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Question 206567: Craig Drives 20 miles per hour faster than sara. in the same time that sara drives 225 miles, craig drives 325 miles. find the speed of each car
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13364)   (Show Source):
You can put this solution on YOUR website!

A quick mental solution....

The difference in distances is 100 miles; the difference in speeds is 20 mph.
Therefore the amount of time is 100/20 = 5 hours.
Therefore Craig's speed is 325/5 = 65 mph and Sara's speed is 225/5 = 45 mph.

ANSWERS:
Craig 65 mph
Sara 45 mph

Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEEDS TIMES DISTANCES Craig r+20 325/(r+20) 325 Sara r 225/r 225

The two time quantities described as equal.
.
Solve for r.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Craig Drives 20 miles per hour faster than sara.
in the same time that sara drives 225 miles, craig drives 325 miles. find the speed of each car
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The answer in the post by @HyperBrain is INCORRECT due to arithmetic error on the way.
        I came to provide a correct solution

d=vt so t=d/v
Let x be the Sara's speed, in miles per hour.
Then the Craig's speed is (x+20) miles per hour.
225/x=325/(x+20)
Multiply both sides by x(x+20)
225(x+20)=325x
225x+4500=325x
100x=4500
x = 45 mi/hr ((Sara' speed)
x+20 = 45 + 20 = 65 mi/hr (Craig' speed).

CHECK the travel time.    = 5 hours.    = 5 hours.     Precisely correct !

Solved correctly.

Question 184851: A bus averages 6 kilometers an hour less than a passenger car. The bus travels 80 kilometers in the same time that the car travels 112 kilometers. Find the rate of each.
Found 3 solutions by greenestamps, josgarithmetic, ikleyn:
Answer by greenestamps(13364)   (Show Source):
You can put this solution on YOUR website!

In the same amount of time, the difference in distances traveled is 32 km and the difference in speeds is 6 km/hr. That means the number of hours that both vehicles traveled is 32/6 = 16/3.

Then the speed of the bus is (80)/(16/3) = 15 km/hr and the speed of the car is (112)/(16/3) = 21 km/hr.

ANSWERS:
bus 15km/h
car 21km/h

Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEEDS IME DISTANCES BUS b 80/b 180 CAR b+6 112/(b+6) 112

