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# The figure above shows a rectangular garden that is comprised of a

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Math Expert
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The figure above shows a rectangular garden that is comprised of a [#permalink]

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19 Nov 2015, 00:09
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The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

A. $$25\sqrt{2}$$

B. $$20\sqrt{2}$$

C. $$20(1-\sqrt{2})$$

D. $$\frac{25\sqrt{2}}{2}$$

E. (25/2)^2

[Reveal] Spoiler:
Attachment:

Untitled.png [ 2.08 KiB | Viewed 1433 times ]
[Reveal] Spoiler: OA

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Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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19 Nov 2015, 01:07
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KUDOS
let length and bredth of central plot be x and y
therefore area =x*y
as per the question 25*20-xy=xy
500=2xy > xy=250
now
25/20=x/y
x/y=5/4 > 5x*4x=250 > x2 =250/20 > x =(25/2)^1/2

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Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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19 Nov 2015, 01:17
Let length of central plot = x
width of central plot = y

Since , the border and the central plot have the same area .
x*y = 25*20 - x*y
=> 2*x*y = 500
=> x*y = 250 --- 1

Length and width of central plot have the same ratio as the length and width of the garden
x/y = 25/20 = 5/4

=> y = (4/5)* x --- 2

Using equation 2 in equation 2 , we get
(4/5) *x *x = 250
=> x^2 = 1250 / 4
=> x = [25*(2^(1/2))]/2

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Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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19 Nov 2015, 02:31
CounterSniper wrote:
let length and bredth of central plot be x and y
therefore area =x*y
as per the question 25*20-xy=xy
500=2xy > xy=250
now
25/20=x/y
x/y=5/4 > 5x*4x=250 > x2 =250/20 > x =(25/2)^1/2

Hi,
you are right upto x=(25/2)^1/2...
x=5/(2^1/2)..
we are looking for 5x=5*5/(2^1/2)=25/(2^1/2)
or 25(2^1/2)/2
ans D
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Senior Manager
Joined: 20 Feb 2015
Posts: 386
Concentration: Strategy, General Management
Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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20 Nov 2015, 00:29
no wonder my answers dont match for half of the questions
Intern
Joined: 09 Feb 2015
Posts: 5
Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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20 Nov 2015, 05:37
1
KUDOS
The length and width of the central plot have the same ratio as the length and width of the garden,
Let the length be 5X and width be 4X for central Plot, so central plot area = $$20X^{2}$$
Area of central Plot = Area of border surrounding, So Area of border surrounding = $$20X^{2}$$
Area of central Plot + Area of border surrounding = Total Area
$$20X^{2}$$ + $$20X^{2}$$ = 20 *25
$$40X^{2}$$ = 500
X = $$\frac{5}{\sqrt {2}}$$
Length of central plot: 5X = 5*$$\frac{5}{\sqrt {2}}$$ = $$\frac{25\sqrt {2}}{2}$$

Ans: D
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The figure above shows a rectangular garden that is comprised of a [#permalink]

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20 Nov 2015, 11:36
I solved it by looking at the answer choices and using logic/educated guessing/Luck.

Because the big rectangle's area is (500 - the little rectangle) = little rectangle. You can guess that the little one going to be about half the size. If it were a perfect square the lengths would be \sqrt{250}. 16^2=256 so we are looking for something that is a little bigger than 15ish.

Answers A, B and E make the length even longer than 25 so cross them out and C makes it 8 which is way too small leaving only D which is 17ish.

This could have been pure luck but I still got it right
Director
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Posts: 961
Re: The figure above shows a rectangular garden that is comprised of a [#permalink]

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17 Jul 2016, 06:39
1
KUDOS
Bunuel wrote:

The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

A. $$25\sqrt{2}$$

B. $$20\sqrt{2}$$

C. $$20(1-\sqrt{2})$$

D. $$\frac{25\sqrt{2}}{2}$$

E. (25/2)^2

[Reveal] Spoiler:
Attachment:
Untitled.png

Length of the central plot must be less than 25
if we analyse every option we see each one is greater than 25 exept (C)& (D)
but option (C) is rather a negative value which is immpossible
leaving option(D) as correct one
Ans D
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Joined: 22 May 2016
Posts: 1418
The figure above shows a rectangular garden that is comprised of a [#permalink]

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30 Sep 2017, 13:01
1
KUDOS
Bunuel wrote:

The figure above shows a rectangular garden that is comprised of a central plot and a border surrounding the central plot; the border and the central plot have the same area. The garden has length 25 feet and width 20 feet. If the length and width of the central plot have the same ratio as the length and width of the garden, what is the length of the central plot, in feet?

A. $$25\sqrt{2}$$

B. $$20\sqrt{2}$$

C. $$20(1-\sqrt{2})$$

D. $$\frac{25\sqrt{2}}{2}$$

E. (25/2)^2

[Reveal] Spoiler:
Attachment:
Untitled.png

General equation
Area of center = LW

Area of border = area of center = LW

Center + border = Total area (25*20)

$$2(LW) = 500$$
$$LW = 250$$

Use the ratio to eliminate W

The ratio of length to width of total area (garden) = ratio of length to width for center

$$\frac{L}{W} = \frac{25}{20} = \frac{5}{4}: 5W = 4L --> W = \frac{4}{5}L$$

Substitute and solve for length

$$LW = 250$$
$$W = \frac{4}{5}L$$

$$\frac{4}{5}L*L = 250$$

$$\frac{4}{5}L^2 = 250$$

$$L^2 = 250 * \frac{5}{4}$$

**$$L^2 =\frac{25*10*5}{4} = \frac{(25)*2*(5)*(5)}{4}$$

$$L =\frac{25\sqrt{2}}{2}$$

**OR
$$L^2 = (250 *\frac{5}{4}) =(\frac{1250}{4}) =(\frac{2*5*5*5*5}{4})$$

$$L =\frac{25\sqrt{2}}{2}$$
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The figure above shows a rectangular garden that is comprised of a   [#permalink] 30 Sep 2017, 13:01
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# The figure above shows a rectangular garden that is comprised of a

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