UPSC Insta–DART (Daily Aptitude and Reasoning Test) 18 Nov 2024

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 21st December of a year was Friday. Then what was the day of the week on July 1 in the same year? (a) Sunday (b) Monday (c) Tuesday (d) Wednesday Correct Solution: Option (a) Explanation: First, we calculate the total number of days from 1st July to 21 December. Days left in July = 31-1 = 30; August= 31; September= 30; October= 31; November= 30; December= 21. So, the total days= 30 + 31 + 30 + 31+ 30 +21 = 173. Odd days in 173 days = 173/7 = 5 (Remainder). So, subtract 5 days from the given day of 21st December, because we are going backward from December to July. So, 1st July will be Friday -5 days, i.e. Sunday. Incorrect Solution: Option (a) Explanation: First, we calculate the total number of days from 1st July to 21 December. Days left in July = 31-1 = 30; August= 31; September= 30; October= 31; November= 30; December= 21. So, the total days= 30 + 31 + 30 + 31+ 30 +21 = 173. Odd days in 173 days = 173/7 = 5 (Remainder). So, subtract 5 days from the given day of 21st December, because we are going backward from December to July. So, 1st July will be Friday -5 days, i.e. Sunday.

#### 1. Question

21st December of a year was Friday. Then what was the day of the week on July 1 in the same year?

• (a) Sunday

• (b) Monday

• (c) Tuesday

• (d) Wednesday

Solution: Option (a)

Explanation:

First, we calculate the total number of days from 1st July to 21 December.

Days left in July = 31-1 = 30; August= 31; September= 30; October= 31; November= 30; December= 21.

So, the total days= 30 + 31 + 30 + 31+ 30 +21 = 173.

Odd days in 173 days = 173/7 = 5 (Remainder).

So, subtract 5 days from the given day of 21st December, because we are going backward from December to July.

So, 1st July will be Friday -5 days, i.e. Sunday.

Solution: Option (a)

Explanation:

First, we calculate the total number of days from 1st July to 21 December.

Days left in July = 31-1 = 30; August= 31; September= 30; October= 31; November= 30; December= 21.

So, the total days= 30 + 31 + 30 + 31+ 30 +21 = 173.

Odd days in 173 days = 173/7 = 5 (Remainder).

So, subtract 5 days from the given day of 21st December, because we are going backward from December to July.

So, 1st July will be Friday -5 days, i.e. Sunday.

• Question 2 of 5 2. Question If instead of four years, leap year follows a cycle of three years, then what day will it be on the 1st January 2020, if year 2013 is a leap year and 1st January 2013 is a Friday? (a) Monday (b) Tuesday (c) Wednesday (d) Thursday Correct Solution: Option (a) Explanation: It is given that instead of every fourth year, Every third year is a leap year. Then number of days from 1st January, 2013 to 1st January, 2020 are Years 2013 2014 2015 2016 2017 2018 2019 Odd days 2 1 1 2 1 1 2 = 10 i.e. 3 days These odd days are till 31st December, 2019. So, 1st January, 2020 will be 3 days after Friday i.e. on Monday. Hence Option (a) is correct answer. Incorrect Solution: Option (a) Explanation: It is given that instead of every fourth year, Every third year is a leap year. Then number of days from 1st January, 2013 to 1st January, 2020 are Years 2013 2014 2015 2016 2017 2018 2019 Odd days 2 1 1 2 1 1 2 = 10 i.e. 3 days These odd days are till 31st December, 2019. So, 1st January, 2020 will be 3 days after Friday i.e. on Monday. Hence Option (a) is correct answer.

#### 2. Question

If instead of four years, leap year follows a cycle of three years, then what day will it be on the 1st January 2020, if year 2013 is a leap year and 1st January 2013 is a Friday?

• (a) Monday

• (b) Tuesday

• (c) Wednesday

• (d) Thursday

Solution: Option (a)

Explanation:

It is given that instead of every fourth year, Every third year is a leap year.

Then number of days from 1st January, 2013 to 1st January, 2020 are

Years 2013 2014 2015 2016 2017 2018 2019

Odd days 2 1 1 2 1 1 2

= 10 i.e. 3 days

These odd days are till 31st December, 2019.

So, 1st January, 2020 will be 3 days after Friday i.e. on Monday. Hence Option (a) is correct answer.

Solution: Option (a)

Explanation:

It is given that instead of every fourth year, Every third year is a leap year.

Then number of days from 1st January, 2013 to 1st January, 2020 are

Years 2013 2014 2015 2016 2017 2018 2019

Odd days 2 1 1 2 1 1 2

= 10 i.e. 3 days

These odd days are till 31st December, 2019.

So, 1st January, 2020 will be 3 days after Friday i.e. on Monday. Hence Option (a) is correct answer.

• Question 3 of 5 3. Question “You must submit your application within 10 days from the date of release of this advertisement.” Question: What is the exact date before which the application must be submitted? S1 : The advertisement was released on 18th February. S2: It was a leap year. Which one of the following is correct in respect of the above Statements and the Question? (a) S1 alone is sufficient to answer the Question. (b) S2 alone is sufficient to answer the Question. (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question. (d) S1 and S2 together are not sufficient to answer the Question. Correct Solution: Option (a) Explanation: S1 => 10 days from the date of release of advertisement (18th February) means application must be submitted before 27 February. Hence, statement 1 alone is sufficient to answer the question. Incorrect Solution: Option (a) Explanation: S1 => 10 days from the date of release of advertisement (18th February) means application must be submitted before 27 February. Hence, statement 1 alone is sufficient to answer the question.