-------speed of the bus.
Car goes 21 kilometers per hour.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A bus averages 6 kilometers an hour less than a passenger car.
The bus travels 80 kilometers in the same time that the car travels 112 kilometers. Find the rate of each.
~~~~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @HyperBrain are irrelevant to the given problem, since
        in his calculations, @HyperBrain uses different numbers from the given in the post.
        It leads him to wrong answer.

        I came to bring a correct/adequate solution.

Step 1: Let x be the rate of the passenger car Then x-6 is the rate of the bus Note: distance=rate*time so time=distance/rate Step 2: Write the algebraic equation 80/(x-6) = 112/x Step 3: multiply both sides by x(x-6) 80x = 112(x-6) 80x = 112x-672 32x = 672 x = 672/32 = 21 Therefore, the passenger car travels at the rate of 21 km/hr. x-6 = 21-6 = 15 Therefore, the bus travels 15 km in 1 hr. CHECK for the travel time: = 5 hours. = 5 hours, the same time. Precisely correct !

Solved correctly.

//////////////////////////////////

The fact that @HyperBrain repeats these words "Power up, @HyperBrain!" from post to post,
tells me that he uses a computer code.

The fact that @HyperBrain makes numerous errors of the same type from post to post,
tells me that his computer code is erroneous and incorrectly reads the input data.

Question 957182: A cruise ship can sail 25 mph in calm water. Sailing with the Gulf Stream, the ship can sail 256 mi in the same amount of time it takes to sail 144 mi against the Gulf Stream. Find the rate of the Gulf Stream.
Found 4 solutions by MathTherapy, josgarithmetic, greenestamps, ikleyn:
Answer by MathTherapy(10855)   (Show Source):
You can put this solution on YOUR website!

A cruise ship can sail 25 mph in calm water. Sailing with the Gulf Stream, the ship can sail 256 mi in the same amount of time it takes to sail 144 mi against the Gulf Stream. Find the rate of the Gulf Stream. ************************************************** Let the stream's rate, be C Then, the DOWNSTREAM rate (rate with the stream) is 25 + C, or 25 + C Also, the UPSTREAM rate (rate against the stream) is: S - C, or 25 - C With the DOWNSTREAM-DISTANCE being 256 mi, time the ship takes to travel said distance is: And, with the UPSTREAM-DISTANCE being 144 mi, time the ship takes to travel said distance is: As the times are the same, we get: 144(25 + C) = 256(25 - C) ------ Cross-multiplying 9(25 + C) = 16(25 - C) ------ Dividing both sides by 8 9(25) + 9C = 16(25) - 16C 9C + 16C = 16(25) - 9(25) 25C = 7(25) Stream's rate, or

Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!
ship speed in still water, 25 mph

SPEED TIME DISTANE with strean 25+s t 256 against stream 25-s t 144

Answer by greenestamps(13364)   (Show Source):
You can put this solution on YOUR website!

Tutor @ikleyn sets the problem up with an equation which says the two times (distance divided by rate) are equal:

That of course is one valid way to set the problem up.

Another valid way to set the problem up is to use a proportion which says that, since the times are equal, the ratio of the distances is equal to the ratio of the speeds:

Clearly the two ways of setting up the problem lead immediately to identical calculations to get the answer -- so it is a matter of personal preference for which method to use.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A cruise ship can sail 25 mph in calm water. Sailing with the Gulf Stream, the ship can sail 256 mi
in the same amount of time it takes to sail 144 mi against the Gulf Stream. Find the rate of the Gulf Stream.
~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @lwsshak3 are incorrect due to arithmetic error.
        His answer "25 mph" for the Gulf Stream rate is absurdist,
        since then the cruise ship could not sail against the current.

        I came to bring a correct solution.

let c = rate of gulf stream
25+c = rate of ship sailing with Gulf Stream
25-c = rate of ship sailing against Gulf Stream
travel time=distance/rate (same in both directions)
...

..
256(25-c) = 144(25+c)
6400 - 256c = 3600 + 144c
400c = 2800
c = 2800/400 = 7
rate of gulf stream = 7 mph

Solved correctly.

Question 955979: Michael drove to work on Monday at 45 mph and arrived one minute early. The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late. How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
Found 3 solutions by josgarithmetic, n2, ikleyn:
Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DISTANCE Monday 45 t-(1/60) d Regularly R t d Tuesday 40 t+(1/60) d

Same distance, d, in the three tabulated situations.

--------this means normal expected trip time would be 17 minutes.

Looking at Tuesday's trip, time was , or 18 minutes.

, drive distance from work, 12 miles.

The question, what speed needed to arrive 5 minutes early?

Answer by n2(91)   (Show Source):
You can put this solution on YOUR website!
.
Michael drove to work on Monday at 45 mph and arrived one minute early.
The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late.
How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let 'd' be the distance from home to work (the same as the returning distance), in miles. The time Michael spent on Monday was hours (the distance divided by the average speed). The time Michael spent on Tuesday was hours (the distance divided by the average speed). These two time intervals differ in 1 + 1 = 2 minutes. So, we write this "time equation" - = . To solve, multiply both sides by 40*45. You will get 45d - 40d = 5d = 60 d = 60/5 = 12. So. the distance from home to work is 12 miles. Thus first question is answered. To answer second question, calculate the travel time on Monday. It is of a hour, which is the same as of an hour, or 16 minutes. +----------------------------------------------------------+ | So, if Michaels leaves his home at the same time | | every day (as the problem states), | | he has 16+1 = 17 minutes to get the office on time. | +----------------------------------------------------------+ To get the office 5 minutes earlier the control time, the travel time should be 17-5 = 12 minutes, or of an hour. Hence, to get the office 5 minutes earlier the control time, his average speed should be = 60 miles per hour. It is the answer to the second question.

At this point, the problem is solved completely.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Michael drove to work on Monday at 45 mph and arrived one minute early.
The employee drove to work on Tuesday, leaving home at the same time driving 40 mph and arriving one minute late.
How far does Michael live from work? What speed does he need to drive to arrive 5 minutes early?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution by @lwsshar3 is incorrect: it has an arithmetic error, which ruins everything and leads to wrong answer.

        I don't like the way how @lwsshar3 construct his solution.
        Actually, this problem is one of standard (and beautiful) problems on Travel and Distance,
        and it has a standard solution, which is equally beautiful as the problem itself is.
        So, anyone involved with this or a similar task. should know its standard method of solution.

        It is why I came to help.

Let 'd' be the distance from home to work (the same as the returning distance), in miles. The time Michael spent on Monday was hours (the distance divided by the average speed). The time Michael spent on Tuesday was hours (the distance divided by the average speed). These two time intervals differ in 1 + 1 = 2 minutes. So, we write this "time equation" - = . To solve, multiply both sides by 40*45. You will get 45d - 40d = 5d = 60 d = 60/5 = 12. So. the distance from home to work is 12 miles. Thus first question is answered. To answer second question, calculate the travel time on Monday. It is of a hour, which is the same as of an hour, or 16 minutes. +----------------------------------------------------------+ | So, if Michaels leaves his home at the same time | | every day (as the problem states), | | he has 16+1 = 17 minutes to get the office on time. | +----------------------------------------------------------+ To get the office 5 minutes earlier the control time, the travel time should be 17-5 = 12 minutes, or of an hour. Hence, to get the office 5 minutes earlier the control time, his average speed should be = 60 miles per hour. It is the answer to the second question.

At this point, the problem is solved completely.

I think you agree with me that this beautiful Travel and Distance problem deserves a perfect solution.

Question 969896: A boat goes downstream 72 miles in 3 hours, and travels upstream 60 miles in 6 hours.
What is the rate of the river, and what is the boat in still water?
Found 2 solutions by josgarithmetic, n2:
Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!
speed of the river, r
speed of boat by itself, b

downstream
upstream

If add the equations then

------boat speed without current

----------river current speed

Answer by n2(91)   (Show Source):
You can put this solution on YOUR website!
.
A boat goes downstream 72 miles in 3 hours, and travels upstream 60 miles in 6 hours.
What is the rate of the river, and what is the boat in still water?
~~~~~~~~~~~~~~~~~~~~~

