Lupe can ride her bike at a rate of 20 mph when there is no wind. On one particular day, she rode 2 miles against the wind and noticed that it took her the same amount of time as it did to ride 3 miles with the wind. How fast was the wind blowing that day?

Answer 1

Let the speed of the wind be \( w \) mph.

When Lupe rides against the wind, her effective speed is \( 20 - w \) mph, and when she rides with the wind, her effective speed is \( 20 + w \) mph.

The time taken to ride against the wind for 2 miles can be expressed as:

\[ \text{Time against the wind} = \frac{2}{20 - w} \]

The time taken to ride with the wind for 3 miles can be expressed as:

\[ \text{Time with the wind} = \frac{3}{20 + w} \]

According to the problem, these two times are equal:

\[ \frac{2}{20 - w} = \frac{3}{20 + w} \]

To solve for \( w \), we cross-multiply:

\[ 2(20 + w) = 3(20 - w) \]

Expanding both sides yields:

\[ 40 + 2w = 60 - 3w \]

Next, we can combine the \( w \) terms:

\[ 2w + 3w = 60 - 40 \]

This simplifies to:

\[ 5w = 20 \]

Now, solving for \( w \):

\[ w = \frac{20}{5} = 4 \]

Thus, the speed of the wind was

\[ \boxed{4} \text{ mph}. \]

To verify this, we can check the times:

1. Against the wind: Effective speed = \( 20 - 4 = 16 \) mph.

   • Time for 2 miles: \[ \frac{2}{16} = \frac{1}{8} \text{ hours}. \]

2. With the wind: Effective speed = \( 20 + 4 = 24 \) mph.

   • Time for 3 miles: \[ \frac{3}{24} = \frac{1}{8} \text{ hours}. \]

Both times are equal, confirming that the wind speed calculation is correct. The wind was indeed blowing at

\[ \boxed{4} \text{ mph}. \]