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# The figure shows a square patio surrounded by a walkway of

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VP
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The figure shows a square patio surrounded by a walkway of [#permalink]  24 Oct 2008, 02:46
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The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters?

A. 56
B. 64
C. 68
D. 81
E. 100
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Re: PS : SQUARE PATIO [#permalink]  24 Oct 2008, 04:16
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64
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Re: PS : SQUARE PATIO [#permalink]  24 Oct 2008, 05:33
LiveStronger wrote:
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64

nice one especially this step ---- x(5+2x) = 33 = 3*11

I was wondering how to reduce the calculation

thanks
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Re: PS : SQUARE PATIO [#permalink]  24 Oct 2008, 08:55
So, x can be 3 and not 11

Why not 11?

(I did this the ultra long way)

amitdgr wrote:
LiveStronger wrote:
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64

nice one especially this step ---- x(5+2x) = 33 = 3*11

I was wondering how to reduce the calculation

thanks
VP
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Re: PS : SQUARE PATIO [#permalink]  24 Oct 2008, 09:07
So, x can be 3 and not 11

Why not 11?

(I did this the ultra long way)

amitdgr wrote:
LiveStronger wrote:
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64

nice one especially this step ---- x(5+2x) = 33 = 3*11

I was wondering how to reduce the calculation

thanks

x > 0 so 5+3x > x

x(5+2x) = 3*11

x has to be the smaller value and (5+2x) the larger.

so x = 3

even I did it the ULTRA lonnggg wayy
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Re: PS : SQUARE PATIO [#permalink]  29 Dec 2008, 14:47
I am confused - if the patio is surounded by the walkway- how is the width of the patio greater than the width of the walkway???
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 01:46
LiveStronger wrote:
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64

can someone please explain this step how did we get this (5+3x)^2 - (5+x)^2 = 132
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 07:27
gurpreet07 wrote:
LiveStronger wrote:
B

(5+3x)^2 - (5+x)^2 = 132
2x(10+4x) = 132
x(5+2x) = 33 = 3*11
So, x can be 3 and not 11

8^2 = 64

can someone please explain this step how did we get this (5+3x)^2 - (5+x)^2 = 132

Width of the patio is 5m greater than the width of the walkway. So, width of the Patio = 5+x (since x is the width of the walkway)

To get the area of the walkway, we need to substract the area of the bigger square minus the patio.

Area of patio = (5+x)^2
Side of the bigger aquare = x+x+5+x = 5+3x
So, area of the bigger square is = (5+3x)^2

Hence, (5+3x)^2 - (5+x)^2 = 132
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 09:17
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Hi LiveStronger,
nice way to solve it. could you please explain me how do you go from step 1 to step 2. It took me 3 lines of operations and you can do it in just one. I definitely need to be faster in the exam so this kind of tips will be really useful.
thks!!

Step 1. (5+3x)^2 - (5+x)^2 = 132
Step 2. 22x(10+4x) = 132
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 09:38
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marcap wrote:
Hi LiveStronger,
nice way to solve it. could you please explain me how do you go from step 1 to step 2. It took me 3 lines of operations and you can do it in just one. I definitely need to be faster in the exam so this kind of tips will be really useful.
thks!!

Step 1. (5+3x)^2 - (5+x)^2 = 132
Step 2. 22x(10+4x) = 132

I used a^2 - b^2 = (a+b)(a-b) formula
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 10:35
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Expert's post
The solutions above subtract the inner square from the outer one, which is a good approach. One other way to get to the same answer: divide up the patio. You have the four corners, measuring x by x, and four rectangles lined up with each side of the inner square which measure x by x+5. So:

4x^2 + 4x(x+5) = 132
2x^2 + 5x = 33

etc.
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 11:13
amitdgr wrote:
The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters?

A. 56
B. 64
C. 68
D. 81
E. 100

width of the walkway = x
width of the patio = x+5 = x+5
width of the whole area = x+5+x = 3x+5

(3x+5)^2 - (x+5)^2 = 132
9x^2 + 30x + 25 - x^2 -10x -25 = 132
8x^2 + 20x = 132
4 (2x^2 + 5x) = 132
2x^2 + 5x - 33 = 0
2x^2 + 11x - 6x - 33 = 0
x (2x+11) -3 (3x +11) = 0
(x-3) (2x+11) = 0
x = 3 or -11/2 but -11/2 is not possible as length/width cannot be in -ve. so x = 30

so area of the patio = (3+5)^2 = 64
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 20:52
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Thanks GMAT TIGER and LiveStronger for d explanation......
Actually i go numb when i see numbers.......

can u please suggest me how should i improve my quant
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Re: PS : SQUARE PATIO [#permalink]  30 Dec 2008, 22:38
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If its convenient, u can also attempt backsolving it:-
On first glance A,C are clearly out.
(E) 100 means side of patio is 10 so width of walkway =5. the walkway has 4 rectangles 2 among them are each 20*5=100 --impossible for total of 132.
(D) 81 means side of patio is 9 so width of walkway =4. the walkway has 4 rectangles 2 among them are each 17*4=68. 2 such rectangles account for 136--hence impossible for total of 132.
(C) 64 means side of patio is 8 so width of walkway =3. the walkway has 4 rectangles 2 among them are each 14*3=42 and rest are 8*24.total= 2(42+24)=132.
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Re: PS : SQUARE PATIO [#permalink]  25 Aug 2011, 22:22
amitdgr wrote:
The figure shows a square patio surrounded by a walkway of width x meters. If the area of the walkway is 132 square meters and the width of the patio is 5 meters greater than the width of the walkway, what is the area of the patio, in square meters?

A. 56
B. 64
C. 68
D. 81
E. 100

Smaller square=Patio
Let the side of Patio be "s".
Width of walkway=x
width of the patio is 5 meters greater than the width of the walkway(Actually it should be "side of the patio is 5 meters greater than the width of the walkway" because patio is a square)

So,
$$s=x+5$$

If we see the figure properly, the outer quadrilateral is also a square with side $$s+x+x$$ OR $$x+5+x+x=3x+5$$

We know, the area of the walkway is 132 square meters:

Area of walkway=Area of outer square-Area of inner square
Area of walkway=(3x+5)^2-(x+5)^2=132

$$(3x+5)^2-(x+5)^2=132$$

$$(3x)^2+(5)^2+2*3x*5-(x^2+5^2+2*5x)=132$$

$$9x^2+25+30x-x^2-25-10x=132$$

$$9x^2+30x-x^2-10x=132$$

$$8x^2+20x=132$$

$$2x^2+5x=33$$

$$2x^2+5x-33=0$$

$$2x^2+11x-6x-33=0$$

$$x(2x+11)-3(2x+11)=0$$

$$(x-3)(2x+11)=0$$

$$x=3 \hspace{3} OR \hspace{3} x=-\frac{11}{2}$$

Width can't be -ve. So, $$x=3$$
$$s=x+5=3+5=8$$

Area of the patio$$=s^2=8^2=64$$

Ans: "B"
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Re: PS : SQUARE PATIO   [#permalink] 25 Aug 2011, 22:22
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