phanendery wrote:1. Working alone, Colleen can paint a house in 2 h less than James. Working together, they can paint the house in 10 h. How long would it take James to paint the house by himself?

Use the standard method for

"work" word problems. Since Colleen's time is defined in terms of James', pick a variable for James' and then

translate Colleen's relationship into an expression in terms of James' variable. Then start the set-up by converting their times to hourly rates.

. . .time to complete job:

. . . . .James: j

. . . . .Colleen: j - h

. . . . .together: 10

. . .completed per hour (rate):

. . . . .James: 1/j

. . . . .Colleen: 1/(j - h)

. . . . .together: 1/10

Since they are assumed to be able to add their labors (they never get in each other's way, so whatever they do individually

adds to what the other has done), add the amounts they each done in an hour, set equal to their total, and

solve.

phanendery wrote:2. The Computer Club went on a field trip to a computer museum. The trip cost $240 to be paid for equally by each club member. The day before the trip, 4 students decided not to go. This increased the cost by $2 per student. How many students went to the computer museum?

Pick a variable for the original number of students. If there were two students, then the per-student cost would be $240/2. If there were three, then the per-student cost would be $240/3. If there were ten, the cost would be $240/10. What expression would then stand for the per-student cost, in terms of the variable you chose?

What expression stands for the number of students which actually went (after the four dropped out)?

What expression then stands for the actual per-student cost?

Using the above, translate "(the actual per-student cost) was (the original per-student cost) plus (another two bucks)" into an equation, and solve.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!