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# M24-10

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Math Expert
Joined: 02 Sep 2009
Posts: 57025

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16 Sep 2014, 01:20
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Difficulty:

75% (hard)

Question Stats:

40% (01:12) correct 60% (01:28) wrong based on 95 sessions

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The 19th of September 1987 was Saturday. What day was the 20th of September 1990 if 1988 was a leap-year?

A. Monday
B. Tuesday
C. Wednesday
D. Thursday
E. Friday

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Joined: 02 Sep 2009
Posts: 57025

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16 Sep 2014, 01:21
Official Solution:

The 19th of September 1987 was Saturday. What day was the 20th of September 1990 if 1988 was a leap-year?

A. Monday
B. Tuesday
C. Wednesday
D. Thursday
E. Friday

There are 365 days in an ordinary year and 366 days in a leap-year.

365 divided by 7 (a week period) yields a remainder of 1, thus the same date after an ordinary year would fall on the next day of a week;

366 divided by 7 (a week period) yields a remainder of 2, thus the same date after a leap-year would fall two days after in a week;

Thus 3 years, 1988, 1989 and 1990, will accumulate $$2+1+1=4$$ days, plus 1-day difference between 19th and 20th, which gives total of 5-day difference (5-day shift). Thus, 20th of September 1990 was$$Saturday+5 days=Thursday$$.

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18 Jun 2015, 09:14
Bunuel wrote:
The 19th of September 1987 was Saturday. What day was the 20th of September 1990 if 1988 was a leap-year?

A. Monday
B. Tuesday
C. Wednesday
D. Thursday
E. Friday

Why is this question also not discussed and solved from other participants? Also found it on the mobile app. Thanks
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Joined: 19 Dec 2015
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14 May 2016, 06:56
I tried to only apply logic on this one:

In a leap year, the Gregorian calendar adds an extra day to the shortest month of the year - February. So instead of February having 28 days as in an ordinary year, in a leap year, February has 29 days.

Thus if 19th of September 1987 was a Saturday, 19th of September 1988 would be a Monday as you add an extra day in February 1988. Using the same logic/pattern, 19th of September 1989 would be a Tuesday, 19th of September 1990 would be a Wednesday and the 20th of September 1990 would be a Thursday.

Thursday is the correct answer, choice D.
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19 Apr 2018, 05:56
+1 for option D. As a rule, same date a year later is always a day ahead. Add to it the fact that leap year will contribute an extra day. We have a total of 4 days. Add another one day because the date is 20 September. That is 20 Sep 1990 will be a Thursday.
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19 Apr 2018, 08:27
364 divisible by 7 for integer .. therefore 365 over 7 has rem 1..

1 leap year (366) rem 2 + 2 years (365) = 2 + 1 + 1 = 4 so 4 days later
And 20 is 1 day later = 5 days

Or 365+365+366+1=1097

1097/7 = rem 5
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Joined: 10 Aug 2014
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02 Aug 2019, 23:24
i solved it this way:

the same day in subsequent years is a day before except leap year that it falls in the same day as the previous year. (always ignore the i day difference between sept 19 and 20)

ordinary year (365 days)
1987 - sept 19 - saturday
1988 - sept 19 - friday (a day before)
1989 - sept 19 thursday ( a day before)
1990- sept 19 wednesday ( a day before)

leap year (assume question is same september 19 1990)
1987 - sept 19 - saturday
1988 - sept 19 - saturday (leap year)
1989 - sept 19 friday ( a day before)
1990- sept 19 thursday ( a day before)

Re: M24-10   [#permalink] 02 Aug 2019, 23:24
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# M24-10

Moderators: chetan2u, Bunuel