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 ? 
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. 
 A: Forget formulas for this one!  
If going forward 1 second the area gets doubled, then going back 1 second the area gets halved. 
So, 1 second before the pond was filled the pond must have been half filled, and 1 second before that it must have been quarter filled.
A: Take a look at the table below: 
\begin{array}{|c|c|}\hline
\text{Area } (A) & 1 & \dfrac12 & \dfrac14 & \dfrac18 & \cdots\\ \hline
\text{Time } (t) & 60 & 59 &  58 & 57 & \cdots\\ \hline
\end{array}
As you might see, you could easily deduce the relationship between $A$ and $t$, 
i.e., $A(t)=\dfrac{1}{2^{60-t}}$.
A: Chronological reasoning might help.


*

*At the beginning or $t=0$, its area is $A(0)$.

*After 1 second elapses or $t=1$, its area becomes $A(1)=2\times A(0)$.

*The next 1 second or $t=2$, its area becomes $A(2)=2 \times A(1) = 2\times2\times A(0)=2^2 A(0)$.

*At $t=60$ the area is $A(60)=2^{60} A(0)$. The pond is fully filled.

*The quarter of the fully filled pond is $A(60)/4 = 2^{60}A(0)/4=2^{58}A(0)$. This area is equal to $A(58)$, so $t=58$.
A: The area of a lily pod can be described by a function:
$$f:[0,60]\to[0,1]\\
f(x)=2^{x-60}$$
Now we have to compute the argument $x$ for which the value of the function $f(x)$ is equal to $\frac{1}{4}$:
$$\frac{1}{4}=2^{x-60}\\
\log_2 \frac{1}{4}=\log_2 2^{x-60}\\
-2 = x-60\\
x=58$$
A: 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.
A: This is exponential rather than linear. If $A$ is the initially covered area, then after one second the covered area will be $2A$, after two second $2\cdot 2A=4A$, after three seconds $2\cdot 4A=8A$. And so on: after $t$ seconds the covered area will be $2^tA$.
After $60$ seconds it will be $2^{60}A$, by assumption this is the whole pond. A quarter of this is
$$
\frac{2^{60}A}{4}=2^{58}A
$$
Of course Carmichael’s answer is slicker.
A: Your 'fraction-like equation' has nothing to do with the problem, because the lily pad doubles every second – its growth is exponential, not linear.
$$Area(t) = 2\cdot Area(t-1)$$
where $t$ is a number of seconds since 'some moment', hence
$$Area(t) = \color{red}{2^t}\cdot Area(0)$$
where zero is an arbitrary 'some moment'.
That implies
$$\frac{Area(60)}{Area(t)} = \frac{2^{60}}{2^t} = 2^{60-t}$$
Then if they ask at what $t$ is
$$Area(t)=1/4\cdot Area(60)$$
you have
$$2^{60-t} = \frac{Area(60)}{1/4\cdot Area(60)} = 4=2^2$$
so
$$60-t = 2$$
and finally
$$t=60-2 = 58$$
– the pond is quater-filled at $58$ seconds.
A: I think it's easiest to work backwards: if the area doubles every second and the pond is totally covered at time $t=60$, then it must be half covered at $t=59$, and therefore one quarter covered at $t=58$.
Alternately, let $f(t)$ be the fraction of the pond's area covered at time $t\leq 60$. Then $f(t)=f(0)2^t$ since the area doubles every second, and since $f(60)=1$ we get $f(0)=2^{-60}$. Therefore $f(t)=2^{-60}2^t=2^{t-60}$. Then setting $ 2^{t-60}=\frac{1}{4}$ and solving for $t$ yields $t=58$.
