UPSC Insta–DART (Daily Aptitude and Reasoning Test) 27 Oct 2025
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 A, B, C are three integers. Two are even and one is odd. Consider the following statements. A + B + C is definitely odd. A × B × C is definitely even. (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Correct Answer: C Solution Statement 1: Two evens add to an even. Even + odd = odd. Hence true. Statement 2: Product includes at least one even. Therefore result is even. Hence true. So both statements are correct. Incorrect Answer: C Solution Statement 1: Two evens add to an even. Even + odd = odd. Hence true. Statement 2: Product includes at least one even. Therefore result is even. Hence true. So both statements are correct.
#### 1. Question
A, B, C are three integers. Two are even and one is odd. Consider the following statements.
• A + B + C is definitely odd.
• A × B × C is definitely even.
• (a) 1 only
• (b) 2 only
• (c) Both 1 and 2
• (d) Neither 1 nor 2
Answer: C Solution Statement 1: Two evens add to an even. Even + odd = odd. Hence true. Statement 2: Product includes at least one even. Therefore result is even. Hence true. So both statements are correct.
Answer: C Solution Statement 1: Two evens add to an even. Even + odd = odd. Hence true. Statement 2: Product includes at least one even. Therefore result is even. Hence true. So both statements are correct.
• Question 2 of 5 2. Question While writing all the numbers from 500 to 999, how many numbers occur in which the digit at hundred’s place is less than the digit at ten’s place, and the digit at ten’s place is less than the digit at unit’s place? (a) 9 (b) 10 (c) 12 (d) 15 Correct Answer: B Solution: Given that, From 500 to 999, we need strictly increasing digits: H < T < U. Now, For each hundred’s digit H, choose two larger digits for (T,U) from {H+1,…,9}. H = 5 → choose two from {6,7,8,9}: 6 ways. H = 6 → from {7,8,9}: 3 ways. H = 7 → from {8,9}: 1 way. H = 8 or 9 → 0 ways. Total = 6 + 3 + 1 = 10. Hence option (b) is correct. Incorrect Answer: B Solution: Given that, From 500 to 999, we need strictly increasing digits: H < T < U. Now, For each hundred’s digit H, choose two larger digits for (T,U) from {H+1,…,9}. H = 5 → choose two from {6,7,8,9}: 6 ways. H = 6 → from {7,8,9}: 3 ways. H = 7 → from {8,9}: 1 way. H = 8 or 9 → 0 ways. Total = 6 + 3 + 1 = 10. Hence option (b) is correct.
#### 2. Question
While writing all the numbers from 500 to 999, how many numbers occur in which the digit at hundred’s place is less than the digit at ten’s place, and the digit at ten’s place is less than the digit at unit’s place?
Given that,
From 500 to 999, we need strictly increasing digits: H < T < U.
For each hundred’s digit H, choose two larger digits for (T,U) from {H+1,…,9}.
H = 5 → choose two from {6,7,8,9}: 6 ways. H = 6 → from {7,8,9}: 3 ways. H = 7 → from {8,9}: 1 way. H = 8 or 9 → 0 ways.
Total = 6 + 3 + 1 = 10.
Hence option (b) is correct.
Given that,
From 500 to 999, we need strictly increasing digits: H < T < U.
For each hundred’s digit H, choose two larger digits for (T,U) from {H+1,…,9}.
H = 5 → choose two from {6,7,8,9}: 6 ways. H = 6 → from {7,8,9}: 3 ways. H = 7 → from {8,9}: 1 way. H = 8 or 9 → 0 ways.
Total = 6 + 3 + 1 = 10.
Hence option (b) is correct.
• Question 3 of 5 3. Question D is a 3-digit number with the hundred’s digit fixed at 2 such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred’s place and the digit at the unit’s place of D? (a) 5 (b) 6 (c) 7 (d) 8 Correct Answer: C Solution: Given that, Hundred’s digit = 2. We want to minimize [number]/[sum of digits]. Now, For a fixed hundred’s digit, to reduce the ratio we maximize the sum while keeping the number as small as possible. Maximize the remaining sum by using 9 and 9; arrange to keep the number minimal → 299. Its sum is 20 and ratio = 299/20, which is smaller than any other 2ab with a+b < 18 or with a>0 increasing the number. Difference between hundred’s and unit’s places = 2 − 9 = 7. Hence option (c) is correct. Incorrect Answer: C Solution: Given that, Hundred’s digit = 2. We want to minimize [number]/[sum of digits]. Now, For a fixed hundred’s digit, to reduce the ratio we maximize the sum while keeping the number as small as possible. Maximize the remaining sum by using 9 and 9; arrange to keep the number minimal → 299. Its sum is 20 and ratio = 299/20, which is smaller than any other 2ab with a+b < 18 or with a>0 increasing the number. Difference between hundred’s and unit’s places = 2 − 9 = 7. Hence option (c) is correct.
