UPSC Insta–DART (Daily Aptitude and Reasoning Test) 23 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 How many three-digit numbers can be formed from numbers 0, 2, 4, 6 and 8 which are divisible by 5 or 2 (repetition of numbers is allowed). A.100 B. 125 C. 80 D. 100 Correct Answer : D Explanation: Among given numbers, only when ‘0’ comes at the unit digit, that particular 3 digit number is divisible by 5. In this condition even that 3 digit number is divisible by 2 also because it ends with 0. Now, we can find out the numbers divisible by 2 units place can be occupied by 0, 2, 4, 6 and 8. Therefore 5 ways Hundred’s place can be occupied by all numbers except ‘0’ i.e., 4 ways Ten’s place can be occupied by any given number that is, 5 ways. Therefore, number of numbers with the given condition are, 5×4×5=100. Incorrect Answer : D Explanation: Among given numbers, only when ‘0’ comes at the unit digit, that particular 3 digit number is divisible by 5. In this condition even that 3 digit number is divisible by 2 also because it ends with 0. Now, we can find out the numbers divisible by 2 units place can be occupied by 0, 2, 4, 6 and 8. Therefore 5 ways Hundred’s place can be occupied by all numbers except ‘0’ i.e., 4 ways Ten’s place can be occupied by any given number that is, 5 ways. Therefore, number of numbers with the given condition are, 5×4×5=100.
#### 1. Question
How many three-digit numbers can be formed from numbers 0, 2, 4, 6 and 8 which are
divisible by 5 or 2 (repetition of numbers is allowed).
Answer : D
Explanation:
Among given numbers, only when ‘0’ comes at the unit digit, that particular 3 digit number is
divisible by 5. In this condition even that 3 digit number is divisible by 2 also because it ends
Now, we can find out the numbers divisible by 2 units place can be occupied by 0, 2, 4, 6 and 8.
Therefore 5 ways Hundred’s place can be occupied by all numbers except ‘0’ i.e., 4 ways
Ten’s place can be occupied by any given number that is, 5 ways.
Therefore, number of numbers with the given condition are,
5×4×5=100.
Answer : D
Explanation:
Among given numbers, only when ‘0’ comes at the unit digit, that particular 3 digit number is
divisible by 5. In this condition even that 3 digit number is divisible by 2 also because it ends
Now, we can find out the numbers divisible by 2 units place can be occupied by 0, 2, 4, 6 and 8.
Therefore 5 ways Hundred’s place can be occupied by all numbers except ‘0’ i.e., 4 ways
Ten’s place can be occupied by any given number that is, 5 ways.
Therefore, number of numbers with the given condition are,
5×4×5=100.
• Question 2 of 5 2. Question A pair of dice is thrown simultaneously, find the probability that the sum is obtained 9 when there is an odd number on the first side. A. 1/6 B. 1/3 C. 1/12 D. 1/9 Correct Answer : D There are 3 ways the first dice reads odd number i.e. 1, 3 or 5. When first dice reads, the second dice can read any of the numbers from 1 to 6. Number of possibilities = 3 × 6 =18. But the sum of 9 can be obtained on two occasions i.e. (3,6) and (5,4). Hence, the probability = 2/18 = 1/9. Incorrect Answer : D There are 3 ways the first dice reads odd number i.e. 1, 3 or 5. When first dice reads, the second dice can read any of the numbers from 1 to 6. Number of possibilities = 3 × 6 =18. But the sum of 9 can be obtained on two occasions i.e. (3,6) and (5,4). Hence, the probability = 2/18 = 1/9.
#### 2. Question
A pair of dice is thrown simultaneously, find the probability that the sum is obtained 9 when there is an odd number on the first side.
Answer : D
There are 3 ways the first dice reads odd number i.e. 1, 3 or 5.
When first dice reads, the second dice can read any of the numbers from 1 to 6.
Number of possibilities = 3 × 6 =18.
But the sum of 9 can be obtained on two occasions i.e. (3,6) and (5,4).
Hence, the probability = 2/18 = 1/9.
Answer : D
There are 3 ways the first dice reads odd number i.e. 1, 3 or 5.
When first dice reads, the second dice can read any of the numbers from 1 to 6.
Number of possibilities = 3 × 6 =18.
But the sum of 9 can be obtained on two occasions i.e. (3,6) and (5,4).
Hence, the probability = 2/18 = 1/9.
• Question 3 of 5 3. Question A principle of Rs.1000 is being charged at 50% per annum. What is the interest for 3rd year at compound interest? A. Rs.1125 B. Rs.2275 C. Rs.2250 D. Rs.3375 Correct Answer – A Since compound interest rate is 50%, then the principle will earn half of it every year. So, simple interest or compound for the first year = 50% of Rs.1000 = Rs.500 Principle for the second year = Rs.1000 + Rs.500 = Rs.1500 Incorrect Answer – A Since compound interest rate is 50%, then the principle will earn half of it every year. So, simple interest or compound for the first year = 50% of Rs.1000 = Rs.500 Principle for the second year = Rs.1000 + Rs.500 = Rs.1500
#### 3. Question
A principle of Rs.1000 is being charged at 50% per annum. What is the interest for
3rd year at compound interest?
