# PE 2 - Calendars | Clocks & Calendars

**PE 2 - Calendars | Clocks & Calendars**

What is the number of odd days in a leap year?

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

Total number of days in a leap year = 366.

∴ Odd days = Remainder (366 ÷ 7) = 2.

Hence, option (b).

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**PE 2 - Calendars | Clocks & Calendars**

Today is Thursday. The day after 59 days will be?

- A.
Monday

- B.
Saturday

- C.
Sunday

- D.
Wednesday

Answer: Option C

**Explanation** :

To find out the day of the week, we need to calculate the odd days i.e., Remainder when 59 is divided by 7.

∴ Odd days = Remainder (59 ÷ 7) = 3.

∴ If it is Thursday today, day of the week after 59 days from today will be same as the day of the week 3 days after Thursday i.e., Sunday.

Hence, option (c).

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**PE 2 - Calendars | Clocks & Calendars**

Today is Friday. The day after 63 days will be?

- A.
Friday

- B.
Wednesday

- C.
Thursday

- D.
Tuesday

Answer: Option A

**Explanation** :

To find out the day of the week, we need to calculate the odd days i.e., Remainder when 63 is divided by 7.

∴ Odd days = Remainder (63 ÷ 7) = 0.

Since, the odd days is 0, the day of the week will be same as it is today i.e., Friday.

Hence, option (a).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

If 3^{rd} Mar of a certain year is a Wednesday, then 29^{th} Mar of the same year will be?

- A.
Tuesday

- B.
Wednesday

- C.
Thursday

- D.
Monday

Answer: Option D

**Explanation** :

From 3^{rd} March till 29^{th} March there are 29 – 3 = 26 days.

To find out the day of the week, we need to calculate the odd days i.e., Remainder when 26 is divided by 7.

∴ Odd days = Remainder (26 ÷ 7) = 5.

∴ Day of the week on 29^{th} march will be 5th day after Wednesday i.e., Monday.

Hence, option (d).

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**PE 2 - Calendars | Clocks & Calendars**

If 7^{th} Aug of a certain year is a Friday, then 30^{th} Dec of the same year will be?

- A.
Saturday

- B.
Wednesday

- C.
Tuesday

- D.
Cannot be determined

Answer: Option B

**Explanation** :

Number of days from 7th Aug till 30^{th} December is:

Aug = 24,

Sep = 30

Oct = 31

Nov = 30

Dec = 30

i.e., total = 24 + 30 + 31 + 30 + 30 = 145 days.

∴ Odd days = Remainder (145 ÷ 7) = 5.

∴ If it is Friday on 7^{th} Aug, day of the week on 30^{th} Dec will be same as the day of the week 5 days after Friday i.e., Wednesday.

Hence, option (b).

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**PE 2 - Calendars | Clocks & Calendars**

If 22^{nd} Jan of a certain year is a Tuesday, then 25^{th} Jul of the same year will be?

- A.
Friday

- B.
Tuesday

- C.
Thursday

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Number of days from 22^{nd} Jan till 25^{th} July is:

Jan = 24,

Feb = 28/29?

Since we do not know if the given year is leap year or not, we cannot calculate the odd days between the dates given.

Hence, the answer cannot be determined.

Hence, option (d).

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**PE 2 - Calendars | Clocks & Calendars**

If 7^{th} Jun 1963 was a Thursday, then which day of the week was 15^{th} Sep 1968?

- A.
Sunday

- B.
Friday

- C.
Saturday

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Let us first calculate the number of odd days after 7^{th} Jun, 1963 till 7^{th} Jun, 1968:

Odd days from

7^{th} Jun, 1963 - 7^{th} Jun, 1964 = 2 (because of 29^{th} Feb, 1964)

7^{th} Jun, 1964 - 7^{th} Jun, 1965 = 1

7^{th} Jun, 1965 - 7^{th} Jun, 1966 = 1

7^{th} Jun, 1966 - 7^{th} Jun, 1967 = 1

7^{th} Jun, 1967 - 7^{th} Jun, 1968 = 2 (because of 29^{th} Feb, 1968)

Total odd days = 2 + 1 + 1 + 1 + 2 = 7 odd days = 0 odd days.

Now, let us calculate the odd days after 7^{th}June, 1968 till 15^{th} Sep, 1968.

Jun = 23

Jul = 31

Aug = 31

Sep = 15

∴ Total days = 23 + 31 + 31 + 15 = 100 day.

∴ Odd days = Remainder (100 ÷ 7) = 2.

⇒ Total odd days from 7^{th }Jun, 1963 till 15^{th} Sep, 1968 = 0 + 2 = 2.

∴ If it is Thursday on 7^{th} June, 1963, day of the week on 15^{th} Sep, 1968 will be same as the day of the week 2 days after Thursday i.e., Saturday.

Hence, option (c).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

If 15^{th} Aug 1947 was a Friday, then which day of the week was 26^{th} Jan 1950?

