# Thread: Word problems (pun intended)

1. ## 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?

2. Originally Posted by Twan
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.

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

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.

3. 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.

4. ..

so glad my major doesn't entail a lot of math

5. 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.