• A. Rs.1125
• B. Rs.2275
• C. Rs.2250
• D. Rs.3375
Answer – A
Since compound interest rate is 50%, then the principle will earn half of it every year.
So, simple interest or compound for the first year = 50% of Rs.1000 = Rs.500
Principle for the second year = Rs.1000 + Rs.500 = Rs.1500
Answer – A
Since compound interest rate is 50%, then the principle will earn half of it every year.
So, simple interest or compound for the first year = 50% of Rs.1000 = Rs.500
Principle for the second year = Rs.1000 + Rs.500 = Rs.1500
• Question 4 of 5 4. Question A, B and C invest in the ratio of 3:4:5. The percentage of return on their investments are 6%, 5% and 4% respectively. Find the total earnings, if B earns Rs.250 more than A A. Rs.6000 B. Rs.7250 C. Rs.5000 D. None of these Correct Answer : B Let the investments of A, B and C be 3x, 4x and 5x respectively. The earnings of A = 6% of 3x = 0.18x The earnings of B = 5% of 4x = 0.2x The earnings of C = 4% of 5x = 0.2x B earns Rs.250 more than A i.e. 0.2x – 0.18x = 0.02x = Rs.250 x = 250/0.02 = 12500. Total earnings = 0.18x + 0.2x + 0.2x = 0.58x = 0.58 = 12500 = Rs.7250. Incorrect Answer : B Let the investments of A, B and C be 3x, 4x and 5x respectively. The earnings of A = 6% of 3x = 0.18x The earnings of B = 5% of 4x = 0.2x The earnings of C = 4% of 5x = 0.2x B earns Rs.250 more than A i.e. 0.2x – 0.18x = 0.02x = Rs.250 x = 250/0.02 = 12500. Total earnings = 0.18x + 0.2x + 0.2x = 0.58x = 0.58 = 12500 = Rs.7250.
#### 4. Question
A, B and C invest in the ratio of 3:4:5. The percentage of return on their investments are 6%, 5% and 4% respectively. Find the total earnings, if B earns Rs.250 more than A
• A. Rs.6000
• B. Rs.7250
• C. Rs.5000
• D. None of these
Answer : B
Let the investments of A, B and C be 3x, 4x and 5x respectively.
The earnings of A = 6% of 3x = 0.18x
The earnings of B = 5% of 4x = 0.2x
The earnings of C = 4% of 5x = 0.2x
B earns Rs.250 more than A i.e. 0.2x – 0.18x = 0.02x = Rs.250
x = 250/0.02 = 12500.
Total earnings = 0.18x + 0.2x + 0.2x = 0.58x = 0.58 = 12500 = Rs.7250.
Answer : B
Let the investments of A, B and C be 3x, 4x and 5x respectively.
The earnings of A = 6% of 3x = 0.18x
The earnings of B = 5% of 4x = 0.2x
The earnings of C = 4% of 5x = 0.2x
B earns Rs.250 more than A i.e. 0.2x – 0.18x = 0.02x = Rs.250
x = 250/0.02 = 12500.
Total earnings = 0.18x + 0.2x + 0.2x = 0.58x = 0.58 = 12500 = Rs.7250.
• Question 5 of 5 5. Question A vendor buys oranges at Rs.2 for 3 oranges and sells then at a rupee each. To make profit of Rs.10, he must sell: A. 10 oranges B. 20 oranges C. 30 oranges D. 40 oranges Correct Answer : C Cost Price of each orange = Rs. 2/3 Selling Price of each orange = Rs.1 Profit on each orange = 1 – (2/3) = Rs. 1/3 To make a profit of Rs.10, the number of oranges = (10)/(1/3) = 30. Incorrect Answer : C Cost Price of each orange = Rs. 2/3 Selling Price of each orange = Rs.1 Profit on each orange = 1 – (2/3) = Rs. 1/3 To make a profit of Rs.10, the number of oranges = (10)/(1/3) = 30.
#### 5. Question
A vendor buys oranges at Rs.2 for 3 oranges and sells then at a rupee each. To make profit of Rs.10, he must sell:
• A. 10 oranges
• B. 20 oranges
• C. 30 oranges
• D. 40 oranges
Answer : C
Cost Price of each orange = Rs. 2/3
Selling Price of each orange = Rs.1
Profit on each orange = 1 – (2/3) = Rs. 1/3
To make a profit of Rs.10, the number of oranges = (10)/(1/3) = 30.
Answer : C
Cost Price of each orange = Rs. 2/3
Selling Price of each orange = Rs.1
Profit on each orange = 1 – (2/3) = Rs. 1/3
To make a profit of Rs.10, the number of oranges = (10)/(1/3) = 30.
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