#### 3. Question

“You must submit your application within 10 days from the date of release of this advertisement.”

Question: What is the exact date before which the application must be submitted?

S1 : The advertisement was released on 18th February.

S2: It was a leap year.

Which one of the following is correct in respect of the above Statements and the Question?

• (a) S1 alone is sufficient to answer the Question.

• (b) S2 alone is sufficient to answer the Question.

• (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

• (d) S1 and S2 together are not sufficient to answer the Question.

Solution: Option (a)

Explanation:

S1 => 10 days from the date of release of advertisement (18th February) means application must be submitted before 27 February.

Hence, statement 1 alone is sufficient to answer the question.

Solution: Option (a)

Explanation:

S1 => 10 days from the date of release of advertisement (18th February) means application must be submitted before 27 February.

Hence, statement 1 alone is sufficient to answer the question.

• Question 4 of 5 4. Question If 2nd and 4th Saturdays, and all the Sundays are taken as holidays for an office, what would be the minimum number of possible working days of any month of any year? (a) 20 (b) 21 (c) 22 (d) 23 Correct Solution: Option (c) Explanation: In any month, in any year there are minimum 28 days (February) Normal year = 28 days in February If every 2nd and 4th Saturday and Sundays are taken as holidays for office, then total working days equal to 22. Incorrect Solution: Option (c) Explanation: In any month, in any year there are minimum 28 days (February) Normal year = 28 days in February If every 2nd and 4th Saturday and Sundays are taken as holidays for office, then total working days equal to 22.

#### 4. Question

If 2nd and 4th Saturdays, and all the Sundays are taken as holidays for an office, what would be the minimum number of possible working days of any month of any year?

Solution: Option (c)

Explanation:

In any month, in any year there are minimum 28 days (February)

Normal year = 28 days in February

If every 2nd and 4th Saturday and Sundays are taken as holidays for office, then total working days equal to 22.

Solution: Option (c)

Explanation:

In any month, in any year there are minimum 28 days (February)

Normal year = 28 days in February

If every 2nd and 4th Saturday and Sundays are taken as holidays for office, then total working days equal to 22.

• Question 5 of 5 5. Question Which of the following year will has the same calendar as that of 2016? (a) 2028 (b) 2024 (c) 2022 (d) 2044 Correct Solution: Option (d) Explanation: Clearly, 2016 was a leap year. So, the calendar can be used again only in a leap year. Between any two leap years, there are 5 odd days (3 odd days for the 3 non leap year and 2 odd days for one leap year). After 7 leap years, the number of odd days will be 7 × 5 = 35 days = 0 as 35 is completely divisible by 7. Thus, the calendar will repeat after 7 leap years = 7 × 4 = 28 years So, 2016 + 28 = 2044 Note: Calendar of a leap year always repeats after 28 years, while that of an ordinary year after 6 or 11 years. Hence, option (d) is the correct answer. Incorrect Solution: Option (d) Explanation: Clearly, 2016 was a leap year. So, the calendar can be used again only in a leap year. Between any two leap years, there are 5 odd days (3 odd days for the 3 non leap year and 2 odd days for one leap year). After 7 leap years, the number of odd days will be 7 × 5 = 35 days = 0 as 35 is completely divisible by 7. Thus, the calendar will repeat after 7 leap years = 7 × 4 = 28 years So, 2016 + 28 = 2044 Note: Calendar of a leap year always repeats after 28 years, while that of an ordinary year after 6 or 11 years. Hence, option (d) is the correct answer.

#### 5. Question

Which of the following year will has the same calendar as that of 2016?

Solution: Option (d)

Explanation:

Clearly, 2016 was a leap year. So, the calendar can be used again only in a leap year. Between any two leap years, there are 5 odd days (3 odd days for the 3 non leap year and 2 odd days for one leap year).

After 7 leap years, the number of odd days will be 7 × 5 = 35 days = 0 as 35 is completely divisible by 7.

Thus, the calendar will repeat after 7 leap years = 7 × 4 = 28 years

So, 2016 + 28 = 2044

Note: Calendar of a leap year always repeats after 28 years, while that of an ordinary year after 6 or 11 years.

Hence, option (d) is the correct answer.

Solution: Option (d)

Explanation:

Clearly, 2016 was a leap year. So, the calendar can be used again only in a leap year. Between any two leap years, there are 5 odd days (3 odd days for the 3 non leap year and 2 odd days for one leap year).

After 7 leap years, the number of odd days will be 7 × 5 = 35 days = 0 as 35 is completely divisible by 7.

Thus, the calendar will repeat after 7 leap years = 7 × 4 = 28 years

So, 2016 + 28 = 2044

Note: Calendar of a leap year always repeats after 28 years, while that of an ordinary year after 6 or 11 years.

Hence, option (d) is the correct answer.

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