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  1. #1
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    word problem

    Two cyclists drive a distance of 15 kilometres. One of the cyclists
    drives
    10 km h faster than the other one and arrives at their destination

    20 minutes earlier. What are the speeds of the cyclists?
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  2. #2
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    Quote Originally Posted by jessmari86 View Post
    Two cyclists drive a distance of 15 kilometres. One of the cyclists
    drives
    10 km h faster than the other one and arrives at their destination

    20 minutes earlier. What are the speeds of the cyclists?
    Solve simultaneously:

    (v + 10)(t - 1/3) = 15 .... (1)

    vt = 15 .... (2)
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    ok so how would i go about doing that because the vt=15 means that v is either 1,15,3,5 and likewise for t but when plugging those numbers into the first equation it does not equal 15 nor can i graph. Also where is the 1/3 from is that for the 20 mins if so why cant you just put t-20 why use a fraction. please help this is really bothering me trying to prepare fopr entrance exams after being out of high school for nearly 9 years is tough. Thank you
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  4. #4
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    Let the time taken by the slower cyclist be x hrs.
    Therefore, the time taken by the faster one is (x - 1/3) hrs.
    The formula for speed is distance/time.
    therefore,
    The speed of the slower cyclist is (15/x) km/h
    The speed of the faster one is [15/(x-1/3)] km/h

    Since the difference between their speeds is 10 km/h,
    the equation is:
    [15/(x-1/3)] - (15/x) = 10

    Solve for x, then calculate their speeds by substituting the value of x into their respective speeds.
    If there are two values, check for both values. The value of x which gives the speed as a negative number should not be considered as an answer.
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  5. #5
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    the fraction 1/3 is used because the time given is 20 MINUTES while the speed has to be calculated in km/hr. therefore, to convert minutes into hours, 20 is divided by 60 ( 1hr= 60 minutes), 20/60 is simplified to give 1/3.
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  6. #6
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    Quote Originally Posted by jessmari86 View Post
    ok so how would i go about doing that because the vt=15 means that v is either 1,15,3,5

    [snip]
    No, it doesn't meant that at all.

    [snip]
    and likewise for t but when plugging those numbers into the first equation it does not equal 15 nor can i graph. Also where is the 1/3 from is that for the 20 mins if so why cant you just put t-20 why use a fraction. please help this is really bothering me trying to prepare fopr entrance exams after being out of high school for nearly 9 years is tough. Thank you
    Units used in the question are km and km/hr. So 20 minutes has to be converted into hours for consistency.

    To solve the two equations I gave you, note that from equation (2) it follows that t = 15/v. Substitute this into equation (1):

    \displaystyle (v + 10)\left( \frac{15}{v} - \frac{1}{3}\right) = 15.

    Expand. Simplify into a quadratic equation for v. Solve for v (only one of the solutions is acceptable). See solve v^2 + 10v - 450 = 0 - Wolfram|Alpha

    and click on Show steps.
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    No, it doesn't meant that at all.


    Units used in the question are km and km/hr. So 20 minutes has to be converted into hours for consistency.

    To solve the two equations I gave you, note that from equation (2) it follows that t = 15/v. Substitute this into equation (1):

    \displaystyle (v + 10)\left( \frac{15}{v} - \frac{1}{3}\right) = 15.

    Expand. Simplify into a quadratic equation for v. Solve for v (only one of the solutions is acceptable). See solve v^2 + 10v - 450 = 0 - Wolfram|Alpha

    and click on Show steps.
    Let s = speed of the slow cyclist
    then
    (s+10) = speed of the faster one
    :
    Change 20 min to hr
    :
    Write a time equation: Time = dist/speed
    :
    slow time - fast time = hr
    - =
    Multiply by 3s(s+10), results:
    3(15)(s-10) - 3(15)s = s(s+10)
    :
    45(s+10) - 45s = s(s+10)
    :
    45s + 450 - 45s = s^2 + 10s
    :
    0 = s^2 + 10s - 450


    Use the quadratic formula to find s:

    In this equation, x=s; a=1, b=10; c=-450

    :


    ;

    :

    Positive solution

    s =
    s = 16.7945 ~ 16.8km/h is the slow cyclist
    and
    26.8 km/h is the fast cyclist
    ;
    :
    Check solution by finding the times
    Slow: 15/16.8 = .89 hrs
    Fast: 15/26.8 = .56 hrs


    and why do you insist on sending infractions to me every time are you really that offended that i questioned your set up? it gets annoying re accepting these rules everytime.
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  8. #8
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    Quote Originally Posted by jessmari86 View Post
    [snip]
    and why do you insist on sending infractions to me every time are you really that offended that i questioned your set up? it gets annoying re accepting these rules everytime.
    When you break the rules you get an Infraction. Avoid getting an Infraction by following these three simple steps:

    Read the rules. Understand the rules. Follow the rules.
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