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Test: Calendars- 1


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Test: Calendars- 1 - Question 1

The century can end with:

Detailed Solution for Test: Calendars- 1 - Question 1
  • 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.

Test: Calendars- 1 - Question 2

Find the leap year?

Detailed Solution for Test: Calendars- 1 - Question 2

Remember the leap year rule:

  • Every year divisible by 4 is a leap year, if it is not a century.
  • Every 4th century is a leap year, but no other century is a leap year.
  • 800,1200 and 2000 comes in the category of 4th century (such as 400,800,1200,1600,2000 etc).

Hence, 800,1200 and 2000 are leap years.

Test: Calendars- 1 - Question 3

What was the day on February 9, 1979?

Detailed Solution for Test: Calendars- 1 - Question 3
  • We know that in 1600 years, there will be 0 odd days. And in the next 300 years, there will be 1 odd day.
  • From 1901 to 1978 we have 19 leap years and 59 non-leap years.
  • So, the total number of odd days up to 31st Dec. 1978 is 19 x 2 + 59 = 97. On dividing 97 by 7 we get 6 as the remainder, which is the total number of odd days in these years.
  • So, till 31st Dec. 1978 we have 1 + 6 = 7 odd days, which forms one complete week. Now, in 1979, we have 3 odd days in January, and 2 odd days in the month of February (up to 9th Feb).
  • So, the total odd days are 3 + 2 = 5.
  • Hence, 9th February 1979 was a Friday.
Test: Calendars- 1 - Question 4

What is the day on July 2 1985?

Detailed Solution for Test: Calendars- 1 - Question 4
  • Every year has one odd day and a leap year has 2 odd days.
  • Though 1984 is a leap year, we don't have Feb 29 in the required period.
  • So, we get only one odd day and as we are moving back we get Tuesday as the answer.
Test: Calendars- 1 - Question 5

India got independence on 15th August 1947. What was the day on that date?

Detailed Solution for Test: Calendars- 1 - Question 5

We shall first calculate the number of odd days till 31 December 1946:

  • Number of odd days in the first 1600 years = 0 odd day
  • Number of odd days in the next 300 years = 1 odd day
  • Now, 46 years had 11 leap years and 35 ordinary years.
  • The number of odd days in 46 years = (2 × 11) + (1 × 35) = 22 + 35 = 57 = 8 weeks and 1 odd day.


Now, we shall calculate the number of odd days in 1947 till 15 August:
Month (Days):

  • January(31)
  • February(28)
  • March(31)
  • April(30)
  • May(31)
  • June(30)
  • July(31)
  • August(15)

Days = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 15 = 227 i.e. 32 weeks and 3 odd days
So, the total number of odd days till 15 August 1947 = 0 + 1 + 1 + 3 = 5
On counting five days from Monday, we get Friday.
Therefore, 15 August 1947 was a Friday.

Test: Calendars- 1 - Question 6

If 10th May, 1997 was a Monday, what was the day on Oct 10, 2001?

Detailed Solution for Test: Calendars- 1 - Question 6

In this question, the reference point is May 10, 1997 and we need to find the number of odd days from May 10, 1997 up to Oct 10, 2001.

► Now, from May 11, 1997 - May 10, 1998 = 1 odd day
► May 11, 1998 - May 10, 1999 = 1 odd day
► May 11, 1999 - May 10, 2000 = 2 odd days (2000 was leap year)
► May 11, 2000 - May 10, 2001 = 1 odd day
► Thus, the total number of odd days up to May 10, 2001 = 5.
► The remaining 21 days of May will give 0 odd days.
► In June, we have 2 odd days; in July, 3 odd days; in August, 3 odd days; in September,2 odd days and up to 10th October, we have 3 odd days. Hence, total number of odd days = 18 i.e. 4 odd days.
Since, May 10, 1997 was a Monday, and then 4 days after Monday would be Friday. So, Oct 10, 2001 was Friday.

Test: Calendars- 1 - Question 7

Second & fourth Saturdays and every Sunday is a holiday. How many working days will be there in a month of 31 days beginning on a Friday ?

Detailed Solution for Test: Calendars- 1 - Question 7

Given that the month begins on a Friday and has 31 days

  • Sundays = 3rd, 10th, 17th, 24th, 31st
    ⇒ Total Sundays = 5
    Every second & fourth Saturday is holiday.
  • 2nd & 4th Saturday in every month = 2
  • Total days in the month = 31
  • Total working days = 31 - (5 + 2) = 24 days
Test: Calendars- 1 - Question 8

Which calendar year will be same as the year 2008?

Detailed Solution for Test: Calendars- 1 - Question 8

For every 28 years, the calendars will same,
so the years 2008,2036 have the same calendar as 1980.

Test: Calendars- 1 - Question 9

Today is Monday. After 61 days, it will be:

Detailed Solution for Test: Calendars- 1 - Question 9
  • 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.
Test: Calendars- 1 - Question 10

The day of the 5th november is equal to the day of the date in the same year?  

Detailed Solution for Test: Calendars- 1 - Question 10

We will show that the number of odd days between the last day of February and the last day of October is zero.

  • March, April, May, June, July, Aug, Sept, Oct , i.e. 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 = 241 days = 35 weeks = 0 odd day.
  • Number of odd days during this period = 0.

Thus, 5th March of a year will be the same day as 5th November of that year.

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