Looking at the solution by @lwsshak3 in his post, I see that the logic
on how the solution is constructed in the post by @lwsshak3 is the same
which @mananth uses in his solutions to similar problems.

From this fact, I make my conclusion that they both use the same computer code.

Their logic is not effective, and their way solving, in my view, is wrong way teaching.

Question 975870: A and B are playing tag . A stay still while B runs from the same spot at a rate 2 meters per second.
After 60 seconds B keeps running . A now runs after him at a rate of 3 meters per second.
How long has A run when he caught up to B
My solution
3 (t-60) = 2t
3t - 180 =2t
3t-2t =180
t = 180
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
A and B are playing tag . A stay still while B runs from the same spot at a rate 2 meters per second.
After 60 seconds B keeps running . A now runs after him at a rate of 3 meters per second.
How long has A run when he caught up to B

My solution
3 (t-60) = 2t
3t - 180 =2t
3t-2t =180
t = 180
~~~~~~~~~~~~~~~~~~~~~~~~

Your error is that you determined the time for B,
while the problem asks about the time for A.

Question 1161528: A boy throws a stone from the top of a building 46.0 m above ground. The
stone is thrown at an angle of 33.0° below the horizontal and strikes the
ground 55.6 m away from the building, find the following: (a) Time of flight.
(b) Initial speed. (c) The magnitude and the direction of the velocity of the
stone just before it strikes the ground.
Answer by KMST(5393)   (Show Source):
You can put this solution on YOUR website!
EDITED for clarity and to correct multiple typos on April 28, 2026:

This is really a Physics class problem,
but from the math point of view it involves algebra, vectors, geometry and trigonometry.
Times will be in seconds from the time the stone left the boy's hand,
distances will be in meters,
speeds will be in m/s, and accelerations in m/s/s,
with the units understood, but not always stated.

initial speed (magnitude of the velocity) of the stone as it leaves the boy's hand

magnitude of the horizontal vector component of that initial velocity
magnitude of the vertical, downward vector component of that initial velocity
Time of flight of the stone, in seconds
We reasonably assume that the effects of air friction or wind do not cause a measurable difference in the motion of the stone,
so that the horizontal velocity is constant,
and the vertical velocity is only affected by gravity.
We know that latitude and altitude affect the acceleration of an object due to gravity,
but not having precise location data, we will use the value of 9.8m/s/s.
The downward velocity will increase linearly with ,
the time in seconds since the stone was thrown,
as
from at to at .
The average downward velocity during that time will be the average of initial and final velocities:

The horizontal and vertical distances traveled by the stone (in meters) will be
and
-->

Part (a) and part (b):
We can find and from the equations above.
From the first of the highlighted equations above, we get ,
and substituting the expression found for into the second equation, we get
-->-->
Using (rounded) , we substitute to get
--> (rounded), and --> (rounded).
Answer for part (a): The stone flies for .
Then, substituting the value found above and into , we get
--> --> (rounded).
The initial speed (magnitude of the initial velocity) of the stone was m/s.

Part(c):
For the stone, the time "just before it strikes the ground" is a minuscule fraction of a second before t=1.42 seconds, which rounds to
We can calculate the magnitude and the direction of the velocity of the
stone at that time from the magnitude of its horizontal and downward components.
As stated early,
the horizontal component of the stone velocity is constant, with magnitude ,
while the vertical component is a function of the time , in seconds, since the stone was thrown.
It was also established that at time , the downwards velocity was ,
Substituting , ,
and the already found values and , we get that
and

Adding those horizontal and vertical components vectors,
we calculate the magnitude of the velocity vector at 1.42 second as
-->
We can calculate its direction as the angle below the horizontal from
(rounded), which corresponds to
.
The calculated magnitude and direction of the velocity of the
stone just before it strikes the ground is m/s at an angle below the horizontal of .

Question 22105: River flows at 2 mph. Phil can paddle in still water at 8 mph. If he is 4 miles downstream from a log floating toward him,
how long will it take him to reach the log?
Found 3 solutions by math_tutor2020, MathTherapy, ikleyn:
Answer by math_tutor2020(3838)   (Show Source):
You can put this solution on YOUR website!

Answer: 1/2 of an hour or 30 minutes

Explanation

Let's say the river flows south.
Define points A, B, and C such that:
A = log's starting location
B = Phil's starting location
C = point in between A and B where Phil reaches the log

The diagram is optional but could be handy.

Phil's speed is 8 mph in still water.
But since Phil has to swim upstream against the 2 mph current, it slows him down to 8-2 = 6 mph.
So he travels 6x miles where x is the length of time in hours.
I'm using the formula:
distance = rate*time
Meanwhile, the log travels at 2 mph for the same amount of time x.
It travels 2x miles.

To recap so far we have these distances
