Calendars: MAH CET, NMAT and SNAP practice questions

Solution

A calendar repeats when the year starts on the same day and is of the same type (leap or ordinary).

Odd days from 2024 to 2051: 35, a multiple of 7, and 2052 is a leap year like 2024.

So the answer is 2052.

Solution

Between the two dates there are 730 days, which leaves 2 odd days (each ordinary year gives 1, and each 29 February crossed adds 1 more).

Saturday + 2 = Monday.

Solution

Use a known date: 1 January 2000 was a Saturday.

From 6 May 1957 to 1 January 2000 there are 15580 days; 15580 ÷ 7 leaves 5, so count 5 days back from Saturday.

So the day is Monday.

Solution

A year is a leap year if it is divisible by 4, but a century year must be divisible by 400.

2000 passes this test, while 2026, 2100, 2200 fail it.

Solution

A year is a leap year if it is divisible by 4, but a century year must be divisible by 400.

1800 fails this test, while 2032, 2024, 2028 pass it.

Solution

Use a known date: 1 January 2000 was a Saturday.

From 12 November 1969 to 1 January 2000 there are 11007 days; 11007 ÷ 7 leaves 3, so count 3 days back from Saturday.

So the day is Wednesday.

Solution

Today is 3 days after Friday: Monday.

The day after tomorrow is 2 days after Monday: Wednesday.

Solution

31 days = 4 weeks + 3 days. Each weekday appears 4 times, and the first 3 weekdays counting from Monday appear once more: Monday, Tuesday, Wednesday.

So there are 5 Mondays.

Solution

Days from 1 January to 25 February = 55 (leap year).

55 ÷ 7 leaves 6, so the day is Sunday + 6 = Saturday.

Solution

31 days = 4 weeks + 3 days. Each weekday appears 4 times, and the first 3 weekdays counting from Sunday appear once more: Sunday, Monday, Tuesday.

So there are 5 Mondays.

Solution

571 ÷ 7 leaves remainder 4.

Monday + 4 days = Friday.

Solution

322 ÷ 7 leaves remainder 0.

Friday + 0 days = Friday.

Solution

A calendar repeats when the year starts on the same day and is of the same type (leap or ordinary).

Odd days from 2005 to 2010: 7, a multiple of 7, and 2011 is an ordinary year like 2005.

So the answer is 2011.

Solution

25 − 5 = 20 days, which leaves 6 after removing whole weeks.

Wednesday + 6 = Tuesday.

Solution

Between the two dates there are 1096 days, which leaves 4 odd days (each ordinary year gives 1, and each 29 February crossed adds 1 more).

Tuesday + 4 = Saturday.

Solution

100 years have 76 ordinary + 24 leap years = 76 + 48 = 124 odd days → 5 odd days. So 100 → 5, 200 → 3, 300 → 1, 400 → 0 (the 400th year is a leap year).

So 200 years have 3 odd days.

Solution

Today is 1 day after Wednesday: Thursday.

The day after tomorrow is 2 days after Thursday: Saturday.

Solution

27 − 8 = 19 days, which leaves 5 after removing whole weeks.

Thursday + 5 = Tuesday.

Solution

Days from 1 January to 10 November = 313.

313 ÷ 7 leaves 5, so the day is Monday + 5 = Saturday.

Solution

100 years have 76 ordinary + 24 leap years = 76 + 48 = 124 odd days → 5 odd days. So 100 → 5, 200 → 3, 300 → 1, 400 → 0 (the 400th year is a leap year).

So 300 years have 1 odd day.