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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(10810)

(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 2-system equation 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(53763)

(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(39799)   (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(53763)   (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(39799)   (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(53763)   (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(10810)   (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(39799)   (Show Source):
You can put this solution on YOUR website!
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(53763)   (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(10810)   (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(155)   (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(39799)   (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(53763)   (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(13334)   (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(155)   (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(53763)   (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(155)   (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(53763)   (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(39799)   (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(53763)   (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(53763)   (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.

Question 263749: How far upstream can a motorboat travel at 8 mi/hr and still return downstream at 12 mi/hr in a total time of 4 hr?

Found 3 solutions by MathTherapy, josgarithmetic, ikleyn:
Answer by MathTherapy(10810)   (Show Source):
You can put this solution on YOUR website!

How far upstream can a motorboat travel at 8 mi/hr and still return downstream at 12 mi/hr in a total time of 4 hr? ********************************* Let distance, back and forth, be D Then time taken to travel upstream, is Also, time taken to travel downstream, is We then get the following TOTAL-TIME equation: 3D + 2D = 96 ---- Multiplying by LCD, 24 5D = 96 Distance, back and forth, or

Answer by josgarithmetic(39799)   (Show Source):
You can put this solution on YOUR website!
Upstream and then downstream
8 and 12 miles per hour
Total time 4 hours.

If t hours going at 12 mph then 4-t going at 8 mph.

distance

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
How far upstream can a motorboat travel at 8 mi/hr and still return downstream at 12 mi/hr in a total time of 4 hr?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is INCORRECT due to arithmetic error.
        His answer "3 hours upstream and 1 hour downstream" does not work, as everybody can check manually.
        I came to bring a correct solution.

let the time taken upstream be x
the down stream time will be 4-x hrs.
speed upstream = 8 mph
distance upstream = 8*x
distance downstream = 12*(4-x)
8x = 48-12x
20x = 48
x = 48/20 = 2.4 hours = 2 hours and 24 minutes ---------------------------------- 24 miles
downstream time = 4 hours - 2.4 hours = 1.6 hours = 1 hour and 36 minutes.

The one way distance is 2.4 hours upstream * 8 mph = 19.2 miles.

ANSWER. How far is 19.2 miles.

Solved correctly.

Question 1027372: Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 am and 9.51 am respectively at the speed of 25 km/hr and 20 km/hr respectively for journey towards Y. Train R leaves station Y at 11.30 am at a speed of 30 km/hr for journey towards X. When and where will P be at equal
distance from Q and R ?
Found 2 solutions by n2, ikleyn:
Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
.
Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 am and 9.51 am respectively
at the speed of 25 km/hr and 20 km/hr respectively for journey towards Y. Train R leaves station Y at 11.30 am
at a speed of 30 km/hr for journey towards X. When and where will P be at equal distance from Q and R ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

P: 8:00 am 25 km/h ---> Q: 9:51 am 20 km/h ---> <--- R: 11:30 am 30 km/h -|------------------------------------------------------|- X (0) Y (220 km) Since the trains start at different time, the whole problem for analyzing is non-linear. We should analyze it step by step separately for different time intervals, as presented in my solution below. (1) At t1 = 9:51 am, the positions relative point A are P(t1) = 25 * 1 = 46.25 km; (train P moved 1 hour and 51 minutes at the rate 25 km/h) Q(t1) = 0; R(t1) = 220 km. So, train P still did not get midpoint between Q and R, and we shall continue our analysis. (2) At t2 = 11:30 am, the positions relative point A are P(t2) = 25 * 3.5 = 87.5 km; (train P moved 3.5 hours at the rate 25 km/h) Q(t2) = 20 * 1 = 33 km; (train Q moved 1 hours at the rate 20 km/h) R(t2) = 220 km. So, train P still did not get midpoint between Q and R, and we shall continue our analysis. (3) After 11:30 am, the positions relative point A are (here 't' is the time after 11:30 am) P(t) = 87.5 + 25*t kilometers; Q(t) = 33 + 20*t kilometers; R(t) = 220 - 30*t kilometers. We want to have P(t) - Q(t) = R(t) - P(t), which is an equation for P(t) to be the midpoint between Q(t) and R(t). It gives us this equation (87.5 + 25*t) - (33 + 20*t) = (220 - 30*t) - (87.5 + 25*t). Simplify it step by step and find 't' 2(87.5 + 25*t) = (220 - 30*t) + (33 + 20*t), 175 + 50*t = 253 - 10*t, 50t + 10t = 253 - 175, 60t = 78, t = of an hour, or 1 hour and 18 minutes. Thus, train P will be at midpoint between trains Q and R in 1 hour and 18 minutes after 11:30 am, i.e. at 12:48 pm. The location of train P will be 4 h 48 min * 25 km/h = = 120 kilometers from point A. (here 4 h 48 min is the travel time for train P from 8:00 am to 12:48 pm). ANSWER. Train P will be at midpoint between trains Q and R at 12:48 pm, i.e. 48 minutes after noon. The position of train P will be 120 km from point A at this time moment.

