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Answers to Puzzle #209
Showing 1 to 50 of 329 answers:
Answer #1:
From: 16576
School: 3107
Tracey lives 12 miles from work.
Extra: Tracey must return instantly.
In my opinion, for this problem it is easier to work with speed
expressed in miles per minute instead of miles per hour, as they are
in the data. I finally worked also with speed expressed in miles per
hour and I obtained the same result.
First, I transformed the miles per hour in miles per minute, because
Tracey is either 1 minute early, or 1 minute late:
miles
40 miles
2 miles
40 ----- = -- ------ = - -----h
60 minutes
3 minute
and
miles
45 miles
3 miles
45 ----- = -- ------ = - -----h
60 minute
4 minute
I know that the distance is the speed multiplied by the time. I noted
with "d" the distance from Tracey's home to work, and with "t" the
time in which Tracey arrives just in time.
So, I take the first affirmation of the problem. The time spent is (t
+ 1) (minutes), because Tracey is one minute late. The equation I
have is:
2
d = - * (t + 1)
3
(1)
where the distance is in miles, the speed in miles/minute and the
time in minutes.
For the second situation, the time is (t - 1), and the equation for
the distance is:
3
d = - * (t - 1)
4
(2)
From equations (1) and (2) I have:
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2(t+1)
3(t-1)
------ = -----3
4
that is an equation in t, that will be naturally expressed in minutes:
2t
2
3t
3
-- + - = -- - 3
3
4
4
Subtracting 2/3 t and adding 3/4 to both members, I obtain:
t
17
-- = -12
12
So, t = 17 (minutes)
Now, I calculate d from equation (1):
d = 2/3 m/min * 18 min
d = 12 miles
I check my result in equation (2):
d = 3/4 m/min * 16 min
d = 12 miles
So, Tracey lives 12 miles from work.
My result is correct because it verifies the system of equations.
Equations (1) and (2) form a system o two equations with two
unknowns, t and d. The problem requires only the value for d.
A sketch of the solution of the problem working with the speed
expressed in the more familiar unit mph is as follows:
Equation (1) is replaced by eq. (3):
1
d = 40 * (t + --)
60
(3)
and eq. (2) has the new form:
1
d = 45 * (t - --)
60
(4)
From eqs. (3) and (4) I obtained successively:
1
1
40(t+ --) = 45(t- --)
60
60
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85
5t = -60
17
t = -60
As specified before, t is now expressed in hours.
Using one of the eq.(3), I obtain:
d = 40 * 18/60
d = 12 (miles)
that is the same as the result obtained working with speed in miles
per minute and the time expressed in minutes. The result is the same
as the value obtained replacing t (in hours) in eq. (4), and is
certainly correct.
Extra
I know that the average speed is the total distance divided by the
total time.
The total distance is 24 miles (12 miles when going and 12 miles when
returning). Now, I calculate the time when going to work, from the
equation of the speed, and supposing the speed constant and equal
with 20 mph. The time is t1, and is expressed in hours:
12
20 = -t1
I do not write again the measure units, because I only use hours and
miles.
t1 = 12/20 = 3/5 (hours)
Now, I have to solve the following equation, that is the equation for
the average speed, along twice the distance home-work for Tracey:
24
------- = 40
t1 + t2
where t2, in hours, is the time Tracey spends on the way back.
I substitute t1 with 3/5, and I have:
24
-------- = 40
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3/5 + t2
or
24 = 24 + 40 * t2
Subtracting 24…
40 * t2 = 0
So,
t2 = 0
The result is surprising, meaning that the speed when returning is
undetermined (divide by 0). Tracey must return instantly.
My result is correct because it is obtained from the formula for the
average speed, and all the calculations are correct.
This means that there is a limit to the average speed. To understand
more, I solve the problem symbolically. Let the distance be “d”, the
speed on the way home-work be v1, and the corresponding time t1. For
the way back, the speed and time are respectively v2 and t2. I note
with v the average speed.
From the definition of the average speed:
2d
2d
v = ------- = ----------t1 + t2
d/v1 + d/v2
that could be written, after simplifying by d and re-arranging:
2
1
1
- = -- + -v
v1
v2
that means that the average speed is the harmonic mean of the two
speeds. The result is somehow surprising, and I thing it could be
generalised: if a mobile travels equal distances with constant (but
different) speeds, than the average speed is the harmonic mean of the
speeds.
For the given problem, substituting the values for v and v1 I obtain:
2
1
1
-- = -- + -40
20
v2
and consequently:
1
-- = 0.
v2
The average speed is calculated as the arithmetic mean of the speeds
for successive distances, if the distances are travelled in equal
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times.
Answer #2:
From: 19248
School: 5252
Tracey lives 12 miles from work.
Extra: There is no answer to the problem.
Let x = time needed to complete trip to work without being early or being late
rateA = 40 mph
rateB = 45 mph
timeA = x + 1/60 (because timeA represents the time spent traveling 40 mph; it has an
increment of 1/60 because of the additional minute she spent)
timeB = x - 1/60 (because timeB represents the time spent traveling 45 mph; it has a
decrease of 1/60 because of the 1 minute she has before arriving on time)
The intuition behind having a decrease and an increment in the times is because when
we travel faster we use up less time that usual in order to get to our destination.
