Example 04 | Age Related Problem in Algebra

Let
A = present age of Albert
B = present age of Bryan
 

	Present	x-yrs hence	y-yrs ago	z-yrs ago	10-yrs hence
Albert	A	A + x	A - y	A - z	A + 10
Bryan	B	B + x	B - y	B - z	B + 10

 

Albert is as old as Bryan will be...
$A = B + x$

$x = A - B$
 

... when Albert is twice as old as Bryan was...
$A + x = 2(B - y)$

$A + (A - B) = 2(B - y)$

$2A - B = 2B - 2y$

$2y = 3B - 2A$
 

... when Albert's age was half the sum of their present ages
$A - y = \frac{1}{2}(A + B)$

$2A - 2y = A + B$

$A - B = 2y$

$A - B = 3B - 2A$

$3A = 4B$

$B = \frac{3}{4}A$
 

Bryan is as old as Albert was...
$B = A - z$

$z = A - B$
 

... when Bryan was half the age he will be ten years from now
$B - z = \frac{1}{2}(B + 10)$

$B - (A - B) = \frac{1}{2}(B + 10)$

$2B - A = \frac{1}{2}(B + 10)$

$4B - 2A = B + 10$

$3B - 2A = 10$

$3(\frac{3}{4}A) - 2A = 10$

$\frac{1}{4}A = 10$

$A = 40 ~ \text{yrs old}$   ← present age of Albert (answer)
 

$B = \frac{3}{4}(40)$

$B = 30 ~ \text{yrs old}$   ← present age of Bryan (answer)