Solved.

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
Distance between two stations X and Y is 220 km. Trains P and Q leave station X at 8 am and 9.51 am respectively
at the speed of 25 km/hr and 20 km/hr respectively for journey towards Y. Train R leaves station Y at 11.30 am
at a speed of 30 km/hr for journey towards X. When and where will P be at equal distance from Q and R ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is conceptually incorrect.
        @mananth assumes from the very beginning that the time spent by trains to get the meeting point
        is the same for all three trains.
        But in reality it is not so, since the trains started at different time moments.

        So, his assumption is inadequate to the problem.

        See my correct solution below.

P: 8:00 am 25 km/h ---> Q: 9:51 am 20 km/h ---> <--- R: 11:30 am 30 km/h -|------------------------------------------------------|- X (0) Y (220 km) Since the trains start at different time, the whole problem for analyzing is non-linear. We should analyze it step by step separately for different time intervals, as presented in my solution below. (1) At t1 = 9:51 am, the positions relative point A are P(t1) = 25 * 1 = 46.25 km; (train P moved 1 hour and 51 minutes at the rate 25 km/h) Q(t1) = 0; R(t1) = 220 km. So, train P still did not get midpoint between Q and R, and we shall continue our analysis. (2) At t2 = 11:30 am, the positions relative point A are P(t2) = 25 * 3.5 = 87.5 km; (train P moved 3.5 hours at the rate 25 km/h) Q(t2) = 20 * 1 = 33 km; (train Q moved 1 hours at the rate 20 km/h) R(t2) = 220 km. So, train P still did not get midpoint between Q and R, and we shall continue our analysis. (3) After 11:30 am, the positions relative point A are (here 't' is the time after 11:30 am) P(t) = 87.5 + 25*t kilometers; Q(t) = 33 + 20*t kilometers; R(t) = 220 - 30*t kilometers. We want to have P(t) - Q(t) = R(t) - P(t), which is an equation for P(t) to be the midpoint between Q(t) and R(t). It gives us this equation (87.5 + 25*t) - (33 + 20*t) = (220 - 30*t) - (87.5 + 25*t). Simplify it step by step and find 't' 2(87.5 + 25*t) = (220 - 30*t) + (33 + 20*t), 175 + 50*t = 253 - 10*t, 50t + 10t = 253 - 175, 60t = 78, t = of an hour, or 1 hour and 18 minutes. Thus, train P will be at midpoint between trains Q and R in 1 hour and 18 minutes after 11:30 am, i.e. at 12:48 pm. The location of train P will be 4 h 48 min * 25 km/h = = 120 kilometers from point A. (here 4 h 48 min is the travel time for train P from 8:00 am to 12:48 pm). ANSWER. Train P will be at midpoint between trains Q and R at 12:48 pm, i.e. 48 minutes after noon. The position of train P will be 120 km from point A at this time moment.

Solved correctly.

Question 1025411: Flying against the wind, an airplane travels 3800km in 5 hours. Flying with the wind, the same plane travels 3660km in 3 hours. What is the rate of the plane in still air and what is the rate of the wind?
Found 3 solutions by josgarithmetic, n2, ikleyn:
Answer by josgarithmetic(39799)   (Show Source):
You can put this solution on YOUR website!
Already asked and answered too many times with different numbers.

r, speed no wind
w, wind's speed
Against:
With:

Simple steps from here...

Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
.
Flying against the wind, an airplane travels 3800km in 5 hours.
Flying with the wind, the same plane travels 3660km in 3 hours.
What is the rate of the plane in still air and what is the rate of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let u be the rate of the plane at no wind (in kilometers per hour) and v be the rate of the wind (in the same units). Then the effective rate of the plane with the wind is u + v and the effective rate of the plane against the wind is u - v. From the problem, the effective rate of the plane against the wind is the distance of 3800 kilometers divided by the time of 5 hours {{3800/5}}} = 760 km/h. The effective rate of the plane with the wind is the distance of 3660 kilometers divided by the time of 3 hours = 1220 mph. So, we have two equations to find 'u' and 'v' u + v = 1220, (1) u - v = 760. (2) To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get 2u = 1220 + 760 = 1980 ---> u = 1980/2 = 990. Now from equation (1) v = 1220 - 990 = 280 - 220 = 230. ANSWER. The rate of the plane in still air is 990 km/h. The rate of the wind is 230 km/h.

Solved.

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
Flying against the wind, an airplane travels 3800km in 5 hours.
Flying with the wind, the same plane travels 3660km in 3 hours.
What is the rate of the plane in still air and what is the rate of the wind?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @mananth is incorrect.
        I came to bring a correct solution.

Let u be the rate of the plane at no wind (in kilometers per hour) and v be the rate of the wind (in the same units). Then the effective rate of the plane with the wind is u + v and the effective rate of the plane against the wind is u - v. From the problem, the effective rate of the plane against the wind is the distance of 3800 kilometers divided by the time of 5 hours {{3800/5}}} = 760 km/h. The effective rate of the plane with the wind is the distance of 3660 kilometers divided by the time of 3 hours = 1220 mph. So, we have two equations to find 'u' and 'v' u + v = 1220, (1) u - v = 760. (2) To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get 2u = 1220 + 760 = 1980 ---> u = 1980/2 = 990. Now from equation (1) v = 1220 - 990 = 280 - 220 = 230. ANSWER. The rate of the plane in still air is 990 km/h. The rate of the wind is 230 km/h.

Solved.

The solution in the post by @mananth is performed in the style of complete loosing of logic,
so you do not try to find a rational idea in his solution - there is no a mathematical accuracy there.

Simply ignore his post.

Question 1107789: Jane took 30 min to drive her boat upstream to water-ski at her favorite spot. Coming back later in the day, at the same boat speed, took her 15 min. If the current in that part of the river is 6 km per hr, what was her boat speed in still water?
Found 3 solutions by greenestamps, timofer, ikleyn:
Answer by greenestamps(13334)   (Show Source):
You can put this solution on YOUR website!

The responses from the other tutors show solutions using an equation that says the distances the two directions are the same.

A different setup makes the algebra needed to solve the problem easier.

Since the distances are the same and the time returning is half the time going, the speed returning is twice the speed going.

Let the boat speed by x; then the speed going is x-6 and the speed returning is x+6:

ANSWER: 18 km/hr

Answer by timofer(155)   (Show Source):
You can put this solution on YOUR website!
Same distance each way.
If b is for the boat speed when no current, then
upstream distance
downstream, same distance

