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- Question 1 of 12
##### 1. Question

What was the day of the week on 15th August, 1947?

##### Hint

15th August, 1947 = (1946 years + Period from 1st Jan., 1947 to 15th Aug., 1947)

Counting of odd days :

1600 years have 0 odd day. 300 years have 1 odd day.

47 years = (11 leap years + 36 ordinary years)

= [(11 × 2) + (36 × 1)] odd days = 58 odd days

⇒ 2 odd days.Jan. Feb. March April May June July Aug. 31 28 31 30 31 30 31 15 = 227 days = (32 weeks + 3 days) = 3 odd days.

Total number of odd days = (0 + 1 + 2 + 3) odd days = 6 odd days.

Hence, the required day was ‘Friday’. - Question 2 of 12
##### 2. Question

The first Republic Day of India was celebrated on 26th January, 1950. It was :

##### Hint

26th Jan., 1950 = (1949 years + Period from 1st Jan., 1950 to 26th Jan., 1950)

1600 years have 0 odd day. 300 years have 1 odd day.

49 years = (12 leap years + 37 ordinary years)

= [(12 × 2) + (37 × 1)] odd days = 61 odd days = 5 odd days.

Number of days from 1st Jan. to 26th Jan. = 26 = 5 odd days

Total number of odd days = (0 + 1 + 5 + 5) = 11 = 4 odd days

∴ The required days was ‘Thursday’ - Question 3 of 12
##### 3. Question

What will be the day of the week on 1st January, 2010?

##### Hint

2000 years have 0 odd days.

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009 Odd days 1 1 1 2 1 1 1 2 1 = 11 odd days (11-7) ⇒ 4 odd days.

1st January, 2010 has 1 odd day.

Total number of odd days = (0 + 4 + 1) = 5.

∴ 5th day of week is Friday.

**Remember:**Number of odd days in 400 years = (5 x 4 + 1) 0 odd day.

Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days. - Question 4 of 12
##### 4. Question

Find the day of the week on 16th July, 1776.

##### Hint

16th July, 1776 mean (1775 years + 6 months + 16 days)

Now, 1600 years have 0 odd days.

100 years have 5 odd days

75 years contain 18 leap years and 57 ordinary years and therefore (36 + 57) or 93 or 2 odd days.

∴1775 years given 0 + 5 + 2 = 7 and so 0 odd days.

Also number of days from 1st Jan. 1776 to 16th July, 1776

Jan. Feb. March April May June July

31 + 29 + 31 + 30 + 31 + 30 + 16

= 198 days = 28 weeks + 2 days = 2 odd days

∴Total number of odd days = 0 + 2 = 2.

Hence the day on 16th July, 1776 was ‘Tuesday’. - Question 5 of 12
##### 5. Question

The calendar for the year 2005 is the same as for the year :

##### Hint

Count the number of days from 2005 onwards to get 0 odd day.

Year 2005 2006 2007 2008 2009 2010 Odd days 1 1 1 2 1 1 = 7 or 0 odd day.

∴ Calender for the year 2005 is the same as that for the year 2011. - Question 6 of 12
##### 6. Question

The year next to 1991 having the same calendar as that of 1990 is :

##### Hint

We go on counting the odd days from 1991 onwards till the sum is divisible by 7. The number of such days are 14 upto the year 2001. So, the calender for 1991 will be repeated in the year 2002.

- Question 7 of 12
##### 7. Question

If 09/12/2001 happens to be Sunday, then 09/12/1971 would have been at

##### Hint

09/12/2001—— Sunday

No. of days between 9/ 12/ 71 & 9 / 12/ 2001

we know every year has 1 odd day and leap year has 2 odd days

Here, No. of normal years = 22And no. of leap years = 8

So odd days = 22 + 16 = 38 i.e., 3odd days

(remainder when 38 is divided by 7, i.e. 3)

Hence it was a Thursday - Question 8 of 12
##### 8. Question

The last day of a century cannot be :

##### Hint

100 years contain 5 odd days. So, last day of 1st century is ‘Friday’

200 years contain (5 × 2) = 10 odd days = 3 odd days.

So, last day of 2nd century is ‘Wednesday’.

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

∴ Last day of 3rd century is ‘Monday’.

400 years contain 0 odd day.

∴ Last day of 4th century is ‘Sunday’

Since the order is continually kept in successive cycles, we see that the last day of a century cannot be Tuesday, Thursday or Saturday. - Question 9 of 12
##### 9. Question

How many times does the 29th day of the month occur in 400 consecutive years ?

##### Hint

In 400 consecutive years there are 97 leap years. Hence, in 400 consecutive years February has the 29th day 97 times and the remaining eleven months have the 29th day 400 × 11 or 4400 times.

Therefore, the 29th day of the month occurs (4400 + 97) or 4497 times. - Question 10 of 12
##### 10. Question

On January 12, 1980, it was Saturday. The day of the week on January 12, 1979 was :

##### Hint

The year 1979 being an ordinary year, it has 1 odd day.

So, the day on 12th January 1980 is one day beyond on the day on 12th January, 1979.

But, January 12, 1980 being Saturday.

∴January 12, 1979 was Friday. - Question 11 of 12
##### 11. Question

Ketan takes casual leave only on first working day of every month. The office has weekly offs on Saturday and Sunday. In a month of 30 days, the first working day happened to be Tuesday. What will be the day for his next casual leave?

##### Hint

If the first working day happened to be Tuesday then 8th, 15th, 22nd and 29th of the month will be Tuesday. Hence, the last day of the month will be Wednesday (since, number of days in the month is 30). Thus, the next casual leave will be on Thursday.

- Question 12 of 12
##### 12. Question

Abhay gave an application for a new ration card to the clerk on Monday afternoon. Next day was a holiday. So the clerk cleared the papers on the next working day on resumption of duty. The senior clerk checked it on the same day but forwarded it to the head clerk on next day. The head clerk decided to dispose the case on the subsequent day. On which of the following days was the case put up to the head clerk by the senior clerk?

##### Hint

(i) Submitted application form: Monday (ii) Holiday: Tuesday (iii) Clearance from clerk: Wednesday (iv) Clearance from senior clerk: Wednesday (v) Submitted to the head clerk: Thursday

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