Minus' Solution to
Posted May 5 - 11, 1997
Minus' Solution to
We have to look at a time when John had lived more
than 9131 minutes. 9131 minutes is 152 hours, and
11 minutes, which is 6 days, 8 hours, and 11 minutes.
When John has lived 9131 minutes, Jane will have lived
approximately 9137 1/3 days.
When John has lived 9137 minutes, Jane will also
have lived approximately 9137 1/3 days, which when rounded
off is 9137 days.
Thus, on January 7, 1995, at 8:17 PM, Jane will have lived as many days
as John has lived minutes.
On January 1, 1995, Jane was 25. There were leap years in 1972,
1976, 1980, 1984, 1988, and 1992.
She had lived 25 x 365 + 6 days = 9131 days.
Minus knew he could count on his old friend, John Moser, to keep him from withering away. And what a gourmet meal John cooked up. John was right on the money with the number of days, but was slightly off with the time due to a small miscalculation. See if you can find the chicken bone in this dish...
Congratulations on being up and running for more than one year!!!A good problem with lots of parts to remember. Leap years & the days of catching up.
First, Jane is 25 x 365 = 9125 + 6 (for leap years) = 9131 days old when John is born.
After six days, John is 60 x 24 x 6 = 8840 minutes old. John still needs 9131 - 8840 = 291 + 6 (the days to catch up to mom) = 297 minutues equals 4 hours and 57 minutes.
Therefore, on January 7 at 4:57 pm they will equal each other minutes to days. Just think in a little more than five more days John will be one million seconds old. Will he live long enough to be a billion seconds old? Is it days, months or years away?
Andrew MacKinlay [andmak@cygnus.uwa.edu.au] proves once again that Australians like math (and Minus!) and they are good at it, too! Andrew concocted this delectable offering through some very sound reasoning. Well done!
John was born exactly 25 years after Jane was born. Jane, not including leap years would be 25*365=9125 days old. But there would be 6 leap years, in 1972, 1976, 1982, 1986, 1988 and 1992. So Jane would 9131 days old at the moment her baby was born.John would reach this age in minutes at 9131/60 = 152+11/60 or 152 hrs 11 min later. This is 152/24 = 6+8/24 or 6 days and 8 hrs. (+ 11 min).
So the date/time would be Jan 7 1995 at 8:11pm and 1 sec. But now Jane is 6 days older. So the time has to be revised to Jan 7 1995 at 8:17pm and 1 sec. This change doesn't make Jane any older so this time works.
Chris Easton [mai@fir.scotch.wa.edu.au] provided Minus with a small side dish. That is, an answer without a solution.
After 9137 days 8 hours and 17 minutes her son will have lived as many minutes as she has days.
Mat J. [mai@fir.scotch.wa.edu.au], perhaps a classmate of Chris, came up with the same answer. Minus was thrilled with the correct answer, but hey..."Where's the beef?"
Jane has been alive 9137 days when at 8:17 in the morning her son John has been alive for the 9137 minutes.
Another side dish came from Daria [dbneal@usit.net]. This one was a little more mysterious than the last. Where is Iaeger? Or, who is Iaeger?
Iaeger Intermediate's 5th Period math class believes the time would be 1 second after 8:17 pm on 1/7/95.
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