View Full Version : Help with word problem Linear Algebra

janviee

07-14-2006, 02:38 PM

Hello

I would really appreciate any help solving the word problem below.

Peter takes 15 minutes longer to mow the lawn by himself than charles does. Together they can mow the lawn in 18 minutes. How long does it take charles to mow the lawn by himself.

Thanks

janviee

janviee

07-14-2006, 02:44 PM

Here is my thinking on how to solve this problem.

I know that i must come up with two equations. I have one so far (P + C = 18). I'm having trouble coming up with the 2nd equation. Once I have the 2nd equation, I will be able to solve.

Thanks

janviee

jonboy

07-14-2006, 04:54 PM

Hello janivee!

Hello

Peter takes 15 minutes longer to mow the lawn by himself than charles does. Together they can mow the lawn in 18 minutes. How long does it take charles to mow the lawn by himself?

I'm sorry but what you have done so far is wrong. But I am very happy that you at least showed some work. :D Whenever you can make ratios like what I' m going to show you that is what you do.

Time it takes Peter = \L c+15

Time it takes Charles = \L c

All time are for one lawn so we can make ratios:

\L \frac{1}{c}+\frac{1}{c+15}=\frac{1}{18}

Now we need to find an LCM. We can do this by multiplying all the denominators together. And whenever we multiply a term times the denominator we need to to that also to the numerator.

So we multiply are denominators: \L c(c+15)18=18c(c+15)

So \L \frac{number}{18c(c+15)}+\frac{number}{18c(c+15)}= \frac{number}{18c(c+15)}

So since our first denominator already has a \L c in it that means I multiply it by \L 18(c+15) to get our common denominator \L {18c(c+15)} so that means we will multiply the first numerator \L 1 by \L 18(c+15) as well. So for \L \frac{1}{c} we have: \L \frac{18(c+15)}{18(c+15)}

Continuing this process you finally reach:

\L \frac{18(c+15)}{18(c+15)}+\frac{18c}{18(c+15)}=\fr ac{c(c+15)}{18(c+15)}

Now we can just ignore the denominator.

Solve \L 18(c+15)+18c=c(c+15)

Can you finish and solve for \L c?

You will get a negative and a positive numer, toss out the negative number since time is not negative.

janviee

07-17-2006, 02:10 PM

Hello jonboy

Thanks alot for your help. I was able to solve follow the reasoning you gave me and was able to solve the equation. I got C =30min an C=-9 min

Thank you so much and i really appreciate your help.

Janviee

jonboy

07-17-2006, 02:16 PM

No problem. But you can't have a negative 9 minutes you gotta throw that one out. Have a great day. :)

TchrWill

07-17-2006, 02:17 PM

Hello

I would really appreciate any help solving the word problem below.

Peter takes 15 minutes longer to mow the lawn by himself than charles does. Together they can mow the lawn in 18 minutes. How long does it take charles to mow the lawn by himself.

Thanks

janviee

Problems of this type are solvable by the following method.

<< If it takes me 2 hours to paint a room and you 3 hours, ow long will it take to paint it together? >>

1--A can paint the house in 5 hours.

2--B can paint the house in 3 hours.

3--A's rate of painting is 1 house per A hours (5 hours) or 1/A (1/5) houses/hour.

4--B's rate of painting is 1 house per B hours (3 hours) or 1/B (1/3) houses/hour.

5--Their combined rate of painting is 1/A + 1/B (1/5 + 1/3) = (A+B)/AB (8/15) houses /hour.

6--Therefore, the time required for both of them to paint the 1 house is 1 house/(A+B)/AB houses/hour = AB/(A+B) = 5(3)/(5+3) = 15/8 hours = 1 hour-52.5 minutes.

Note - T = AB/(A + B), where AB/(A + B) is one half the harmonic mean of the individual times, A and B.

I'll let you work out your numbers.

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