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The area of a square garden is A square feet and the

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Manager
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The area of a square garden is A square feet and the [#permalink]

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25 Jul 2010, 23:56
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The area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?

A. 28
B. 36
C. 40
D. 56
E. 64
[Reveal] Spoiler: OA
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Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 00:39
Hi,

The perimeter of the square is 4*length of one side.
The area of the square = Side*Side

Let x = side of the square
a = 2p+9
x*x = 2(x/4) +9
x*x - x/2 - 9 = 0
Solve for x and the area is x*x.

regards,
Jack
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Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 00:49
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xmagedo wrote:
the area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?
28
36
40
56
64

thanks
tell me about the formula !

The above solution is not right.

Let the side of garden be $$x$$ feet, then: $$area=a=x^2$$ and $$perimeter=p=4x$$. Given: $$a=2p+9$$ --> $$x^2=2*4x+9$$ --> solving for $$x$$: $$x=-1$$ (not a valid solution as $$x$$ represents the length and therefore must be positive) or $$x=9$$ --> $$perimeter=p=4x=36$$.

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Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 00:54
Pleas can you explain how x=9 ? I didnt get this point
thanks
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Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 01:30
xmagedo wrote:
Pleas can you explain how x=9 ? I didnt get this point
thanks

You'll have quadratic equation $$x^2-8x-9=0$$ and you should solve it for $$x$$ --> $$x=-1$$ or $$x=9$$.
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Re: Gemotery complex, pleas help [#permalink]

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26 Jul 2010, 17:24
If side = s
A=s*s
P= 4s
A = 2(p)+9
= 2(4s)+9
= 8s +9
s*s = 8s+9
s*s-8s-9=0
s*s-9s+s-9=0
s(s-9)+1(s-9)=0
(s-9) (s+1)=0
s=9 or -1
s=9
P=4(s) = 4(9)=36
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Re: Gemotery complex, pleas help [#permalink]

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27 Jul 2010, 02:42
xmagedo wrote:
the area of a square garden is A square feet and the perimeter is p feet. If a=2p+9, what is the perimeter of the garden, in feet?
28
36
40
56
64

thanks
tell me about the formula !

You can also solve this via using the given numbers in the answer choices!

Of course you need to be aware of the basic properties as outlined by the other posts above (a = x^2 and p = 4x)

Starting with D you will notice that x=14 is way too big for your area (14^2) and will not satisfy: a=2p+9

--> Eliminate D and E

Now pick B (its either too big, then its A, or too small then you know its C or it is B itsself)

And picking B indeed solves the problem! (36/4 --> 9; a= 9^2 = 81 and 81=2x36+9)

Cheers,
André
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Re: GMAT Prep CAT PS: Area of Square Garden [#permalink]

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20 Dec 2010, 21:03
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$$a^2=A$$
$$4*a=P$$
$$a^2=8*a+9$$
$$a^2-8a-9=0$$
$$(a-9)(a+1)=0$$
$$a=9$$
$$P=4*a=36$$
Re: GMAT Prep CAT PS: Area of Square Garden   [#permalink] 20 Dec 2010, 21:03
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