AC = 2x
BC = 6x

These two distance subtotals must add to the 4 mile gap between the points A and B.
AC+BC = AB
2x+6x = AB
2x+6x = 4
8x = 4
x = 4/8
x = 1/2 of an hour aka 30 minutes

Answer by MathTherapy(10855)   (Show Source):
You can put this solution on YOUR website!

River flows at 2 mph. Phil can paddle in still water at 8 mph. If he is 4 miles downstream from a log floating toward him, how long will it take him to reach the log? ******************************************* Two of the persons who responded don't have a clue, and that's why their answers don't make sense. Let the time he takes to get to the log, be T Going downstream, his total average speed is his speed in still water, plus the speed of the current, or 8 + 2 = 10 mph Of the 4 miles between he and the log, Phil will cover 10T miles when he gets to the log Going upstream towards Phil, the log's speed is its speed in still water (0 mph), less the speed of the current, or 0 - 2 = - 2 mph Of the 4 miles between the log and Phil, the log will cover - 2T miles when Phil catches up to it. A speed of - 2 mph, and a distance of - 2T miles might seem strange and non-sensical, but this actually means that the log is actually travelling/drifting BACKWARDS, or AWAY from Phil. We now get the following DISTANCE equation: 10T + - 2T = 4 8T = 4 Time it takes Phil to get to the log, or ANECDOTE** Phil actually traveled 10T, or 10() = 5 miles to get to the log, although he started out just 4 miles from it. With the log travelling backwards, or drifting away from him, he had to travel an extra mile to get to it. Incidentally, the log travelled/drifted - 2() = - 1 mile, or 1 mile, BACKWARDS, which is the extra distance Phil ended up making up/travelling to get to it. Quite INTERESTING, isn't it?

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
River flows at 2 mph. Phil can paddle in still water at 8 mph. If he is 4 miles downstream from a log floating toward him,
how long will it take him to reach the log?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solutions in the posts by @Paul and by @JWG both are incorrect.
        For the correct solution see that follows in my post.

According to the problem's context, Phil paddles upstream, while the log is floating toward him, i.e. downstream. The Phil effective rate relative the river's banks is 8-2 = 6 miles per hour. The log floats toward Phil with the rate 2 miles per hour, so their approaching rate is 6+2 = 8 miles per hour. Hence, the time when Phil will reach the log is = of an hour, or 30 minutes. ANSWER

Solved.

Question 299160: a plane flew 500 km from atlanta to louisville. when returning to atlanta the flight took 1/2 hour less time.if the rate to Louisville was 50km/hr faster than the rate returning, find the two rates in km/hr.
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DIST. A to L r+50 500/(r+50) 500 L to A r 500/r 500 DIFF. 1/2

...
Suspecting that part of the description wording is mixed-up.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
a plane flew 500 km from Atlanta to Louisville. when returning to Atlanta the flight took 1/2 hour less time.
if the rate to Louisville was 50km/hr faster than the rate returning, find the two rates in km/hr.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

After reading this problem and its solution in the post by @mananth, I have two notices.

Firstly, the problem's formulation in the post is absurdist and self-contradictory.
Indeed, it says that the flight to Atlanta took 1/2 hour less time
and the rate to Louisville was 50 km/h faster than the rate returning.

So, the problem's creator was so hurry that missed circumstances.

I will edit the problem as it is shown below, to make sense from nonsense:

A plane flew 500 km from Atlanta to Louisville. When returning to Atlanta the flight took 1/2 hour time. If the rate to Louisville was 50km/h faster than the rate returning, find the two rates in km/hr.

Secondly, in the post by @mananth, both the solution and the answer are incorrect due to arithmetic errors.

I came to bring a correct solution to this my edited formulation.

Let the rate to Louisville be x km/h. Time of this flight is hours. The rate to Atlanta (returning) will be (x-50) km/h. The time of this returning flight is hours. Time equation is - = of an hour. <<<---=== It says literally what the problem says Simplify and find x 2*500*x - 2*500*(x-50) = x*(x-50), 1000x - 1000x + 50000 = x^2 - 50x, x^2 - 50x + 50000 = 0, (x-250)*(x+200) = 0, x = 250 or x = -200. We choose the positive root and reject the negative one. ANSWER. The rate from Atlanta to Louisville was 250 km/h. The rate of the returning trip from Louisville to Atlanta was 250-50 = 200 km/h. CHECK. The tome flying from Atlanta to Louisville was = 2 hours. The time flying from Louisville to Atlanta (returning) was = 2.5 hours, in 0.5 hours longer. Precisely correct.

The problem edited and converted from self-contradictory to self-consistent
and then solved in a right way.

Question 296160: a trip that takes 3 hours by car takes 30 minutes by plane. At these rates, how many minutes would be required for a plane trip that takes 4 hours by car?
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
a trip that takes 3 hours by car takes 30 minutes by plane. At these rates,
how many minutes would be required for a plane trip that takes 4 hours by car?
~~~~~~~~~~~~~~~~~~~~~~~~~~

