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

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

##### Hint

15th Aug 1947 ⇒ (1946 years + Period from 1st Jan 1947 to 15th Aug 1947)

**Counting of odd days:**

1600 years have 0 odd days.

300 years have 1 odd day.47 years contain 11 leap years and 36 ordinary years.

⇒ (11×2)+(36×1) odd days

⇒ 58 odd days

⇒ 2 odd days.No. of the odd day from 1st Jan 1947 to 15th Aug 1947:

31+28+31+30+31+30+31+16=227 days

⇒ 227 days =(32 weeks+3 days)

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

Hence, the 6th day of the week is**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)

**Counting of odd days:**

1600 years have 0 odd days.

300 years have 1 odd day.49 years contain 12 leap years and 37 ordinary years.

⇒ (12×2) + (37×1) odd days

⇒ 61 odd days

⇒ 5 odd days.Number of odd days from 1st Jan to 26th Jan (26 days)

⇒ 5 odd days

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

Hence, the 4th day of the week is**Thursday.** - Question 3 of 12
##### 3. Question

What will be the day of the week on 1st January 2030?

##### Hint

1st January 2030 ⇒ (2030 years + 1st Jan)

**Counting of odd days:**

2000 years have 0 odd days.

29 years contain 7 leap years and 22 ordinary years.

⇒ (7×2)+(22×1)=36 odd days

⇒ 1 odd day

1st January 2030 has 1 odd day.Total number of odd days =(0+1+1)=2.

Hence, 2nd day of the week is**Tuesday**.**Remember:**Number of odd days in 400 years =(5×4+1)=0

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

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

##### Hint

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

**Counting of odd days:**

Now, 1600 years have 0 odd days.

100 years have 5 odd days.75 years contain 18 leap years and 57 ordinary years.

⇒ (18×2)+(57×1)=93 days

⇒ 2 odd dayNo. of the odd day from 1st Jan 1976 to 16th Jul 1976:

31+29+31+30+31+30+16=198 days

= 28 weeks + 2 days

⇒ 2 odd daysTotal number of odd days

= (0+5+2+2)=9.

⇒ 2 i.e., 2nd day of the week

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 days.

Year–odd day: 2005–

**1**, 2006–**1**, 2007–**1**, 2008–**2**, 2009–**1**, 2010–**1**

= 7 or 0 odd days.Thus, the calendar 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 1990 onwards till the sum is divisible by 7.

The number of such days are 14 up to the year 2000.

Year–odd day: 1990–**1**, 1991–**1**, 1992–**2**, 1993–**1**, 1994–**1**, 1995–**1**, 1996–**2**, 1997–**1**, 1998–**1**, 1999–**1**, 2000–**2**The year next to 1991 having the same calendar as that of 1990 is 2001.

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

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

##### Hint

No. of years from 9/12/71 to 9/12/2001=30

We know every ordinary year has 1 odd day and a leap year has 2 odd days.Here, no. of ordinary years are 22, and no. of leap years are 8.

So total counting of odd days =(22×1)+(8×2)=38

i.e., 3 odd days (3 is remainder when 38 is divided by 7)So, the day on 09/12/1971 is 3 days before the day on 09/12/2001.

But, 09/12/2001 is Sunday.

Thus, 09/12/1971 will be 3 days before Sunday.

Hence, it was**Thursday**. - Question 8 of 12
##### 8. Question

The last day of a century cannot be:

##### Hint

100 years contain 5 odd days.

So, the last day of the 1st century is Friday.

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

So, the last day of the 2nd century is Wednesday.

300 years contain (5×3=15)⇒1 odd day.

So, the last day of the 3rd century is Monday.

400 years contain 0 odd days.

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

##### Hint

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

So, the day on 12th January 1979 is one day before the day on 12th January 1980.But, 12 January 1980 is Saturday.

Thus, 12 January 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 the 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. The 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 the 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

1. Submitted application form Monday 2. Holiday Tuesday 3. Clearance from the clerk Wednesday 4. Clearance from the senior clerk Wednesday 5. Submitted to the head clerk Thursday