-----------------------------speed of boat without a current

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
Jane took 30 min to drive her boat upstream to water-ski at her favorite spot.
Coming back later in the& day, at the same boat speed, took her 15 min.
If the current in that part of the river is 6 km per hr, what was her boat speed in still water?
~~~~~~~~~~~~~~~~~~~~~

Let boat speed in still water be x The speed of the current = 6 km/h upstream speed = (x-6) km/h. downstream speed = (x+6) km/h distance = time * speed Distance upstream is 1/2 of an hour * (x-6) km/h d = kilometers. Distance downstream is 1/4 of an hour * (x+6) km/h d = kilometers. The distance is the same in both directions - - - so, we have this equation = (1) Multiply equation (1) by 4 (both sides). We get 2(x-6) = x + 6, 2x - 12 = x + 6, 2x - x = 12 + 6, x = 18. ANSWER. The speed of boat in still water is 18 km/h.

Question 1031863: frank went 16 miles at one speed and then caame back going 4mph faster. if the return trip took 40mins less time find the 2 speeds.
Found 3 solutions by josgarithmetic, n2, ikleyn:
Answer by josgarithmetic(39799)   (Show Source):
You can put this solution on YOUR website!

SPEEDS TIME DISTANCE WENT r 16/r 16 BACK r+4 16/(r+4) 16 DIFFERENCE 2/3 40 minutes is hour.

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Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
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frank went 16 miles at one speed and then came back going 4mph faster. if the return trip took 40 mins
less time find the 2 speeds.
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Let x be the rate going to there, in miles per hour. Then the rate going back is (x+4) miler per hour. The time going at one speed is hours. The time going back is hours. The time equation is - = . To solve, multiply both sides by LCD 3x*(x+4). You will get 48(x+4) - 48x = 2x(x+4) 192 = 2x(x+4) 96 = x(x+4) x^2 + 4x - 96 = 0 (x+12)*(x-8) = 0 The roots are -12 and 8. We reject negative root and accept the positive one x = 8. ANSWER. The speeds are 8 mph (going to there) and 12 mph (going back). CHECK. The time going to there is = 2 hours. The time going back is = = = 1 hours. The difference is of an hour, which is PRECISELY CORRECT.

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
frank went 16 miles at one speed and then came back going 4mph faster. if the return trip took 40 mins
less time find the 2 speeds.
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        The solution in the post by @mananth is incorrect.
        It has an arithmetic error on the way, which leads to wrong answer.
        I came to bring a correct solution.

Let x be the rate going to there, in miles per hour. Then the rate going back is (x+4) miler per hour. The time going at one speed is hours. The time going back is hours. The time equation is - = . To solve, multiply both sides by LCD 3x*(x+4). You will get 48(x+4) - 48x = 2x(x+4) <<<---=== note it is DIFFERENT equation than that in the post by @mananth 192 = 2x(x+4) 96 = x(x+4) x^2 + 4x - 96 = 0 (x+12)*(x-8) = 0 The roots are -12 and 8. We reject negative root and accept the positive one x = 8. ANSWER. The speeds are 8 mph (going to there) and 12 mph (going back). CHECK. The time going to there is = 2 hours. The time going back is = = = 1 hours. The difference is of an hour, which is PRECISELY CORRECT.

Solved correctly.

Question 1155527: two brothers leave their house at the same time. alex runs due east at a constant speed of 10km per hour, and Boris walks due south at a constant speed of 4km per hour. How far has each brother travelled after 30 minutes?

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
two brothers leave their house at the same time. alex runs due east at a constant speed of 10km per hour,
and Boris walks due south at a constant speed of 4km per hour. How far has each brother travelled after 30 minutes?
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Regarding the solution in the post by @mananth, I'd like to notice that the problem
does not require to calculate the distance between the brothers after 30 minutes.

So, this calculation is unnecessary and performed by @mananth without any real necessity.

Question 1179660: A motorboat traveling with the current went 42 mi in 3.5 h. Traveling against the current, the boat went 18 mi in 3 h. Find the rate of the boat in calm water and the rate of the current.
Found 4 solutions by josgarithmetic, greenestamps, n2, ikleyn:
Answer by josgarithmetic(39799)   (Show Source):
You can put this solution on YOUR website!
ust a different example of one of the frequent forms of travel-problems.

r, speed of boat if no current
c, speed of current
D, the large distance in T, the large time
d, the small distance in t, the small time
The unknown variables for this example are r and c.
The other variables are given.

add corresponding parts

subtract corresponding parts

Problem description gives D=42, T=3.5, d=18, and t=3.
Using these to evaluate r and c,

and

Answer by greenestamps(13334)   (Show Source):
You can put this solution on YOUR website!