The correct answer to this problem is 2/3 of an hour, or 40 minutes by the plane
(not 45 minutes, as in the post by @mananth).

Question 626102: The question is......Two jets leave at the same time,one is flying eat, the other west. the speed of the jet that is flying west is 75km/h faster than the speed of the jet flying east. in 3 hr they are 3075km apart. let w represent the speed of the plane travelling west and e represent the speed of plane travelling east. you need to know the speed of each plane. write a system of equation to model this situation
Answer by MathTherapy(10855)   (Show Source):
You can put this solution on YOUR website!

The question is......Two jets leave at the same time,one is flying eat, the other west. the speed of the jet that is flying west is 75km/h faster than the speed of the jet flying east. in 3 hr they are 3075km apart. let w represent the speed of the plane travelling west and e represent the speed of plane travelling east. you need to know the speed of each plane. write a system of equation to model this situation ****************************************** The other "PERSON" totally ignored the given instructions, regarding the assigned variables. It/He/She even came up with a solution, although one wasn't requested. As requested, speeds of westbound and eastbound jets are w, and e, respectively Since the westbound jet is travelling 75 km/h faster than the eastbound, we get the following SPEED equation: w = e + 75 ---- eq (i) Also, it takes the jets 3 hrs to be 3,075 km apart, at which time, the westbound had travelled "3w" km, and the eastbound, "3e" km, resulting in the following DISTANCE equation: 3w + 3e = 3,075 3(w + e) = 3(1,025) w + e = 1,025 --- eq (ii) w = e + 75 -- eq (i) The above two-equation system can be easily solved using SUBSTITUTION. It can also be solved using ELIMINATION, but written as follows: w - e = 75 ----- eq (i) w + e = 1,025 -- eq (ii) That's IT!. That's ALL that's needed.

