UPSC Insta–DART (Daily Aptitude and Reasoning Test) 30 Dec 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 If 1st March is a Friday in a leap year, then what is the day of the week on 1st September of the same year? (a) Sunday (b) Monday (c) Tuesday (d) Wednesday Correct Answer: (a) Solution: Count number of days from 1st March to 1st September (leap year): Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug (31) = 184 days 184 ÷ 7 = 26 weeks + 2 odd days ⇒ 1st September = Friday + 2 = Sunday Incorrect Answer: (a) Solution: Count number of days from 1st March to 1st September (leap year): Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug (31) = 184 days 184 ÷ 7 = 26 weeks + 2 odd days ⇒ 1st September = Friday + 2 = Sunday
#### 1. Question
If 1st March is a Friday in a leap year, then what is the day of the week on 1st September of the same year?
• (a) Sunday
• (b) Monday
• (c) Tuesday
• (d) Wednesday
Answer: (a)
Solution: Count number of days from 1st March to 1st September (leap year): Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug (31) = 184 days
184 ÷ 7 = 26 weeks + 2 odd days
⇒ 1st September = Friday + 2 = Sunday
Answer: (a)
Solution: Count number of days from 1st March to 1st September (leap year): Mar (31), Apr (30), May (31), Jun (30), Jul (31), Aug (31) = 184 days
184 ÷ 7 = 26 weeks + 2 odd days
⇒ 1st September = Friday + 2 = Sunday
• Question 2 of 5 2. Question Consider the following two statements and a question: Statement-1: The last day of the month is a Thursday. Statement-2: The third Saturday of the month is the 21st. Question: What day of the week is the 18th of the same month? (a) Statement-1 alone is sufficient to answer the question. (b) Statement-2 alone is sufficient to answer the question. (c) Both Statement-1 and Statement-2 are required to answer the question. (d) Neither Statement-1 nor Statement-2 alone is sufficient to answer the question. Correct Answer: (b) Solution: From Statement-1: We know only the last day is Thursday. But without knowing the total number of days (30/31), or a fixed date like 1st, we cannot trace back to the 18th. So, Statement-1 alone is not sufficient. From Statement-2: If 21st is the third Saturday → then Saturdays are: 7 (1st Saturday), 14 (2nd), 21 (3rd), 28 (4th) So, 21st = Saturday ⇒ 20th = Friday ⇒ 19th = Thursday ⇒ 18th = Wednesday Hence, Statement-2 alone is sufficient Incorrect Answer: (b) Solution: From Statement-1: We know only the last day is Thursday. But without knowing the total number of days (30/31), or a fixed date like 1st, we cannot trace back to the 18th. So, Statement-1 alone is not sufficient. From Statement-2: If 21st is the third Saturday → then Saturdays are: 7 (1st Saturday), 14 (2nd), 21 (3rd), 28 (4th) So, 21st = Saturday ⇒ 20th = Friday ⇒ 19th = Thursday ⇒ 18th = Wednesday Hence, Statement-2 alone is sufficient
#### 2. Question
Consider the following two statements and a question:
Statement-1: The last day of the month is a Thursday. Statement-2: The third Saturday of the month is the 21st.
Question: What day of the week is the 18th of the same month?
• (a) Statement-1 alone is sufficient to answer the question.
• (b) Statement-2 alone is sufficient to answer the question.
• (c) Both Statement-1 and Statement-2 are required to answer the question.
• (d) Neither Statement-1 nor Statement-2 alone is sufficient to answer the question.
Answer: (b)
Solution:
From Statement-1: We know only the last day is Thursday. But without knowing the total number of days (30/31), or a fixed date like 1st, we cannot trace back to the 18th. So, Statement-1 alone is not sufficient.
From Statement-2: If 21st is the third Saturday → then Saturdays are: 7 (1st Saturday), 14 (2nd), 21 (3rd), 28 (4th) So, 21st = Saturday ⇒ 20th = Friday ⇒ 19th = Thursday ⇒ 18th = Wednesday
Hence, Statement-2 alone is sufficient
Answer: (b)
Solution:
From Statement-1: We know only the last day is Thursday. But without knowing the total number of days (30/31), or a fixed date like 1st, we cannot trace back to the 18th. So, Statement-1 alone is not sufficient.
