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Math Help - word problem.

  1. #1
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    word problem.

    hi ! im having a difficult time answering this word problem.

    "Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."

    I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???


    Thank you very much !!
    Last edited by mr fantastic; December 1st 2010 at 04:59 AM. Reason: Deleted begging in title.
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  2. #2
    MHF Contributor harish21's Avatar
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    Whats wrong with getting 5 as your answer?

    You can check:

    In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms

    In 5 hours, Ben travels: 4+5+6+7+8=30kms

    that means they were together after 5 hours.
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  3. #3
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    Nope. Ben started 2 hours after
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  4. #4
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    It should be 10 hours. But I don't know how to prove it using arithmetic progression.
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  5. #5
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    Here's what I did. We know these two things about arithmetic progressions:
    a_n = a_1 + (n - 1)d where d is the distance between numbers in the sequence. We also know that:

    S_n = \frac{n}{2}(a_1 + a_n) where S_n is the sum of the first n terms.

    We also know that the distance traveled by the first is just 6t.

    I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
    So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:

    6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))
    6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)

    Hope you see what I did there. After that, rearrange terms so you get:
    t^2 - 9t - 10 = 0

    For that, I get one positive root of t = 10 (also t = -1 which is obviously not useful). Voila! Thanks for the problem, I had fun with that.
    Last edited by Grep; December 1st 2010 at 12:28 AM. Reason: Small clarification
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  6. #6
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    Hello, dugongster!

    Both of you forgot about the two-hour headstart.


    Alvin set out from a certain point and travelled at the rate of 6 kph.
    After Alvin had gone for 2 hours, Ben set on to overtake him and went 4km the first hour,
    5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour.

    After many hours were they together?

    Alvin had a two-hour headstart; he is already 12 km ahead.
    . . In the next \,h hours, he travels another 6h km.
    Alvin has traveled 6h + 12 km in the first \,h hours.

    During the same \,h hours, Ben has travelled:
    . .  4 + 5 + 6 + \hdots + (h+3) \:=\:\dfrac{h(h+7)}{2} km. .**

    The two distances are equal: . \dfrac{h(h+7)}{2} \:=\:6h + 12

    . . h^2 + 7h \:=\:12h + 24 \quad\Rightarrow\quad h^2 - 5h - 24 \:=\:0

    . . (h - 8)(h + 3) \:=\:0 \quad\Rightarrow\quad h \:=\:8,\:-3


    Answer: . \text{8 hours}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    We want the sum of the numbers from \,4 to h\!+\!3.

    The sum of the numbers from \,1 to h\!+\!3 is: . \dfrac{(h+3)(h+4)}{2}

    The sum of the numbers from \,1 to \,3 is: . 1+2+3 \:=\:6


    Hence,the sum of the numbers from \,4 to h\!+\!3 is:

    . . \dfrac{(h+3)(h+4)}{2} - 6 \;\;=\;\; \dfrac{h^2 + 7x + 12 - 12}{2} \;\;=\;\;\dfrac{h(h+7)}{2}

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  7. #7
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    great. thanks to both of you. grep, you rock !!!
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  8. #8
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    Quote Originally Posted by Soroban View Post
    Hello, dugongster!

    Both of you forgot about the two-hour headstart.
    Hi Soroban! With due respect (which is considerable, I might add, you being one of the most helpful and, let's face it, awesome people here), I took into account the 2 hour headstart by using (t - 2) as the time in the arithmetic progression (n = (t - 2)), and starting at 4 ( a_1 = 4).

    The answer is indeed 10 hours from when Alvin sets out. I do like the way you solved it, however. You just had to add back the 2 hours head start to get the total time of 10, so your calculations are essentially correct. I think the head start is part of the time it took for them to meet, though, and should be counted.

    And dugongster, you're quite welcome.
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