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Christopher’s youth lasted 1/6 of his life. He had his first beard in the next 1/12 of his life. At the end of the following 1/7 of his life Christopher got married. Five years from then his son was born. His son lived exactly 1/2 of Christopher’s life. Christopher died 4 years after the death of his son. How long did Christopher live?
[contact-form 8 “brain-teaser”]
Weighing 24 coins | Back – Next | 101-102 = 1 Can you solve this equality?
Yeah, what they said! Pretty straightforward linear equation.
84 Years. Son’s life is then 42 years.
After 1/6 (14 years) + 1/12(7 years) + 1/7(12 years) = 33 Years old, gets married.
5 Years later son is born – 38 Years.
Plus son’s like; 38+42 = 80 Years.
4 Years later Christopher dies at 84 years old.
Allan is right. Christopher’s life is 84 years. His son’s life is 42 years, as given in the question. The following mathematical explanation justifies the answer:
Suppose, life of Christopher = L,
Life of son = S.
S= L/2 ………………..eq 1
1/6 L (youth) + 1/12 L (beard appearance) + 1/7 L (marriage) + 5 yrs (son’s birth) + S (life of son) + 4 yrs (Christopher’s death) = L (Christopher’s life)
1/6 L + 1/12 L +1/7 L + 5 + S + 4 = L ……………eq 2
By solving these 2 simple equations, we get L = 84 and S = 42.
Hence, Christopher’s life is 84 years.
Thanks for that explanation.
Ken.
Christopher was 84 when he died.
14 yrs of youth (1/6 of 84)
7 yrs to grow beard (1/2 of 84)
12 yrs before marriage (1/7 of 84)
5 yrs before child is born
42 yrs child lives (1/2 of 84)
4 yrs to death.
1/6x + 1/12x + 1/7x + 1/2x + (4 + 5) = x (life span)
Of course, as a father, I never like to see parents outlive the children. Make it fish next time. 😉
Good job.
not sure, but quickly…
1/6x + 1/12x + 1/7x + 5 + 1/2x + 4 = x
x = 84, son = 42?
84 years … Life = youth (L/6) + beard (L/12) + single (L/7) + birth (5) + son’s life (L/2) + after son’s death (4) … L = [(14 + 7 + 12 + 42) / 84] x Life +9 … 9 = [(14 + 7 + 12 + 42 – 84) / 84] x Life … 9 = 9/84 Life … Life = 84