# Word problems

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• Jun 11th 2008, 02:21 PM
King_nic
Word problems
1) Person A runs around a circular track in 40 seconds. He meets person B coming the other way every 15 seconds. How fast does Person B run around the track in seconds?

2) Five kilometers upstream from his starting point,
a rower passed a raft flowing with the current.
He rowed upstream for one more hour and then rowed back
and reached his starting point at the same time as the raft.
Find the speed of the current.

(Edit: I could not reconstruct the escalator problem. - Dan)
• Jun 11th 2008, 04:02 PM
TheEmptySet
Quote:

Originally Posted by King_nic
1) Person A runs around a circular track in 40 seconds. He meets person B coming the other way every 15 seconds. How fast does Person B run around the track in seconds?

If anyone could post the steps in finishing these, it would be appreciated.

Since the track is circular the distance around it is $2\pi r$ units

So runner A runs $\frac{2 \pi r}{40}=\frac{\pi r}{20}$ units per second.

So in 15 seconds he runs $15 \cdot \frac{\pi r}{20}=\frac{3 \pi r}{4}$ units so runner B ran $\frac{5 \pi r}{4}$ units in 15 seconds

$\frac{5 \pi r}{4}=15 s$ now multiply both sides by $\frac{8}{5}$ to get $2 \pi r = 24s$

I hope this helps.
• Jun 11th 2008, 04:11 PM
Soroban
Hello, King_nic!

Quote:

1) Person A runs around a circular track in 40 seconds.
He meets person B coming the other way every 15 seconds.
How fast does Person B run around the track in seconds?

$A$ runs 360° in 40 seconds.
. . He runs: $\frac{360}{40} = 9$ degrees per second.

$B$ runs 360° in $x$ seconds.
. . He runs: $\frac{360}{x}$ degrees per second.

They approach each other at a combined speed of: . $9 + \frac{360}{x}$ degrees per second.

Together, they cover 360° in: . $\frac{360}{9 + \frac{360}{x}} \,=\,\frac{360x}{9x + 360}$ seconds.

But we are told that this happens every 15 seconds.

There is our equation! . . . $\frac{360x}{9x+360} \:=\:15$

. . $360x \:=\:135x + 5400\quad\Rightarrow\quad 225x \:=\:5400\quad\Rightarrow\quad x \:=\:24$

Therefore, $B$ runs a lap in 24 seconds.

Edit: Too fast for me, $\emptyset$ !
• Jun 11th 2008, 04:14 PM
King_nic
Thanks
Thanks alot for the posts guys, just two more to unravel
• Jun 11th 2008, 06:17 PM
Soroban
Hello again, King_nic!

Quote:

2) Five kilometers upstream from his starting point,
a rower passed a raft flowing with the current.
He rowed upstream for one more hour and then rowed back
and reached his starting point at the same time as the raft.
Find the speed of the current.

Let $b$ = speed of the boat in still water (km/hr).
Let $c$ = speed of the current (km/hr).
Code:

      C    b-c    B        5          A       * - ← - ← - * - ← - ← - ← - ← - - *       * - → - → - * - → - → - → - → - - *                   * - → - → - → - → - - *                   B          5          A

Going upstream, the boat's speed is $(b-c)$ km/hr.
It started at $A$, went 5 km to $B$, where it met the floating log.
It continued upstream for an hour, traveling $(b-c)$ km to $C.$

Going downstream, the boat's speed is $(b+c)$ km/hr.
It went downstream $(b-c)$ km to $B.$
. . This took: . $\frac{b-c}{b+c}$ hours.
Then it went 5 km to $A.$
. . This took: . $\frac{5}{b+c}$ hours.
Since it met the log, the boat traveled for: . $1 + \frac{b-c}{b+c} + \frac{5}{b+c}$ hours. .[1]

During this same time, the log went 5 km at $c$ km/hr.
. . This took: . $\frac{5}{c}$ hours. .[2]

Equate [1] and [2]: . $1 + \frac{b-c}{b+c} + \frac{5}{b+c} \;=\;\frac{5}{c}$

Multiply by $c(b-c)(b+c)\!:\;\;c(b-c)(b+c) + c(b-c)^2 + 5c(b-c) \;=\;5(b-c)(b+c)$

. . which simplifies to: . $2b^2c - 2bc^2 + 5bc - 5b^2 \:=\:0$

Since $b \neq 0$, divide by $b\!:\;\;2bc - 2c^2 - 5b + 5c\:=\:0$

Factor: . $2c(b - c) -5(b-c) \:=\:0$

Factor: . $(b-c)(2c-5) \:=\:0$

If $b-c \:=\:0$, then: . $b = c$
The speed of the current equals the speed of the boat.
. . Then the boat could not have gone upstream at all.

Therefore: . $2c - 5 \:=\:0 \quad\Rightarrow\quad c \:=\:\frac{5}{2}$

The speed of the current is 2.5 km/hr.

• Jun 12th 2008, 06:37 AM
galactus
For the escalator problem we could let s=the number of steps in the escalator standing still. It is moving against the woman so she moves s-28 steps. Let r be her rate down, so we have:

$\frac{s-28}{r}=16$

and

$\frac{s-22}{r}=24$

Solve the sytem and we see s=40 steps.
• Jun 12th 2008, 08:45 PM
mr fantastic
Quote:

Originally Posted by King_nic
Thank you for the help guys.
Last edited by King_nic : Today at 07:43 PM.

Please don't delete questions once they've been answered. The questions will be useful to others.
• Jun 13th 2008, 03:24 AM
galactus
Yes, why did you delete them?. They were fun little algebra problems. A wee bit tougher than most of those kinds of problems.