I'm sort of struggling with this riddle told to me by a friend:
Hank owns a car. He has been taking good care of his car;
In fact, he has been taking such good care of it that the age of Hank, and the age his car combined is 56 years!Coincidentally his car is twice as old as Hank was when his car was as old as Hank is now.
How old are Hank and his car?
I'm having trouble figuring out what is easiest way of unraveling the 2nd sentence ("Coincidentally his car(…)").
For example, when representing the age of Hank, his car, and their combined age as $x$, $y$ and $c$ respectively, I get $x+y=c=56$ from the 1st sentence.
Then when trying do the same, the 2nd sentence I get $y = 2(?)$
So what I guess it boils down to is; I simply don't know how to express the 2nd sentence to define $y$.
Please excuse bad grammar and syntax, as english is not my primary language.
Best Answer
We can translate the statements about Hank and his car thusly:
$$\begin{align} \text{hank}_{\text{now}} + \text{car}_{\text{now}} &= 56 &(1)\\ \text{hank}_{\text{now}} &= \text{car}_{\text{then}} &(2)\\ \text{car}_{\text{now}} &= 2\cdot \text{hank}_{\text{then}} &(3) \end{align}$$
And we can use the passage of time to convert from "now" to "then":
$$\begin{align} \text{hank}_{\text{now}} &=\text{time} + \text{hank}_{\text{then}} \\ \text{car}_{\text{now}} &= \text{time} + \text{car}_{\text{then}} \end{align}$$
Writing $h := \text{hank}_{\text{then}}$, $c := \text{car}_{\text{then}}$, $t := \text{time}$, we have $$\begin{align} (t+h) + (t+c) = 2 t + h + c &= 56 &(1^\prime)\\ t+h &= c &(2^\prime)\\ t + c &= 2h &(3^\prime) \end{align}$$ Solving gives $$t = 8 \qquad h = 16 \qquad c = 24$$
So,
$$\text{hank}_{\text{then}} = 16 \qquad \text{hank}_{now} = 16 + 8 = 24$$ $$\text{car}_{\text{then}} = 24 \qquad \text{car}_{now} = 24 + 8 = 32$$
Let's check:
That's it!