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Source: — Data Sufficiency |
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by prindaroy » Fri Jul 24, 2009 10:25 am
The answer is 64. It's pretty algebra intensive so I might cut through a few steps.

First width = length of square = x + 5

First let's find x in the walkway;

Area of bottom most side is ((x+5)+x+x)x

Area of top most side is the same so total area of bottom and top sides =

2(3x+5)x.

Area of two sides of walkway = (x+5)x + x(x+5)

Total area = 2(3x+5)x + 2x(x+5) = 132

in short

2x^2 + 5x - 33 = 0

So x = 3 or x = -5.5. x can only be positive. So we know width = length of square = 3 + 5 = 8. So area = 8^2 = 64
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by raghavsarathy » Fri Jul 24, 2009 10:30 am
IMO -B

Length of each side of patio = (x+5)

Length of each side of the outer square = 3x + 5

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

(9x^2 + 25+ 30x ) - (x^2 + 10x + 25) = 132
8x^2 + 20x - 132 =0
2x^2 + 5x - 33 = 0
2x^2 -6x +11x - 33 = 0
2x(x-3) + 11(x-3)= 0

Hence x=3

Side of patio = x+5 = 8
Area = 8*8 = 64
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by gmat740 » Fri Jul 24, 2009 6:14 pm
Length of each side of patio = (x+5)

Length of each side of the outer square = 3x + 5
How you both are arriving at the above equation?

IMO : Length of each side of patio = 5

Length of each side of the outer square = 2x + 5

Let me know where I am wrong.
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by raghavsarathy » Sat Jul 25, 2009 7:31 am
gmat740 wrote:
Length of each side of patio = (x+5)

Length of each side of the outer square = 3x + 5
How you both are arriving at the above equation?

IMO : Length of each side of patio = 5

Length of each side of the outer square = 2x + 5

Let me know where I am wrong.
Hi Karan

It is mentioned that " side of patio is 5 metres greater than the width of the walkway(x) "

Hence we get side of patio = x+5
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