Solved Problems

A stranger asked about his age replied :
If I have to reach 100 years, the half of the $\frac{5}{6}$ of the $\frac{9}{4}$ of my age is 12 years more than the $\frac{3}{4}$ of the $\frac{11}{13}$ of my remaining time to live. Find my age.

Solution:

Let $x$ be the stranger’s age.
The time remaing to reach 100 years in years is $100-x$

We have the equation:
$\frac{1}{2}\cdot \frac{5}{6}\cdot \frac{9}{4}\cdot x=\frac{3}{4}\cdot \frac{11}{13}\cdot (100- x)$
After simplification (see video how)
$\frac{5}{4}x=\frac{11}{13}(100-x)+16$
After multiplying both sides by $4\times 13$
It yields:
$13 \times 5x=4 \times 11(100-x)+16 \times 4 \times 13$
$65x=44 \times 100-44x+832$
$109x=5232$
$\frac{109x}{109}=\frac{5232}{109}$
$x=48$
The stranger’s age is $48$
Remaining time in years : $100-48=52$

Let’s check it out:

$\frac{9}{4}\; of\;48=\frac{9 \times 48}{4}$
$\frac{9}{4}\; of\;48=9 \times 12$
$\frac{9}{4}\; of\;48=108$

$\frac{5}{6}\; of\;108=\frac{5 \times 108}{6}$
$\frac{5}{6}\; of\;108=5 \times 18$
$\frac{5}{6}\; of\;108=90$
$\frac{1}{2}\; of\;90=\frac{90}{2}$
$\frac{1}{2}\; of\;90=45$