Question 288985: A uniform-width boardwalk is built around the inside edge of a rectangular parkland that is 10m by 15m. If the boardwalk takes 20% of the lot, how wide is the boardwalk to the nearest centimetre?
Found 3 solutions by mananth, ikleyn, n2:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! A uniform-width boardwalk is built around the inside edge of a rectangular parkland that is 10m by 15m. If the boardwalk takes 20% of the lot, how wide is the boardwalk to the nearest centimetre?
The area of parkland = 10* 15 = 150 sq. meters
Boardwalk takes 20% of 150
=20/100 * 150
=30 sq. mtrs.
let the width of the boardwalk be x
the length of parkland excluding width of board walk = 15-2x
width will be 10-2x
150 - (15-2x)(10-2x)=30
150 -{ 150 -30x-20x+4x^2}=30
150-150+50x-4x^2-30=0
4x^2-50x-30=0
taking the roots of the equation
x1= 50+sqrt(2500+480) / 8
=50+sqrt2980 /8
=13.07 Or 13
x2 will be negative
so 13 meters is the width of the board walk
Answer by ikleyn(53777)  (Show Source):
You can put this solution on YOUR website! .
A uniform-width boardwalk is built around the inside edge of a rectangular parkland that is 10m by 15m.
If the boardwalk takes 20% of the lot, how wide is the boardwalk to the nearest centimetre?
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In his post, @mananth produces the answer for the wide of the boardwalk: 13.07, or 13 meters.
To me, it means only one thing: he even did not read his solution and did not compare
his answer with the problem. To me, there is nothing unexpected in it: by observing his numerous solutions,
I just learned it long time ago and firmly: this person never reads his own solutions.
It is simply non-interesting to him.
Below is my correct solution to this problem.
Let x be the uniform wide of the boardwalk, in meters.
The area inside the boardwalk is 0.8*10*15 = 120 square meters.
Therefore, our equation for this area is
(10-2x)*(15-2x) = 120.
Simplify and reduce it to the standard form quadratic equation
150 - 50x + 4x^2 = 120,
4x^2 - 50x + 30 = 0,
2x^2 - 25x + 15 = 0.
 =  .
One root  is about 11.86 meters, and it is too big to be a solution
to the problem, so we reject it.
The other root  is about 0.63 meters, and it makes sense as a solution
to the problem, so we accept it.
ANSWER. The width of the boardwalk is about 63 centimeters.
Solved correctly to teach you in a right way.
Answer by n2(82)  (Show Source):
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