SOLUTION: The perimeter of a rectangular fields is 200m. If the length were reduced 20m and the width reduced by 25%, the perimeter would be reduced by 30%. The original length of the field

Algebra ->  Finance -> SOLUTION: The perimeter of a rectangular fields is 200m. If the length were reduced 20m and the width reduced by 25%, the perimeter would be reduced by 30%. The original length of the field       Log On


   



Question 1170798: The perimeter of a rectangular fields is 200m. If the length were reduced 20m and the width reduced by 25%, the perimeter would be reduced by 30%. The original length of the field is:
Found 2 solutions by VFBundy, MathTherapy:
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
Original length = L
Original width = W

2L + 2W = 200

2(L - 20) + 2(0.75W) = 0.70(200)

2L - 40 + 1.5W = 140

2L + 1.5W = 180

1.5W = 180 - 2L

W = 180/1.5 - 2L/1.5

W = 120 - 2L/1.5

W = 120 - 4L/3

Plug W = 120 - 4L/3 into the first equation:

2L + 2W = 200

2L + 2(120 - 4L/3) = 200

2L + 240 - 8L/3 = 200

2L - 8L/3 = -40

6L/3 - 8L/3 = -40

-2L/3 = -40

-2L = -120

L = 60

Original length = L = 60m

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!

The perimeter of a rectangular fields is 200m. If the length were reduced 20m and the width reduced by 25%, the perimeter would be reduced by 30%. The original length of the field is:
Let original length, be L
Then
We then get:
Original length, or highlight_green%28matrix%281%2C6%2C+L%2C+%22=%22%2C+30%2F.5%2C+%22=%22%2C+60%2C+m%29%29