## Infosys 3 Sample Puzzles On Dates

Dear Reader, you can expect few puzzles dealing with dates and days of years during Infosys placement tests. Below are three useful questions with answers for your practice.

Question 1

If Jeya celebrated her 18th birthday on 28 February 2009, Saturday. Then what day will come on her 25th birthday?

a)Monday b) Sunday c) Friday d) Saturday

Solution:

Jeya Celebrated her 18th birthday on 28 February 2009, Saturday.
Then her 19th birthday will be on 28 February 2010, Sunday. (This is because 2010 is not a leap year and hence contains 365 days. 365 divided by 7 gives remainder 1 which means any date on 2009 corresponds to the same date on 2010 advanced by one day.)
Her 20th birthday will be on 28 February 2011, Monday.
Her 21st birthday will be on 28 February 2012, Tuesday.

Her 22nd birthday will be on 28 February 2013, Thursday. (Since, 2012 is a leap year hence contains 366 days. 366 divided by 7 gives remainder 2 which means any date on 2013 corresponds to the same date on 2012 advanced by two days.)
Her 23rd birthday will be on 28 February 2014, Friday.
Her 24th birthday will be on 28 February 2015, Saturday.
Her 25th birthday will be on 28 February 2016, Sunday.
So, her 25th birthday will be on Sunday.

Question 2

Ravi went to the hospital on Monday. Doctors made some check up and they asked to come three days before the two days after the next day of day after tomorrow. When will Ravi go to Hospital?

a) Friday b) Saturday c) Tuesday d) Wednesday

Solution:

Ravi went to hospital on Monday.
Then, three days before Monday is Friday and the two days after Friday is Sunday.
And the next day of Sunday is Monday and the day after tomorrow of Monday is Wednesday.
So, Ravi will go to hospital on Wednesday.

Question 3

The Calendar for the year 2013 is the same as for the year?

a) 2020 b) 2018 c) 2017 d) 2019

Solution:

For calendar to repeat exactly, the dates and days have to match perfectly.
Consider 2014 :
Any date on 2014 will correspond to same date on 2013 advanced by one day. (same logic used in first question.) For example if Jan 1 is Tuesday on 2013, then Jan 1 will be Wednesday on 2014.

```Year		2014	2015	2016 (leap year)    2017   2018    2019
Advanced  days	 1	 1	 2  	              1	     1       1
```

In 2019 the total number of advancements will be 1 + 1 + 1 + 2 + 1 + 1 = 7 . Any week has seven days. Hence advancement of 7 days also means the days are going to be the same for any dates. That is if 1st Jan on 2013 is Tuesday, then 1st Jan on 2019 will also be Tuesday.
Hence the calendar for the year 2013 and 2019 is the same.

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