Question 290678: Plane A and Plane B are 500 kilometres apart and flying towards each other. If plane A is travelling at a constant speed of 240 k/m and plane B is travelling at a constant speed of 360 k/m when will the planes meet?
Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Plane A and Plane B are 500 kilometres apart and flying towards each other. If plane A is travelling
at a constant speed of 240 k/m and plane B is travelling at a constant speed of 360 k/m when will the planes meet?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @mananth both are incorrect.
        I came to bring a correct answer.

The approaching rate of the planes is 240 + 360 = 600 kilometers per hour. Therefore, the time before they meet each other is = of an hour, or 50 minutes. ANSWER

Solved correctly and right answer is obtained, disproving the solution by @mananth.

Question 287905: Jan and Tariq took a canoeing trip, traveling 6 mi upstream against a 2 mi/h current. They then returned to the same point downstream. If their entire trip took 4 h, how fast can they paddle in still water? [Hint: If r is their rate (in miles per hour) in still water, their rate upstream is r - 2 and their rate downstream is r + 2.]

Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DISTANCE UP r-2 6/(r-2) 6 DOWN r+2 6/(r+2) 6 TOTAL 4

.
.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!

Question 284297: Fred and Desi left Steamtown Mall at 9:00 am and began walking in opposite directions. At 1:00 pm that same day they were 20 miles apart. If Fred walks 0.5 mph slower than Desi, what is Desi's speed of walking?
A. 2.25 mph B. 2.75 mph C. 3.5 mph D. 5 mph E. None of these.
Found 3 solutions by mccravyedwin, josgarithmetic, ikleyn:
Answer by mccravyedwin(421)   (Show Source):
You can put this solution on YOUR website!

Let Desi's speed be R, then Fred's speed is then R-0.5 Their speed of separation is the sum of their speeds or R+R-0.5 or 2R-0.5 From 9:00 am to 1:00 pm is 4 hours. Distance = Rate x Time 20 = (2R-0.5)T 20 = (2R-0.5)(4) 20 = 8R-2 22 = 8R 22/8 = R 2.75 = R Answer: Desi's speed of walking is 2.75 mph Edwin

Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DISTANCE Fred r-0.5 4 (r-0.5)(4) Desi r 4 (r)(4) TOTAL 20

