Calendar & Clock

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1. (b); The year 2007 is an ordinary year. So, it has 1 odd day. 

1st day of the year 2007 was Monday. 

1st day of the year 2008 will be 1 day beyond Monday. 

Hence, it will be Tuesday.

2. (c); The year 2008 is a leap year. So, it has 2 odd days. 

1st day of the year 2008 is Tuesday (Given) 

So, 1st day of the year 2009 is 2 days beyond Tuesday. 

Hence, it will be Thursday.

3. (d); The year 2006 is an ordinary year. So, it has 1 odd day.

So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.

But, 8th Dec, 2007 is Saturday. 8th Dec, 2006 is Friday.

4. (a); The year 2004 is a leap year. So, it has 2 odd days.

But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.

The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.

Given that, 6th March, 2005 is Monday. 

6th March, 2004 is Sunday (1 day before to 6th March, 2005).

5. (d); Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.

Year: 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 2017

Odd day: 1 2 1 1 1 2 1 1 1 2 1

Sum = 14 odd days 0 odd days.

Calendar for the year 2018 will be the same as for the year 2007.

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6. (d); We shall find the day on 1st April, 2001.

1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)

Odd days in 1600 years = 0  

Odd days in 400 years = 0

Jan. Feb. March April  

(31 + 28 + 31 + 1) = 91 days 0 odd days.  

Total number of odd days = (0 + 0 + 0) = 0

On 1st April, 2001 it was Sunday.  

In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

7. (c); 17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)

Odd days in 1600 years = 0  

Odd days in 300 years = (5 x 3) = 1 odd day

97 years has 24 leap years + 73 ordinary years.  

Number of odd days in 97 years (24 x 2 + 73) = 121 = 2 odd days.

Jan. Feb. March April May June  

(31 + 28 + 31 + 30 + 31 + 17) = 168 days  

168 days = 24 weeks = 0 odd day.

Total number of odd days = (0 + 1 + 2 + 0) = 3.  

Given day is Wednesday.

8. (d); 28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)

Odd days in 1600 years = 0  

Odd days in 400 years = 0

5 years = (4 ordinary years + 1 leap year)  

(4 x 1 + 1 x 2) = 6 odd days

Jan. Feb. March April May  

(31 + 28 + 31 + 30 + 28) = 148 days  

148 days = (21 weeks + 1 day) 1 odd day.  

Total number of odd days  

= (0 + 0 + 6 + 1) = 7 odd days.

Given day is Sunday.

9. (a); 15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)

Odd days in 1600 years = 0  

Odd days in 400 years = 0

9 years = (2 leap years + 7 ordinary years)  

(2 x 2 + 7 x 1) = 4 odd days.

Jan. Feb. March April May June July Aug.  

(31 + 28 + 31 + 30 + 31 + 30 + 31 + 15) = 227 days  

227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 0 + 4 + 3) = 7 odd days.  

Given day is Sunday.

10. (b); Each day of the week is repeated after 7 days.

So, after 63 days, it will be Monday.  

After 61 days, it will be Saturday.

11. (c); 100 years contain 5 odd days.

Last day of 1st century is Friday.  

200 years contain (5 x 2) = 3 odd days.  

Last day of 2nd century is Wednesday.  

300 years contain (5 x 3) = 15 = 1 odd day.  

Last day of 3rd century is Monday.  

400 years contain 0 odd day.  

Last day of 4th century is Sunday.  

This cycle is repeated.

Last day of a century cannot be Tuesday or Thursday or Saturday.

12. (a); The century divisible by 400 is a leap year.

The year 700 is not a leap year.

13. (b); x weeks x days = (7x + x) days = 8x days.

14. (c); On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.  

On 31st December, 2009 it was Thursday.  

Thus, on 1st Jan, 2010 it is Friday.

15. (c); The year 2004 is a leap year. It has 2 odd days.

The day on 8th Feb, 2004 is 2 days before Sunday.  

Hence, this day is Sunday.

16. (b); Angle traced by the hour hand in 5 hours = (360/12 x 5) = 150°

17. (c); In 12 hours, they are at right angles 22 times.

In 24 hours, they are at right angles 44 times.

18. (a); The hands of a clock coincide 11 times in every 12 hours (Since between 11 and 1, they coincide only once, i.e. at 12 o’clock).

AM: 12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55  

PM: 12:00, 1:05, 2:11, 3:16, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:55  

The hands coincide 22 times in a day.

19. (c); 15 x 1/2 = 7.5°

20. (a); 30° + 15/2 = 37.5°

21. (c); Angle traced by hour hand in 12 hrs = 360°.  

Angle traced by hour hand in 5 hrs 10 min = (360/12 x 31/6) = 155°

22. (c); 22 times the hands of the clock will be opposite directions in a day.

Here is the extracted content from questions 23 to 30

23. (b); Angle traced by hour hand in 17/2 hrs

image

Required angle = (255 – 180)° = 75°.

24. (a); The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours.
(Because between 5 and 7 they point in opposite directions at 6 o’clock only.)

So, in a day, the hands point in the opposite directions 22 times.

25. (b); Angle traced by hour hand in (9 1/2) hrs

image 1

Required angle = (180 – 135)° = 45°.

26. (c); All century years divisible by 400 will be leap years.
Year 1600 is divisible by 400. So, 1600 is a leap year.

27. (a); Thursday + 9 days = Saturday

28. (b); In ordinary year 1 Jan. and 31 Dec. will be the same day.
If Jan first is Saturday, then 31 Dec. will be Saturday also.

29. (c); In leap year 1 Jan. is Sunday, then the last day or 31 Dec. will be the next day from 1 Jan.
If here Jan 1 is Sunday, then 31 Dec. will be Monday. And on 1 Jan. 2013 will be next day means Tuesday

30. (c);

 


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Dhalendra Kothale

Dhalendra Kothale

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