I am very much in need of solution for this problem. I can't figure out the answer for this. does anybody know about this problem below?
thanks,,
Problem:
A man's boyhood lasted for 1/6 of his life, he played soccer for the next 1/12 of his life, and he married after 1/7 more of his life. a daughter was born 5 years after his marriage & the daughter lived 1/2 as many years as her father did. If the man died four years after his daughter did, how old was the man when he died?
Others have explained how to translate to the equation below. Here's a nice way to solve it.
$$\begin{eqnarray} L &\,=\,&\ \,\frac{L}6 + \frac{L}{12} + \frac{L}7\ +\ \frac{L}2\, +\, 9\\ \\ \\ \smash{\overset{\large\times\ \color{#c00}{84}}\iff}\ \ 84 L &\,=\,& 14L + 7L + 12L + 42L + 9\cdot 84\\ \\ &\,=\,& 75L + 9\cdot 84\\ \\ \iff\ \ \ 9L &\,=\,& 9\cdot 84\\ \\ \iff\ \ \ \ \ L &\,=\,& 84\end{eqnarray}\qquad$$
Note on lcm $= \color{#c00}{84}$: $\rm\,\ \color{#0a0}{2}\mid \color{#0a0}6\mid 12\,\Rightarrow\, lcm(\color{#0a0}{2,6},12, 7) = lcm(12,7) = 12\cdot 7= \color{#c00}{84},\,$ by $\,\gcd(12,7)=1.$
We can set up this problem by adding all the fractions of his life so they equal $1$ (his whole life). Also, let's say that he lived to $x$ years old.
Now,
$\frac{1}{6}$ of his life was boyhood.
$\frac{1}{12}$ of his life was soccer.
$\frac{1}{7}$ of his life was end of soccer to marriage.
After $5$ years, his daughter was born so
$\frac{5}{x}$ of his life was marriage to daughter's birth.
His daughter lived half the amount of years that he will have lived so
$\frac{1}{2}$ of his life was his daughter's life.
Then he died $4$ years later so
$\frac{4}{x}$ of his life was from his daughter's death to his death.
And all of these should equal to $1$. Thus
$\frac{1}{6}+\frac{1}{12}+\frac{1}{7}+\frac{5}{x}+\frac{1}{2}+\frac{4}{x}=1$
Now we simply solve for $x$. We get
$\frac{9}{x}=1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}$
then $9=x(1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2})$
then $x=\frac{9}{1-\frac{1}{6}-\frac{1}{12}-\frac{1}{7}-\frac{1}{2}}$.
By simplifying, we get $x=84$. So he lived to $84$ years old.
Let $L_M$ denote the length of the man's life and $L_D$ denote the length of the daughter's life.
Literally writing down what the problem tells us, we have $L_M = L_M\left(\frac16 + \frac1{12}+\frac17\right) + 5 + \frac12L_M + 4$.
Noting that the LCM of $12$ and $7$ is $84$, we then have $L_M = L_M\left(\frac{14}{84}+\frac{7}{84}+\frac{12}{84}\right) + \frac{42}{84}LM + 9$
So $L_M = \frac{75}{84}L_M + 9$. So $9=L_M - \frac{75}{84}L_M$, i.e. $9 = L_M\left(1-\frac{75}{84}\right)$, i.e. $9=\frac{9}{84}L_M$, so $1 = \frac{1}{84}L_M$, so $\frac{1}{L_M} = \frac{1}{84}$, so $L_M = 84$. As his daughter reached half his age, $L_D = \frac12 \cdot 84 = 42$.
Let
X = Man's Age
M = Man's Age at Marriage.
B = Man's Age at Daughter's Birth.
D = Daughter's Age.
We have 4 equations in 4 unknowns.
M + 5 = B
B + D + 4 = X
D = X / 2
M = 11/28 X
Recasting this as a linear system, we get
M + 5 - B = 0
B + D + 4 - X = 0
D - X / 2 = 0
M - 11/28 X = 0
Solving for this linear system, gives us that X = 84.
i will give you a more conceptual answer. let the age of the father when he died be $x$. then his age when he had the child is $$x(\frac{1}{6}+\frac{1}{12}+\frac{1}{7})+5$$ his age when his daughter died is $x-4$.
it is given that her age when she died is $\frac{x}{2}$. $$(x-4)-(x( \frac{1}{6}+\frac{1}{12}+\frac{1}{7} )+5)=\frac{x}{2}$$ solving this equation gives the age of the father when he died as $84$.
