Rikataka

i'm having trouble with this math question from the official sat study guide by the collegeboard. it's #18 on the 6th section

Esther drove to work in the morning at an average 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 in the morning?

i'm not sure why I couldn't get this problem. i usually do well on math algebra questions.

i can get to the point where I see it takes 24 minutes in the morning and 36 minutes at night, i just get confused multiplying the 45 miles per hour by the 24 minutes. also, i see the answer is 18 and i can get that by logically figuring 45/60 = 3/4 a mile per hour, and multiplying that rate by 24 minutes, but is there another way that won't seem so trivial when i'm actually sitting for my test?

cujoe

no no no... that's the simplest way you can do it... just stick to taht... and 45/60 is 3/4 of a mile per minute*

setzwxman

Here's how I did it (and I only came up with it after I played around with numbers using dimensional analysis)...

In general, R = d/t

So we know this...
[R1 = 45 mi/h, R2 = 30 mi/h]

(d/R1) + (d/R2) = t ...d in the morning = d in the afternoon

Combining that and then factoring for d, we have...

(dR2 + dR1)/(R1*R2) = t

d(R2 + R1)/(R1*R2) = t

d = t[(R1*R2)/(R1 + R2)]

Plugging in,

d = 1[(45*30)/(45 + 30)]
d = 1[(1350)/(75)]
d = 18 mi

optimizerdad

Using the same logic as Setxwxman, but in perhaps simpler fashion:

(d/45) + (d/30) = morning_travel_hours + evening_travel_hours = 1
(2d/90) + (3d/90) = 5d/90 = 1
d = 90/5 = 18 miles

little john

yep, did it the same way as optimizerdad... its probly the simplest way to do it

EllenF

To add to optimizerdad's solution, be sure to show your units. If your units don't come out correctly, your answer is usually wrong.

d/45 mph + d/30 mph = 1 h.
Hours cancel out.
d = 18 miles

PeteSAT

There is another (easier, I think) way to tackle this kind of problem! It also used the Rate*Time=Distance formula, but the trick is to use a variable for the time. For the trip to work, use t for the time:

45mph * t hours = 45t miles

For the trip home, use (1-t) to represent the time LEFT (If total time was 1 hour, and the trip to work was t hours, the trip home was (1-t) hours):

30mph * (1-t) hours = 30(1-t) miles

Since the distances to and from work are the same, just set the distances equal to each other:

45t = 30(1-t)
75t=30
t=2/5

Then just solve for the distance using t=2/5

45 * (2/5) =18.

Use whichever method makes more sense to you.

xiggi

As I have posted countless times on CC, this type problem is only difficult on the surface. It only requires a bit of attention to the problem statement, a bit of logic, and the use of a simple formula that should be in everyone's SAT repertoire.

One of the big problems is that our good friends at the PR and Kaplan of the world, as well as arrogant and pretentious authors such as Adam Robinson (from the error filled RocketReview SAT book) and Bill Drucker, prefer to unload a healthy dose of balderdash, flimflam, and tomfoolery onto hordes of poor students who are told NOT to trust their capabilities. Those "experts" send their readers in a tailspin of doubt with half-baked theories of finding the common denominators that require lengthy calculations, or even worse in a solution that ends up with a wild guess between two answers. They'll give easy examples that are supposed to work. I have little doubt that they would not be able to solve their own problems if the speeds happened to be 23 and 27, as opposed to the nice round numbers of 30 and 20.

Okies, off my soapbox for a few ... seconds. The reality is that this problem only requires you to know the formula for average speeds and pay attention to the problem statement.

How long SHOULD it take me to solve this problem?

Start your stopwatch.

Calculate (2*30*45)/(30+45) => 2700/75 equals 36.
Now I have the average speed of 36 mph for one hour of travel. Since the distance IS the same both ways (duh), the distance is 36/2 or 18 miles.

Stop you stopwatch. 12 seconds!

PS The formula is 2 * speed1 * speed2 / speed 1 + speed 2. I could explain the formula, but it is not necessary since the formula is easy to remember and ALWAYS works. Astute students will also notice that the formula presents a variance: multiply both speeds and divide by the straight average of the speeds given. For people who use their head before jumping to the calculator, this may allo for a quicker number manipulation. For students interested in super shorcuts, you should also know that if this type of problem appears on a MC, the answer is ALWAYS slighly below the straight average. The straight average is ALWAYS the trick question.

EllenF

Seven seconds to get an answer on the calculator for your problem, Xiggi, although that time is probably a bit skewed because we've been discussing this. Normally, I'd have to think a few seconds before starting to punch buttons.

gcf101

Rikataka, check out a related question on vinny380 06-28-2005 "Math Help =/" thread.
Just wonder - why is it's always women who drive?

dualityim

If this forum allows hotlinking of images, I would post that pictoral warning of Old Threads.

EllenF

Gcf101 - women drive in there problems because men crash and don't make it home. For the latter, see physics problems involving force, impulse, etc.

EllenF

"those" not "there"

gcf101

EllenF - two days passed by, and I still can't come up with a decent retort. At least I am driving slower now. On the way home.

gcf101

in post #10:
Math Help =/