UPSC Insta–DART (Daily Aptitude and Reasoning Test) 10 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 train travels 25% faster than a car. Both start together to cover 80 km. The train halts at stations for 12 minutes, yet they reach together. The speed of the car is: (a) 72 kmph (b) 75 kmph (c) 80 kmph (d) 84 kmph Correct Answer – C Solution: Given that, Train is 25% faster than car. Let car speed be v kmph ⇒ train speed = 1.25v = 5v/4 kmph. Now, Car time = 80/v. Train running time = 80/(5v/4) = 64/v. Halt time = 12 minutes = 12/60 = 1/5 hour. So, 80/v = 64/v + 1/5 ⇒ 16/v = 1/5 ⇒ v = 16 × 5 = 80 kmph. Hence option (c) is correct Incorrect Answer – C Solution: Given that, Train is 25% faster than car. Let car speed be v kmph ⇒ train speed = 1.25v = 5v/4 kmph. Now, Car time = 80/v. Train running time = 80/(5v/4) = 64/v. Halt time = 12 minutes = 12/60 = 1/5 hour. So, 80/v = 64/v + 1/5 ⇒ 16/v = 1/5 ⇒ v = 16 × 5 = 80 kmph. Hence option (c) is correct
#### 1. Question
A train travels 25% faster than a car. Both start together to cover 80 km. The train halts at stations for 12 minutes, yet they reach together. The speed of the car is:
• (a) 72 kmph
• (b) 75 kmph
• (c) 80 kmph
• (d) 84 kmph
Answer – C Solution:
Given that,
Train is 25% faster than car.
Let car speed be v kmph ⇒ train speed = 1.25v = 5v/4 kmph.
Car time = 80/v. Train running time = 80/(5v/4) = 64/v. Halt time = 12 minutes = 12/60 = 1/5 hour.
80/v = 64/v + 1/5 ⇒ 16/v = 1/5 ⇒ v = 16 × 5 = 80 kmph.
Hence option (c) is correct
Answer – C Solution:
Given that,
Train is 25% faster than car.
Let car speed be v kmph ⇒ train speed = 1.25v = 5v/4 kmph.
Car time = 80/v. Train running time = 80/(5v/4) = 64/v. Halt time = 12 minutes = 12/60 = 1/5 hour.
80/v = 64/v + 1/5 ⇒ 16/v = 1/5 ⇒ v = 16 × 5 = 80 kmph.
Hence option (c) is correct
• Question 2 of 5 2. Question Seema plans to reach office by 5 P.M. if she drives at 20 kmph; she would reach by 3 P.M. if she drives at 30 kmph. At what speed must she drive to reach by 4 P.M.? (a) 22 kmph (b) 24 kmph (c) 25 kmph (d) 26 kmph Correct Answer – B Solution: Given that, At 20 kmph arrival is 5 P.M., at 30 kmph arrival is 3 P.M. (2 hours earlier). Let the distance be D. Now, D/20 − D/30 = 2 D(1/20 − 1/30) = 2 ⇒ D(1/60) = 2 ⇒ D = 120 km At 20 kmph, time = 120/20 = 6 hours (5 P.M.). To reach at 4 P.M. (1 hour earlier), required time = 5 hours. Required speed = 120 / 5 = 24 kmph Hence option (b) is correct Incorrect Answer – B Solution: Given that, At 20 kmph arrival is 5 P.M., at 30 kmph arrival is 3 P.M. (2 hours earlier). Let the distance be D. Now, D/20 − D/30 = 2 D(1/20 − 1/30) = 2 ⇒ D(1/60) = 2 ⇒ D = 120 km At 20 kmph, time = 120/20 = 6 hours (5 P.M.). To reach at 4 P.M. (1 hour earlier), required time = 5 hours. Required speed = 120 / 5 = 24 kmph Hence option (b) is correct
#### 2. Question
Seema plans to reach office by 5 P.M. if she drives at 20 kmph; she would reach by 3 P.M. if she drives at 30 kmph. At what speed must she drive to reach by 4 P.M.?