Similarly, we can travel slower and still arrive at the same destination but spending a
longer period of time.
Since she is traveling to the same place of work, both distances involving the two
scenarios are equal.
40(x + 1/60) = 45(x - 1/60)
40x + 2/3 = 45x - 3/4
-5x = -17/12
x = 17/60 hr.
Substituting to any of the times and multiply by their corresponding rates,
40(17/60 + 1/60) = 40(18/60) = 12 miles
Similarly,
45(17/60 - 1/60) = 45(16/60) = 12 miles
As previously stated, the distance between Tracey's home to her workplace is the
same even with different rates.
Extra:
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There is a need to have the correct answer in the previous problem to make the
solution of the current problem a lot easier. Let us change the notation a bit so that it
would be easier to handle.
Let t 1 = the time needed to complete the trip from home to work (the time spent while
stuck in traffic)
t 2 = the time needed to complete the trip from work to home
Therefore, t 1 + t 2 is the total time for the round trip.
Let r 1 = 20 mph
r 2 = the speed she will have to average on the trip home so that her average speed
for the entire round trip is 40 miles per hour
Therefore, 20t 1 and r 2 t 2 are both the distances from home to work and vice versa.
They are both equal to 12 as seen in the previous problem.
20t 1 = 12
t 1 = 12/20 = 3/5 hr.
We now know the time needed to complete the trip from home to work, however be
reminded that this is not necessarily equal to the time she spends to go from work to
home.
To continue we have to use the given average rate for the round trip which is 40 mph.
Let us look deeper in to the meaning of average rate for round trip. It refers to the
ratio of the total distance traveled for the round trip to the time needed to complete
the round trip.
The total distance for the round trip is 12 + 12 = 24 miles. The total time is t 1 + t 2 as
derived earlier. So the solution will be:
24 / (t 1 + t 2 ) = 40
However, it leaves us with two unknowns. But we already know t 1 which is 3/5.
Substituting to the previous equation, we get:
24 / (3/5 + t 2 ) = 40
&nbsp24 = 40*(3/5 + t 2 )
&nbsp24 = 24 + 40t 2
t2 = 0
The answer seems implausible because it takes Tracey no time to reach home which
means she was traveling at infinite speed.
Clearly, there must be some mistake in the problem. But how can she average a rate
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of 40 miles per hour for the entire trip if the other part of the trip is done at infinite
speeds?
Answer #3:
From: 17985
School: 1354
Tracy lives 17 miles from work.
If Tracy travles at 40 miles per hour and is a minute late. Her drive
can be represented by the equation 40X+1. The 40 represents the 40
Miles per hour that she drove and the number 1 represens the one
minute that she is late to work. If she drives at 45 miles per hour
she arrives one minute early. That drive can be represented with the
equation 45X-1. Because the equations are equal distance they can be
set equal eachother. Therefore 40X+1=45X-1. Add one to both sides
and subtract 40 from both sides and you have 2=5X. You divided 2 by
5 to get X. X=.4. When you put X back into the equation you get 17
on both sides.
Answer #4:
From: 18622
18687
School: 62
Tracy Lives 12 miles away from her work.
The strategy I used was making an organized list.
Answer #5:
From: 18619
18620
School: 62
Tracey lives 12 miles from her work.
We found out this answer by finding how fast Tracey will be going to
get to work on time. Then we found out about how long it will take
her to get to work. Finally we used that as our answer.
Answer #6:
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From: 18383
School: 1354
I really Do not understand this problem. First, I thought what about
what would work.. What if the distance from home to work was 40
miles, it would then take her 60 minutes to get to work going 40 mph
and 53.33 minutes when going 45 mph... But now I am stuck! How do I
then determine the exact milage from home to work.. Do I take the
average of these two times and then divide by the 40 miles again?
Please help! Thank you!
Is there a starting point to this problem, or some type of ratio that
I should be using?
Answer #7:
From: 18528
18529
School: 62
Tracey's work is 12 miles from home.
We figured this out because when she goes 40 mph makes her 1 minute
late and when she goes 45 mph she is 1 minute early. So we took the
number inbetween 40 and 45, and got 42.5mph. So our equation was
42.5m=d. m=mph d=distance. We used this equation to get 12.
Answer #8:
From: 16625
School: 2922
Tracey liives 42.5 miles from work.
First I found the middle of 40 and 45 which is 42.5. Then I got
the miles per minute that Tracey travels at: 40mph=2.4 45mph=2.7.
Then I checked my work. 42.5 - 40 (which is the speed she goes) = 2.5
and since she goes 2.4mph she would be just over 1 min. late. At
45mph she arrives with just over 1 min. to spare.
Answer #9:
From: 16679
School: 25
Tracy lives 12 miles from work.
