# Intermediate Algebra Problems

• Jun 20th 2011, 08:08 AM
Aravsion
Intermediate Algebra Problems
Can someone pleace explain how to solve these problems?
The problem statement, all variables and given/known data
Sally can paint a room in 9 hours while it takes Steve 3 hours to paint the same room. How long would it take them to paint the room if they worked together?
The problem statement, all variables and given/known data
One hose can fill a goldfish pond in 18 minutes, and two hoses can fill the same pond in 14 minutes. Find how long it takes the second hose alone to fill the pond.
• Jun 20th 2011, 08:17 AM
Prove It
Re: Intermediate Algebra Problems
1. Set these up as ratios "How much of the room painted : Hours".

Sally
$\displaystyle \displaystyle 1 : 9 = \frac{1}{9} : 1$

Steve
$\displaystyle \displaystyle 1 : 3 = \frac{1}{3} : 1$

Together
\displaystyle \displaystyle \begin{align*} \frac{1}{9} + \frac{1}{3} &: 1 \\ \frac{4}{9} &: 1 \\ 4 &: 9 \\ 1 &: \frac{9}{4} \end{align*}.

So together they will take $\displaystyle \displaystyle 2\frac{1}{4}$ hours.
• Jun 20th 2011, 08:33 AM
IBstudent
Re: Intermediate Algebra Problems
For the second one, lets call the hoses hose A and hose B. A takes 18 minutes so the ratio is 18:1. A+B would be 14:1, so 1/14-1/18=1/63. Hence it is 63 minutes...
• Jun 20th 2011, 08:33 AM
HallsofIvy
Re: Intermediate Algebra Problems
A slightly different point of view: when people or things "work together", their rates of work add.

"Sally can paint a room in 9 hours"
so Sally's rate of work is 1/9 "room per hour".
"it takes Steve 3 hours to paint the same room"
so Steve's rate of work is 1/3 "room per hour".

Together they work at $\displaystyle \frac{1}{9}+ \frac{1}{3}= \frac{1}{9}+ \frac{3}{9}= \frac{4}{9}$ "room per hour". Invert that to get "hours per room".

"One hose can fill a goldfish pond in 18 minutes"
so that hose works at 1/18 "pond per minute". Since we want to find how long it would take the second hose to fill the pond, call that time "x" minutes. The
second hose works at 1/x "pond per minute".
" and two hoses can fill the same pond in 14 minutes."
so $\displaystyle \frac{1}{18}+ \frac{1}{x}= \frac{1}{14}$

Solve that for x. I recommend multiplying both sides of the equation by the "least common denominator", 2(7)(9)x= 126x.
• Jun 20th 2011, 09:31 AM
Aravsion
Re: Intermediate Algebra Problems
thanks everybody for the explanations! i understand it now!