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

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Director
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The figure shows a square patio surrounded by a walkway of width x [#permalink]

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30 Jun 2005, 20:03
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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

[Reveal] Spoiler:
Attachment:

patio.JPG [ 8.03 KiB | Viewed 6499 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 05 Jul 2015, 08:05, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: The figure shows a square patio surrounded by a walkway of width x [#permalink]

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30 Jun 2005, 20:21
Clearly the patio is a square, since all of its sides of off 2x from the bigger square.

Let the patio's side be a. => x=a-5

Side of bigger square = a + 2(a-5) = 3a - 10

The area of the walkway = Area of big square - Area of patio

132 = (3a-10)^2 - a^2

132 = 9a^2 -60a + 100 -a^2

This reduces to:

8a^2 - 60a - 32 = 0
2a^2 -15a - 8 = 0
2a^2 + a -16a - 8 = 0
a(2a + 1) -8(2a+1) = 0
(a-8)*(2a+1) = 0
a = 8 or -1/2, But a is positive => a=8

Patio's area = a^2 = 64

VP
Joined: 30 Jun 2008
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The figure shows a square patio surrounded by a walkway of [#permalink]

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24 Oct 2008, 03:46
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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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square patio.jpg [ 22.65 KiB | Viewed 12189 times ]

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Re: PS : SQUARE PATIO [#permalink]

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24 Oct 2008, 05:16
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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
VP
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Re: PS : SQUARE PATIO [#permalink]

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24 Oct 2008, 06:33
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]

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24 Oct 2008, 09:55
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
Joined: 30 Jun 2008
Posts: 1034
Re: PS : SQUARE PATIO [#permalink]

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24 Oct 2008, 10:07
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]

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30 Dec 2008, 02:46
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]

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30 Dec 2008, 08:27
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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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Joined: 19 Aug 2008
Posts: 3
Re: PS : SQUARE PATIO [#permalink]

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30 Dec 2008, 10:17
1
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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]

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30 Dec 2008, 10:38
1
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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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Joined: 24 Jun 2008
Posts: 1179
Re: PS : SQUARE PATIO [#permalink]

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30 Dec 2008, 11:35
1
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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]

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30 Dec 2008, 12:13
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]

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30 Dec 2008, 21:52
1
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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]

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30 Dec 2008, 23:38
1
KUDOS
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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26 Aug 2010, 08:23
the answer is B. here's how i did it...

the width of the patio is X + 5 and the width of the big square is 3X +5, therefore the are of the big square is equal to the area of the walkway plus the area of the patio, the equation is
(3X +5)^2 -132 = (X + 5)^2,
X = 3 or X = -11/2
since the width of the patio = X+5 = 3 + 5 =8
the area equals to 8^ 2 =64
hope it helps!!
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Re: PS : SQUARE PATIO [#permalink]

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25 Aug 2011, 23:22
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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: The figure shows a square patio surrounded by a walkway of width x met [#permalink]

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10 Nov 2011, 12:48
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The algebraic solution to the problem is given below:

Given, Width of walkway = x
Width of Patio = Width of Walkway + 5 meters = x+5 meters
Area of walkway = 132 square meters

From the given figure, we know that the width of the square figure i.e. Patio + Walkway will be:
2 times the width of walkway + width of Patio i.e 2(x) + (x+5) = 3x+5

Therefore, the area of this combined figure (square) will be (3x+5)^2

Now, the area of the walkway will be equal to the difference between the area of the combined figure and the square Patio i.e. (3x+5)^2 - (x+5)^2
This is given to be 132 square meters.

Therefore, expanding and simplifying the equation, (3x+5)^2 - (x+5)^2 = 132
We get,
8x^2 + 20x - 132 = 0
i.e. 2x^2 + 5x - 33 = 0
Solving the above quadratic equation will yield the value of x as -6.5 and 3. Since the width of the walkway can't be negative, the value of x is 3 meters.

Using this, we can calculate the area of the Patio, which is (x+5)^2 i.e. (3+5)^2 = 8^2 = 64 square meters.

This is option B.

Hope this helps

Cheers!
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Re: The figure shows a square patio surrounded by a walkway of [#permalink]

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04 Jul 2015, 06:57
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

Hi bunuel,

i didnt get it.

Thanks.
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Re: The figure shows a square patio surrounded by a walkway of [#permalink]

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04 Jul 2015, 07:31
PathFinder007 wrote:
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

Hi bunuel,

i didnt get it.

Thanks.

It is given that width of the patio is 5 more than width of the walkway. Now, walkway's 'extension on either side of the patio' is 'x' and thus width of the of the patio = $$W_P$$ = 5+x

Now, total width of the walkway = $$W_W$$ = width of the patio + 2*x = 5+x+2x = 5+3x
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Re: The figure shows a square patio surrounded by a walkway of   [#permalink] 04 Jul 2015, 07:31

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