• (a) 22 kmph
• (b) 24 kmph
• (c) 25 kmph
• (d) 26 kmph
Answer – B Solution:
Given that,
At 20 kmph arrival is 5 P.M., at 30 kmph arrival is 3 P.M. (2 hours earlier).
Let the distance be D.
D/20 − D/30 = 2 D(1/20 − 1/30) = 2 ⇒ D(1/60) = 2 ⇒ D = 120 km
At 20 kmph, time = 120/20 = 6 hours (5 P.M.). To reach at 4 P.M. (1 hour earlier), required time = 5 hours.
Required speed = 120 / 5 = 24 kmph
Hence option (b) is correct
Answer – B Solution:
Given that,
At 20 kmph arrival is 5 P.M., at 30 kmph arrival is 3 P.M. (2 hours earlier).
Let the distance be D.
D/20 − D/30 = 2 D(1/20 − 1/30) = 2 ⇒ D(1/60) = 2 ⇒ D = 120 km
At 20 kmph, time = 120/20 = 6 hours (5 P.M.). To reach at 4 P.M. (1 hour earlier), required time = 5 hours.
Required speed = 120 / 5 = 24 kmph
Hence option (b) is correct
• Question 3 of 5 3. Question A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 2 km/h and Y walks with a uniform speed of 1 km/h in the first hour, 1.25 km/h in the second hour, 1.5 km/h in the third hour, 1.75 km/h in the fourth hour and 2 km/h in the fifth hour and so on. Which of the following is/are correct? They take 5 hours to meet. They meet midway between A and B. (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Correct Answer – D Solution: Given that, Total distance between A and B = 15 km X walks with a uniform speed = 2 km/h Y’s hourly speeds = 1, 1.25, 1.5, 1.75, 2, … Now, After 4 hours: X covers = 2 × 4 = 8 km Y covers = 1 + 1.25 + 1.5 + 1.75 = 5.5 km Total = 13.5 km; remaining = 15 − 13.5 = 1.5 km In the 5th hour, speeds are X = 2 km/h and Y = 2 km/h ⇒ combined = 4 km/h. Time to finish remaining 1.5 km = 1.5/4 = 0.375 hour = 22.5 minutes. Therefore, they meet in 4 hours 22.5 minutes (not 5 hours), and distances at meet: X total = 8 + (2 × 0.375) = 8.75 km; Y total = 5.5 + (2 × 0.375) = 6.25 km (not midpoint 7.5 km). They take 5 hours to meet. Hence statement 1 is incorrect They meet midway between A and B. Hence statement 2 is incorrect Hence option (d) is correct Incorrect Answer – D Solution: Given that, Total distance between A and B = 15 km X walks with a uniform speed = 2 km/h Y’s hourly speeds = 1, 1.25, 1.5, 1.75, 2, … Now, After 4 hours: X covers = 2 × 4 = 8 km Y covers = 1 + 1.25 + 1.5 + 1.75 = 5.5 km Total = 13.5 km; remaining = 15 − 13.5 = 1.5 km In the 5th hour, speeds are X = 2 km/h and Y = 2 km/h ⇒ combined = 4 km/h. Time to finish remaining 1.5 km = 1.5/4 = 0.375 hour = 22.5 minutes. Therefore, they meet in 4 hours 22.5 minutes (not 5 hours), and distances at meet: X total = 8 + (2 × 0.375) = 8.75 km; Y total = 5.5 + (2 × 0.375) = 6.25 km (not midpoint 7.5 km). They take 5 hours to meet. Hence statement 1 is incorrect They meet midway between A and B. Hence statement 2 is incorrect Hence option (d) is correct
#### 3. Question
A person X from a place A and another person Y from a place B set out at the same time to walk towards each other. The places are separated by a distance of 15 km. X walks with a uniform speed of 2 km/h and Y walks with a uniform speed of 1 km/h in the first hour, 1.25 km/h in the second hour, 1.5 km/h in the third hour, 1.75 km/h in the fourth hour and 2 km/h in the fifth hour and so on.
