Sign Up For Free

**Join for FREE**,
and start talking with other members, weighing in on community polls,
and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one!)

- Reply to threads, and start your own
- Create reports of your
**campus visits** - Share college
**photos**and**videos** **Find your dream college**, save your search and share with friends- Receive our
**monthly newsletter**

College Confidential’s “Dean,” Sally Rubenstone, put together 25 of her best tips. Get your free copy of the "25 Tips from the Dean" eBook and get helpful advice on how to choose a college, get in, and pay for it: http://goo.gl/9zDJTM

tweeeek
Posts: **94**Registered User Junior Member

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?

Post edited by tweeeek on

## Replies to: Hard Math Problem =(

1,334Registered User Senior Member45(time she took in the morning, or p) = x

30(time she took in the evening, or q) = x

Therefore 45(p) = 30(q), or divide both sides by 5 and get 9(p) = 6(q). I know you can divide it further, but these numbers are small enough and it's not worth the time.

Since the whole trip took an hour, (p + q) = 60min, and so, p = 60-q.

Therefore 9(60-q) = 6q or 540-9q = 6q. So 540 = 15q, which makes q = 36. If q = 36, then by (p+q)=60, p (the time she took in the morning) must equal 24.

45 miles per hour, her speed in the morning, times (24/60) hours, her time, makes 18 miles travelled in the morning. If you check, 30 miles per hour times (36/60) hours also makes 18 miles in the evening.

Hope that makes a little sense. And I also hope it's right

356Registered User Memberr = d/t

In this case, we want to rearrange that equation to show that the total time = 1...

t = d/r

Since the distance is the same each way, we can use ONE variable to represent the distance one way...

t = (d/r1) + (d/r2)

r1 and r2 represent the two different rates, of course.

getting a common denominator and solving for d, we have...

t = [(dr2)/(r1r2)] + [(dr1)/(r1r2)]

t = (dr2 + dr1)/(r1r2)

t = [d(r2 + r1)]/(r1r2)

t[(r1r2)/(r2 + r1)] = d

Plugging in,

1[(45*30)/(45 + 30)] = 18 miles

Since it is only asking for the distance ONE WAY (ie, the morning), this is the answer. If it wanted the total trip distance, it'd be 36 mi.

139Registered User Junior Memberfor morning: 30/(30+45) = 0.6 of total time

for evening: 45/(30+45) = 0.4 of total time

relating back to distance = rate*time, mile/hour*h = mile, 30*.6 = 18. this is all there is!

the above solution basically is this using systems without telling you and the one before is perfect in every way, but i have found this one to be helpful and easy. for times it is also very easy, just 0.4 or .6 *60.

517Registered User Member(Speed 1 * Speed 2) / (Speed 1 + Speed 2)

That always gives you the distance traveled (for one path, not the total distance back and forth) if you the total time is one hour. Interestingly enough, I came accross a problem where the total time traveled was 10 hours (not 1); I used this formula and multiplied the answer by 10 and it ended up corresponding perfectly with the right answer. I'm not sure WHY this formula works at all...I just know it does (try for it for yourself).

2,642Registered User Senior Member517Registered User Member74- Junior MemberRusen Meylani.

736Registered User MemberThe distance going to work is 45t, where t is the time driving to work. The distance coming from work is 30(1 - t), because she spent one hour total driving.

45mph(t) = 30mph(1 - t)

t = 6/15 hours

To find the one-way distance: 45 mph(6/15 hr) = 18 miles

24,333Registered User Senior MemberFirst, there is an obvious contradiction in your position.

1. You don't need special formulae for this type of problem.

2. You know that speed x time = distance. = formula isn't it?

It is obvious that one does not to know a formula! SO, why not memorize the correct formula since it is fast and elegant!

Following your approach is exactly what the TCB expects a student who is NOT well prepared to do. Following your approach is not inherently wrong but takes too much time to arrive at the correct answer, not to mention the HIGH change of messing up in the simplifications. That is why the SAT is different from high school math. On the SAT, only two things matter: picking the right answer AND not wasting time.

FWIW, I have posted in DETAILS how to solve this problem in anywhere between 10 seconds to twenty seconds, without the possibility of errors.

736Registered User MemberIf you can think and reason, you can do any problem on the test no matter how it is asked - and do it quickly. I can easily do that problem in 10 seconds. There are no tricky questions on the SAT for those who truly understand mathematical concepts.

