TITLE: A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond ?
QUESTION [34 upvotes]: A lily pad doubles in area every second. After one minute, it fills the pond. How long would it take to quarter fill the pond?

To me this seems like we can set up a fraction-like equation:
$$\frac{60 \ \text{seconds}}{1} = \frac{x \ \text{seconds}}{1/4}$$ then $x = 15$ seconds. But the answer is $58$ seconds which really makes no sense to me. Any suggestions are greatly appreciated.

REPLY [5 votes]: We want a function relating the current time to the size of the lily pad; let's call it $f$. From the problem, we know that
$f(60\ \mathrm{seconds})=1\ \mathrm{pond}$
$f(x\ \mathrm{seconds})=1/4\ \mathrm{pond}$
so we can indeed write an equation like the one you want by solving for $1\ \mathrm{pond}$ in each equation; then:
$\frac{f(60\ \mathrm{seconds})}1=\frac{f(x\ \mathrm{seconds})}{1/4}$
But note the mediating $f$ that you are missing in your equation! If we assumed $f(x)=x$, then we would get your equation; but the problem also tells us that the lily pad doubles in size every second, that is, that:
$f((x+1)\ \mathrm{seconds}) = 2f(x\ \mathrm{seconds})$
If we choose $f(x)=x$, then this equality is not validated, since $(x+1)\ \mathrm{seconds}=2x\ \mathrm{seconds}$ is not validated.
Luckily we can make progress even without assuming $f(x)=x$. Simplifying the corrected version of your equation, we have:
$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$
Now we can apply the other equation given in the problem twice:
$f(60\ \mathrm{seconds})=4f(x\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=2(2f(x\ \mathrm{seconds}))$
$\phantom{f(60\ \mathrm{seconds})}=2f((x+1)\ \mathrm{seconds})$
$\phantom{f(60\ \mathrm{seconds})}=f((x+2)\ \mathrm{seconds})$
Then we can conclude that $60=x+2$ would be sufficient to validate this equation, so $x=58$ is one possible solution.