After using the given information to find that the speed with the current is 12 mph and the speed against the current is 6 mph, the other tutors solved the problem with formal algebra by using a pair of equations involving the boat speed and the current speed.

Of course, a formal algebraic solution is possibly what was required.

However, this kind of problem is so common that, if formal algebra is not required, it can be solved informally with very little effort using logical reasoning.

If the current speed added to the boat speed is 12 mph and the current speed subtracted from the boat speed is 6 mph, then logical reasoning says that the boat speed is halfway between 12 mph and 6mph -- i.e., 9 mph -- and the current speed is then the difference between 9 mph and either 6 mph of 12 mph.

ANSWERS:
boat speed: 9 mph (halfway between 6 mph and 12 mph)
current speed: 12-9 = 3 mph; or 9-6 = 3 mph

Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
.
A motorboat traveling with the current went 42 mi in 3.5 h. Traveling against the current, the boat went 18 mi in 3 h.
Find the rate of the boat in calm water and the rate of the current.
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Let x be the rate of the motorboat in still water (in miles per hour) and y be the rate of the current (in the same units). Then the effective rate of the motorboat downstream is x + y and the effective rate of the motorboat upstream is x - y. From the problem, the effective rate of the motorboat downstream is the distance of 42 miles divided by the time of 3.5 hours = 12 mph. The effective rate of the motorboat upstream is the distance of 18 miles divided by the time of 3 hours = 6 mph. So, we have two equations to find 'k' and 'c' x + y = 12, (1) x - y = 6. (2) To solve, add equations (1) and (2). The terms 'y' and '-y' will cancel each other, and you will get 2x = 12 + 6 = 18 ---> x = 18/2 = 9. Now from equation (1) v = 12 - u = 12 - 9 = 3. ANSWER. The rate of the motorboat in still water is 9 mph. The rate of the current is 3 mph km/h.

Solved.

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
A motorboat traveling with the current went 42 mi in 3.5 h. Traveling against the current, the boat went 18 mi in 3 h.
Find the rate of the boat in calm water and the rate of the current.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the rate of the motorboat in still water (in miles per hour) and y be the rate of the current (in the same units). Then the effective rate of the motorboat downstream is x + y and the effective rate of the motorboat upstream is x - y. From the problem, the effective rate of the motorboat downstream is the distance of 42 miles divided by the time of 3.5 hours = 12 mph. The effective rate of the motorboat upstream is the distance of 18 miles divided by the time of 3 hours = 6 mph. So, we have two equations to find 'k' and 'c' x + y = 12, (1) x - y = 6. (2) To solve, add equations (1) and (2). The terms 'y' and '-y' will cancel each other, and you will get 2x = 12 + 6 = 18 ---> x = 18/2 = 9. Now from equation (1) v = 12 - u = 12 - 9 = 3. ANSWER. The rate of the motorboat in still water is 9 mph. The rate of the current is 3 mph km/h.

Solved.

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This solution produces the same answer as in the post by @mananth, but has an advantage
that it does not contain excessive calculations that the solution by @mananth has.

We, the tutors, write here our solutions not only to get certain numerical answer.
We write to teach - and, in particular, to teach solving in a right style.

This style solving presented in my post, is straightforward with no logical loops.
The solution presented in the post by @mananth has two logical loops.

One loop in the @mananth post is writing 42/(x+y) = 3.5 ---> divide by 3.5 12/(x+y) = 1 ---> x+y = 12, while in my solution I simply write for the effective rate x + y = 42/3.5 = 12. Second loop in the @mananth post is writing 18/(x-y) = 3 ---> divide by 3 6/(x-y) = 1 ---> x-y = 6, while in my solution I simply write for the effective rate upstream x - y = 18/3 = 6.

It is why I present my solution here and why I think it is better than the solution by @mananth:
- because it teaches students to present their arguments in a straightforward way, without logical zigzags.

@mananth repeats his construction of solution with no change for all similar problems on floating
with and against the current simply because his COMPUTER CODE is written this way.
But this way is not pedagogically optimal - in opposite, it is pedagogically imperfect.

Question 1179659: Flying with the wind, a plane flew 1,120 mi in 4 h. Against the wind, the plane required 7 h to fly the same distance. Find the rate of the plane in calm air and the rate of the wind.
Found 2 solutions by n2, ikleyn:
Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
.
Flying with the wind, a plane flew 1,120 mi in 4 h.
Against the wind, the plane required 7 h to fly the same distance.
Find the rate of the plane in calm air and the rate of the wind.
