1. ## Help (story problems)

1. The length of a rectangular playing field is 5 ft less than twice its width. If the perimeter of the playing field is 230 ft, find the length and width of the field.

2. A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi apart?

2. Originally Posted by Patience
1. The length of a rectangular playing field is 5 ft less than twice its width. If the perimeter of the playing field is 230 ft, find the length and width of the field....

Hi,

let l be the length of the rectangle and
let w be the width of the rectangle. Then
the perimeter of a rectangle is calculated by:

p = 2l + 2w

You know that l = 2w - 5 and that p = 230'
Now substitute these term into the first equation:

2(2w - 5) + 2w = 230. Expand the bracket

6w - 10 = 230
6w = 240

w = 40 feet and l = 75 feet

p = 2*40 + 2*75

p = 80 + 150 = 230 feet. Seems to be OK

EB

3. Originally Posted by Patience
...
2. A bus leaves a station at 1 P.M., traveling west at an average rate of 44 mi/h. One hour later a second bus leaves the same station, traveling east at a rate of 48 mi/h. At what time will the two buses be 274 mi apart?

Hi,

at 2:00 pm the distance between the 2 buses is 44 mi. From now on the distance increases at (44 + 48) mi/h.

44 + (44 + 48)*t = 274

92t = 230

t = 230/92 h

After 230/92 h after 2:00 pm the buses have a distance of 274 mi

That is at 4:23:28.7 pm

EB