Question

What was the day of the week on 28th May, 2006?

A) Thursday B) Friday C) Saturday D) Sunday

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

→ 28May :- 31jan + 28feb + 31march + 30April + 28may

= 31 + 28 + 31 + 30 + 28

= 3 + 0 + 3 + 2 + 0

= 8

= 1 odd day.

Similarly,

→ 2006 = 2000 + 5 + 1 Leap Year.

= 0 + 6

= 6 odd day.

→ Total Odd day = 1 + 6 = 7 = 0

That Means , on 28th May, 2006 it was Sunday. (D)

____________________

Odd Day :- Total Number of days divide by 7 , the remainder we get, is called odd day.

• 0 Remainder = sunday.
• 1 Remainder = Monday.
• 2 Remainder = Tuesday.
• 3 Remainder = Wednesday.
• 4 Remainder = Thursday .
• 5 Remainder = Friday.
• 6 Remainder = Saturday.

___________________________

Answer 2

${ \huge{ \bold\red{ { \underline{ \: question}}}}}$

• What was the day of the week on 28th May, 2006??

a) Thursday b) Friday

c) Saturday d) Sunday

${ \bold{ \red{ \huge{ \underline{ solution}}}}}$

for solving these type of questions , we use the concept of odd days .....

${ \pink{ \underline{ \underline{counting \: of \: odd \: days}}}}$

28th May , 2006 =

2005 + period from 1.1.2006 to 28.5.2006

• May 28 = 31 Jan +. 28 Feb + 31 Mar + 30 Apr + 28 may

= 148 days

=. ( 21 weeks + 1 day). = 1 odd day

• 1 ordinary year = 365 days=(52 weeks+1 day)

1 ordinary year has 1 odd day.

• 1 leap year = 366 days =(52 weeks +2 day)

1 leap year has 2 odd days

${ { \red{5years = 4ordinary \: year + 1 \: leap \: year}}}$

= ( 4 × 1 odd day) + ( 1 × 2 odd day)

=. 4 + 2 = 6 odd days

${ \blue{ \bold{ \underline{total \: no. \: of \: odd \: days = 6 + 1 = 7}}}}$

dividing the total number of odd days by 7.....

remainder is 0 then , the day. is Sunday

therefore,. option (d) is correct....

${ \bold{ \red{ \underline{ \underline{sunday}}}}}$