UPSC Insta–DART (Daily Aptitude and Reasoning Test) 13 Jan 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 Simple interest on a certain sum for 4 years at 5% per annum is one-third of the compound interest on ₹3000 for 2 years at 10% per annum. What is the sum placed on simple interest? (a) ₹900 (b) ₹1000 (c) ₹1050 (d) ₹1200 Correct Answer: (c) Solution: CI = 3000 × [(1 + 10/100)² – 1] = 3000 × (1.21 – 1) = 3000 × 0.21 = ₹630 SI = (1/3) × 630 = ₹210 Let principal be P SI = (P × 5 × 4)/100 = 210 P = (210 × 100)/(5 × 4) = ₹1050 Incorrect Answer: (c) Solution: CI = 3000 × [(1 + 10/100)² – 1] = 3000 × (1.21 – 1) = 3000 × 0.21 = ₹630 SI = (1/3) × 630 = ₹210 Let principal be P SI = (P × 5 × 4)/100 = 210 P = (210 × 100)/(5 × 4) = ₹1050
#### 1. Question
Simple interest on a certain sum for 4 years at 5% per annum is one-third of the compound interest on ₹3000 for 2 years at 10% per annum. What is the sum placed on simple interest?
Answer: (c)
Solution: CI = 3000 × [(1 + 10/100)² – 1] = 3000 × (1.21 – 1) = 3000 × 0.21 = ₹630
SI = (1/3) × 630 = ₹210 Let principal be P SI = (P × 5 × 4)/100 = 210 P = (210 × 100)/(5 × 4) = ₹1050
Answer: (c)
Solution: CI = 3000 × [(1 + 10/100)² – 1] = 3000 × (1.21 – 1) = 3000 × 0.21 = ₹630
SI = (1/3) × 630 = ₹210 Let principal be P SI = (P × 5 × 4)/100 = 210 P = (210 × 100)/(5 × 4) = ₹1050
• Question 2 of 5 2. Question A cube is painted on opposite faces with orange, pink, and grey. The cube is then sliced into 36 smaller cubes: 32 are small (identical), and 4 are larger (identical). None of the large cubes is adjacent to the pink face. How many cubes have no face painted? (a) 0 (b) 1 (c) 2 (d) None Correct Answer: (d) Solution: Cubes that have no face painted must lie completely inside the big cube, not touching any surface In this case: → All small and big cubes are either on or adjacent to the surface → Cube is too small to contain an inner cube untouched by any face → So no cube is completely unpainted Incorrect Answer: (d) Solution: Cubes that have no face painted must lie completely inside the big cube, not touching any surface In this case: → All small and big cubes are either on or adjacent to the surface → Cube is too small to contain an inner cube untouched by any face → So no cube is completely unpainted
#### 2. Question
A cube is painted on opposite faces with orange, pink, and grey. The cube is then sliced into 36 smaller cubes: 32 are small (identical), and 4 are larger (identical). None of the large cubes is adjacent to the pink face. How many cubes have no face painted?
Answer: (d)
Solution: Cubes that have no face painted must lie completely inside the big cube, not touching any surface
In this case: → All small and big cubes are either on or adjacent to the surface → Cube is too small to contain an inner cube untouched by any face → So no cube is completely unpainted
Answer: (d)
Solution: Cubes that have no face painted must lie completely inside the big cube, not touching any surface
In this case: → All small and big cubes are either on or adjacent to the surface → Cube is too small to contain an inner cube untouched by any face → So no cube is completely unpainted
• Question 3 of 5 3. Question In a company, only the first and last Sundays of each month are holidays. All Saturdays are working. What is the maximum number of possible working days in a 30-day month? (a) 28 (b) 27 (c) 26 (d) 25 Correct Answer: (a) Solution: 30 days = 4 full weeks + 2 days → So, at most 5 Sundays in the month But only first and last Sundays are holidays. Maximum number of Sundays = 5, so 2 Sundays off → Total holidays = 2 ⇒ Working days = 30 − 2 = 28 Incorrect Answer: (a) Solution: 30 days = 4 full weeks + 2 days → So, at most 5 Sundays in the month But only first and last Sundays are holidays. Maximum number of Sundays = 5, so 2 Sundays off → Total holidays = 2 ⇒ Working days = 30 − 2 = 28
#### 3. Question
In a company, only the first and last Sundays of each month are holidays. All Saturdays are working. What is the maximum number of possible working days in a 30-day month?
