Question

Sean went to a concert 120 miles away from his home. Sean’s house is located on the top of a mountain. On the way back from the concert, it started to snow, so he drove 10 mph slower than on the way to the concert, making his trip home two hours longer. How fast did he drive on the way to the concert?(1 point)
Responses

30 mph
30 mph

10 mph
10 mph

60 mph
60 mph

20 mph

Answer 1

Let's denote Sean's speed on the way to the concert as \( x \) mph. According to the problem, the trip to the concert, which is 120 miles away, can be expressed as:

\[
\text{Time to concert} = \frac{120}{x}
\]

On the way back, his speed is \( x - 10 \) mph, and the time taken to return home is:

\[
\text{Time back home} = \frac{120}{x - 10}
\]

We know that the return trip took 2 hours longer than the trip to the concert. Thus, we can set up the equation:

\[
\frac{120}{x - 10} = \frac{120}{x} + 2
\]

Now, let's solve this equation. First, we'll multiply both sides by \( x(x - 10) \) to eliminate the fractions:

\[
120x = 120(x - 10) + 2x(x - 10)
\]

Expanding both sides gives:

\[
120x = 120x - 1200 + 2x^2 - 20x
\]

Now, simplify the equation:

\[
0 = -1200 + 2x^2 - 20x
\]

Rearranging the equation:

\[
2x^2 - 20x - 1200 = 0
\]

We can simplify this by dividing the entire equation by 2:

\[
x^2 - 10x - 600 = 0
\]

Next, we can use the quadratic formula to solve for \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -10 \), and \( c = -600 \):

\[
b^2 - 4ac = (-10)^2 - 4(1)(-600) = 100 + 2400 = 2500
\]

So, we find:

\[
x = \frac{10 \pm \sqrt{2500}}{2} = \frac{10 \pm 50}{2}
\]

This gives us two potential solutions:

\[
x = \frac{60}{2} = 30 \quad \text{and} \quad x = \frac{-40}{2} = -20
\]

Since speed cannot be negative, we take the positive solution:

\[
x = 30
\]

Thus, Sean drove **30 mph** on the way to the concert.