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Fred and Desi left Steamtown Mall at 9:00 am and began walking in opposite directions.
At 1:00 pm that same day they were 20 miles apart. If Fred walks 0.5 mph slower than Desi,
what is Desi' speed of walking?
A. 2.25 mph B. 2.75 mph C. 3.5 mph D. 5 mph E. None of these.
~~~~~~~~~~~~~~~~~~~~~~~~

        In the post by @mananth, the treatment is not adequate to the problem.
        Due to this reason, the solution and the answer are incorrect.

        I came to bring a correct solution.

let desi's speed be x mph
Fred's speed = x-0.5 mph
In 4 hours Desi has walked 4x miles
In 4 hours Fred has walked 4(x-0.5) miles = 4x-2
4x + (4x -2) = 20
8x = 22
x = = = 2.75 mph     Desi' speed        <<<---=== ANSWER

CHECK.   The distance apart in 4 hours, at 1:00 pm, is 4*2.75 + 4*(2.75 - 0.5) = 20 miles.     ! Precisely correct !

Solved correctly.

Question 282326: Each day, Amit runs nine miles and then walks one mile. He runs 10 mph faster than he walks. If his total time is 75 minutes, then what is Amit's running speed?
Found 3 solutions by MathTherapy, josgarithmetic, ikleyn:
Answer by MathTherapy(10855)   (Show Source):
You can put this solution on YOUR website!

Each day, Amit runs nine miles and then walks one mile. He runs 10 mph faster than he walks. If his total time is 75 minutes, then what is Amit's running speed? ******************************************************************* Let Amit's running-speed be S Since his running-speed is 10 mph faster than his walking-speed, then Amit's walking-speed is "S - 10" mph With his running distance being 9 miles, Amit's time to run these 9 miles is And, with his walking distance being 1 mile, Amit's time to cover this mile is It's stated that Amit's total time to run and walk "....is 75 minutes," or . This gives us the following total TIME equation: , with S > 10 9(4)(S - 10) + 4S = 5S(S - 10) --- Multiplying by LCD, 4S(S - 10) (S - 12)(S - 6) = 0 S - 12 = 0 or S - 6 = 0 ----- Setting FACTORS equal to 0 S = 12 mph or 6 mph. However, 6 is NOT > 10 (see above constraint). So, Amit's running-speed is 12 mph

Answer by josgarithmetic(39835)   (Show Source):
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Seventy-five minutes is one and quarter hours.

SPEED TIME DISTANCE WALK r-10 1/(r-10) 1 RUN r 9/r 9 SUM 1.25

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Each day, Amit runs nine miles and then walks one mile. He runs 10 mph faster than he walks.
If his total time is 75 minutes, then what is Amit's running speed?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @mananth both are incorrect due to arithmetic errors on the way.
        I came to bring a correct solution.

let him walk at x mph he runs x+10 mph time taken for running = 9/x+10 time he walks = 1/x Time equation 1/x + 9/(x+10) = 5/4 hours Simplify and find x 4(x+10) + 36x = 5x(x+10) 4x + 40 + 36x = 5x^2 + 50x 40x + 40 = 5x^2 + 50x 5x^2 + 10x - 40 = 0 x^2 + 2x - 8 = 0 (x+4)*(x-2) = 0. The roots are x= -4 and x= 2. Since we look for the speed, we accept the positive root x= 2 and reject the negative one. ANSWER. Amit's walking speed is 2 miles per hour and his running speed is 2+10 = 12 mph. CHECK for the travel time. = = = of an hour. ! precisely correct !

Solved correctly.

Question 277004: it takes a plane 40 min longer to fly from Boston to LA at 525 mi/h than it does to return at 600 mi/h. how far apart are the cities?
Found 4 solutions by MathTherapy, timofer, josgarithmetic, ikleyn:
Answer by MathTherapy(10855)   (Show Source):
You can put this solution on YOUR website!

it takes a plane 40 min longer to fly from Boston to LA at 525 mi/h than it does to return at 600 mi/h. how far apart are the cities? **************************************************** Let distance between the 2 cities, be D Then, time the plane takes to fly from Boston to LA is And, time the plane takes to return from LA to Boston is: The plane takes 40 minutes, or longer to fly from Boston to LA. So, we get the following TIME equation: 8D - 7D = 2(1,400) ---- Multiplying by LCD, 4,200 Distance between the 2 cities, or D = 2,800 miles

Answer by timofer(159)   (Show Source):
You can put this solution on YOUR website!
Two different online webpages showed about 2600 miles between LA and Boston
by air. If we did not know and use d for this distance, then

Time for Boston to LA,
Time for LA to Boston,
and difference was given,

MAYBE.

Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DISTANCE Boston to LA 525 x+2/3 525(x+2/3) LA to Boston 600 x 600x Diff. 2/3

Distance each way, the same

----MISTAKE HERE

--the rest removed---

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
it takes a plane 40 min longer to fly from Boston to LA at 525 mi/h
than it does to return at 600 mi/h. how far apart are the cities?
~~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect, leading to wrong answer.
        I came to bring a correct solution.

Let the time taken to return be x hours (LA to Boston)
The time to go will be x+2/3 hours
600x = 525(x+2/3)
600x = 525x + 350
75x = 350
x = 350/75 = 4 2/3 hours