- A.
Thursday

- B.
Friday

- C.
Saturday

- D.
None of these

Answer: Option A

**Explanation** :

Let us first calculate the number of odd days after 15^{th} Aug, 1947 till 15^{th} Aug, 1949:

Odd days from

15^{th} Aug, 1947 - 15^{th} Aug, 1948 = 2 (because of 29^{th} Feb, 1948)

15^{th} Aug, 1948 - 15^{th} Aug, 1949 = 1

Total odd days = 2 + 1 = 3 odd days.

Now, let us calculate the odd days after 15^{th} Aug, 1949 till 26^{th} Jan, 1950.

Aug = 16

Sep = 30

Oct = 31

Nov = 30

Dec = 31

Jan = 26

∴ Total days = 16 + 30 + 31 + 30 + 31 + 26 = 164 day.

∴ Odd days = Remainder (164 ÷ 7) = 3.

⇒ Total odd days from 15^{th} Aug, 1947 till 26^{th} Jan, 1950 = 3 + 3 = 6.

∴ If it is Friday on 15^{th} Aug, 1947, day of the week on 26^{th} Jan, 1950 will be same as the day of the week 6 days after Friday i.e., Thursday.

Hence, option (a).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

If a year starts on a Saturday, then what is the maximum possible number of Saturdays in that year?

- A.
53

- B.
52

- C.
51

- D.
49

Answer: Option A

**Explanation** :

**Case 1:** Non-Leap year

In a leap year there are 52 complete weeks and 1 odd day.

Since the year starts with Saturday, 365th day will also be Saturday.

Hence, there will be 53 Saturdays.

**Case 2:** Leap year

In a leap year there are 52 complete weeks and 2 odd days.

Since the year starts with Saturday, 365th day will also be Saturday.

Hence, there will be 53 Saturdays.

In any case, the number of Saturday will be 53.

Hence, option (a).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

If a year starts on a Saturday, then what is the maximum possible number of Sundays in that year?

- A.
53

- B.
52

- C.
51

- D.
49

- E.
Cannot be determined

Answer: Option D

**Explanation** :

**Case 1:** Non-Leap year

In a leap year there are 52 complete weeks and 1 odd day.

Since the year starts with Saturday, 365^{th} day will also be Saturday and there won’t be a 53^{rd} Sunday.

Hence, there will be 52 Sundays.

**Case 2:** Leap year

In a leap year there are 52 complete weeks and 2 odd days.

Since the year starts with Saturday, 365^{th} day will also be Saturday and 366^{th} day will be a Sunday.

Hence, there will be 53 Sundays.

Since we do not know whether the year is a leap or non-leap year, we cannot determine the number of Sundays.

Hence, option (d).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

Which among the following years is a leap year?

- A.
2600

- B.
2700

- C.
2800

- D.
3000

Answer: Option C

**Explanation** :

For a year to be leap year there are 2 conditions

- If it is a non-century year, it should be divisible by 4.

- If it is a century year, it should be divisible by 400.

Here all the given years are century years, and only 2800 is divisible by 400.

Hence, 2800 is a leap year.

Hence, option (c).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

What was the day on 1^{st} Jan 1901?

- A.
Wednesday

- B.
Thursday

- C.
Tuesday

- D.
Saturday

Answer: Option C

**Explanation** :

Here we need to break days till 1^{st} Jan 1901.

**Complete years = 1900**

Odd days in first 1600 years = 0

Odd days in next 300 years = 1

∴ Odd days in complete years = 0 + 1 = 1.

**Incomplete year (1901)**

Just 1 day i.e., 1^{st} of January 1901.

∴ Odd days in incomplete year = 1

∴ Total odd days till 1^{st} Jan 1901 = 1 + 1 = 2.

∴ Day of the week on 1st Jan 1901 will be 2 days after Sunday i.e., Tuesday.

Hence, option (c).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

What was the day of the week on 2^{nd} July 1184?

- A.
Thursday

- B.
Wednesday

- C.
Saturday

- D.
Monday

Answer: Option D

**Explanation** :

Here we need to break days till 2^{nd} July 1184.

**Complete years = 1183**

Odd days in first 800 years = 0

Odd days in next 300 years = 1

Odd days in next 83 years (20 leap and 63 non-leap years) = 20 × 2 + 63 × 1 = 103.

∴ Odd days in complete years = 0 + 1 + 103 = 104 odd days.

∴ Odd days in complete years = Remainder (104 ÷ 7) = 6 odd days.

**Incomplete year (1184, a leap year)**

Jan + Feb + Mar = 0 odd days,

Apr + May + Jun = 0 = odd days,

Jul = 2

∴ Odd days in incomplete year = 0 + 0 + 2 = 2 odd days.

∴ Total odd days till 2^{nd} July 1184 = 6 + 2 = 8 = 1 odd day.

∴ Day of the week on 2^{nd} July 1184 will be 1 day after Sunday i.e., Monday.

Hence, option (d).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

What was the day on 31^{st} Oct 1984?

- A.
Sunday

- B.
Wednesday

- C.
Monday

- D.
Thursday

Answer: Option D

**Explanation** :

Here we need to break days till 31^{st} Oct 1984.