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We need to find the distance from her house that Tracy works. We know
the rate that she is driving both days. we know that her time, which
we will represent by t, is one minute more one day and one minute less
the next. Finally, we know that the distance from her home to work
does not change.
we can now relate this information by knowing that rate * time
=distance. So, on the first day her rate is 40mph. Her time is t+1,
since she is one minute late. On the 2nd day, her rate is 45mph, but
her time is t-1, since she is one minute early. Multiplying rate*
time, you get 40t+40 for the first day, and 45t-45 for the second day.
Since the distance is constant, you can set the 2 equal to each other
and solve for t. You get t=17min. So, t+1=18, and 18 min=.3 hrs. So in
.3 hrs, traveling at 40mph speeds, you travel 12 miles.
Answer #10:
From: 19253
School: 5256
well for her to recieve the same MPH as the same what she did before.
well you use the problem and realize the answer with common sense.
Answer #11:
From: 16539
School: 4440
My answer is that she starts 12 miles away from her work.
Let x = the distance that she has to travel to work.
First, I made an equation that used what we have. To do this I
thought about what I would do to check a result. I would find the
time that it takes her at 40 miles/hour and the time at 45
miles/hour. I would subtract the time it takes at 45 miles/hour from
the time it takes at 40 miles/hour
because the time it takes to drive a certain distance at 40
miles/hour is bigger. The difference would have to be 2
minutes,
1 minute from being late, and 1 from early. To make what I will
explain simpler, let's pretend
that she lives 15
miles away. Then, the equation would be:
15
y
-- = -45
60
In this equation I am divding the distance she travels by her speed
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and comparing it to minutes out of an hour. Here, y = the number of
minutes. You're basically asking yourself, if
15 is a third of 45, what is a third of 60? To solve these kinds of
equations, you multiply both sides by
60, in order
to isolate the variable on one side. This leaves you with:
15 * 60
------- = y
45
I'm doing this because if I were to leave out the 60, the right side
would equal minutes. However, the left side contains miles/hour.
These two units don't match, so you have to multiply by 60 to have
the same term on both sides.
Only in the problem itself, I don't know the distance,
so I label it as x. So, the equation becomes:
60x
60x
--- - --- = 2
40
45
We can simplify the two fractions because 60/40 = 3/2 and 60/45 =
4/3. Therefore:
3x
4x
-- - -- = 2
2
3
Next, we have to get a common denominator so that we can subtract the
two fractions. The common denominator here will be 6:
9x
8x
-- - -- = 2
6
6
Now we subtract:
x
- = 2
6
Now, all that's left is to multiply both sides by 6 to get rid of the
fraction and isolate the variable on one side:
x = 12
That means that she lives 12 miles away from work. To check, put this
into the first equation:
60 * (12)
60 * (12)
--------- - --------- = 2
40
45
The first part equals 18, the second - 16, which gives a difference
of 2. So it works.
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Extra: Since she lives 12 miles away, she will have to drive a total
of 24 miles there and back. Therefore, if she wants to drive at an
average speed of 40 miles an hour, her total time driving has to be
36 minutes. To get this, use the formula I used and explained in the
main part of the problem:
24
x
-- = -40
60
At 20 miles/hour, her speed to work, she needs to drive 12 miles. It
will take her 36
minutes (12/20 = 36/60, get this the same way as above). She's
already traveled all her alloted time,
therefore, no matter how fast she drives, she won't be able to go at
an average of 40 miles/hour.
Answer #12:
From: 19254
School: 2518
Tracy lives 12 miles far from work
Change:
40 miles/hour=2/3 mile/minute
45 miles/hour=3/4 mile/minute
Let t = the right time she gets to work(minute)
s = the distance between her house and her work(miles)
because she is 1 minute late when she drives at 40 mph so
t= s/ 2/3miles/minute - 1
because she is 1 minute early when she drives at 45 mph
t= s/ 3/4miles/minute + 1
so s/ 2/3 - 1
so
3/2s -1
so 3/2s-4/3s
so
1/6s
so
s
=
=
=
=
=
s/ 3/4 + 1 = t
4/3s + 1
1+1
2
12
So she lives 12 miles far from work
Answer #13:
From: 19064
School: 2062
Tracey lives 12 miles from work.
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Given:
12/2/08 5:26 PM
40 mph and 45 mph
Formula for distance:
d=rt or t=d/r
Let d=distance in miles
Let r=rate in mph
Let t=time in hours
Using t=d/r in which r=40mph and r=45mph.
At 40 mph, Tracey is one minute late or 1/60 hr late.
Thus:
t=(d/40mph)-(1/60hr)
t is equal to the time that it takes to get to work when she is
neither late nor early, however, since
she is driving slowly, she is one minute late. This is equivalent to
the subtraction of 1/60 hr from t.
Driving at 45 mph, Tracey is one minute early
Thus:
t=(d/45mph)+(1/60hr)
t again is equal to the time that it takes to
time she is one minute earlier because she is
faster. Now 1/60 hr needs to be added to the
or 1/60 hr early.
get to work, but this
driving 5 miles
total.
