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Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 01:27
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eGMAT Question of the Week #17Bill celebrated his 12th birthday on April 1, 1898, which was a Friday. On which day of the week did Bill celebrate his 18th birthday? A. Tuesday B. Wednesday C. Thursday D. Friday E. Saturday
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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 03:36
Bill celebrated his 12th birthday on April 1, 1898, which was a Friday. On which day of the week did Bill celebrate his 18th birthday? Few facts 1) each year has 365 days,so when divided by 7 remainder is 1.. 2) every 4th year is leap so one day extra, and when divided by 7, remainder is 2.. 3) the century year is leap once in 4 years, so 1600 will be leap but 1700,1800 and 1900 will not be. ( here this concept is tested but i doubt you will be checked for your knowledge of leap year specially this point of century year18th birthday will be on 1898+1812=1904 So in 1898 Friday 1899  Saturday 1900( not a leap year)  sunday 1901 monday 1902  tuesday 1903  Wednesday 1904( leap year) as april 1 falls after 29th feb, so Friday Also you have to add 6 days for 6 years from 12 to 18th birthday. Add 1 to 6 for the leap year 1904., So total days above a multiple of 7 = 6+1=7 So day on 01: apr:1898 and 01 : apr : 1904 will be same that is Friday D A. Tuesday B. Wednesday C. Thursday D. Friday E. Saturday
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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 04:59
chetan2u leap year comes after every 4 yrs.. and a century leap year is the one which os divisible by 400. By this logic.. 1896 is leap but 1900 is not leap right? Posted from my mobile device



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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 05:06
saurabh9gupta wrote: chetan2u leap year comes after every 4 yrs.. and a century leap year is the one which os divisible by 400. By this logic.. 1896 is leap but 1900 is not leap right? Posted from my mobile device Yes you are correct... Every 4th year is leap but when it comes to century year, it is once in 4, that is the one divisible by 400, so 1600,2000 and 2400
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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 05:14
chetan2u wrote: Bill celebrated his 12th birthday on April 1, 1898, which was a Friday. On which day of the week did Bill celebrate his 18th birthday? Few facts 1) each year has 365 days,so when divided by 7 remainder is 1.. 2) every 4th year is leap so one day extra, and when divided by 7, remainder is 2.. 3) the century year is leap once in 4 years, so 1600 will be leap but 1700,1800 and 1900 will not be. ( here this concept is tested but i doubt you will be checked for your knowledge of leap year specially this point of century year18th birthday will be on 1898+1812=1904 So in 1898 Friday 1899  Saturday 1900( not a leap year)  sunday 1901 monday 1902  tuesday 1903  Wednesday 1904( leap year) as april 1 falls after 29th feb, so Friday Also you have to add 6 days for 6 years from 12 to 18th birthday. Add 1 to 6 for the leap year 1904., So total days above a multiple of 7 = 6+1=7 So day on 01: apr:1898 and 01 : apr : 1904 will be same that is Friday D A. Tuesday B. Wednesday C. Thursday D. Friday E. Saturday I marked answer C that is Thursday...i didn't count leap year in my calculation as i was not sure which year is a leap year in this particular problem as is 6 years there could be 2 leap years or just one leap year...
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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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05 Oct 2018, 22:50
Here is the standard formula to determine whether a year is a leap year or not, 1. If the year is evenly divisible by 4, go to step 2. Otherwise, go to step 5. 2. If the year is evenly divisible by 100, go to step 3. Otherwise, go to step 4. 3. If the year is evenly divisible by 400, go to step 4. Otherwise, go to step 5. 4. The year is a leap year (it has 366 days). 5. The year is not a leap year (it has 365 days).
Coming to the question, since we are being said Bill celebrated his 12th birthday on April 1, 1898, which was Friday, we need to determine how many number of days are there from today till his 18th birthday. If it is a multiple of 7, then it would be Friday. If not, the remainder would be the day of the week it is, starting from Friday.
Since the month of April has 30 days in it, April 2nd to April 30th would be 29 days. Similarly, for the remaining months, we can determine there would be 5 months which are having 31 days each and 3 remaining months in the year with 30 days.
So, the total number of days would be.
1898  29 + (31 * 5) + (30 * 3) = 274
Now, utilizing the formula to determine whether or not a year is leap year, we can determine the subsequent years 1899, 1900, 1901, 1902, 1903 are not leap year.
So, the total number of days would be (365 * 5) = 1825
As for 1904, since it is a leap year there would be 366 days in it, and the month of February would be having 29 days. We know, January and March would be having 31 days each, and February with 29 days.
So, the total number of days in 1904, till 31st March would be (31 * 2) + 29 = 91
So, adding all the required values, we get the total number of days  274 + 1825 + 91
If we were to divide it by 7, we will see the remainder is 1 for 274, 5 for 1825 and 0 for 91. So, 31st March, 1904 is Thursday. Which implies, April 1st, 1904 when Bill will be 18 years old, is Friday.
Answer choice would be 'D'.



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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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09 Oct 2018, 23:08
Solution Given:• Bill celebrated his 12th birthday on April 1, 1898, which was a Friday To find:• The day of the week, on which Bill celebrated his 18th birthday Approach and Working: The year in which Bill celebrated his 18th birthday = 1898 + 6 = 1904 • We know that, in a normal year there are 365 days, that is 365 = 7*52 + 1, which when divided by 7 leaves a remainder of 1.
o Thus, the number of odd days in a normal year = 1. • We also know that, in a leap year there are 366 days, that is 366 = 7*52 + 2, which when divided by 7 leaves a remainder of 2.
o Thus, the number of odd days in a leap year = 2. • So, the total number odd days from April 1, 1898 to April 1, 1904 = the number of normal years from 1899 to 1904 + 2 * (the number of leap years from 1899 to 1904)
o The only leap year in the given period is 1904, thus, the total number of odd days = 5 + 2*1 = 7
• Thus, April 1, 1904 = Friday + 7 days, which is again a Friday. Therefore, Bill celebrated his 18th birthday on a Friday. Hence, the correct answer is option D. Answer: D
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Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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27 Nov 2018, 12:57
365*6 = 2190 days
Add 1 day for the 1 leap year: 2191 days
2191/7 = 713 remainder 0, so 18th birthday will fall on Friday again
Answer: D




Re: Question of the Week 17 (Bill celebrated his 12th birthday on Apr...)
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27 Nov 2018, 12:57






