Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    question is attached.
    answer in the back of the textbook is 553.6 kmph
    dont know how to work it out, help would be so much appreciated!
    Attached Images
     
    • Study Helper
    Offline

    0
    ReputationRep:
    (Original post by wrightarya)
    question is attached.
    answer in the back of the textbook is 553.6 kmph
    dont know how to work it out, help would be so much appreciated!
    If the speed is x km/h then the time taken is \frac{1000}{x} hours.

    If the speed is x-120 then the time taken is \frac{1000}{x-120} hours and this is equal to the other time plus 0.5 hours.

    Write and solve an equation.
    • Thread Starter
    Offline

    0
    ReputationRep:
    i tried solving it like this:
    1000/x-120 + 1/2= 1000/x
    1000/x-120 = 2000 + x
    1000= (2000+x) (x-120)
    1000= x^2 + 1880x - 240000
    = x^2 + 1880x - 241000
    substituted into quadratic formula
    x=120 or -2000

    dont know if i did that right, but its not what the answer says in the back of the book...
    • TSR Support Team
    • Clearing and Applications Advisor
    Offline

    20
    ReputationRep:
    Spoiler:
    Show

    let d = distance
    let t = time
    let u = initial speed (what we're after)

    d = 1000km = ut

    u = \frac{1000}{t}

    then we reduce u by 120km/h and increase t by half an hour

    d = (u-120)(t+0.5)

    1000 = (\frac{1000}{t}-120)(t+0.5)

    1000 = 1000+\frac{500}{t}-120t-60

    0 = \frac{500}{t}-120t-60

    multiply t to get rid of the fraction

    0 = 120t^2+60t-500

    divide through by 20

    0 = 6t^2+3t-25

    solving the quadratic gets you t = 1.8065 (5 s.f) (I'm discarding the negative result because obviously time cannot be negative)

    then we just substitute our t value to get the answer.

    u = \frac{1000}{1.8065}

    u = 553.56km/h


    Hope that helps.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by The Financier)
    Spoiler:
    Show

    let d = distance
    let t = time
    let u = initial speed (what we're after)

    d = 1000km = ut

    u = \frac{1000}{t}

    then we reduce u by 120km/h and increase t by half an hour

    d = (u-120)(t+0.5)

    1000 = (\frac{1000}{t}-120)(t+0.5)

    1000 = 1000+\frac{500}{t}-120t-60

    0 = \frac{500}{t}-120t-60

    multiply t to get rid of the fraction

    0 = 120t^2+60t-500

    divide through by 20

    0 = 6t^2+3t-25

    solving the quadratic gets you t = 1.8065 (5 s.f) (I'm discarding the negative result because obviously time cannot be negative)

    then we just substitute our t value to get the answer.

    u = \frac{1000}{1.8065}

    u = 553.56km/h


    Hope that helps.
    Thank You so so so much!
    if you dont mind, i also worked out another way to do it, sort of like how i tried to do it in the previous post.

    i did

    1000/x-120 = 1000/x + 1/2
    1000/x-120 = 2000+x/2x
    2000x = (2000+x)(x-120)
    and i solved for x which this time (accidentally) gave me x=553.56km/h

    im not sure why the 1/2 had to be added on to the 1000/x but not on the 1000/x-120
    would you know why this is so?

    thanks again
    • Study Helper
    Offline

    7
    ReputationRep:
    (Original post by wrightarya)
    i tried solving it like this:
    1000/x-120 + 1/2= 1000/x
    1000/x-120 = 2000 + x
    1000= (2000+x) (x-120)
    1000= x^2 + 1880x - 240000
    = x^2 + 1880x - 241000
    substituted into quadratic formula
    x=120 or -2000

    dont know if i did that right, but its not what the answer says in the back of the book...
    You have put the +\frac{1}{2} on the wrong side of the equation
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by The Financier)
    ..
    Please don't post full solutions. It's against the rules of the maths forum plus it would have been much more beneficial to point out the mistake in the OPs working.
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by wrightarya)
    Thank You so so so much!
    if you dont mind, i also worked out another way to do it, sort of like how i tried to do it in the previous post.

    i did

    1000/x-120 = 1000/x + 1/2
    1000/x-120 = 2000+x/2x
    2000x = (2000+x)(x-120)
    and i solved for x which this time (accidentally) gave me x=553.56km/h

    im not sure why the 1/2 had to be added on to the 1000/x but not on the 1000/x-120
    would you know why this is so?

    thanks again
    The 1000/(x-120) time is 0.5 hrs more than the 1000/x time.

    So 1000/(x-120) = (1000/x) + 0.5
    • TSR Support Team
    • Clearing and Applications Advisor
    Offline

    20
    ReputationRep:
    (Original post by notnek)
    Please don't post full solutions. It's against the rules of the maths forum plus it would have been much more beneficial to point out the mistake in the OPs working.
    My apologies. I thought the spoiler tag would indicate it would be a full solution (and so would be seen as a last resort) and it seemed like it was only just an expansion of what BabyMaths had originally posted. I'll bear this in mind in future.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by notnek)
    The 1000/(x-120) time is 0.5 hrs more than the 1000/x time.

    So 1000/(x-120) = (1000/x) + 0.5
    I have attached a very similar question to the one I originally posted.
    and i worked out the solution by making this equation:

    40/x = 40/(x+40) + 1/3
    and solved for x to get
    x= 52.11 which is the correct answer.

    im not sure why in this question the 1/3 hour is added on to 40/(x+40)
    but for the question I originally posted, the 1/2 is added onto the 1000/5 instead of the 1000/(x-120)

    could anyone tell me why this is so? thanks
    Attached Images
     
    • TSR Support Team
    • Study Helper
    Offline

    20
    ReputationRep:
    (Original post by wrightarya)
    I have attached a very similar question to the one I originally posted.
    and i worked out the solution by making this equation:

    40/x = 40/(x+40) + 1/3
    and solved for x to get
    x= 52.11 which is the correct answer.

    im not sure why in this question the 1/3 hour is added on to 40/(x+40)
    but for the question I originally posted, the 1/2 is added onto the 1000/5 instead of the 1000/(x-120)

    could anyone tell me why this is so? thanks
    The main difference here is that your x in the first question is the faster speed and x in the second question is the slower speed.

    If in the second question, you called the speed of Jack x instead of Mark then you'll find that your equations for both questions are similar.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.