**Complete years = 1983**

Odd days in first 1600 years = 0

Odd days in next 300 years = 1

Odd days in next 83 years (20 leap and 63 non-leap years) = 20 × 2 + 63 × 1 = 103.

∴ Odd days in complete years = 0 + 1 + 103 = 104 odd days.

∴ Odd days in complete years = Remainder (104 ÷ 7) = 6 odd days.

**Incomplete year (1984, a leap year)**

Jan + Feb + Mar = 0 odd days,

Apr + May + Jun = 0 = odd days,

Jul + Aug + Sep = 1 odd day

Oct = 31

∴ Odd days in incomplete year = 0 + 0 + 1 + 31 = 32 odd days = 4 odd days.

∴ Total odd days till 31^{st} Oct 1984 = 6 + 4 = 10 odd days = 3 odd days.

∴ Day of the week on 31^{st} Oct 1984 will be 3 days after Sunday i.e., Wednesday.

Hence, option (d).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

What was the day on 14^{th} Mar 2393?

- A.
Tuesday

- B.
Sunday

- C.
Friday

- D.
Saturday

Answer: Option B

**Explanation** :

Here we need to break days till 14^{th} Mar 2393.

**Complete years = 2392**

Odd days in first 2000 years = 0

Odd days in next 300 years = 1

Odd days in next 92 years (23 leap and 69 non-leap years) = 23 × 2 + 69 × 1 = 115.

∴ Odd days in complete years = 0 + 1 + 115 = 116 odd days.

∴ Odd days in complete years = Remainder (116 ÷ 7) = 4 odd days.

**Incomplete year (2393, a non-leap year)**

Jan = 31,

Feb = 28

Mar = 14

∴ Odd days in incomplete year = 31 + 28 + 14 = 73 odd days = 3 odd days.

∴ Total odd days till 14^{th} Mar 2393 = 3 + 4 = 7 odd days = 0 odd days.

∴ Day of the week on 14^{th} Mar 2393 will be 0 days after Sunday i.e., Sunday.

Hence, option (b).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

On what dates of Aug 1980 did Monday fall?

- A.
4

^{th}, 11^{th}, 18^{th}, and 25^{th} - B.
3

^{rd}, 10^{th}, 17^{th}and 24^{th} - C.
6

^{th}, 13^{th}, 20^{th}and 27^{th} - D.
9

^{th}, 16^{th}, 23^{rd}and 30^{th}

Answer: Option A

**Explanation** :

Let us first calculate the day of the week on 1^{st} August 1980.

Here we need to break days till 1^{st} August 1980.

**Complete years = 1979**

Odd days in first 1600 years = 0

Odd days in next 300 years = 1

Odd days in next 79 years (19 leap and 60 non-leap years) = 19 × 2 + 60 × 1 = 98.

∴ Odd days in complete years = 0 + 1 + 98 = 99 odd days.

∴ Odd days in complete years = Remainder (99 ÷ 7) = 1 odd days.

**Incomplete year (1980, a leap year)**

Jan + Feb + Mar = 0 odd days,

Apr + May + Jun = 0 odd days,

Jul = 31

Aug = 1

∴ Odd days in incomplete year = 0 + 0 + 31 + 1 = 32 odd days = 4 odd days.

∴ Total odd days till 1^{st} August 1980 = 1 + 4 = 5 odd days.

∴ Day of the week on 1^{st} August 1980 will be 5 days after Sunday i.e., Friday.

Since 1^{st} August 1980 is Friday, next Monday will be on 4^{th} August.

Hence Mondays in August 1980 will be on 4^{th}, 11^{th}, 18^{th} and 25^{th}.

Hence, option (a).

Workspace:

**PE 2 - Calendars | Clocks & Calendars**

The calendar for year 2007 is same as calendar of which of these years?

- A.
2011

- B.
2013

- C.
2018

- D.
2020

Answer: Option D

**Explanation** :

For the calendar to be same the year should be a non-leap year and the number of odd days should be 0.

Here, we will calculate the total odd days from 1st Jan of 2007 till 1^{st} Jan of the year.

2007 – 2008 = 1 odd day

2008 – 2009 = 2 odd days (Total 3 odd days)

2009 – 2010 = 1 odd day (Total 4 odd days)

2010 – 2011 = 1 odd day (Total 5 odd days)

2011 – 2012 = 1 odd day (Total 6 odd days)

2012 – 2013 = 2 odd day (Total 8 odd days)

2013 – 2014 = 1 odd day (Total 9 odd days)

2014 – 2015 = 1 odd day (Total 10 odd days)

2015 – 2016 = 1 odd day (Total 11 odd days)

2016 – 2017 = 2 odd day (Total 13 odd days)

2017 – 2018 = 1 odd day (Total 14 odd days = 0 odd days)

Now, 2018 is also a non-leap year like 2007 and odd days from 2007 till 2018 are 0.

∴ Calendar of 2018 will be same as calendar of 2007.

Hence, option (d).

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