UPSC Insta–DART (Daily Aptitude and Reasoning Test) 12 Sep 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 The houses of A and B face each other on a road going east–west; A’s being on the southern side. A comes out of his house, turns left, travels 5 km, turns right, travels 12 km to the front of D’s house. B does exactly the same and reaches the front of C’s house. In this context, which one of the following statements is correct? (a) C and D live on the same street. (b) C’s house faces west. (c) The houses of C and D are 26 km apart. (d) None of the above Correct Answer: c Solution: Given that, The houses of A and B face each other on a road going east–west; A’s being on the southern side. A comes out, faces north; turns left (west) 5 km; turns right (north) 12 km to D’s house. B does exactly the same (from northern side: facing south; left = east; right = south) and reaches C’s house. Now, The direction concluded: From A to D is a right‑triangle of legs 5 km and 12 km. From B to C is the symmetric right‑triangle in the opposite quadrant. Distance of CA 52+122=CA2 CA2=25+144=169⇒CA=13 Similarly, DA2=52+122=169⇒DA=13 CA + DA = CD So, CD = 26 km Hence option (c) is correct. Incorrect Answer: c Solution: Given that, The houses of A and B face each other on a road going east–west; A’s being on the southern side. A comes out, faces north; turns left (west) 5 km; turns right (north) 12 km to D’s house. B does exactly the same (from northern side: facing south; left = east; right = south) and reaches C’s house. Now, The direction concluded: From A to D is a right‑triangle of legs 5 km and 12 km. From B to C is the symmetric right‑triangle in the opposite quadrant. Distance of CA 52+122=CA2 CA2=25+144=169⇒CA=13 Similarly, DA2=52+122=169⇒DA=13 CA + DA = CD So, CD = 26 km Hence option (c) is correct.
#### 1. Question
The houses of A and B face each other on a road going east–west; A’s being on the southern side. A comes out of his house, turns left, travels 5 km, turns right, travels 12 km to the front of D’s house. B does exactly the same and reaches the front of C’s house. In this context, which one of the following statements is correct?
• (a) C and D live on the same street.
• (b) C’s house faces west.
• (c) The houses of C and D are 26 km apart.
• (d) None of the above
Answer: c
Solution:
Given that,
The houses of A and B face each other on a road going east–west; A’s being on the southern side.
A comes out, faces north; turns left (west) 5 km; turns right (north) 12 km to D’s house. B does exactly the same (from northern side: facing south; left = east; right = south) and reaches C’s house.
The direction concluded:
From A to D is a right‑triangle of legs 5 km and 12 km. From B to C is the symmetric right‑triangle in the opposite quadrant.
Distance of CA
52+122=CA2
CA2=25+144=169⇒CA=13
Similarly,
DA2=52+122=169⇒DA=13
CA + DA = CD
So, CD = 26 km
Hence option (c) is correct.
Answer: c
Solution:
Given that,
The houses of A and B face each other on a road going east–west; A’s being on the southern side.
A comes out, faces north; turns left (west) 5 km; turns right (north) 12 km to D’s house. B does exactly the same (from northern side: facing south; left = east; right = south) and reaches C’s house.
The direction concluded:
From A to D is a right‑triangle of legs 5 km and 12 km. From B to C is the symmetric right‑triangle in the opposite quadrant.
Distance of CA
52+122=CA2
CA2=25+144=169⇒CA=13
Similarly,
DA2=52+122=169⇒DA=13
CA + DA = CD
So, CD = 26 km
Hence option (c) is correct.
• Question 2 of 5 2. Question The average score of a student in the first 4 subjects was 70. The average in the last 4 subjects (2nd, 3rd, 4th, and 5th) was 74. If the ratio of marks in 1st and 5th subjects was 4:5, what was the score in the 1st subject? (a) 76 (b) 68 (c) 64 (d) 60 Correct Answer: C Explanation: Let scores be A, B, C, D, E A + B + C + D = 70 × 4 = 280 B + C + D + E = 74 × 4 = 296 Subtract: E − A = 296 − 280 = 16 A : E = 4 : 5 ⇒ Let A = 4x, E = 5x E − A = x = 16 ⇒ x = 16 So A = 4x = 64 Incorrect Answer: C Explanation: Let scores be A, B, C, D, E A + B + C + D = 70 × 4 = 280 B + C + D + E = 74 × 4 = 296 Subtract: E − A = 296 − 280 = 16 A : E = 4 : 5 ⇒ Let A = 4x, E = 5x E − A = x = 16 ⇒ x = 16 So A = 4x = 64
#### 2. Question
The average score of a student in the first 4 subjects was 70. The average in the last 4 subjects (2nd, 3rd, 4th, and 5th) was 74. If the ratio of marks in 1st and 5th subjects was 4:5, what was the score in the 1st subject?
