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Math Help - Some Problems...

  1. #1
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    Some Problems...

    Problem 1
    A commercial jet took off from city 1 for city 2, a distance of 2550 km, at a speed of 800 km/h. At the same time a private jet, traveling 900 km/h, left City 2 for City 1. How long after takeoff will the jet pass each other?

    Problem 2
    Wendy took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h and the train 60 mi/h. The entire trip took 5 and a half hour. How long did Wendy spend on the train?

    Problem 3
    Kiran drove from Tortula to Cactus, a distance of 250 mi. She increaseed her speed by 10 mi/h for 360-mi trip from cactus to Dry Junction. If the total trip took 11 hr, what was her speed from Tortula to Cactus?

    Problem 4
    A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 mins more time than the first leg, how fast was he driving between Ajax and Barrington

    sidenote:
    mi = miles

    reply fast please!!
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  2. #2
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    Quote Originally Posted by DrSimple View Post
    its very urgent please!
    bumping is against the rules. don't rush, you should have known better than to post last minute.

    it seems obvious to me that this is for an assignment or something, so i won't give you full solutions. that would be cheating. anyway

    Quote Originally Posted by DrSimple View Post
    Problem 1
    A commercial jet took off from city 1 for city 2, a distance of 2550 km, at a speed of 800 km/h. At the same time a private jet, traveling 900 km/h, left City 2 for City 1. How long after takeoff will the jet pass each other?
    speed = distance/time. this means time = distance/speed

    now, lets say they pass each other at a distance x from city 1. so the 800km/h plane travels x km. that would mean the 900km/h plane traveled (2550 - x) km.

    so, for the 800-plane, time = x/800

    for the 900-plane, time = (2550 - x)/900

    this is a pair of simultaneous equations with two unknowns. you can solve for time

    do that while i work on the others.
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  3. #3
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    sorry for bumping, wont happen again, i dont know what u mean with simaltanous equations, could u put it really simple please?
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    Quote Originally Posted by DrSimple View Post
    Problem 2
    Wendy took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h and the train 60 mi/h. The entire trip took 5 and a half hour. How long did Wendy spend on the train?
    again, speed = distance/time, so time = distance/speed

    lets say Wendy traveled x mi by bus. then she traveled (300 - x) mi by train.

    the time Wendy spent on the bus is therefore, t_1 = x/40

    the time Wendy spent on the train is, t_2 = (300 - x)/60

    and we know that t_1 + t_2 = 5.5

    so now we can find x using the last equation, then we will be able to find t_2, the time by train
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DrSimple View Post
    sorry for bumping, wont happen again, i dont know what u mean with simaltanous equations, could u put it really simple please?
    time = x/800 and time = (2550 - x)/900, so it means x/800 = (2550 - x)/900

    solve for x then you can find time
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  6. #6
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    I dont get his, its to hard..i was hoping for answers so i would understand how ur getting this all , taking my time.
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    Quote Originally Posted by DrSimple View Post
    Problem 3
    Kiran drove from Tortula to Cactus, a distance of 250 mi. She increaseed her speed by 10 mi/h for 360-mi trip from cactus to Dry Junction. If the total trip took 11 hr, what was her speed from Tortula to Cactus?
    and yet again, we will use speed = distance/time

    let s be Kiran's average speed from Tortula to Cactus. and let t_1 be the time for the trip. so,

    s = 250/t_1 ....................(1)

    she increased her average speed by 10 to drive to Dry Junction, so her speed is (s + 10). and so,

    s + 10 = 360/t_2 ..............(2)

    you can solve those for t_1 and t_2. then set t_1 + t_2 = 11. then you can solve for s and find the required answer from that
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    Quote Originally Posted by DrSimple View Post
    I dont get his, its to hard..i was hoping for answers so i would understand how ur getting this all , taking my time.
    i'm sorry, but i cannot give you what you're hoping for. i told you why. i have already done more for you than i should. i gave you the equations you must use, all you have to do is solve them. you can literally forget about the word problem and just treat it as solving equations. i cannot give you any more help than this
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    Quote Originally Posted by DrSimple View Post
    arent you only making this harder for me..lol
    i think i would make it harder for you by not answering at all, wouldn't you agree? just solve the equations
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  10. #10
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    Problem 3
    Kiran drove from Tortula to Cactus, a distance of 250 mi. She increaseed her speed by 10 mi/h for 360-mi trip from cactus to Dry Junction. If the total trip took 11 hr, what was her speed from Tortula to Cactus?