Answer: (a)
Solution: 30 days = 4 full weeks + 2 days → So, at most 5 Sundays in the month
But only first and last Sundays are holidays. Maximum number of Sundays = 5, so 2 Sundays off → Total holidays = 2 ⇒ Working days = 30 − 2 = 28
Answer: (a)
Solution: 30 days = 4 full weeks + 2 days → So, at most 5 Sundays in the month
But only first and last Sundays are holidays. Maximum number of Sundays = 5, so 2 Sundays off → Total holidays = 2 ⇒ Working days = 30 − 2 = 28
• Question 4 of 5 4. Question Two brothers buy bus tickets. The elder one has ₹8, which he realizes is 40% of the total fare for both. The younger one gives him ₹5. How much more money do they still need to buy the tickets? (a) ₹2 (b) ₹7 (c) ₹1 (d) None, they already have enough Correct Answer: (b) Solution: Let total fare = F. 40% of F = 8 ⇒ F = (8 × 100) / 40 = ₹20. Total money = 8 + 5 = ₹13. Still needed = 20 – 13 = ₹7. Incorrect Answer: (b) Solution: Let total fare = F. 40% of F = 8 ⇒ F = (8 × 100) / 40 = ₹20. Total money = 8 + 5 = ₹13. Still needed = 20 – 13 = ₹7.
#### 4. Question
Two brothers buy bus tickets. The elder one has ₹8, which he realizes is 40% of the total fare for both. The younger one gives him ₹5. How much more money do they still need to buy the tickets?
• (d) None, they already have enough
Answer: (b)
Solution: Let total fare = F. 40% of F = 8 ⇒ F = (8 × 100) / 40 = ₹20. Total money = 8 + 5 = ₹13. Still needed = 20 – 13 = ₹7.
Answer: (b)
Solution: Let total fare = F. 40% of F = 8 ⇒ F = (8 × 100) / 40 = ₹20. Total money = 8 + 5 = ₹13. Still needed = 20 – 13 = ₹7.
• Question 5 of 5 5. Question If every year 2/5th of a community’s population migrates to another region, find the leftover population after the 3rd year, assuming no growth during this period. (a) 27/125 part of the population. (b) 64/125 part of the population. (c) 8/25 part of the population. (d) 16/25 part of the population. Correct Answer: (a) Solution: Given that, there is a policy that 2/5th of the population migrates every year to another region. There is no further growth in the population within three years. Now, Let the population be p. Leftover population after one year = p − (2/5)×p = (3/5)p. After every passing year population will become 3/5th of previous population. Population after 2nd year = (3/5)(3/5)×p = (9/25)p. Population after 3rd year = (27/125)p. Hence option (a) is correct. Incorrect Answer: (a) Solution: Given that, there is a policy that 2/5th of the population migrates every year to another region. There is no further growth in the population within three years. Now, Let the population be p. Leftover population after one year = p − (2/5)×p = (3/5)p. After every passing year population will become 3/5th of previous population. Population after 2nd year = (3/5)(3/5)×p = (9/25)p. Population after 3rd year = (27/125)p. Hence option (a) is correct.
#### 5. Question
If every year 2/5th of a community’s population migrates to another region, find the leftover population after the 3rd year, assuming no growth during this period.
• (a) 27/125 part of the population.
• (b) 64/125 part of the population.
• (c) 8/25 part of the population.
• (d) 16/25 part of the population.
Answer: (a)
Solution:
Given that,
there is a policy that 2/5th of the population migrates every year to another region. There is no further growth in the population within three years.
Let the population be p.
Leftover population after one year = p − (2/5)×p = (3/5)p.
After every passing year population will become 3/5th of previous population.
Population after 2nd year = (3/5)(3/5)×p = (9/25)p. Population after 3rd year = (27/125)p.
Hence option (a) is correct.
Answer: (a)
Solution:
Given that,
there is a policy that 2/5th of the population migrates every year to another region. There is no further growth in the population within three years.
Let the population be p.
Leftover population after one year = p − (2/5)×p = (3/5)p.
After every passing year population will become 3/5th of previous population.
Population after 2nd year = (3/5)(3/5)×p = (9/25)p. Population after 3rd year = (27/125)p.
Hence option (a) is correct.
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