#### 3. Question
D is a 3-digit number with the hundred’s digit fixed at 2 such that the ratio of the number to the sum of its digits is least. What is the difference between the digit at the hundred’s place and the digit at the unit’s place of D?
Given that,
Hundred’s digit = 2. We want to minimize [number]/[sum of digits].
For a fixed hundred’s digit, to reduce the ratio we maximize the sum while keeping the number as small as possible. Maximize the remaining sum by using 9 and 9; arrange to keep the number minimal → 299. Its sum is 20 and ratio = 299/20, which is smaller than any other 2ab with a+b < 18 or with a>0 increasing the number.
Difference between hundred’s and unit’s places = 2 − 9 = 7.
Hence option (c) is correct.
Given that,
Hundred’s digit = 2. We want to minimize [number]/[sum of digits].
For a fixed hundred’s digit, to reduce the ratio we maximize the sum while keeping the number as small as possible. Maximize the remaining sum by using 9 and 9; arrange to keep the number minimal → 299. Its sum is 20 and ratio = 299/20, which is smaller than any other 2ab with a+b < 18 or with a>0 increasing the number.
Difference between hundred’s and unit’s places = 2 − 9 = 7.
Hence option (c) is correct.
• Question 4 of 5 4. Question A certain species of fish in a lake triples in population every 15 years. If the population at the beginning of 2020 was 300, which one of the following equations best models the population P after n years? (a) P = 300 × (3)^(n/15) (b) P = 300 + 15n (c) P = 15 + 300n (d) P = 300 × (2)^(n/15) Correct Answer: A Solution: Given that, Fish population triples every 15 years. At 2020, population = 300. Now, growth is exponential with tripling time = 15 years. When n = 0, P = 300. When n = 15, P should be 900. So, P = 300 × (3)^(n/15). Hence option (a) is correct. Incorrect Answer: A Solution: Given that, Fish population triples every 15 years. At 2020, population = 300. Now, growth is exponential with tripling time = 15 years. When n = 0, P = 300. When n = 15, P should be 900. So, P = 300 × (3)^(n/15). Hence option (a) is correct.
#### 4. Question
A certain species of fish in a lake triples in population every 15 years. If the population at the beginning of 2020 was 300, which one of the following equations best models the population P after n years?
• (a) P = 300 × (3)^(n/15)
• (b) P = 300 + 15n
• (c) P = 15 + 300n
• (d) P = 300 × (2)^(n/15)
Given that,
Fish population triples every 15 years. At 2020, population = 300.
Now, growth is exponential with tripling time = 15 years.
When n = 0, P = 300. When n = 15, P should be 900.
So, P = 300 × (3)^(n/15).
Hence option (a) is correct.
Given that,
Fish population triples every 15 years. At 2020, population = 300.
Now, growth is exponential with tripling time = 15 years.
When n = 0, P = 300. When n = 15, P should be 900.
So, P = 300 × (3)^(n/15).
Hence option (a) is correct.
• Question 5 of 5 5. Question In a college, seventy percent are undergraduates. Forty-five percent are hostellers. Forty percent are in sports clubs. Fifty-five percent have laptops. Which one of the following statements is certainly correct? (a) All undergraduates are in sports clubs. (b) Some hostellers have laptops. (c) Some undergraduates are in sports clubs. (d) No undergraduate has a laptop. Correct Answer: C Solution: Given that, Undergraduates → 70%. Hostellers → 45%. In sports clubs → 40%. Have laptops → 55%. Now, from the given statements 70% are undergraduates and 40% are in sports clubs, so some undergraduates are in sports clubs (overlap ≥ 70+40−100 = 10%). Hence option (c) is correct. Incorrect Answer: C Solution: Given that, Undergraduates → 70%. Hostellers → 45%. In sports clubs → 40%. Have laptops → 55%. Now, from the given statements 70% are undergraduates and 40% are in sports clubs, so some undergraduates are in sports clubs (overlap ≥ 70+40−100 = 10%). Hence option (c) is correct.
#### 5. Question
In a college, seventy percent are undergraduates. Forty-five percent are hostellers. Forty percent are in sports clubs. Fifty-five percent have laptops. Which one of the following statements is certainly correct?
• (a) All undergraduates are in sports clubs.
• (b) Some hostellers have laptops.
• (c) Some undergraduates are in sports clubs.
• (d) No undergraduate has a laptop.
Answer: C
Solution:
Given that,
Undergraduates → 70%. Hostellers → 45%. In sports clubs → 40%. Have laptops → 55%.
Now, from the given statements
70% are undergraduates and 40% are in sports clubs, so some undergraduates are in sports clubs (overlap ≥ 70+40−100 = 10%). Hence option (c) is correct.
Answer: C
Solution:
Given that,
Undergraduates → 70%. Hostellers → 45%. In sports clubs → 40%. Have laptops → 55%.
Now, from the given statements
70% are undergraduates and 40% are in sports clubs, so some undergraduates are in sports clubs (overlap ≥ 70+40−100 = 10%). Hence option (c) is correct.
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