From Statement-2: If 21st is the third Saturday → then Saturdays are: 7 (1st Saturday), 14 (2nd), 21 (3rd), 28 (4th) So, 21st = Saturday ⇒ 20th = Friday ⇒ 19th = Thursday ⇒ 18th = Wednesday
Hence, Statement-2 alone is sufficient
• Question 3 of 5 3. Question Ananya goes to yoga class every 6th day, Bhavna every 8th day, and Charu every 12th day. If they all met on a Monday, on which day will they all meet again? (a) Tuesday (b) Wednesday (c) Thursday (d) Friday Correct Answer: (c) Solution: Ananya – every 6 days Bhavna – every 8 days Charu – every 12 days LCM of 6, 8, 12 = 24 → They will meet every 24 days They met on Monday 24 ÷ 7 = 3 weeks + 3 odd days → Monday + 3 days = Thursday Incorrect Answer: (c) Solution: Ananya – every 6 days Bhavna – every 8 days Charu – every 12 days LCM of 6, 8, 12 = 24 → They will meet every 24 days They met on Monday 24 ÷ 7 = 3 weeks + 3 odd days → Monday + 3 days = Thursday
#### 3. Question
Ananya goes to yoga class every 6th day, Bhavna every 8th day, and Charu every 12th day. If they all met on a Monday, on which day will they all meet again?
• (a) Tuesday
• (b) Wednesday
• (c) Thursday
• (d) Friday
Answer: (c)
Solution: Ananya – every 6 days Bhavna – every 8 days Charu – every 12 days
LCM of 6, 8, 12 = 24 → They will meet every 24 days
They met on Monday 24 ÷ 7 = 3 weeks + 3 odd days
→ Monday + 3 days = Thursday
Answer: (c)
Solution: Ananya – every 6 days Bhavna – every 8 days Charu – every 12 days
LCM of 6, 8, 12 = 24 → They will meet every 24 days
They met on Monday 24 ÷ 7 = 3 weeks + 3 odd days
→ Monday + 3 days = Thursday
• Question 4 of 5 4. Question A cube has all its faces painted. It is cut into 64 smaller cubes of equal size. How many of the smaller cubes will have exactly one face painted? (a) 16 (b) 24 (c) 32 (d) 36 Correct Answer: (b) Solution: Since 64 = , the original cube was cut into 4 × 4 × 4 small cubes → Small cubes with one face painted are center cubes on each face, i.e., not touching edges → Each face has inner such cubes → 6 faces × 4 = 24 Incorrect Answer: (b) Solution: Since 64 = , the original cube was cut into 4 × 4 × 4 small cubes → Small cubes with one face painted are center cubes on each face, i.e., not touching edges → Each face has inner such cubes → 6 faces × 4 = 24
#### 4. Question
A cube has all its faces painted. It is cut into 64 smaller cubes of equal size. How many of the smaller cubes will have exactly one face painted?
Answer: (b)
Solution: Since 64 = , the original cube was cut into 4 × 4 × 4 small cubes
→ Small cubes with one face painted are center cubes on each face, i.e., not touching edges
→ Each face has inner such cubes → 6 faces × 4 = 24
Answer: (b)
Solution: Since 64 = , the original cube was cut into 4 × 4 × 4 small cubes
→ Small cubes with one face painted are center cubes on each face, i.e., not touching edges
→ Each face has inner such cubes → 6 faces × 4 = 24
• Question 5 of 5 5. Question The outer surface of a 5 cm × 5 cm × 5 cm cube is painted completely in blue. It is sliced parallel to the faces to yield 125 small cubes of 1 cm × 1 cm × 1 cm. How many of the small cubes do not have any face painted? (a) 8 (b) 16 (c) 27 (d) 64 Correct Answer: (c) Solution: The large cube is of size 5 × 5 × 5 = 125 small cubes. Cubes that are not painted at all are those that are completely inside — i.e., not on the surface. This forms a smaller cube of side = 5 − 2 = 3 ⇒ Total unpainted cubes = 3³ = 27 Incorrect Answer: (c) Solution: The large cube is of size 5 × 5 × 5 = 125 small cubes. Cubes that are not painted at all are those that are completely inside — i.e., not on the surface. This forms a smaller cube of side = 5 − 2 = 3 ⇒ Total unpainted cubes = 3³ = 27
#### 5. Question
The outer surface of a 5 cm × 5 cm × 5 cm cube is painted completely in blue. It is sliced parallel to the faces to yield 125 small cubes of 1 cm × 1 cm × 1 cm. How many of the small cubes do not have any face painted?
Answer: (c)
Solution: The large cube is of size 5 × 5 × 5 = 125 small cubes. Cubes that are not painted at all are those that are completely inside — i.e., not on the surface.
This forms a smaller cube of side = 5 − 2 = 3 ⇒ Total unpainted cubes = 3³ = 27
Answer: (c)
Solution: The large cube is of size 5 × 5 × 5 = 125 small cubes. Cubes that are not painted at all are those that are completely inside — i.e., not on the surface.
This forms a smaller cube of side = 5 − 2 = 3 ⇒ Total unpainted cubes = 3³ = 27
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