How Old in 1900?Date: 11/09/2002 at 17:34:14 From: Josh Subject: Word Problem In 1930, a correspondent proposed the following question: A man's age at death was one twenty-ninth of the year of his birth. How old was the man in 1900? Date: 11/10/2002 at 12:47:46 From: Doctors Greenie and Ian Subject: Re: Word Problem Hi Josh, Since this problem was posed in 1930, the man must have died in 1930 or earlier. The other date given in the problem is also significant. By asking what the man's age was in 1900, we know that he didn't die before 1900. Suppose the man's age at death was x. Then, according to the given information, the man's year of birth was 29x. He lived for x years after that, so the year of his death was (29x + x), or 30x. Only one year between 1900 and 1930 is divisible by 30, which is 1920. So he must have died in 1920. Again, we're told that "the man's age at death was 1/29th of the year of his birth". If he died in 1920, and he lived x years, then he must have been born in the year (1920 - x). So (age at death) = (1/29)(year of birth) x = (1/29)(1920 - x) This will let you figure out his age when he died, which will let you figure out his age in 1900. Can you take it from here? - Doctors Greenie and Ian, The Math Forum http://mathforum.org/dr.math/ Date: 11/11/2002 at 05:29:13 From: Confused Subject: Word Problem Hi Dr. Math, I was able to follow you up to getting the year of the man's death, but I can't see how you came up with his age at death, unless you just tried guessing a lot of different values. Is there a way to get it directly? Date: 11/11/2002 at 16:41:36 From: Doctor Ian Subject: Re: Word Problem Hi, Let's review what we know: 1) The man lived for x years. 2) He died in 1920. 3) His age at death was 1/29th of the year of his birth. From (1) and (2) we can conclude: 4) He was born in (1920-x). And now we're ready to make use of (3): (age at death) = (1/29)(year of birth) x = (1/29)(1920 - x) 29x = 1920 - x 30x = 1920 x = 1920/30 = 64 So he died in 1920 at age 64. That makes him 44 years old in 1900. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/