Explanation:
Let scores be A, B, C, D, E
• A + B + C + D = 70 × 4 = 280
• B + C + D + E = 74 × 4 = 296
Subtract: E − A = 296 − 280 = 16 A : E = 4 : 5 ⇒ Let A = 4x, E = 5x E − A = x = 16 ⇒ x = 16 So A = 4x = 64
Explanation:
Let scores be A, B, C, D, E
• A + B + C + D = 70 × 4 = 280
• B + C + D + E = 74 × 4 = 296
Subtract: E − A = 296 − 280 = 16 A : E = 4 : 5 ⇒ Let A = 4x, E = 5x E − A = x = 16 ⇒ x = 16 So A = 4x = 64
• Question 3 of 5 3. Question A question is given followed by two statements: Question: What is the selling price of the article? Statement I: The article was bought for ₹500 and the loss incurred was 10%. Statement II: The article was sold at 90% of the cost price. Which of the following is correct with respect to above? (a) Statement I alone is sufficient (b) Statement II alone is sufficient (c) Both the statements together are not sufficient (d) Either statement alone is sufficient Correct Answer: A Solution: Statement I: CP = ₹500, Loss = 10% SP = 90% of 500 = ₹450 ⇒ Sufficient Statement II: SP = 90% of CP, but CP not given ⇒ Cannot find SP value ⇒ Not sufficient Hence, option a is correct. Incorrect Answer: A Solution: Statement I: CP = ₹500, Loss = 10% SP = 90% of 500 = ₹450 ⇒ Sufficient Statement II: SP = 90% of CP, but CP not given ⇒ Cannot find SP value ⇒ Not sufficient Hence, option a is correct.
#### 3. Question
A question is given followed by two statements:
Question: What is the selling price of the article?
Statement I: The article was bought for ₹500 and the loss incurred was 10%. Statement II: The article was sold at 90% of the cost price.
Which of the following is correct with respect to above?
• (a) Statement I alone is sufficient
• (b) Statement II alone is sufficient
• (c) Both the statements together are not sufficient
• (d) Either statement alone is sufficient
Solution: Statement I:
• CP = ₹500, Loss = 10%
• SP = 90% of 500 = ₹450 ⇒ Sufficient
Statement II:
• SP = 90% of CP, but CP not given ⇒ Cannot find SP value ⇒ Not sufficient
Hence, option a is correct.
Solution: Statement I:
• CP = ₹500, Loss = 10%
• SP = 90% of 500 = ₹450 ⇒ Sufficient
Statement II:
• SP = 90% of CP, but CP not given ⇒ Cannot find SP value ⇒ Not sufficient
Hence, option a is correct.
• Question 4 of 5 4. Question A biker rides to college at an average speed of 36 km/h. The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining distance. How far is his college? (a) 18 km (b) 20 km (c) 24 km (d) 30 km Correct Answer: C Solution: Given that, Biker rides at an average speed of 36 km/h. The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining 60%. Now, Let total distance = D. Time for first 40% = (0.4D)/36 hours; time for remaining 60% = (0.6D)/36 hours. Difference = [(0.6D − 0.4D)/36] hours = (0.2D)/36 hours = 8 minutes = 8/60 hours. So, (0.2D)/36 = 8/60 ⇒ D = 24 km. Hence option (c) is correct. Incorrect Answer: C Solution: Given that, Biker rides at an average speed of 36 km/h. The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining 60%. Now, Let total distance = D. Time for first 40% = (0.4D)/36 hours; time for remaining 60% = (0.6D)/36 hours. Difference = [(0.6D − 0.4D)/36] hours = (0.2D)/36 hours = 8 minutes = 8/60 hours. So, (0.2D)/36 = 8/60 ⇒ D = 24 km. Hence option (c) is correct.
#### 4. Question
A biker rides to college at an average speed of 36 km/h. The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining distance. How far is his college?
Solution:
Given that,
Biker rides at an average speed of 36 km/h.
The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining 60%.
Let total distance = D.
Time for first 40% = (0.4D)/36 hours; time for remaining 60% = (0.6D)/36 hours.
Difference = [(0.6D − 0.4D)/36] hours = (0.2D)/36 hours = 8 minutes = 8/60 hours.
