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?
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 days184 ÷ 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 days184 ÷ 7 = 26 weeks + 2 odd days
⇒ 1st September = Friday + 2 = Sunday
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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?
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 = WednesdayHence, 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 = WednesdayHence, Statement-2 alone is sufficient
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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?
Correct
Answer: (c)
Solution:
Ananya – every 6 days
Bhavna – every 8 days
Charu – every 12 daysLCM of 6, 8, 12 = 24
→ They will meet every 24 daysThey 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 daysLCM of 6, 8, 12 = 24
→ They will meet every 24 daysThey met on Monday
24 ÷ 7 = 3 weeks + 3 odd days→ Monday + 3 days = Thursday
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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?
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 = 24Incorrect
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?
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 = 27Incorrect
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
