GATE Civil 2020 Set 2 | GA | Question: 7

Question

For the year $2019$, which of the previous year’s calendar can be used?

• A. $2011$
• B. $2012$
• C. $2013$
• D. $2014$

Answer 1

A calendar repeats when the number of days between $2$ years is a multiple of $7.$ Since, $365\mod 7=1$ a non leap year adds $1$ and a leap year adds $2$ to the extra days after an exact multiple of $7.$ So, $2019-7+1=2013$ $(+1$ for leap year $2016)$ must be the answer here.

Correct Answer: C

---

More general approach. 

Calendars repeat after certain cycles:

• If it is a leap year, after $28$ years.
• If it is leap year $+ 1,$ after $6$ years, then $11,$ and again $11.$
• If it is a leap year $+ 2,$ after $11$ years, then $6,$ and again $11.$
• If it is a leap year $+ 3,$ after $11$ years, then $11,$ and finally $6.$

These cycles are valid unless the years span a non-leap century year $(e.g. 2100,$ which is a century year but not a leap year$).$

Since $2019$ is a leap year $+ 3,$ it will repeat in $2030, 2041, 2047, 2058, 2069, 2075,$ etc.

Checking all the options, for $2013$ it will repeat after $6$ years.

So, the correct answer is $(C).$

Comments

@Lakshman Patel RJIT From where did you learn this topic . Can u share any reference!!!

---

I watch some videos on youtube and read the concepts from the Internet.

---

Okay Thanks !!

---

@Lakshman Patel RJIT , @DebRC 

A calendar repeats when the number of days between $2$ years is a multiple of $7.$ Since, $365\mod 7=1$ a non leap year adds $1$ and a leap year adds $2$ to the extra days after an exact multiple of $7.$ So, $2019-7+1=2013$ $(+1$ for leap year $2016)$ must be the answer here.

is this also true if asked for leap years? 

like for 2016,

according to the general approach you mentioned it will repeat after 28 years.?

Answer 2

Let us go backward from 2019.

2019—1 odd days

2018-1 odd days

2017 -1 odd days

2016- 2 odd days (leap year)

2015 -1 odd days

2014- 1 odd day

2013- 1 odd days    

Now from 2013 onwards year 2019 will have the same calendar as 2013. As no of odd days from 2013 to 2019=Rem(1+1+2+1+1+1)/7=0

So our answer is 2013. You need to check the concept of no of odd days for these types of questions

Answer 3

DO REVERSE ENGINEERING AND FIND THE ANSWE

EX. 2019-0

       18-1, 17-2, 16-4, 15-5, 14-6, 13-7

2013 IS THE ANSWER.

OPTION C.)