Which of the following is/are correct?
• They take 5 hours to meet.
• They meet midway between A and B.
• (a) 1 only
• (b) 2 only
• (c) Both 1 and 2
• (d) Neither 1 nor 2
Answer – D Solution:
Given that,
Total distance between A and B = 15 km
X walks with a uniform speed = 2 km/h
Y’s hourly speeds = 1, 1.25, 1.5, 1.75, 2, …
After 4 hours: X covers = 2 × 4 = 8 km Y covers = 1 + 1.25 + 1.5 + 1.75 = 5.5 km Total = 13.5 km; remaining = 15 − 13.5 = 1.5 km
In the 5th hour, speeds are X = 2 km/h and Y = 2 km/h ⇒ combined = 4 km/h. Time to finish remaining 1.5 km = 1.5/4 = 0.375 hour = 22.5 minutes.
Therefore, they meet in 4 hours 22.5 minutes (not 5 hours), and distances at meet: X total = 8 + (2 × 0.375) = 8.75 km; Y total = 5.5 + (2 × 0.375) = 6.25 km (not midpoint 7.5 km).
• They take 5 hours to meet. Hence statement 1 is incorrect
• They meet midway between A and B. Hence statement 2 is incorrect
Hence option (d) is correct
Answer – D Solution:
Given that,
Total distance between A and B = 15 km
X walks with a uniform speed = 2 km/h
Y’s hourly speeds = 1, 1.25, 1.5, 1.75, 2, …
After 4 hours: X covers = 2 × 4 = 8 km Y covers = 1 + 1.25 + 1.5 + 1.75 = 5.5 km Total = 13.5 km; remaining = 15 − 13.5 = 1.5 km
In the 5th hour, speeds are X = 2 km/h and Y = 2 km/h ⇒ combined = 4 km/h. Time to finish remaining 1.5 km = 1.5/4 = 0.375 hour = 22.5 minutes.
Therefore, they meet in 4 hours 22.5 minutes (not 5 hours), and distances at meet: X total = 8 + (2 × 0.375) = 8.75 km; Y total = 5.5 + (2 × 0.375) = 6.25 km (not midpoint 7.5 km).
• They take 5 hours to meet. Hence statement 1 is incorrect
• They meet midway between A and B. Hence statement 2 is incorrect
Hence option (d) is correct
• Question 4 of 5 4. Question Two cars start towards each other from two places A and B which are at a distance of 150 km. They start at the same time 06:40 am. If the speeds of the cars are 40 km/h and 35 km/h respectively, they will meet each other at: (a) 08:30 am (b) 08:40 am (c) 08:50 am (d) 09:00 am Correct Answer – B Solution: Given that, Two cars start towards each other from A and B, 150 km apart. They start at the same time 06:40 am. The speeds of the cars are 40 km/h and 35 km/h respectively. Now, Relative speed = 40 + 35 = 75 km/hr Distance = 150 km Time taken = 150/75 = 2 hrs 06:40 am + 2 hours = 08:40 am Hence option (b) is correct Incorrect Answer – B Solution: Given that, Two cars start towards each other from A and B, 150 km apart. They start at the same time 06:40 am. The speeds of the cars are 40 km/h and 35 km/h respectively. Now, Relative speed = 40 + 35 = 75 km/hr Distance = 150 km Time taken = 150/75 = 2 hrs 06:40 am + 2 hours = 08:40 am Hence option (b) is correct
#### 4. Question
Two cars start towards each other from two places A and B which are at a distance of 150 km. They start at the same time 06:40 am. If the speeds of the cars are 40 km/h and 35 km/h respectively, they will meet each other at:
• (a) 08:30 am
• (b) 08:40 am
• (c) 08:50 am
• (d) 09:00 am
Answer – B Solution:
Given that,
Two cars start towards each other from A and B, 150 km apart.
