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# Having a difficult time with Algebra word problems!!

Sun 8 Oct, 2006 11:11 am
Here are the questions...

1) If Sam can do a job in 4 days that Lisa can do in 6 days and Tom can do in 2 days, how long would the job take if they all work together to complete it?

2) Two cyclists start biking from a trails start 3 hours apart. The second cyclists travels at 10 mph and starts 3 hours after the first cyclists who is traveling at 6 mph. How much time will pass before the second cyclists catches up with the first from the time the second cyclists started biking?

I'm having trouble setting up the equations, I understand equations once they are in an equation form. Thanks for your help!!
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Type: Discussion • Score: 3 • Views: 6,472 • Replies: 6
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fresco

1
Mon 9 Oct, 2006 09:34 am
1. Let the job be x units. So per day, Sam does x/4 units , Lisa does x/6 etc. Add the units per day and divide the total into x to give the number of days.

2. Distance = Speed x Time.
When they meet distances travelled are equal so
6xT= 10 (T-3)
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Brandon9000

1
Mon 9 Oct, 2006 10:42 pm
Re: Having a difficult time with Algebra word problems!!
daphne592 wrote:
2) Two cyclists start biking from a trails start 3 hours apart. The second cyclists travels at 10 mph and starts 3 hours after the first cyclists who is traveling at 6 mph. How much time will pass before the second cyclists catches up with the first from the time the second cyclists started biking?

D1 = distance covered by 1st cyclist
D2 = distance covered by 2nd cyclist

D1 = 6t
D2 = 10(t-3)

At the time in question, D1 = D2
6t = 10(t-3)
6t = 10t - 30
0 = 4t - 30
4t = 30
t = 7.5 hours

Check:
At 7.5 hours, cyclist 1 has travelled 6 mph (7.5 hrs) = 45 miles
At 7.5 hours, cyclist 2 has travelled 10 mph (4.5 hrs) = 45 miles
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mebemikeyc

1
Sun 16 Nov, 2008 11:14 am
These solutions are actually incorrect, because you're subtracting from the faster cyclist's rate, based on the amount of time that the slower cyclist is ahead.

The ratio should be:
6(t+3) = 10t
6t + 18 = 10t
18=4t
18/4 = t
4.5 = t
flyboy804

1
Sun 16 Nov, 2008 02:36 pm
@mebemikeyc,
mebemikeyc wrote:

These solutions are actually incorrect, because you're subtracting from the faster cyclist's rate, based on the amount of time that the slower cyclist is ahead.

The ratio should be:
6(t+3) = 10t
6t + 18 = 10t
18=4t
18/4 = t
4.5 = t
You're a bit late. The homework was due two years ago.
Foxfyre

1
Sun 16 Nov, 2008 03:22 pm
@flyboy804,
Hmm. Wonder what kind of grade she got
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Fountofwisdom

1
Sat 3 Jan, 2009 08:13 am
Well: lets not be too down hearted: congratulations everyone: some clear algebra logically and clearly presented: I'm impressed.
0 Replies

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