens can be 1, 2, 3 If tens = 1 → units = 0 → 1 number If tens = 2 → units = 0, 1 → 2 numbers If tens = 3 → units = 0, 1, 2 → 3 numbers Total = 1 + 2 + 3 = 6 Considering 500 to 599 Hundreds digit = 5 tens can be 1, 2, 3, 4 Total = 1 + 2 + 3 + 4 = 10 Considering 600 to 699 Hundreds digit = 6 tens can be 1, 2, 3, 4, 5 Total = 1 + 2 + 3 + 4 + 5 = 15 Total possible numbers = 6 + 10 + 15 = 31 Incorrect Answer: (c) Solution: Given that, From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place. Now, Considering 400 to 499 Hundreds digit = 4 So we need: 4 > tens > units Possible pairs (tens, units): tens can be 1, 2, 3 If tens = 1 → units = 0 → 1 number If tens = 2 → units = 0, 1 → 2 numbers If tens = 3 → units = 0, 1, 2 → 3 numbers Total = 1 + 2 + 3 = 6 Considering 500 to 599 Hundreds digit = 5 tens can be 1, 2, 3, 4 Total = 1 + 2 + 3 + 4 = 10 Considering 600 to 699 Hundreds digit = 6 tens can be 1, 2, 3, 4, 5 Total = 1 + 2 + 3 + 4 + 5 = 15 Total possible numbers = 6 + 10 + 15 = 31
#### 1. Question
While writing all the numbers from 400 to 699, how many numbers occur in which the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place?
Answer: (c)
Solution:
Given that,
From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.
Considering 400 to 499
Hundreds digit = 4 So we need: 4 > tens > units
Possible pairs (tens, units): tens can be 1, 2, 3
If tens = 1 → units = 0 → 1 number If tens = 2 → units = 0, 1 → 2 numbers If tens = 3 → units = 0, 1, 2 → 3 numbers
Total = 1 + 2 + 3 = 6
Considering 500 to 599
Hundreds digit = 5 tens can be 1, 2, 3, 4
Total = 1 + 2 + 3 + 4 = 10
Considering 600 to 699
Hundreds digit = 6 tens can be 1, 2, 3, 4, 5
Total = 1 + 2 + 3 + 4 + 5 = 15
Total possible numbers = 6 + 10 + 15 = 31
Answer: (c)
Solution:
Given that,
From 400 to 699, the digit at hundred’s place is greater than the digit at ten’s place, and the digit at ten’s place is greater than the digit at unit’s place.
Considering 400 to 499
Hundreds digit = 4 So we need: 4 > tens > units
Possible pairs (tens, units): tens can be 1, 2, 3
If tens = 1 → units = 0 → 1 number If tens = 2 → units = 0, 1 → 2 numbers If tens = 3 → units = 0, 1, 2 → 3 numbers
Total = 1 + 2 + 3 = 6
Considering 500 to 599
Hundreds digit = 5 tens can be 1, 2, 3, 4
Total = 1 + 2 + 3 + 4 = 10
Considering 600 to 699
Hundreds digit = 6 tens can be 1, 2, 3, 4, 5
Total = 1 + 2 + 3 + 4 + 5 = 15
Total possible numbers = 6 + 10 + 15 = 31
• Question 2 of 5 2. Question One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 144. The torn page contains which of the following numbers? (a) 3, 4 (b) 4, 5 (c) 7, 8 (d) 9, 10 Correct Answer: (b) Solution: Given that, One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 144. Now, Let the torn page contain consecutive numbers P and (P + 1) So, sum of torn page numbers = P + (P + 1) = 2P + 1 Total sum of pages = 144 + (2P + 1) Sum of first n natural numbers = (n/2)(n + 1) So, 144 + (2P + 1) = (n/2)(n + 1) Hit and trial method The value of n must be such that the sum should be near 144 n = 17, sum = 153 So, 144 + (2P + 1) = 153 2P + 1 = 9 2P = 8 P = 4 Torn page contains 4 and 5 Incorrect Answer: (b) Solution: Given that, One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 144. Now, Let the torn page contain consecutive numbers P and (P + 1) So, sum of torn page numbers = P + (P + 1) = 2P + 1 Total sum of pages = 144 + (2P + 1) Sum of first n natural numbers = (n/2)(n + 1) So, 144 + (2P + 1) = (n/2)(n + 1) Hit and trial method The value of n must be such that the sum should be near 144 n = 17, sum = 153 So, 144 + (2P + 1) = 153 2P + 1 = 9 2P = 8 P = 4 Torn page contains 4 and 5
#### 2. Question
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 144. The torn page contains which of the following numbers?
