Question

On a ruler's tombstone, it is said that one sixth of his life was spent in childhood and one twelfth as a teenager. One seventh of his life passed between the time he became an adult and the time he married; five years later, his son was born. Alas, the son died four years before he did. He lived to be twice as old as his son did. How old did the ruler live to be ?

Answer 1

Hey mate ^_^

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Answer:
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Let the age of ruler is x so that of son = $\frac{x}{2}$

(Given)

Now according to the given condition:

$x = ( \frac{x}{6} ) + ( \frac{x}{12} ) + ( \frac{x}{7} ) + 5 + ( \frac{x}{2} ) + 4$

$x = ( \frac{14x}{84} ) + ( \frac{7x}{84} ) + ( \frac{12x}{84} ) + ( \frac{42x}{84} ) + 9$

$x = ( \frac{75x}{84} ) + 9$

$9 = ( \frac{84x}{84} ) - ( \frac{75x}{84} )$

$9 = ( \frac{9x}{84} )$

$1 = ( \frac{x}{84} )$

$84 = x$

Therefore,

Answer: 84 Years.

#Be Brainly❤️

Answer 2

Heyy mate ❤✌✌❤

Here's your Answer.....

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Let the age of ruler is x so that of son = x/2(given)

Now according to the given condition 

(x/6) + (x/12) + (x/7) + 5 + (x/2) + 4 = x

=> x = 84
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