# CRE 1 - Odd Days | LR - Calendar

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**CRE 1 - Odd Days | LR - Calendar**

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

- (a)
1

- (b)
2

- (c)
3

- (d)
4

Answer: Option B

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**Explanation** :

Total number of days in a leap year = 366.

∴ Odd days = Remainder (366 ÷ 7) = 2.

Hence, option (b).

Workspace:

**CRE 1 - Odd Days | LR - Calendar**

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

- (a)
Monday

- (b)
Saturday

- (c)
Sunday

- (d)
Wednesday

Answer: Option C

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**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).

Workspace:

**CRE 1 - Odd Days | LR - Calendar**

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

- (a)
Friday

- (b)
Wednesday

- (c)
Thursday

- (d)
Tuesday

Answer: Option A

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**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:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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).

Workspace:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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).

Workspace:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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).

Workspace:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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:

**CRE 1 - Odd Days | LR - Calendar**

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

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**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:

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