    Let Kiran's usual speed from Tortula to Cactus = x mi/h

    Time taken by Kiran from Tortula to Cactus =  \frac{Distance}{speed} = \frac{250}{x}

    Now, Kiran's new speed after increase (from Cactus to Dry Junction) = (x + 10) mi/h

    Time taken from Cactus to Dry Junction = \frac{360}{x + 10}

    Now, total time = 11 h

    so, \frac{250}{x} + \frac{360}{x + 10} = 11

    Now, solve for x.

    Did you get it now??? If not, pleaase ask agaain.....
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    Quote Originally Posted by DrSimple View Post
    Problem 4
    A salesman drives from Ajax to Barrington, a distance of 120 mi, at a steady speed. He then increases his speed by 10 mi/h to drive the 150 mi from Barrington to Collins. If the second leg of his trip took 6 mins more time than the first leg, how fast was he driving between Ajax and Barrington
    No surprise here, speed = distance/time is again our main formula. (i hope you are picking up the pattern of how to do these now)

    there is a snag in this question though. note that the speed is given with hours as the unit of time, while the trip, they told you about time in minutes. we have to work in one kind of unit. lets change 6 minutes to hours. note that 6 mins = 0.1 hours.

    now, let s be the speed he drove between Ajax and Barrington. then (s + 10) is the speed he drove from Barrington to Collins. let t be the time he took to drive from Ajax to Barrington. then (t + 0.1) is the time he took to drive from Barrington to Collins.

    for the first trip then, s = 120/t .................(1)

    for the second trip, (s + 10) = 150/(t + 0.1) .................(2)

    again, simultaneous equations. you want to solve for s

    and here is where i get off. good luck!
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  12. #12
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    Can you guys give me just the simple equations i need to solve for the questions?

    simple calculations to finish it off?

    please!
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  13. #13
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    DrSimple - Your question should be answered shortly. I know you are new here, but please treat all members, especially staff members with respect.
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  14. #14
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    Quote Originally Posted by DrSimple View Post
    Problem 1
    A commercial jet took off from city 1 for city 2, a distance of 2550 km, at a speed of 800 km/h. At the same time a private jet, traveling 900 km/h, left City 2 for City 1. How long after takeoff will the jet pass each other?
    The two planes are approaching one another at a speed of 1700 km/hr. They start 2550 km apart, how long untill they are 0 km apart?

    RonL
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  15. #15
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    Quote Originally Posted by DrSimple View Post
    Problem 1
    A commercial jet took off from city 1 for city 2, a distance of 2550 km, at a speed of 800 km/h. At the same time a private jet, traveling 900 km/h, left City 2 for City 1. How long after takeoff will the jet pass each other?

    Problem 2
    Wendy took a trip from Davenport to Omaha, a distance of 300 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h and the train 60 mi/h. The entire trip took 5 and a half hour. How long did Wendy spend on the train?

    ...
    to #1:

    The distance between the 2 planes decreases by (900+800)km/h. The distance between the planes is zero after 2550km / 1700 km/h = 1.5 h

    to #2:

    Let x denote the distance travelled by bus and y the distance travelled by train.
    You are supposed to know that

    \frac{distance}{speed} = travelled\ time

    You can set up a system of simultaneous equations:

    \left| \begin{array}{l}x+y=300 \\ \frac x{40} + \frac y{60} = \frac{11}2 \end{array} \right.

    I've got x= 60 mi and y = 240 mi
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