Since in both equations time (t) is equal to the time it takes to get
to work, both equations can be set equal to each other:
(d/40mph)-(1/60hr)=(d/45mph)+(1/60hr)
Solve for d:
(d/40)-(1/60)
(d/40)-(1/60)-(d/45)
(d/40)-(d/45)-(1/60)
(d/40)-(d/45)-(1/60)+(1/60)
(d/40)-(d/45)
(9d/360)-(8d/360)
d/360
(d/360)*(360)
d
=
=
=
=
=
=
=
=
=
(d/45)+(1/60)
(d/45)+(1/60)-(d/45)
1/60
(1/60)+(1/60)
(1/60)+(1/60)
2/60
1/30
(1/30)*(360)
12 miles
Check:
d/r=t
d=12 miles just calculated
r=40mph when a minute late and 45mph when a minute early
thus:
(12mi/40mph)*60=18 min
(12mi/45mph)*60=16 min
18min-16min=2min
Substitiute the distance back in the formula t=d/r to find the time
it takes to get to work at 40mph when a minute early and 45mph when a
minute late.
Multiply by 60 to get to minutes.
Since Tracey was one minute late and one minute early, the time
should come to a combined total of a two minute difference.
The time for 40mph subtracted from the time for 45 mph gives the same
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difference:
12/2/08 5:26 PM
2min so d=12 miles is correct.
Answer #14:
From: 18842
School: 5103
Tracey lives 12 miles from work.
Given that Tracey drives to work at 40 mph and is 1 minute late, and
drives 45 mph and is 1 minute early. Let x be the amount of time
that Tracey would take to be right on time for work.
I believe that it is easier to work with one unit. Therefore, I
would convert "miles-per-hour" to "miles-per-minute" since the time
would be in minutes.
To convert 40 mph, I know that there are 60 minutes in an hour.
Using a conversion factor, I can say:
40 miles
1 hr
40 miles
-------- x ------- = -------1 hr
60 min
60 min
The reason that my answer is in "miles-per-minutes" is that the
hours cancel each other out. I can simplify this fraction to 2/3
miles/min.
I can do the same thing to 45 mph.
45 miles
1 hr
45 miles
--------- x ---------- = ----------1 hr
60 min
60 min
Simplifying this, I would get 3/4 miles/min.
We know that the formula for distance is:
distance = speed x times
Let d be the distance, s be the speed, and t be the time.
For the 2/3 miles/min being one minute late, we can say:
d=st
d=(2/3)(x+1)--------> remember x is the exact time plus 1 minute
that was late
For the 3/4 miles/min being one minute early, we can say:
d=(3/4)(x-1)-------> remember x is the exact time minus 1 minute
that was early
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"S" and "T" are changed in the two equations because Tracey drives
at different speed and takes different amount of times. However,
she still traveled the same distance therefore it is unchanged.
Since d equals to both (2/3)(x+1) and (3/4)(x-1) we can use the
transitive property.
(2/3)(x+1)=(3/4)(x-1)
We can use the distributive property.
(2/3)x+(2/3)=(3/4)x-(3/4)
We try to get the variable on one side of the equation.
this by subtracting (2/3)x from both sides.
We can do
(2/3)x-(2/3)x+(2/3)=(3/4)x-(2/3)x-(3/4)
(2/3)=(1/12)x-(3/4)
Using inverse operation, we need to add 3/4 to both sides.
(17/12)=(1/12)x
Multiply both sides by the reciprocal of 1/12 which is 12.
I end up with x=17. This indicates that the exact time for Tracey
to arrive at work is 17 minutes. 1 minute late is 18 minutes, and 1
minute early is 16 minutes.
Go back to solve for the distance.
d=(2/3)(x+1)
d=(2/3)(17+1)
d=(2/3)(18)
d=12
Substitute x in the other equation to make sure.
d=(3/4)(x-1)
d=(3/4)(17-1)
d=(3/4)(16)
d=12
I can conclude that Tracey lives 12 miles from work.
Answer #15:
From: 18641
School: 5002
Tracey lives 12 miles from work.
Extra: You would have to be going at a infinite speed
First let's declare our variables:
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Let D = The distance to Tracey's office in miles
Let T = The time from when she leaves to when she needs to get to
work to be on time.
I decided to start with the formula Distance = Rate * Time
D = RT
Next I adapted this formula for use in this problem. I looked at
the first situation given "Tracey drives to work at 40 mph and is 1
minute late." So from this information I can change the formula D =
RT into D = 40 mph. * (T hours + (1/60) hours)
(1/60 hours being the 1 minute) giving me my first equation:
D = 40 (T + (1/60))
Using the distributive property we can distribute 40 over (T +
(1/60)) to get:
D = 40T + 40 * (1/60)
1
D = 40T + 40 * -60
40
1
D = 40T + -- * -1
60
40
D = 40T + -60
40 / 20
D = 40T + ------60 / 20
2
D = 40T + 3
D = 40T + (2/3)
So now we have taken the first situation that we are given and
created the equation D = 40T + (2/3). So let's start on the next
situation that we are given "If she leaves at the same time and
drives 45 mph, she'll be 1 minute early." Let's start with the same
formula as last time D = RT. If we modify it in the same way as
before we get the distance D she drives to work = 45 Mph. * (T
hours - (1/60) hours (1 minute) ) So now we have the equation:
D = 45 * (T - (1/60))
We can use the distributive property to distribute 45 over T (1/60) giving us
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D = 45T - 45 * (1/60)
1
D = 45T - 45 * -60
45
1
D = 45T - -- * -1
60
45
D = 45T - -60
45 / 15
D = 45T - ------60 / 15
3
D = 45T - 4
D = 45T - (3/4)
We now have a equation describing the seconded situation. Next
let's reverse the sides of this equation (giving us 45T - (3/4) = D)
and add that equation to our previous one.