The distance will be   4 2/3 * 600 = 2800 miles.         ANSWER

Question 275984: a cyclist travels 30mi in 3 hours going against the wind and 80mi in 5 hours with wind. What is the rate of the cyclist in still air and what is the rate of the wind?
Found 3 solutions by greenestamps, timofer, ikleyn:
Answer by greenestamps(13364)   (Show Source):
You can put this solution on YOUR website!

These kinds of problems are intended as mathematical exercises, without concern for the physics.

In this problem, the speed of the cyclist against the wind is 30/3 = 10 mph and with the wind is 80/5 = 16 mph.

You can solve the problem using formal algebra:
c the speed of the cyclist in still air
w the speed of the wind
c+w=16; c-w=10 --> c=13, w=3

But an informal solution using logical reasoning is much easier: the speed of the cyclist is halfway between his speed with the wind and his speed against the wind:
(10+16)/2 = 13
and the speed of the wind is the difference between that speed and either of the other speeds:
16-13 = 3 mph, or 13-10 = 3 mph

ANSWERS: cyclist 13 mph; wind 3 mph

Answer by timofer(159)   (Show Source):
You can put this solution on YOUR website!
The stated problem does not make much sense. How exactly wind speed affects a bicyclist on the ground is not clear.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
a cyclist travels 30mi in 3 hours going against the wind and 80mi in 5 hours with wind.
What is the rate of the cyclist in still air and what is the rate of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~

The solution in the post by @mananth is incorrect and makes no sense.

The fact that the rate of the cyclist is 'x' miles per hour at no wind and the fact
that the rate of the wind is 'y' miles per hour DO NOT IMPLY that the effective rate
of the cyclist with the wind is (x+y) mph and the effective rate of the cyclist against
the wind is (x-y) miles per hour.

Only a person absolutely illiterate in Physics could create/compose such a nonsense.

No one peer-reviewed Math textbook or Physics textbook would publish such a non-sensical
gibberish.

My advise to a reader is to ignore the problem itself as a kind of nonsense
and ignore the solution by @mananth since it is WRONG TEACHING.

The creator of this "problem" should be ashamed of himself for composing such nonsense
and distributing it on the Internet.

Question 267480: Ramesh crosses a street 600 m long in 5 minutes. His speed in km/
hr is
Found 2 solutions by timofer, ikleyn:
Answer by timofer(159)   (Show Source):
You can put this solution on YOUR website!
If he crosses the street, that is through the width, and not through how long is the street. Not enough information until the actual width is given.

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Ramesh crosses a street 600 m long in 5 minutes. His speed in km/hr is
~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect.
        See below my correct solution below.

600 meters is = 0.6 of a kilometer.

5 minutes = 5/60 hrs = 1/12 of an hour

speed = distance / time = = 0.6*12 = 7.2 km/h.

his speed is 7.2 km/ h         ANSWER

Solved correctly.

Question 264398: Two hikers started from opposite ends of a 29-mile trail. One hiker walked 1.25 mph slower than the other hiker, and they met after 4 hours. How fast did each hiker walk?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39835)   (Show Source):
You can put this solution on YOUR website!

SPEED TIME DISTANCE one hiker r-1.25 4 4(r-1.25) other hiker r 4 4r TOTAL 29

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
Two hikers started from opposite ends of a 29-mile trail. One hiker walked 1.25 mph slower than the other hiker,
and they met after 4 hours. How fast did each hiker walk?
~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution and the answer in the post by @mananth both are incorrect.
        The fact that his answer values are incorrect, everybody can check manually.

        Below is my correct solution.

Let x be the rate of the slower hiker, in miles per hour.
Then the rate of the faster hiker is (x+1.25) miles per hour.

Write the total distance equation

        4x + 4*(x+1.25) = 29 miles.

Simplify and find x

        4x + 4x + 4*1.25 = 29,
        8x + 5 = 29,   --->   8x = 29 - 5 = 24,   --->   x = 24/8 = 3.

ANSWER.   The slower hiker rate is 3 mph;   the faster hiker rate is 3+1.25 = 4.25 mph.

CHECK.     4*(3 + 4.25) = 4*7.25 = 29 miles, the total distance.     ! correct !

Solved correctly.

Question 264376: an aircraft covers a distance of 42km in 2minutes and 30 seconds what is the average speed?

Answer by ikleyn(53937)   (Show Source):
You can put this solution on YOUR website!
.
an aircraft covers a distance of 42 km in 2 minutes and 30 seconds. what is the average speed?
~~~~~~~~~~~~~~~~~~~~~~

        Calculations in the post by @mananth are incorrect/inaccurate.
        I came to bring a correct accurate solution.

2 minutes and 30 seconds = 2 and 1/2 minutes = 5/2 minutes = (5/2 )/60 = 5/120 of an hour.

speed = d/t = 42 / (5/120) = 1008 kilometers per hour.         ANSWER

Solved correctly.