(0.2D)/36 = 8/60 ⇒ D = 24 km.
Hence option (c) is correct.
Solution:
Given that,
Biker rides at an average speed of 36 km/h.
The time taken to cover the first 40% of the distance is 8 minutes less than the time taken to cover the remaining 60%.
Let total distance = D.
Time for first 40% = (0.4D)/36 hours; time for remaining 60% = (0.6D)/36 hours.
Difference = [(0.6D − 0.4D)/36] hours = (0.2D)/36 hours = 8 minutes = 8/60 hours.
(0.2D)/36 = 8/60 ⇒ D = 24 km.
Hence option (c) is correct.
• Question 5 of 5 5. Question A man started from home at 14:00 hours and drove to a village, arriving there when the village clock indicated 14:45 hours. After staying for 10 minutes, he drove back by a different route of length 1.5 times the first route at a rate twice as fast, reaching home at 15:20 hours. As compared to the clock at home, the village clock is (a) 10 minutes slow (b) 5 minutes slow (c) 10 minutes fast (d) 5 minutes fast Correct Answer: D Solution: Given that, He started at 14:00 and reached home at 15:20. He stayed in the village for 10 minutes. The return route is 1.5 times the onward route, and the return speed is twice the onward speed. Now, Let onward speed be S km/h and onward distance be D km. Total time taken = 15:20 − 14:00 = 1 h 20 min = 80 min. Total travelling time = 80 − 10 = 70 min. Return time = (1.5D)/(2S) = 0.75 × (D/S). So, total travelling time = (D/S) + 0.75(D/S) = 1.75(D/S) = 70 min ⇒ D/S = 40 min. Exact arrival time on home clock = 14:00 + 40 min = 14:40. Village clock showed 14:45, i.e., 5 minutes fast. Hence option (d) is correct. Incorrect Answer: D Solution: Given that, He started at 14:00 and reached home at 15:20. He stayed in the village for 10 minutes. The return route is 1.5 times the onward route, and the return speed is twice the onward speed. Now, Let onward speed be S km/h and onward distance be D km. Total time taken = 15:20 − 14:00 = 1 h 20 min = 80 min. Total travelling time = 80 − 10 = 70 min. Return time = (1.5D)/(2S) = 0.75 × (D/S). So, total travelling time = (D/S) + 0.75(D/S) = 1.75(D/S) = 70 min ⇒ D/S = 40 min. Exact arrival time on home clock = 14:00 + 40 min = 14:40. Village clock showed 14:45, i.e., 5 minutes fast. Hence option (d) is correct.
#### 5. Question
A man started from home at 14:00 hours and drove to a village, arriving there when the village clock indicated 14:45 hours. After staying for 10 minutes, he drove back by a different route of length 1.5 times the first route at a rate twice as fast, reaching home at 15:20 hours. As compared to the clock at home, the village clock is
• (a) 10 minutes slow
• (b) 5 minutes slow
• (c) 10 minutes fast
• (d) 5 minutes fast
Solution:
Given that,
He started at 14:00 and reached home at 15:20. He stayed in the village for 10 minutes. The return route is 1.5 times the onward route, and the return speed is twice the onward speed.
Let onward speed be S km/h and onward distance be D km.
Total time taken = 15:20 − 14:00 = 1 h 20 min = 80 min.
Total travelling time = 80 − 10 = 70 min.
Return time = (1.5D)/(2S) = 0.75 × (D/S).
So, total travelling time = (D/S) + 0.75(D/S) = 1.75(D/S) = 70 min ⇒ D/S = 40 min.
Exact arrival time on home clock = 14:00 + 40 min = 14:40.
Village clock showed 14:45, i.e., 5 minutes fast.
Hence option (d) is correct.
Solution:
Given that,
He started at 14:00 and reached home at 15:20. He stayed in the village for 10 minutes. The return route is 1.5 times the onward route, and the return speed is twice the onward speed.
Let onward speed be S km/h and onward distance be D km.
Total time taken = 15:20 − 14:00 = 1 h 20 min = 80 min.
Total travelling time = 80 − 10 = 70 min.
Return time = (1.5D)/(2S) = 0.75 × (D/S).
So, total travelling time = (D/S) + 0.75(D/S) = 1.75(D/S) = 70 min ⇒ D/S = 40 min.
Exact arrival time on home clock = 14:00 + 40 min = 14:40.
Village clock showed 14:45, i.e., 5 minutes fast.
Hence option (d) is correct.
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