Question

Carly walks $30$ feet in seven second. At this rate, how many minutes will it take for Carly to walk a mile if there are $5280$ feet in one mile?

Answer 1

Hint: In the above question we have given a rate of walking of Carly is$\dfrac{30ft}{7\sec }$ and Carly, at this rate will take one mile which is also given $5280$ feet. We have to find that, how many minutes will Carly take? It means we need answers in minutes.

Complete step by step solution:
Here we have given a rate in $\dfrac{ft}{\sec }$ , but since we want to end up with minutes.
$\Rightarrow $ In $7\sec $ the Carly covered distance $30feet$
$\Rightarrow $ In $1\sec $, the Carly covered $\dfrac{30ft}{7}$
$\Rightarrow $ In$60\sec $or in $1\min $ , the rate of Carly become\[=\dfrac{30ft\cdot 60}{7}\]
Now,
$\Rightarrow rate=\dfrac{30ft\times 60}{7}$
 Now, by according to the question we have to find time in minutes, the formula to find time is: $\Rightarrow time\left( \min \right)=\dfrac{dis\tan ce}{rate}$
Now we put our given values and the value of rate in min.
In question the distance is equals to $5280ft$ and rate or speed which we find is $\Rightarrow rate=\dfrac{30ft\times 60}{7}$
We will put those two values in the $time\left( \min \right)=\dfrac{dis\tan ce}{rate}$, we get
$\begin{align}
  & \Rightarrow time\left( \min \right)=\dfrac{dis\tan ce}{rate} \\
 & \Rightarrow time=\dfrac{5280ft}{\dfrac{30ft\times 60}{7}} \\
 & \Rightarrow time=\dfrac{5280\times 7}{30\times 60} \\
\end{align}$
Now multiplying $5280\times 7$ and $30\times 60$ in calculator we get,
$\Rightarrow time=\dfrac{36960}{1800}$
Now dividing these we get,
$\Rightarrow time=20.5333...$
Where $\Rightarrow 20.533\approx 20\dfrac{1}{2}\min $
Hence Carly takes $20\dfrac{1}{2}\min $ to cover $5280feet$.

Note:
We can also find the solution of the above question by another way.
Since we have given a rate $\dfrac{ft}{\sec }$ but we want to end up with time\minutes as a result, we need time term in numerator:
$\Rightarrow \min =\dfrac{\sec }{ft}\times ?$
We know the number of feet to be walked:
$\Rightarrow \min =\dfrac{\sec \times ft}{ft}\times ?$
And now we have convert seconds into minutes:
$\begin{align}
  & \Rightarrow \min =\dfrac{\sec \times ft}{ft}\times \dfrac{\min }{\sec }=\min \\
 & \\
\end{align}$
Now write the values given in question:
$\begin{align}
  & \Rightarrow \min =\dfrac{7}{30}\times 5280\times \dfrac{1}{60} \\
 & \Rightarrow \min =\dfrac{36960}{1800} \\
 & \Rightarrow \min =20.533...\approx 20\dfrac{1}{2} \\
\end{align}$
Hence we get the same answer as above.