View Full Version : Word problems - perhaps about rate?
zqueez
Apr 11, 2007, 12:44 PM
Machine A and B can do a job in 6/5 hrs.
Machine A and C can do a job in 3/2 hrs.
Machine B and C can do a job in 2 hrs.
How long would it take machines ABC to do the job?
galactus
Apr 11, 2007, 02:53 PM
A way to do it is to solve the system of equations.
A+B=\frac{5}{6}
A+C=\frac{2}{3}
B+C=\frac{1}{2}
Solve for A, B, and C.
Add them up to find the amount of time for all three.
It'll have to be less than the lowest time given, which is 1.2 hours. That's with two machines(A and B) together. 3 will be even faster.
zqueez
Apr 11, 2007, 02:56 PM
May I ask you why you reversed the fractions?
galactus
Apr 11, 2007, 03:15 PM
It is convenient to introduce the part of the job done in one hour.
For instance, suppose a problem said, "two pumps are used to fill a pool. Pump A, by itself, can fill it in 3 hours. Pump B, by itself, can fill it in 4 hours. How long if both work together?."
Well, we could say:
\frac{1}{3}+\frac{1}{4}=\frac{1}{t}
Solve for t=12/7 hours. Working together they fill it in 1 hour and 43 minutes.
Same principle here.
zqueez
Apr 11, 2007, 03:36 PM
thank You Again...you've Been Great!
galactus
Apr 11, 2007, 03:38 PM
You're welcome. Feel free to post any other problems you are stuck on.
zqueez
Apr 11, 2007, 04:10 PM
The answer I got was 1/2 but it says that was incorrect. It says the answer is 1 but I don't know how how I'm doing it wrong :(
vrooje
Apr 11, 2007, 07:13 PM
zqueez:
Try adding up all three equations to form one big equation with 2A + 2B + 2C = ?? (whatever the number ends up being).
Then I would suggest trying to solve for A + B + C and asking yourself if that's the quantity you're looking for (or is it 1/(A+B+C)?)...
galactus
Apr 11, 2007, 07:23 PM
The answer is indeed 1. If you correctly solve the system I gave you and add up the 3 solutions, you will arrive at 1.
Can you solve:
\left[\begin{array}{ccc|c}1&1&0&\frac{5}{6}\\1&0&1&\frac{2}{3}\\0&1&1&\frac{1}{2}\end{array}\right]
Or, if not, solve the first equation for A and sub into the second equation:
A+B=\frac{5}{6}
A=\frac{5}{6}-B
Sub into second equation:
\frac{5}{6}-B+C=\frac{2}{3}
Solve third for C and sub into the second:
B+C=\frac{1}{2}
C=\frac{1}{2}-B
\frac{5}{6}-B+(\frac{1}{2}-B)=\frac{2}{3}
Solve for B=\frac{1}{3}
There's one of them for free. Now, find A and C.