Most people recognize that, if they travel 30 mph for 2 hours, they will travel 60 miles. So they don't have to study and remember that speed x time = distance. They *know* it. If you know this basic concept and you can reason, you don't have to remember a formula. You can apply what you know to any speed/distance/time problem, no matter what the format is. Furthermore, you can generalize your knowledge to other problems.

Xiggi, your method may work for some people. Mine may work for others. People don't all learn using the same method. I'm merely letting people know there are alternatives.

2,214Super Moderator Senior MemberOn the other hand, it's hard to argue that on many SAT II occasions a formula could be a time/life saver - say, Heron's formula.

On the first hand again, if somebody remembers a formula, but is not very good at reasoning, the formula could be of little help, and sometimes even counterproductive.

A minor modification to setzwxman's solution.

t = (d/r1) + (d/r2),

t = d (1/r1 + 1/r2)

d = t / (1/r1 + 1/r2)

This formula is less elegant (it looks neater if you write it multi-storied), but it opens a way to expansions.

If Esther drove equal distances to work, then to the movies, and then home with r1=45, r2=54, r3=30 mph, and all this driving took her 1hour, then one leg of her travels is

d = 1 / (1/r1 + 1/r2 + 1/r3).

Don't forget to use [x^(-1)] key on your TI-83/84:

d = 45[x^(-1)] + 54[x^(-1)] + 30[x^(-1)] [ENTER] [x^(-1)] [ENTER] = 13.5

(I hope I did not offend anybody with this keys' playing).

Let's go back to the original question and add one detail:

on a way home Esther had a 10min. coffee break. What was her morning work commute?

Mayday! Is there a formula?!

Just stay calm, we don't need one (remember reasoning?):

d miles - distance between home and work

d/45 hr takes Esther to cover it going to work

d/30 hr takes Esther to cover it going home

10/60 hr Esther wasted on coffee, but it still counts in her total 1 hour of commute time:

1 = d/45 + d/30 + 10/60

d (1/45 + 1/30) = 5/6, ...

d = 15 miles.

Yeah, you could say, it still looks like setzwxman's formula, where "t" is driving only time.

I'd stick with a sure-fire equation in this case, rather then feel almost sure.

Besides, questions very rarely exactly repeat on SATs, and the next time Esther might be stuck in the traffic, trudging 30mph half a distance to work, then rushing 60mph for the rest of her morning commute, and cruising 40mph on a way home, still totaling 1 hour.

Same difference:

(d/2) / 30 + (d/2) / 60 + d/40 = 1

d(1/60 + 1/120 + 1/40) = 1

d = 20.

I forgot what I meant to prove here, but one thing I wanted to say for sure was that I liked boethian's solution the best. No formulas or equations (almost) - hard to beat. If only his approach could be explained simply...

Let's try:

(morning rate) * (morning time) = (evening rate) * (evening time),

or

(morning time) / (evening time) = (evening rate) / (morning rate) =

30/45= 2/3.

So, the ratio of two things is 2:3, or 2 equal parts to 3 equal parts.

Their total is 5 equal parts.

The smaller thing is 2 out of 5 parts, or 2/5 of their total.

The bigger thing is 3 out of 5 parts, or 3/5 of their total.

Frankly, a formula (or an equation) is hidden here.

Anyway, since total of morning and evening times is 1 hour,

morning time = 2/5 of 1 hour, or .4 h.

Therefore, distance = (morning rate) * (morning time) = 45 * .4 = 18.

This "parts" method helps in may other questions, especially on solutions (I mean, chemical), but that's for another time.

One more comment though: I can not agree with EllenF that there are no tricky questions on the SAT for "mathters". As xiggi rightly said, SAT ain't no high school math.

736Registered User MemberIf your math course didn't teach you how to think and reason, but you learn this skill in the course of preparing for the SAT, you have gained much more than a high score.

24,333Registered User Senior MemberTry your method with this problem:

Esther drove to work in the morning at an average speed of 12.75 miles per hour. She returned home in the evening along the same route and averaged 18.25 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?

It takes more than ten seconds to write down the first step of your solution. I still do not understand why anyone would bother remembering how to do compute the various steps of your approach when a solution that uses a simplified formula is faster, more elegant, and idiot proof.

There are indeed different approaches to every problem, but not all the methods are applicable to the SAT because they are circular or waste time.

People who are truly aware of the demands of the SAT recognize the difference easily.

517Registered User Member859Registered User MemberEasy.