What was Andrew's average walking speed in mph?

Question

Andrew drove one and a half miles to work and then walked back home at a tenth of his average driving speed. The two journeys lasted a total of 33 minutes. What was Andrew's average walking speed in mph?

Would be grateful if you could walk me through it.

Answer 1

This is a distance rate time problem

 

D = RxT

 

-----------------------------------------------------------    

|               | Distance |      Rate     |  Time   |

|---------------------------------------------------------|

| driving    | 1.5        |   r               |   t₁      |

| --------------------------------------------------------|

| walking   | 1.5         | r/10           |   t₂      |

----------------------------------------------------------

 

We can develop equations out of this chart

 

1.5 = rt₁

1.5 = (r/10) t₂

So

rt₁ = (r/10) t₂

or

rt₁ - (r/10) t₂ = 0    call this equation 1

 

We also know from the narrative that the total trip lasted 33 minutes

t₁ + t₂ = 33         call this equation 2

 

rt₁ - (r/10) t₂ = 0

t₁ + t₂ = 33

 

We can use elimination or substitution to solve this.

 

Let's use elimination.

To do so, we need to multiply the 2nd equation by -r to that we can eliminate the t₁ term.

rt1 - (r/10) t2 = 0
-rt₁ - rt₂ = -33r

 

Add the two equations together.

-(r+r/10)t₂ = -33r

 

Divide both sides by -r

(1 + 1/10)t₂ = 33

11/10 t₂ = 33

 

Divide both sides by 11/10

t₂ = 30

 

This tells us it took 30 minutes to walk, leaving 3 minutes for the drive.

 

But we're not done because we need to find the walker's speed.  To do this, let's go back to one of our earlier equations

 

1.5 = (r/10) t₂

Substituting the values for t₂,

1.5 = (r/10) 30

r/10 = 1/20 mph