Answer: (b)
Given that,
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.
The sum of the numbers on the remaining pages is 144.
Let the torn page contain consecutive numbers P and (P + 1)
So, sum of torn page numbers = P + (P + 1) = 2P + 1
Total sum of pages = 144 + (2P + 1)
Sum of first n natural numbers = (n/2)(n + 1)
So, 144 + (2P + 1) = (n/2)(n + 1)
Hit and trial method
The value of n must be such that the sum should be near 144
n = 17, sum = 153
So, 144 + (2P + 1) = 153 2P + 1 = 9 2P = 8 P = 4
Torn page contains 4 and 5
Answer: (b)
Given that,
One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1.
The sum of the numbers on the remaining pages is 144.
Let the torn page contain consecutive numbers P and (P + 1)
So, sum of torn page numbers = P + (P + 1) = 2P + 1
Total sum of pages = 144 + (2P + 1)
Sum of first n natural numbers = (n/2)(n + 1)
So, 144 + (2P + 1) = (n/2)(n + 1)
Hit and trial method
The value of n must be such that the sum should be near 144
n = 17, sum = 153
So, 144 + (2P + 1) = 153 2P + 1 = 9 2P = 8 P = 4
Torn page contains 4 and 5
• Question 3 of 5 3. Question . What is the sum of all digits which appear in all the integers from 10 to 99? (a) 810 (b) 820 (c) 855 (d) 900 Correct Answer: (c) Solution: Given that, Sum of all digits which appear in all the integers from 10 to 99 Now, For tens digit 10 to 19 tens digit = 1 20 to 29 tens digit = 2 30 to 39 tens digit = 3 … 90 to 99 tens digit = 9 Each tens digit appears 10 times. So, (1 + 2 + 3 + … + 9) × 10 = 45 × 10 = 450 For unit digits 0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99) So, (0 + 1 + 2 + … + 9) × 9 = 45 × 9 = 405 Total sum of digits = 450 + 405 = 855 Hence option (c) is correct. Incorrect Answer: (c) Solution: Given that, Sum of all digits which appear in all the integers from 10 to 99 Now, For tens digit 10 to 19 tens digit = 1 20 to 29 tens digit = 2 30 to 39 tens digit = 3 … 90 to 99 tens digit = 9 Each tens digit appears 10 times. So, (1 + 2 + 3 + … + 9) × 10 = 45 × 10 = 450 For unit digits 0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99) So, (0 + 1 + 2 + … + 9) × 9 = 45 × 9 = 405 Total sum of digits = 450 + 405 = 855 Hence option (c) is correct.
#### 3. Question
. What is the sum of all digits which appear in all the integers from 10 to 99?
Answer: (c)
Given that,
Sum of all digits which appear in all the integers from 10 to 99
For tens digit
10 to 19 tens digit = 1 20 to 29 tens digit = 2 30 to 39 tens digit = 3 … 90 to 99 tens digit = 9
Each tens digit appears 10 times.
So, (1 + 2 + 3 + … + 9) × 10 = 45 × 10 = 450
For unit digits
0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99)
So, (0 + 1 + 2 + … + 9) × 9 = 45 × 9 = 405
Total sum of digits = 450 + 405 = 855
Hence option (c) is correct.
Answer: (c)
Given that,
Sum of all digits which appear in all the integers from 10 to 99
For tens digit
10 to 19 tens digit = 1 20 to 29 tens digit = 2 30 to 39 tens digit = 3 … 90 to 99 tens digit = 9
Each tens digit appears 10 times.
