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

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

So, x can be 3 and not 11

Why not 11?

(I did this the ultra long way)

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

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

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

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.

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

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

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.

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!!

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"

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!

Hi bunuel,

Could you please provide your comments on how we are getting 5+3x.

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