Results 1 to 5 of 5

Math Help - Word problems (pun intended)

  1. #1
    Newbie
    Joined
    Sep 2009
    Posts
    7

    Word problems (pun intended)

    I'm having issues setting up a correct equation to solve these 2 problems. I have the answers to the problems I'm just not sure how to get there.

    Beth can walk six miles in the same time she can bike 32 miles. If her biking rate is one more than five
    times her walking rate, how fast does she walk?

    Zack is standing near an 8-foot street lamp. His distance from the base of the lamppost is two feet more
    than his height. The length of his shadow is four times his height. How tall is Zack?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7
    Quote Originally Posted by Twan View Post
    I'm having issues setting up a correct equation to solve these 2 problems. I have the answers to the problems I'm just not sure how to get there.

    Beth can walk six miles in the same time she can bike 32 miles. If her biking rate is one more than five
    times her walking rate, how fast does she walk?

    Zack is standing near an 8-foot street lamp. His distance from the base of the lamppost is two feet more
    than his height. The length of his shadow is four times his height. How tall is Zack?
    Let Beth's walking rate be x.
    Her biking rate is 5x+1.

    \frac{6}{x}=\frac{32}{5x+1}

    \frac{3}{x}=\frac{16}{5x+1}

    16x=15x+3

    x=3

    ---------------------------------------------------------------------------------------------------------------------------------------------

    Refer diagram.

    AB is Zack.
    LO is lamp.
    BS is Zack's shadow.

    \frac{AB}{LO}=\frac{BS}{OS}

    Let Zack's height be h feet.

    \frac{h}{8}=\frac{4h}{4h+2+h}

    \frac{h}{8}=\frac{4h}{5h+2}

    \frac{1}{8}=\frac{4}{5h+2}

    5h+2=32

    5h=50

    h=6 feet

    Zack is 6 feet tall.
    Attached Thumbnails Attached Thumbnails Word problems (pun intended)-zack.jpg  
    Last edited by alexmahone; November 11th 2009 at 02:55 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2009
    Posts
    7
    Ah I see now, thank you.

    edit: the proportion i get with that problem leaves me with an exponent and i dunno how to solve that one out.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2009
    Posts
    7
    ..

    so glad my major doesn't entail a lot of math
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,551
    Thanks
    542
    Hello, Twan!

    Beth can walk six miles in the same time she can bike 32 miles.
    If her biking rate is one more than five times her walking rate, how fast does she walk?

    We will use: . \text{Distance} \:=\:\text{Rate} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{R}

    Let: . \begin{array}{ccc} r &=& \text{walking rate} \\ 5r+1 &=& \text{biking rate} \end{array}

    She walks 6 miles at r mph.
    . . This takes her: . \frac{6}{r} hours.

    She bikes 32 miles at 5r+1 mph.
    . . This takes her: . \frac{32}{5r+1} hours.

    These times are equal: . \frac{6}{r} \:=\:\frac{32}{5r+1}

    Now solve for r.



    Zack is standing near an 8-foot street lamp.
    His distance from the base of the lamppost is two feet more than his height.
    The length of his shadow is four times his height.
    How tall is Zack?
    Code:
        C *
          |   *
          |       *   A
          |           o
        8 |           |   *
          |          h|       *
          |           |           *
          * - - - - - o - - - - - - - *
          D   h+2     B       4h        E

    Zack is AB = h.

    The street lamp is CD = 8.

    We are told that: . DB \:=\:h+2,\;BE \:=\:4h.
    We see that: DE \:=\:5h+2

    From the similar right triangles: . \frac{DE}{CD} \,=\,\frac{BE}{AB} \quad\Rightarrow\quad \frac{5h+2}{8} \:=\:\frac{4h}{h}

    We have: . \frac{5h+2}{8} \:=\:4 \quad\Rightarrow\quad 5h+2 \:=\:32 \quad\Rightarrow\quad 6h \:=\:30

    Therefore: . h \:=\:6 . . . Zack is 6 feet tall.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Word problems
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 6th 2011, 02:09 PM
  2. Word Problems
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 10th 2011, 02:17 PM
  3. Problems with integration word problems
    Posted in the Calculus Forum
    Replies: 5
    Last Post: April 25th 2010, 05:39 PM
  4. Matlab not returning variable as intended
    Posted in the Math Software Forum
    Replies: 7
    Last Post: April 18th 2010, 03:35 PM
  5. Check please! (pun intended)
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: April 9th 2009, 11:08 AM

Search Tags


/mathhelpforum @mathhelpforum