45T - (3/4) = D
D = 40T + (2/3)
+
--------------------------45T - (3/4) + D = 40T + (2/3) + D
let's now get rid of D by subtracting it from both sides
45T - (3/4) + D - D = 40T + (2/3) + D - D
45T - (3/4) = 40T + (2/3)
Let's now subtract 40T from both sides
45T - (3/4) - 40T = 40T + (2/3) - 40T
45T - 40T - (3/4) = 40T - 40T + (2/3)
5T - (3/4) = (2/3)
5T + -(3/4) = (2/3)
Let's now move the -(3/4) from the left side to the right side by
adding (3/4) to both sides
5T + -(3/4) + (3/4) = (2/3) + (3/4)
2
3
5T = - + Page 16 of 33
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3
12/2/08 5:26 PM
4
2 * 4
3 * 3
5T = ----- + ----3 * 4
4 * 3
8
9
5T = -- + -12
12
8 + 9
5T = ----12
17
5T = -12
Now let's turn the 5T into T by multiplying both sides by (1/5)
(this is the same as dividing by 5)
1
17
1
5T * - = -- * 5
12
5
5T
1
17
1
-- * - = -- * 1
5
12
5
5T
17
1
-- = -- * 5
12
5
17
1
T = -- * 12
5
17 * 1
T = -----12 * 5
17
T = -60
We now know that the amount of time it takes her to drive to work
when she is not late T = (17/60) hours or 17 minutes. We can now
replace (17/60) for T in our first equation ( D = 40T + (2/3) ) to
find the value of D
D = 40T + (2/3)
D = 40 * (17/60) + (2/3)
40
17
2
D = -- * -- + 1
60
3
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12/2/08 5:26 PM
40 * 17
2
D = ------- + 1 * 60
3
40 * 17
2
D = ------- + 60
3
680
2
D = --- + 60
3
D = (680/60) + (2/3)
D = 11 + (1/3) + (2/3)
D = 11 + 1
D = 12
So we now know that the distance Tracey 12 miles lives from work and
this is our answer.
We can check our answer using the second equation we found.
D = 45 * (17/60) - (3/4)
D = (765/60) - 3/4
D = 12 + (3/4) - (3/4)
D = 12
Yep, it checks so we know our answer, 12 Miles, is correct.
Extra:
Let X = her speed on the way back from her office
Let's start with the formula from before D = RT. The total distance
there and back is 24 Miles so D = 24. The average rate for her trip
40 Mph. so we know the rate (R) = 40. The time (T) is = the sum of
the time of the two trips (there and back). We can take D = RT
divide both sides by R (D/R = RT/R), simplify that (D/R = T) and
then reverse the sides to get T = D/R. So we know that the time of
the trip = the distance / the rate. Using this we can say that in
the first half of the trip we are traveling 12 miles at 20 Mph so
the time of the first half of the trip = 12/20 = 6/10 = .6 . On the
seconded half of the trip we are traveling 12 miles at X Mph so the
time of the seconded half of the trip = 12/X. Knowing this we can
say that 24 (the total distance) = 40 (the average speed) * (.6 +
(12/X) ) (the sum of the times of the 2 trips(there and back).)
24 = 40 (.6 + (12/X))
24 = 40 * .6 + 40 * 12 / X
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12/2/08 5:26 PM
24 = 24 + 40 * 12 / X
24 = 24 + 480 / X
24 - 24 = 24 + 480 / X - 24
0 = 480 / X
X * 0 = 480 / X * X
X * 0 = 480
We know that 0 times anything = 0 but, our equation says that 0
times something = 480 so there is no answer. We know that the
faster you are traveling the greater your average speed, so we can
infer that you would have to be traveling at a infinite speed.
Answer #16:
From: 19255
School: 442
Tracy lives 12 miles away from work.
X = Time it takes Tracy to get to work on time (hours).
Y = How far away Tracy lives (miles).
So.....two equations...
40 miles/hour (X hours + 1/60 hours) = Y miles
45 miles/hour (X hours - 1/60 hours) = Y miles
Both are equal to Y miles, so....
40(X + 1/60) = 45(X - 1/60)
40x + 2/3 = 45x + 3/4
17/12 = 5x
x = 17/60
(Takes 17 minutes to get to work on time)
Plug into an equation....