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Let u be the rate of the plane at no wind (in miles per hour) and v be the rate of the wind (in the same units). Then the effective rate of the plane with the wind is u + v and the effective rate of the plane against the wind is u - v. From the problem, the effective rate of the plane with the wind is the distance of 1120 miles divided by the time of 4 hours {{1120/4}}} = 280 mph. The effective rate of the plane against the wind is the distance of 1120 miles divided by the time of 7 hours = 160 mph. So, we have two equations to find 'u' and 'v' u + v = 280, (1) u - v = 160. (2) To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get 2u = 280 + 160 = 440 ---> u = 440/2 = 220. Now from equation (1) v = 280 - u = 280 - 220 = 60. ANSWER. The rate of the plane in still air is 220 mph. The rate of the wind is 60 mph km/h.

Solved.

Answer by ikleyn(53763)   (Show Source):
You can put this solution on YOUR website!
.
Flying with the wind, a plane flew 1,120 mi in 4 h.
Against the wind, the plane required 7 h to fly the same distance.
Find the rate of the plane in calm air and the rate of the wind.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let u be the rate of the plane at no wind (in miles per hour) and v be the rate of the wind (in the same units). Then the effective rate of the plane with the wind is u + v and the effective rate of the plane against the wind is u - v. From the problem, the effective rate of the plane with the wind is the distance of 1120 miles divided by the time of 4 hours {{1120/4}}} = 280 mph. The effective rate of the plane against the wind is the distance of 1120 miles divided by the time of 7 hours = 160 mph. So, we have two equations to find 'u' and 'v' u + v = 280, (1) u - v = 160. (2) To solve, add equations (1) and (2). The terms 'v' and '-v' will cancel each other, and you will get 2u = 280 + 160 = 440 ---> u = 440/2 = 220. Now from equation (1) v = 280 - u = 280 - 220 = 60. ANSWER. The rate of the plane in still air is 220 mph. The rate of the wind is 60 mph km/h.

Solved.

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My solution produces the same answer as in the post by @mananth, but makes it in clean, clear and correct form,
while calculations in the post by @mananth contradict to each other literally in every line.

We, the tutors, write here our solutions not only to get certain numerical answer.
We write to teach - and, in particular, to teach writing solutions in accurate and clear style.

@mananth repeats his construction of solution with no change for all similar problems on flies
with and against the wind simply because his COMPUTER CODE is written this way.
But his computer code is written in a bad style. It is simply DEFECTIVE.
But @mananth does not change it just during several ( five or seven or even more ) years.
My impression of his writing is that he does not read what his code produces, because it is not interesting to him.

To me, it is monstrous style posting to the forum.

Question 166202: a plane which can fly 520mph in still air flies for 3 hours against a wind and for 2 hours with the same wind. The total distance it covers is 2565 miles. find rate of the wind.
Answer by MathTherapy(10810)   (Show Source):
You can put this solution on YOUR website!

a plane which can fly 520mph in still air flies for 3 hours against a wind and for 2 hours with the same wind. The total distance it covers is 2565 miles. find rate of the wind. ====================================== Speed, in still air: 520 mph Let wind-speed be W Average speed, against the wind: 520 - W Average speed, with the wind: 520 + W Distance travelled against the wind, in 3 hours: 3(520 - W) = 1,560 - 3W Distance travelled with the wind, in 2 hours: 2(520 + W) = 1,040 + 2W With the TOTAL DISTANCE travelled being 2,565 miles, we get the following TOTAL DISTANCE equation: 1,560 - 3W + 1,040 + 2W = 2,565 - 3W + 2,600 + 2W = 2,565 - 3W + 2W = 2,565 - 2,600 - W = - 35 Speed of wind, or

Question 1179658: A plane traveling with the wind flew 1312.5 mi in 5.25 h. Against the wind, the plane required 6.25 h to fly the same distance. Find the rate of the plane in calm air and the rate of the wind.
Answer by n2(82)   (Show Source):
You can put this solution on YOUR website!
.
A plane traveling with the wind flew 1312.5 mi in 5.25 h.
Against the wind, the plane required 6.25 h to fly the same distance.
Find the rate of the plane in calm air and the rate of the wind.
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Let x be the rate of the plane in calm air, in miles per hour. Let y be the rate of the wind. Then the effective speed of the plane with the wind is (x+y) mph, while the effective speed of the plane against the wind is (x-y) mph, From the problem, the effective rate of the plane with the wind is the distance divided by the travel time = 250 miles per hour, and the effective rate of the plane against the wind is the distance divided by the travel time = 210 miles per hour. So, we have these two equations x + y = 250 (1) x - y = 210 (2) To find x, add the equation. You will get 2x = 250 + 210 = 460, x = 460/2 = 230. Now express y from equation (1) and calculate y = 250 - x = 250 - 230 = 20. At this point the problem is solved completely. ANSWER. The speed of the plane in calm air is 230 mph. The rate of the wind is 20 mph.

Solved.