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# S95-22

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:49
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Difficulty:

55% (hard)

Question Stats:

67% (03:06) correct 33% (02:59) wrong based on 61 sessions

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Note: Figure not drawn to scale.

A rectangular yard (white area) is surrounded by a hedge (shaded region) with a length of 15 meters and a width of 12 meters, as shown in the figure above. The area of the yard is equal to the area of the hedge. If the ratio of the length of the hedge to the width of the hedge is the same as the ratio of the length of the yard to the width of the yard, what is the width, in meters, of the yard?

A. $$6\sqrt{2}$$
B. $$\frac{5}{4}$$
C. $$\frac{6}{\sqrt{2}}$$
D. $$7\frac{\sqrt{2}}{2}$$
E. $$\frac{6}{2}$$

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16 Sep 2014, 01:49
Official Solution:

Note: Figure not drawn to scale.

A rectangular yard (white area) is surrounded by a hedge (shaded region) with a length of 15 meters and a width of 12 meters, as shown in the figure above. The area of the yard is equal to the area of the hedge. If the ratio of the length of the hedge to the width of the hedge is the same as the ratio of the length of the yard to the width of the yard, what is the width, in meters, of the yard?

A. $$6\sqrt{2}$$
B. $$\frac{5}{4}$$
C. $$\frac{6}{\sqrt{2}}$$
D. $$7\frac{\sqrt{2}}{2}$$
E. $$\frac{6}{2}$$

Let the length of the yard be $$x$$, and let the width of the yard be $$y$$.

We are told that the length of the hedge is 15 meters and that the width of the hedge is 12 meters, so we can find the area of the yard and hedge combined using the formula for area of a rectangle: $$A_{\text{yard and hedge}} = \text{length} \times \text{width} = 15 \times 12 = 180$$. Similarly, we can find the area of the yard: $$A_{yard} = xy$$.

We are told that the area of the yard and the hedge are equal, so the area of the hedge must also be $$A_{hedge} = A_{yard} = xy$$. We can use this information to write $$xy + xy = 2xy = 180$$, or $$xy = 90$$.

We know that the length and width of the yard are in the same ratio as the length and width of the hedge. We express this with a proportion: $$\frac{15}{12} = \frac{x}{y}$$. Cross-multiply to get $$15y = 12x$$. Simplify to get $$5y = 4x$$.

Now we have two equations for $$x$$ and $$y$$. We will solve for these unknowns using substitution. Since we want to find the width, $$y$$, we first solve for $$x$$ in terms of $$y$$. From the equation $$xy = 90$$, we get $$x = \frac{90}{y}$$. We plug this into $$5y = 4x$$ and get $$5y = 4 \times \frac{90}{y} = \frac{360}{y}$$. Dividing both sides by 5 and multiplying by $$y$$, we find $$y^2 = 72$$. We now take the square root of both sides to get $$y = \sqrt{72}$$.

To simplify this square root, we look for a perfect square that is a factor of 72. We note that $$72 = 36 \times 2$$, and so $$y = \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$$.

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25 Oct 2016, 09:01
any easy explanation....
Math Expert
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25 Oct 2016, 09:06
VIJAYTHAPLIYAL wrote:
any easy explanation....

Hope it helps.
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03 Jun 2017, 04:47
I don't agree with the explanation. you have assumed 15/12 = x/y, which means the ratio of length to width for whole yard and hedge = length to width ratio of the yard. To my mind this is not what the problem reads. Please correct me if my interpretation is not correct.
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03 Jun 2017, 17:58
Ganeshsrinivasan wrote:
I don't agree with the explanation. you have assumed 15/12 = x/y, which means the ratio of length to width for whole yard and hedge = length to width ratio of the yard. To my mind this is not what the problem reads. Please correct me if my interpretation is not correct.

Discussion here might help: http://gmatclub.com/forum/a-rectangular ... 88513.html

Hope it helps.
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Joined: 18 Mar 2015
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Schools: ISB '19
GMAT 1: 600 Q47 V26
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18 Aug 2017, 05:03
I have used expression for area of hedge = area of yard = (15-2x)(12-2x)=15*12
can anyone point out the mistake ? I feel this equation would be valid had square was given in the question. Am I correct?
Re: S95-22   [#permalink] 18 Aug 2017, 05:03
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# S95-22

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