Insta–DART (Daily Aptitude and Reasoning Test) 2020 - 21
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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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
If in a particular year 12th January is sunday a then which one of the following is correct?
Correct
Sol. (c)
Number of days between 12th January to 12 July in a leap year = 19 + 29 + 31 + 30 +31 + 30 + 12 * 182 = 26 weeks
Hence, there is no remainder, therefore 12th July shall be same as 12th January in a leap year. So,
it shall be sunday.
Incorrect
Sol. (c)
Number of days between 12th January to 12 July in a leap year = 19 + 29 + 31 + 30 +31 + 30 + 12 * 182 = 26 weeks
Hence, there is no remainder, therefore 12th July shall be same as 12th January in a leap year. So,
it shall be sunday.
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Question 2 of 5
2. Question
Mr. ‘X’ has three children. The birthday of the first child falls on the 5th Monday of April that of the second one falls on the 5th Thursday of November. On which day is the birthday of his third child, which falls on 20th December?
Correct
Sol. (b)
According to the question the birthday of first child of Mr. X is on fifth Monday of April and birthday of second child is on fifth Thursday of November. So, fifth Monday of April 29″ or 30 April and 29 or 30th nov.
No of odd days between Monday and Thursday=3 days
So, the birth of first child is on 30th April .
So the number of odd days between the birthdays of two child=3.
The birth of first child is on 30th April.so no of odd days=31+30+31+31+30+31+29=213=3 odd days.so the birth of first child is on 30th april and the birth of second child is on 29th November. Now, number of odd days between 29th November to 20th December = 1 + 20 = 21 = 0 odd day.Hence, the birthday of third child is on 20th December which is Thursday.
Incorrect
Sol. (b)
According to the question the birthday of first child of Mr. X is on fifth Monday of April and birthday of second child is on fifth Thursday of November. So, fifth Monday of April 29″ or 30 April and 29 or 30th nov.
No of odd days between Monday and Thursday=3 days
So, the birth of first child is on 30th April .
So the number of odd days between the birthdays of two child=3.
The birth of first child is on 30th April.so no of odd days=31+30+31+31+30+31+29=213=3 odd days.so the birth of first child is on 30th april and the birth of second child is on 29th November. Now, number of odd days between 29th November to 20th December = 1 + 20 = 21 = 0 odd day.Hence, the birthday of third child is on 20th December which is Thursday.
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Question 3 of 5
3. Question
A wall clock moves 10 min fast in every 24 hr. The clock was set right to show the correct time at 8:00 am on Monday. When the clock shows the time 6:00 pm on Wednesday, what is the correct time?
Correct
Sol. (a)
Clock to move fast in 24 hr = 10 min. Time from 8:00 am on Monday to till 6:00 pm on Wednesday 58 hr. So, clock to move fast in 58 hr=10/24*58
=24.1
= 24.1
24 min
So, correct time 6:00pm- 24 min= 5:36pm
Incorrect
Sol. (a)
Clock to move fast in 24 hr = 10 min. Time from 8:00 am on Monday to till 6:00 pm on Wednesday 58 hr. So, clock to move fast in 58 hr=10/24*58
=24.1
= 24.1
24 min
So, correct time 6:00pm- 24 min= 5:36pm
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Question 4 of 5
4. Question
Which year has the same calendar as that of 2009?
Correct
Sol. (d)
In one normal year we have 1 odd day and in one leap year we have 2 odd days. So, the repetition of same calendar happens after every 7 odd days. Now, 7 odd days will come in 6 years (5 normal years + 1 leap year). So, the next year which will have same calendar as 2009 = 2009 + 6 =2015.
Incorrect
Sol. (d)
In one normal year we have 1 odd day and in one leap year we have 2 odd days. So, the repetition of same calendar happens after every 7 odd days. Now, 7 odd days will come in 6 years (5 normal years + 1 leap year). So, the next year which will have same calendar as 2009 = 2009 + 6 =2015.
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Question 5 of 5
5. Question
If second and fourth Saturdays and all the Sundays are taken as only holidays for an office, what would be the minimum number of possible working days of any month of
Correct
Sol. (b)
February is the month in which there are minimum number of working days, i.e., 28 days. There will
be maximum four Sundays and two Saturdays off, i.e., 6 day .So, minimum number of working days = 28 – 6 =
22 days. Hence, (b) is the correct option.
Incorrect
Sol. (b)
February is the month in which there are minimum number of working days, i.e., 28 days. There will
be maximum four Sundays and two Saturdays off, i.e., 6 day .So, minimum number of working days = 28 – 6 =
22 days. Hence, (b) is the correct option.
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