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The length (in feet) of a rectangle is 4 more than twice its width. Th

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The length (in feet) of a rectangle is 4 more than twice its width. Th  [#permalink]

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New post 29 Mar 2018, 00:45
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

83% (01:34) correct 17% (02:13) wrong based on 52 sessions

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Re: The length (in feet) of a rectangle is 4 more than twice its width. Th  [#permalink]

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New post 29 Mar 2018, 00:53
Bunuel wrote:
The length (in feet) of a rectangle is 4 more than twice its width. The area of the rectangle is 70 feet^2. What is the length (in feet) of the rectangle?

(A) 5
(B) 7
(C) 10
(D) 12
(E) 14


Area of rectangle is length * width.

Let the length be L.
It is given to be 2*W + 4 , where W is width.
\(L = 2W+4\)
\(W = (L-4)/2\)

\(L*W = 70\)
\(L*(L-4)/2 = 70\) ...(Substitute W in)
\(L^2 - 4L - 140 = 0\)

140 = 7*2*2*5 ... hence 14*10 we can use this to factorize...

\(L^2 - 14L + 10L - 140 = 0\)
\(L (L- 14)+ 10(L - 14) = 0\)
\((L + 10)(L - 14) = 0\)
L= +14, -10 ... since L >0
\(L=14\)

Hence Option (E) is correct.

Best,
Gladi
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The length (in feet) of a rectangle is 4 more than twice its width. Th  [#permalink]

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New post 29 Mar 2018, 19:30
Bunuel wrote:
The length (in feet) of a rectangle is 4 more than twice its width. The area of the rectangle is 70 feet^2. What is the length (in feet) of the rectangle?

(A) 5
(B) 7
(C) 10
(D) 12
(E) 14

\(l * w = Area, A\)
\(l= 2w + 4\)
\(A = 70\)
\((2w + 4) * w = 70\)
\(2w^2 + 4w - 70 = 0\)
\(w^2 + 2w - 35=0\)
\((w + 7)(w - 5) = 0\)
\(w = 5\) (length cannot be negative)
\(l = (2w + 4) = (10 + 4) = 14\)


Answer E
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The length (in feet) of a rectangle is 4 more than twice its width. Th   [#permalink] 29 Mar 2018, 19:30
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