So, (1 + 2 + 3 + … + 9) × 10 = 45 × 10 = 450
For unit digits
0 to 9 appears 9 times (once in each block 10–19, 20–29, …, 90–99)
So, (0 + 1 + 2 + … + 9) × 9 = 45 × 9 = 405
Total sum of digits = 450 + 405 = 855
Hence option (c) is correct.
• Question 4 of 5 4. Question Two Statements are given followed by two Conclusions: Statements: All prime numbers are natural numbers. All natural numbers are integers. Conclusions: Conclusion-I: All prime numbers are integers. Conclusion-II: All integers are prime numbers. Which of the above Conclusions logically follows/follow from the two given Statements? (a) Only Conclusion-I (b) Only Conclusion-II (c) Neither Conclusion-I nor Conclusion-II (d) Both Conclusion-I and Conclusion-II Correct Answer: (a) Solution: Given that, Statements: All prime numbers are natural numbers. All natural numbers are integers. Now, Conclusion-I: All prime numbers are integers. This logically follows as prime numbers ⟶ natural numbers ⟶ integers. Hence Conclusion-I is correct. Conclusion-II: All integers are prime numbers. This does not follow. Integers include composite numbers and zero. Hence Conclusion-II is incorrect. Therefore, option (a) is correct. Incorrect Answer: (a) Solution: Given that, Statements: All prime numbers are natural numbers. All natural numbers are integers. Now, Conclusion-I: All prime numbers are integers. This logically follows as prime numbers ⟶ natural numbers ⟶ integers. Hence Conclusion-I is correct. Conclusion-II: All integers are prime numbers. This does not follow. Integers include composite numbers and zero. Hence Conclusion-II is incorrect. Therefore, option (a) is correct.
#### 4. Question
Two Statements are given followed by two Conclusions:
Statements: All prime numbers are natural numbers. All natural numbers are integers.
Conclusions: Conclusion-I: All prime numbers are integers. Conclusion-II: All integers are prime numbers.
Which of the above Conclusions logically follows/follow from the two given Statements?
• (a) Only Conclusion-I
• (b) Only Conclusion-II
• (c) Neither Conclusion-I nor Conclusion-II
• (d) Both Conclusion-I and Conclusion-II
Answer: (a)
Solution:
Given that,
Statements:
All prime numbers are natural numbers. All natural numbers are integers.
Conclusion-I: All prime numbers are integers.
This logically follows as prime numbers ⟶ natural numbers ⟶ integers.
Hence Conclusion-I is correct.
Conclusion-II: All integers are prime numbers.
This does not follow. Integers include composite numbers and zero.
Hence Conclusion-II is incorrect.
Therefore, option (a) is correct.
Answer: (a)
Solution:
Given that,
Statements:
All prime numbers are natural numbers. All natural numbers are integers.
Conclusion-I: All prime numbers are integers.
This logically follows as prime numbers ⟶ natural numbers ⟶ integers.
Hence Conclusion-I is correct.
Conclusion-II: All integers are prime numbers.
This does not follow. Integers include composite numbers and zero.
Hence Conclusion-II is incorrect.
Therefore, option (a) is correct.
• Question 5 of 5 5. Question Each digit of an 8-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number? (a) 64 (b) 72 (c) 80 (d) 81 Correct Answer: (a) Solution: Given that, Each digit of an 8-digit number is 1. It is multiplied by itself. Now, 11111111 × 11111111 = 123456787654321 Thus sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 64 Hence option (a) is correct. Incorrect Answer: (a) Solution: Given that, Each digit of an 8-digit number is 1. It is multiplied by itself. Now, 11111111 × 11111111 = 123456787654321 Thus sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 64 Hence option (a) is correct.
#### 5. Question
Each digit of an 8-digit number is 1. It is multiplied by itself. What is the sum of the digits of the resulting number?
Answer: (a)
Solution:
Given that,
Each digit of an 8-digit number is 1.
It is multiplied by itself.
11111111 × 11111111 = 123456787654321
Thus sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 64
Hence option (a) is correct.
Answer: (a)
Solution:
Given that,
Each digit of an 8-digit number is 1.
It is multiplied by itself.
11111111 × 11111111 = 123456787654321
Thus sum of digits = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 64
Hence option (a) is correct.
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