Algebra Distance Question

Question

Dara drove from home to the airport to pick up her friend and then came back home on the same route. Her average speed on the way to the airport was 55 miles per hour (mph), and her average speed on the way back home was 45 mph. If the total driving time was 1 hour and 20 minutes, how far, in miles, is the airport from Dara's home?

A) 29.7 B) 33 C) 100/3 D) 66

How do you solve such problems relating to distance given time and rate? Previously seen on SAT.

I understand d=rt but I can't get the set up to be correct. I learned this before but I can't recall how to do it.

Answer 1

Let $t_1$ be the time it took Dara to drive from home to the airport, and $t_2$ be the time it took her to drive back home. Then $$t_1 + t_2 = \left (1 + \frac{20}{60} \right)\, \mathrm h. \tag{1}$$ Let $d$ be the distance from home to the airport. Then $$d = 55\, \text{mi}/\text{h} \cdot t_1$$ but also $$d = 45\, \text{mi}/\text{h} \cdot t_2$$ and so $$55\, \text{mi}/\text{h} \cdot t_1 = 45\, \text{mi}/\text{h} \cdot t_2. \tag{2}$$

From equations $(1)$ and $(2)$ you should be able to find $t_1$ and $t_2$, and thus also $d$.

Answer 2

Lets let

$x = $ Distance from Dara's home to airport

$t_{1} =$ Time it took Dara to get from her home to the airport

$t_{2} =$ Time it took Dara to get from the airport to her home

$s_{1} =$ Average speed one her way to the airport

$s_{2} =$ Average speed coming back to the airport

So we know that

$$Speed = \frac{distance}{time}$$

So when Dara was going on her way to the airport

$$s_{1} = \frac{x}{t_{1}}$$

We know that $s_{1} = 55\, mph$

We can rearrange this formula to get

$$t_{1} = \frac{x}{s_{1}} = \frac{x}{55}$$

Similarly when Dara is coming back from the airport

$$s_{2} = \frac{x}{t_{2}}$$

We know that $s_{2} = 45\, mph$

We can rearrange this formula to get

$$t_{2} = \frac{x}{s_{2}} = \frac{x}{45}$$

We are also given that the total time is 1 hour and 20 minutes.

This means $t_{total} = t_{1} + t_{2}=$ 80 minutes

Now by plugging in the equations we found for $t_{1}$ and $t_{2}$ we end up with

$$80 = \frac{x}{55} + \frac{x}{45}$$

By finding a common denominator we get

$$80 = \frac{45x +55x}{45*55}$$

Now we can solve for $x$

$$x= \frac{80*45*55}{45+55}=1980\,\frac{min*mile}{h}*\frac{1\,h}{60\, min}=33\,miles$$