40(17/60 + 1/60) = Y
Y = 12 miles (Distance from home to work)
Answer #17:
From: 19256
School: 5257
The distance from traceys home to her work place is 12miles
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12/2/08 5:26 PM
Let the normal time she's to take to the office be X
when she drives at 40mph, she's a minute late i.e she spends (x+1min)
or (x+1/60)hours
when she drives at 45mph , she's a minute early, i.e she spends (x1/60)mins.
distance = speed x time
Since the distance for both speeds is the same
therefore 40(x+1/60) = 45(x-1/60)
40x+2/3= 45x-3/4
5x =17/12
x = 17/60.
distance from her house to work = speed x time
=40(x+1/60)
40 x(17/60+1/600
= 12miles
Answer #18:
From: 19257
School: 3949
Tracey lives 12 miles from home.
Because we have two unknowns, 2 equations are required to solve the
problem.
Using the distance equation x=vt, I came up with 2 equations based on
the conditions that at 40 mph she is a minute late, and at 45 mph she
is a minute early.
The equations are as follows:
x=40*(t+1) and x=45*(t-1). Solving the first equation for t, yields
the result t=x/40 - 1. Plugging this into x=45(t-1) yields the
equation x=45*(x/40-1)-45. Solving for x gives you the result of 720.
Dividing this number by 60 because the speed used in the equations
was miles per hour yields a result of 12 miles.
Answer #19:
From: 19258
School: 5261
Tracey drove 12 miles.
We used our bwains
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12/2/08 5:26 PM
Answer #20:
From: 19259
School: 5263
Tracy Lives 12 Miles From Work
We know that the distance is the same in each example so we need to
solve for time - Y=Time. We know that at 40mpl it takes her 1 min
longer or 1/60 of an hour longer. We knoe that at 45mpl it takes her
1 min less or 1/60 of an hour shorter. Thus, we can come up with the
following expression:
40(Y+1/60)=45(Y-1/60) Solve for Y and
Y=17/60 of an hour or 17min
40(17/60+1/60)=12miles & 45(17/60-1/60)=12miles
Extra:
She can't - A distance of 24 miles round trip at 40mpl takes
(18/60+18/60)or 36min. Her trip in already took 36min 20(y/60)=12 or
y=36min
Answer #21:
From: 19260
School: 4527
la
la
Answer #22:
From: 17313
School: 2922
Let x equal half of 40 and 45 and y equal the solution to this
problem. The answer would be 40+x=y or 45-x=y.
Extra: Let x equal the average solution to this problem, so the
answer would be 20*x=40.
How I got the answer to this solutin is because Tracy drives to work
at 40 mph and gets there 1 minute late and drives home at 45 mph and
gets there 1 minute early.Because of that half of 40 and 45 would
tell you how many mph it takes Tracey to get home.
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12/2/08 5:26 PM
Extra: The way I got this average solution is because Tracey got to
work at only an average of 20 mph. It asks to tell you algebracly
the average speed for the entire round trip to equal 40 mph. So I
subtracted 40 from 20 and I got 20 as I used for x.
Answer #23:
From: 18533
School: 62
I'm not sure how exacty to get the right answer to this problem. But
I did figure one thing out and I was curious if it was right.
I know that at the rate that Tracey is driving she is going about 1
mile per 1 and a half minutes. Will that help me solve the problem?
Answer #24:
From: 16516
School: 3177
Tracey drives 12 miles to work.
I know that the formula for distance is rt = d where r is rate, t is
time, and d is distance. We know the rate for the two different days
of work - 40 mi/hr and 45 mi/hr. We know that when she went at 45
mi/hr, it took here two less minutes to drive (because the difference
between -1 or one minute late and 1 or one minute early is 2). So,
Let t = the time it took her to get to work when she was late
Let d = the distance to work
40 and 45 are rates, miles per hour. I'm multiplying by time, so the
answer - d - will be in miles. The only problem is - the time is in
minutes, since you must subtract two _minutes_ to get the time for
when she is early. To change the rates from miles per hour to miles
per minute, we must divide the rate by 60 minutes (you probably drive
less in a minute than you do in an hour, and there are 60 minutes in
an hour). You get
40 / 60 = 2/3 of a mile per minute
45 / 60 = 3/4 of a mile per minute
We have
d = t * 2/3
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d = (t - 2) * 3/4
12/2/08 5:26 PM
It's -2 because it took her less time to get
to work
We can now say that
(2/3)t = 3/4(t - 2)
(2/3)t = 3/4t - 1 1/2
0 = (1/12)t - 1 1/2
1 1/2 = (1/12)t
t = 18
First, I'll distribute
Get the t's on one side - subract (2/3)t
on both sides.
Add 1 1/2.
Divide by 1/12, or multiply by 12
This is in minutes because when you divide
miles (d) by mi/minute, you get that t is
in minutes.
Now we can figure out the distance by substituting this value for t
in the first two equations.
d = 18 * 2/3 = 12
d = 16 * 3/4 = 12 (!)
They both work, so, there are 12 miles separating Tracey from work.
