SOLUTION: A man born in the first half of the 19th century was x years old in the year x2. He was born in: (A) 1849 (B) 1806 (C) 1812 (D) 1825

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Question 1132177: A man born in the first half of the 19th century was x years old in the year x2. He was born in: (A) 1849 (B) 1806 (C) 1812 (D) 1825
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39618) About Me  (Show Source):
Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
The 19-th century is the time period from 1801 to 1900, inclusively.


sqrt%281801%29 = 42.44 (approximately);


sqrt%281900%29 = 43.59 (approximately).


Therefore, the only positive integer number x, whose square  x%5E2 lies between 1801 and 1900 is 43.


43%5E2 = 1849.


So, the problem says that the man was 43 years old in the year 1849.


Hence, he was born in the year 1849 - 43 = 1806.    ANSWER

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This problem is very well known,  and it was solved at this forum  MANY  TIMES  (about  10  times).

See,  for example,  this link

https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.1126192.html

https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.1126192.html


as well as many other links

https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.493026.html

https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.829890.html

https://www.algebra.com/algebra/homework/word/age/Age_Word_Problems.faq.question.16844.html

https://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.982465.html