The last time he counted, Father William had three children: Noah, Arthur, and John. The combined ages of the three children was half of Father William’s.
Five years later, during which time Joyce was born, Father William’s age equaled the total of all of his children’s ages.
A decade has since passed, with Phyllis appearing during that interval. When Phyllis was born, Noah was as old as John and Joyce together. The combined ages of all of the children is now double Father William’s age, which is only equal to that of Noah and Arthur together. Noah’s age also equals that of the two daughters together.
What are the current ages of Father William and all of his children? Note: Only Joyce and Phyllis are girls.
Here are some of my thoughts so far: Let $w$ be the original age of father William, $n$ be the original age of Noah, $a$ be the original age of Arthur, $j_b$ be John $j_g$ be Joyce, and $p$ be Phyllis.
Then originally we have $\frac{1}{2}w= n+a+ j_b$.
5 years later, we have $w=15+n+a+j_b$.
Now this is the part I got stuck, since we don't know when Phyillis was born.
Could someone please help me with this one?
Thank you!
When the first time we counted, we write John's age as $j$, then
$2 (a + j + n) = w \tag1$
If Joyce was born $x$ years after, then $5$ years after last count,
$(a + 5) + (j + 5) + (n + 5) + (5 - x) = w + 5 \tag2$
Say Phyllis was born $y$ years after second time we counted,
$n + 5 + y = (j + 5 + y) + (5 - x + y) \tag3$
After a decade after second count,
$(a + 15) + (n + 15) + (j + 15) + (15 - x) + (10-y) = 2 (w + 15) \tag4$
Also,
$(a + 15) + (n + 15) = w + 15 \tag5$
$n + 15 = (15 - x) + (10-y) \tag6$
Now you can solve for $a, j, n, x, y$ and $w$.
Their ages are now $w + 15, a + 15, n + 15, j + 15, 15 - x$ and $10- y$.