Another way to solve
there is no variable
equations equal each
they are both in the
divided by 60, since
d
----45/60
60d
--45
60d
--40
this problem is to introduce equations where
for time - only for distance. Here, the
other because they both equal the same time form d/r = t. The rate, as we said before is
we need minutes, not hours. The times are:
This is for when she is early. Also, the 60 is divided twice,
once because of the vinculum, and once because of the division
sign in the denominator, so it is multiplied:
Then, there's the time when she is late:
Now, to form these into an equation, they have to be equal.
And for them to be equal, you must include the 2 minute
difference between the times. Here, you add 2 to the early
time, since she took 2 more minutes getting to work when she
was late. So, add 2, and make an equation:
60d
60d
--- + 2 = --45
40
4
3
- d + 2 = - d
3
2
Now, simplify the fractions:
Find the LCD, 6, and make the fractions have the same
denominator
8
9
- d + 2 = - d
6
6
Multiply _everything_ by 6
8d + 12 = 9d
Subtract 8d
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12 = d
Extra:
12/2/08 5:26 PM
Once again, the distance is 12 miles.
First, let's define "average speed". It is
total distance
-------------total time
We have: We are trying to make Tracey have an average speed of 40
miles/hour over the 24 mile trek to work and back, considering that
she has already done 12 miles of that (to work) at 20 miles. So, it
took her
12
-20
hours to get to work, and that's 6/10 of an hour, or 36 minutes.
She needs to take
24
-40
hours to get to work _and_ back. This also simplifies to 6/10, or
36 minutes. It is impossible to get home in 0 minutes, since
Tracey has already used up all 36 minutes she needs to take to
get to work and back to have an average speed of 40 miles per
hour. If she tried, she would be _really_ speeding, and not be
driving safely.
In other words, it is impossible to go at 20 miles per hour to work
and go at an average speed of 40 miles per hour there and back.
Answer #25:
From: 19263
School: 4351
44-1=43 41+1= 42 she needs to drive 42.5 miles per hour to get there
on time
i figured it out by using equations and differents processes. Using
linear functions and other thing
Answer #26:
From: 17249
School: 2965
sAda
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12/2/08 5:26 PM
asfdas
Answer #27:
From: 18193
School: 293
i got 77
i addded 44 +33
Answer #28:
From: 18320
School: 2414
dont know
i have no idea
Answer #29:
From: 19265
School: 4527
The distance from Tracey's house to work is 12 miles.
First I stet up these two equations:
40x+(2/3),45x-(3/4)
Then I set them equal to each other and solved for x. x=17/60
So the distance is
40(17/60)+(2/3) = 12
Answer #30:
From: 19266
School: 1380
gertet
10 pluse 2 plus 3
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12/2/08 5:26 PM
Answer #31:
From: 19269
School: 5270
she lives 24 miles away
45-40=5 (there is a difference of 5m travelled each hour). 60 (# of
minutes in hour) /5=12 (# of miles travelled for each minute
difference) 12*2(# of minutes different) = 24
Answer #32:
From: 16776
School: 9468
My answer would be that Tracy lives 12 miles away from her work place.
I got that answer by first getting down the information on a piece of
paper. Then I figured out that it takes 1 minuite to drive from one
exit to the next on the freeway. Since Tracy drives 40 mph and she is
one minute late and 45 mph and is 1 minute early I figured that she
lives 12 miles away from her work place.
Answer #33:
From: 19270
School: 2
1.5 miles
becuase i said so
Answer #34:
From: 19271
School: 1520
Tracy lives 42.5 miles from work.
EXTRA: Tracy must average 60mph on the way back from work
i figured that if she averaged 40mph and was a minute late, and if she averaged 45
miles an hour and was a minute early than it must be exactly in between.42.5 miles is
halfway EXTRA:i figured this out this way.20mph+60mph=80mph 80mph/2=40mph
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12/2/08 5:26 PM
Answer #35:
From: 19272
School: 4474
Tracy lives 42.5 miles from work.
I took the average of the two speeds and I used that as my
answer.
Answer #36:
From: 19273
School: 5271
Tracey lives 42.5 miles from work.
The equation d=r/t
d= distance
r= rate
t= time
45+40 = 55 divide by (-1-1) = 2
55 over 2 is 42.5
Answer #37:
From: 18689
School: 5040
Tracey lives 29 miles away from work
i plugged in variables and concluded
Answer #38:
From: 19210
School: 9468
42.5 mph
the time would be 42.5 mph
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12/2/08 5:26 PM
Answer #39:
From: 19274
School: 5271
Tracy lives 6.5 miles from her work.
40/61*x/60
2400=61x
x=39.3
45/59*x/60
2700=59x
x=45.8
45.8-39.3=6.5
Answer #40:
From: 19016
School: 9468
The distance between home and work is 12 miles.
d=rt so at 40 mph it would take 1 1/2 minutes per mile at 45 miles
per hour it only takes 1 1/3 minutes.
1 1/2d = 1 1/3d+2
1/6d = 2
d = 12 miles
The answer to the extra problem is that you can not average 40 mph
for the round trip because using the formula of total time divided by
total distance tells you that all of the time was used up on the
first half. If it is true that time stands still at the speed of
light perhaps that would be a theoretical answer.
Answer #41:
From: 17732
School: 1084
Tracey lives 12 miles from work.
