Question

At 10:00, Angela starts from her home and runs at a constant pace to Kathleen’s house, which is exactly 2 miles away. Immediately, she and Kathleen turn around and walk back to Angela’s house exactly at 4 miles an hour slower than Angela ran. They arrive at the house at 10:45. Find the speed at which Angela ran.
[a] 6 miles per hour
[b] 6.5 miles per hour
[c] 7 miles per hour
[d] 7.5 miles per hour
[e] 8 miles per hour.

Answer 1

Hint: Let the speed at which Angela ran be x miles per hour. Hence the speed at which Angela and Kathleen ran together is equal to x-4 miles per hour. Using $\text{time}=\dfrac{\text{Distance}}{\text{Speed}}$, calculate in terms of x, the time spent by Angela in covering the distance to Kathleen’s and the time spent by Angela and Kathleen together on the way back. Equate the total time to 45 mins. Hence form an equation in x. Solve for x and hence find the speed of Angela.

Complete step-by-step answer:
Let the speed of Angela be x miles per hour.
Hence the speed at which she and Kathleen ran on the way back is x-4 mile per hour.
Distance between Angela and Kathleen’s house is equal to 2 miles.
Hence the time taken by Angela to reach Kathleen’s house $=\dfrac{2}{x}hrs$ and the time taken by Kathleen and Angela on the way back $=\dfrac{2}{x-4}hrs$
Time spent in the whole trip is equal to $\left( \dfrac{2}{x}+\dfrac{2}{x-4} \right)hrs=120\left( \dfrac{1}{x}+\dfrac{1}{x-4} \right)\min $
But the total time spent in the trip is equal to 45 min
Hence, we have
$120\left( \dfrac{1}{x}+\dfrac{1}{x-4} \right)=45$
Dividing both sides by 15, we get
$8\left( \dfrac{1}{x}+\dfrac{1}{x-4} \right)=3$
Multiplying both sides by x(x-4), we get
$8\left( x-4+x \right)=3\left( x \right)\left( x-4 \right)$
Hence, we have
$\begin{align}
  & 8\left( 2x-4 \right)=3\left( {{x}^{2}}-4x \right) \\
 & \Rightarrow 16x-32=3{{x}^{2}}-12x \\
\end{align}$
Subtracting 16x from both sides, we get
$3{{x}^{2}}-28x=-32$
Adding 32 on both sides, we get
$3{{x}^{2}}-28x+32=0$
We have $32\times 3=96$
Also, we have $4\times 24=96$ and $4+24=28$
Hence, we have
$\begin{align}
  & 3{{x}^{2}}-28x+32=0 \\
 & \Rightarrow 3{{x}^{2}}-24x-4x+32=0 \\
\end{align}$
Taking 3x common from the first two terms and 4 common from the last two terms, we get
$3x\left( x-8 \right)-4\left( x-8 \right)=0$
Taking x-8 common from the two terms, we get
$\left( x-8 \right)\left( 3x-4 \right)=0$
Using zero product property, we get
$x-8=0\ \text{ or }3x-4=0$
Hence, we have
$x=8\text{ or }x=\dfrac{4}{3}$
But since the speed of Angela and Kathleen together on the way back is non-negative, we have
$x-2\ge 0\Rightarrow x\ge 2$
Hence, we have $x=\dfrac{4}{3}$ rejected.
Hence, we have x =8
Hence the speed of Angela is 8 miles per hour.
Hence option [e] is correct.

Note: Verification:
Time taken by Angela to reach Kathleen’s house is equal to $\dfrac{2}{8}\times 60\min =15\min $
Time taken by Anela and Katleen together on the way back is equal to $\dfrac{2}{4}\times 60\min =30\min $
Hence the time taken in the whole trip is equal to 15+30 = 45 min.
Hence our answer is verified to be correct.