Worksheet Solutions: Calendar | Mental Mathematics for Class 8 PDF Download

Q1: It was Sunday on Jan 1, 2006. What was the day of the week Jan 1, 2010?
(a) Sunday
(b) Saturday
(c) Friday
(d) Wednesday
Ans: (c)
On 31st December, 2005 it was Saturday.
Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.
∴ On 31st December 2009, it was Thursday.
Thus, on 1st Jan, 2010 it is Friday.

Q2: What was the day of the week on 17th June, 1998?
(a) Monday
(b) Tuesday
(c) Wednesday
(d) Thursday
Ans: (c)
17th June, 1998 = (1997 years + Period from 1.1.1998 to 17.6.1998)
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) ≡ 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.

Therefore 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.

Q3: Today is Monday. After 61 days, it will be:
(a) Wednesday
(b) Saturday
(c) Tuesday
(d) Thursday
Ans: (b)
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.

Q4: On what dates of April, 2001 did Wednesday fall?
(a) 1st, 8th, 15th, 22nd, 29th
(b) 2nd, 9th, 16th, 23rd, 30th
(c) 3rd, 10th, 17th, 24th
(d) 4th, 11th, 18th, 25th
Ans: (d)
We shall find the day on 1st April, 2001.
1st April, 2001 = (2000 years + Period from 1.1.2001 to 1.4.2001)
Odd days in 1600 years = 0
Odd days in 400 years = 0
Jan. Feb. March April
(31 + 28 + 31 + 1)     = 91 days ≡ 0 odd days.
Total number of odd days = (0 + 0 + 0) = 0
On 1st April, 2001 it was Sunday.
In April, 2001 Wednesday falls on 4th, 11th, 18th and 25th.

Q5: The last day of a century cannot be
(a) Monday
(b) Wednesday
(c) Tuesday
(d)  Friday
Ans: (c)
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.

Q6: What was the day of the week on 28th May, 2006?
(a) Thursday
(b) Friday
(c) Saturday
(d) Sunday
Ans: (d)
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) ≡ 6 odd days

∴ 148 days = (21 weeks + 1 day) ≡ 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 ≡ 0 odd day.
Given day is Sunday.

Q7: What will be the day of the week 15th August, 2010?
(a) Sunday
(b) Monday
(c) Tuesday
(d) Friday
Ans: (a)
15th August, 2010 = (2009 years + Period 1.1.2010 to 15.8.2010)
Odd days in 1600 years = 0
Odd days in 400 years = 0
9 years = (2 leap years + 7 ordinary years) = (2 x 2 + 7 x 1) = 11 odd days ≡ 4 odd days.

∴ 227 days = (32 weeks + 3 days) ≡ 3 odd days.
Total number of odd days = (0 + 0 + 4 + 3) = 7 ≡ 0 odd days.
Given day is Sunday.

Q8: If 6th March, 2005 is Monday, what was the day of the week on 6th March, 2004?
(a) Sunday
(b) Saturday
(c) Tuesday
(d) Wednesday
Ans: (a)
The year 2004 is a leap year. So, it has 2 odd days.
But, Feb 2004 not included because we are calculating from March 2004 to March 2005. So it has 1 odd day only.
∴ The day on 6th March, 2005 will be 1 day beyond the day on 6th March, 2004.
Given that, 6th March, 2005 is Monday.
∴ 6th March, 2004 is Sunday (1 day before to 6th March, 2005).

Q9: How many days are there in x weeks x days?
(a) 7x²
(b) 8x
(c) 14x
(d) 7
Ans: (b)
x weeks x days = (7x + x) days = 8x days.

Q10: On 8th Feb, 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004?
(a) Tuesday
(b) Monday
(c) Sunday
(d) Wednesday
Ans: (c)
The year 2004 is a leap year. It has 2 odd days.
∴ The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.