60 miles on her way home.
For the extra, she has to average
The formula to find distance is d = rt.
mph at first.
I know that her rate is 40
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d = 40(t + 1/60)
A minute is 1/60 of an hour and we're dealing with hours.
minute late.
12/2/08 5:26 PM
She was a
The rate home was 45 mph.
d = 45(t - 1/60)
Since the distance to and from work is the same, it would mean:
40(t + 1/60) = 45(t - 1/60)
I used the distributive property.
40t + 40/60 = 45t - 45/60
I subtract 40t from both sides:
(40t -40t) + 40/60 = (45t - 40t) - 45/60
Which equals:
40/60 = 5t -45/60
Next I add 45/60 to both sides of the equation:
40/60 + 45/60 = 5t
85/60 = 5t
I divide 5 from both sides of the equation.
85/60(1/5) = t
t = 17/60 hour
Which equals 17 minutes.
To get the distance I plug it in.
I will plug in the 40 which means that I have to add another minute
(18/60) because she is one minute late.
d = 40(18/60)
d = 12 miles
If it were 45 miles, she is a minute early so I subtracted a minute
(16/60).
d = 45(16/60)
d = 12 miles
For the extra:
r1= rate to work
r2= rate to go back home
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r3= average speed (40)
(r1 + r2): 2 = r3
(20 + r2) : 2 = 40
20 + r2 = 40(2)
20 + r2 = 80
r2 = 80 - 20 = 60.
Answer #42:
From: 19275
School: 5273
i donno....
i donno
Answer #43:
From: 19277
School: 5271
Tracey lives 12 miles from work.
To find how long it takes to get somewhere, you take the number of
miles, this being x since we don't know, and dividing that by the
mph and multiplying the quotient by 60, the number of minutes in an
hour since that's what we're looking for.
Basically, ((# of miles(x)/ mph)) times 60.
But you have two speeds, so we have two quotients, i.e.
60x
and
60x
45mph
40mph
Now, the area of time it takes her to get to work is 2 minutes, one
minute that is too early and one minute that is too late.
So the difference in the two has to be 2. The equation is
60x
_
60x =
2
45
40
Which simplifies to
(4/3)X - (3/2)X = 2
solve for x, and you get 12.
Answer #44:
From: 19278
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12/2/08 5:26 PM
School: 1520
<html>
extra: 60 mph<br>
</html>
dont know answer to original problem
Extra:
20+x=40
2
*2 *2
20+x = 80
-20 -20
x=60
60 Miles Per Hour.
~Parker Finan
Answer #45:
From: 19279
School: 4434
Tracey lives 12 miles from work.
EXTRA: She will have to average 60 mph on the trip home.
I set up an equation of 40mph times Y divided by 60 minutes plus 1 minute = 45mph
times Y divided by 60 minutes minus 1 minute. This gave me the answer of 24 miles.
I then plugged 24 miles back into the above equation: 40 mph times 24 miles divided
by 60 minutes plus 1 minte = 45 mph times 24 miles divided by 60 mintes minus 1
minute. This gave me the answer of 17 minutes. I then plugged 17 mintes into the
equation 40 mph times 17 minutes divided by 60 minutes plus 1 minute which = 12
miles and 45 mph times 17 minutes divided by 60 minutes minus 1 minute which = 12
miles. EXTRA: I set up the equation 20 mph plus X mph divided by 2= 40 mph. This
gave me the anwser of 60 mph.
Answer #46:
From: 19128
School: 9468
12 miles
40
45
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__
60
*
x+1=
(x+1) =
45
-40
-60
+
12/2/08 5:26 PM
(x-1)
+ (x-1)
x+1
18
--- = -x-1
16
x
x
18-1
= ---16+1
x=17
18/60=3/10
3
----10
*
40
=
120/10 = 12
Answer #47:
From: 19281
19283
19282
School: 3074
We have calculated that Tracey must live 12 miles from her employment
office.
First we read the problem. Then we made an equation using
exponents. The equation was x5=60. We then solved the equation and
found out that x=12.
Answer #48:
From: 19284
School: 5276
Driving to work: 42.5 mile
Extra: 60 mph
On Driving to work, she was 1 min late at 40 and 1 min early at 45.
because it's a min exactly differant the differance in the distance
would be the differance between her change in speed (5 mph) the exact
differance between 0 and 5 is 2.5 so it would be 42.5 mph
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12/2/08 5:26 PM
on the extra. The differant was 20 mph (she went 20 there and 40 is
normal) so you add the differance to the trip she needs to make it up
on. 40+20=60
Right? oh yeah I still have allot of trouble with algebra I use my
own formulas
Answer #49:
From: 17868
School: 3048
Tracey lives 12 miles far from work.
x=time(minute)
so
(x+1).(40/60)=(x-1).(45/60)
when we multiply ech side by 60
(x+1).40=(x-1).45
40x+40=45x-45
85=5x
17=x
y=distance between work and home
(17+1).(40/60)=y
18.2/3=y
12=y
so
Answer #50:
From: 19285
School: 896
answer is 14...
40x+1=41y
Solve...
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