They start at the same time 06:40 am.
The speeds of the cars are 40 km/h and 35 km/h respectively.
Relative speed = 40 + 35 = 75 km/hr
Distance = 150 km
Time taken = 150/75 = 2 hrs
06:40 am + 2 hours = 08:40 am
Hence option (b) is correct
Answer – B Solution:
Given that,
Two cars start towards each other from A and B, 150 km apart.
They start at the same time 06:40 am.
The speeds of the cars are 40 km/h and 35 km/h respectively.
Relative speed = 40 + 35 = 75 km/hr
Distance = 150 km
Time taken = 150/75 = 2 hrs
06:40 am + 2 hours = 08:40 am
Hence option (b) is correct
• Question 5 of 5 5. Question A motorboat’s speed in still water is 15 km/h. It goes from P to Q and back. The round trip takes 12 hours, and the upstream journey takes 2 hours longer than the downstream journey. What is the speed of the stream? (a) 2 km/h (b) 2.5 km/h (c) 3 km/h (d) 3.5 km/h Correct Answer – B Solution: Given that, Still-water speed = 15 km/h; upstream time = (downstream time) + 2 h; total = 12 h. Now, Let downstream time = t hours ⇒ upstream time = t + 2; so 2t + 2 = 12 ⇒ t = 5, upstream = 7. Let stream speed = u. For one-way distance D: D = (15 + u)×5 and also D = (15 − u)×7 (15 + u)×5 = (15 − u)×7 ⇒ 75 + 5u = 105 − 7u ⇒ 12u = 30 ⇒ u = 2.5 km/h Hence option (b) is correct Incorrect Answer – B Solution: Given that, Still-water speed = 15 km/h; upstream time = (downstream time) + 2 h; total = 12 h. Now, Let downstream time = t hours ⇒ upstream time = t + 2; so 2t + 2 = 12 ⇒ t = 5, upstream = 7. Let stream speed = u. For one-way distance D: D = (15 + u)×5 and also D = (15 − u)×7 (15 + u)×5 = (15 − u)×7 ⇒ 75 + 5u = 105 − 7u ⇒ 12u = 30 ⇒ u = 2.5 km/h Hence option (b) is correct
#### 5. Question
A motorboat’s speed in still water is 15 km/h. It goes from P to Q and back. The round trip takes 12 hours, and the upstream journey takes 2 hours longer than the downstream journey. What is the speed of the stream?
• (a) 2 km/h
• (b) 2.5 km/h
• (c) 3 km/h
• (d) 3.5 km/h
Answer – B
Given that,
Still-water speed = 15 km/h; upstream time = (downstream time) + 2 h; total = 12 h.
Let downstream time = t hours ⇒ upstream time = t + 2; so 2t + 2 = 12 ⇒ t = 5, upstream = 7.
Let stream speed = u. For one-way distance D:
D = (15 + u)×5 and also D = (15 − u)×7
(15 + u)×5 = (15 − u)×7 ⇒ 75 + 5u = 105 − 7u ⇒ 12u = 30 ⇒ u = 2.5 km/h
Hence option (b) is correct
Answer – B
Given that,
Still-water speed = 15 km/h; upstream time = (downstream time) + 2 h; total = 12 h.
Let downstream time = t hours ⇒ upstream time = t + 2; so 2t + 2 = 12 ⇒ t = 5, upstream = 7.
Let stream speed = u. For one-way distance D:
D = (15 + u)×5 and also D = (15 − u)×7
(15 + u)×5 = (15 − u)×7 ⇒ 75 + 5u = 105 − 7u ⇒ 12u = 30 ⇒ u = 2.5 km/h
Hence option (b) is correct
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