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?
The key is Esther travelled the same distance - x - in both her morning and evening commute.
45(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
ok. i found a less complicated way to look at these. if looking for distance, you should identify this with distance = rate*time. you can find the ratio of time (which adds up to 1) by dividing the other speed by total speed, because the higher speed corresponds to less time.
for 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.
I don't know where I got this formula, but this is what I do when I come accross this type of problem.
(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).
George - thats essentially the same formula that Xiggi has posted (t[(r1r2)/(r2 + r1)] = d, with r = rate/speed ) Look up two posts for an explanation.
You don't need special formulae for this type of problem. All you need is common sense. The common sense you need for this specific problem is recognizing that the distance driving to work equals the distance returning from work. You know that speed x time = distance. (If you forget which numbers to multiply, just make sure your units cancel. If you don't know how to cancel units, you might want to spend some time learning that. Cancelling units is very easy and can prevent many mistakes.)
The 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
It seems that we have discussed this particular question several times. I have no idea why you insist on coming back with an erroneous tactic.
First, 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.
There are different approaches to taking the test. For me, thinking works better than formulae. Formulae may work better for people who are good at memorization. I'm not.
If 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.
I agree with EllenF that you should do math and other things the way you can do them best. If I could solve a problem fast and errorless without formulas, I'd rather use that route.
On 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.
I agree that SAT math isn't *standard* high school math, but that's because most standard high school math courses don't teach students how to think. Many students can do a problem only if it is asked in one specific way. Ask it a bit differently and they're lost. SAT math is a cinch for students whose math courses teach them how to reason.
If 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.
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.
xiggi, your formula obviously is superior when it comes to time management. Do you have any other interesting formulas that are uncommon to share with others? To note, I had never heard of that formula except from a post I read from you earlier this month (without even realizing it was you who wrote it). It's this little stuff that can give you sooo much extra time to go back and check for stupid mistakes (the difference between a 700 and an 800).
Replies to: Hard Math Problem =(
45(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
r = 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.
for 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.
(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).
Rusen Meylani.
The 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
First, 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.
If 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.
On 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.
If 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.
Try 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.
Easy.