Mile per Hour SAT Questions!

[wlkid9]

Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

I had absolutely no idea what to do so I looked the question up on Google and sure enough I was able to find the answer and a formula. What I found became this:
(Speed 1 * Speed 2) / (Speed 1 + Speed 2), which is pretty much the same as (t[(r1r2)/(r2 + r1)] = d, with r = rate/speed ).

The answer is 18 Miles. The amount of miles going back home should still be 18 Miles correct?

Apparently these formulas only give you one distance. When I'm saying one distance I mean, like just Esther going to work, but not going to work and back home. Can anyone explain this formula or can you give me a link to FreeMathHelp.com that has a lesson on these types of problems?

 

[wjm11]

Esther drove to work in the morning at an average speed of 45 miles per hour. She returned home in the evening along the same route and averaged 30 miles per hour. If Esther spent a total of one hour commuting to and from work, how many miles did Esther drive to work in the morning?

You show good initiative in tracking down some formulas and a solution to your problem. Let’s look at some of the underlying logic to see if it all makes sense.

Distance equals velocity times time: d = vt
Rearranging this equation, we get: t = d/v

We always want to check our answers, so
t = d/v
t1 = (18mi)/(45mi/hr) = .4 hr
t2 = (18mi)/(30mi/hr) = .6 hr
Our check looks good: t1 + t2 = .4 +.6 = 1.0 hr.

...a lesson on these types of problems?

How would we tackle the problem initially? Let’s assign variables to our unknown quantities:

d = distance from home to work (one way)
t1 = time from home to work
t2 = time from work to home

Plugging this into “d = vt”, we get two equations (which we refer to as a “system of equations”):

d = (45mi/hr)(t1)
d = (30mi/hr)(t2)

Since we have only two equations to go with our three unknowns, we still need one more equation in order to have sufficient information to solve the problem. Our third equation comes from the problem statement about time. The total commuting time was one hour:

t1 + t2 = 1 hr, which can also be rearranged to
t2 = 1 hr – t1

We now have all the information needed to solve the problem.

In looking at our three equations, one thing we should recognize is that if we can find the value of any one of our three variables (d, t1, t2), the other two variables are very easily found. This gives us flexibility in our approach. (There are numerous ways to proceed from this point.) We don’t necessarily have to solve for d first!

Since the first two equations are both “d =”, we can set the right hand sides of those equations equal to each other (by substitution):

(45mi/hr)(t1) = (30mi/hr)(t2)

Substituting for t2, this becomes

(45mi/hr)(t1) = (30mi/hr)(1 hr – t1) = 30 mi – (30mi/hr)(t1)

Rearranging and simplifying:

(45mi/hr)(t1) + (30mi/hr)(t1) = 30 mi
(75mi/hr)(t1) = 30 mi
(t1) = (30 mi)/(75mi/hr) = .4 hr

Now that we have t1, the solutions for d and t2 are easily found.

The “big picture” is simply this: assign variables to unknown quantities, write equations using those variables, then combine those equations (often by substitution) to solve.

 

[wlkid9]

Thanks for all the help. I remember in Middle School we learned about this and we always had to use equations that looked like this: T and then T minus something. I never understood why we had to include the minus into one of the two equations. Could you have used (1 Hr - 1 ) for the side equation that has 45 mph?

 

[wjm11]

I remember in Middle School we learned about this and we always had to use equations that looked like this: T and then T minus something. I never understood why we had to include the minus into one of the two equations. Could you have used (1 Hr - 1 ) for the side equation that has 45 mph?

We could not have used (1 Hr - 1 ). What would that –1 represent? There is nothing in the problem statement that says one of the times is “1” less than another time.

The “T and then T minus something” would only apply if the problem were stated differently – if, for example, it had stated that the time coming home had taken 12 minutes longer than the time going to work (and leaving out one of the speeds). Then we could have used T for one equation and T + 12 for the next equation. Make sense?