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A larger playground has a rectangular-shaped track shaded shown as abo

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A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 26 Apr 2017, 01:16
1
4
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

52% (01:58) correct 48% (02:12) wrong based on 137 sessions

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A larger playground has a rectangular-shaped track shaded shown as above figure such that the track has a uniform d as its width. What is the track’s area?

1) The perimeter of the smaller playground is 100.
2) The perimeter of the larger playground is 200.

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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 26 Apr 2017, 09:25
MathRevolution wrote:
Attachment:
1.png


A larger playground has a rectangular-shaped track shaded shown as above figure such that the track has a uniform d as its width. What is the track’s area?

1) The perimeter of the smaller playground is 100.
2) The perimeter of the larger playground is 200.


1) The perimeter of the smaller playground is 100. - Not sufficient
2) The perimeter of the larger playground is 200 - Not sufficient

1) + 2) Let length and breadth of smaller rectangle be l and b.

2(l+b) = 100
2(l+2d+b+2d) = 200.

On solving the above two equations, we'll be able to find out the the width of the track d but won't be able to find the area of the same as we need the values of l and b.

Hence E.

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New post 28 Apr 2017, 01:31
==> In the original condition, if you set the width and the height of the playground as a and b, there are 3 variables (a,b,d). In order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), from 2(a+b)=100 and 2(a+b)+8d=200, you cannot find a and b in a unique way, hence it is not sufficient.

Therefore, the answer is E.
Answer: E
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 12 May 2017, 01:37
I'm getting C. Can someone please check my math and see if there's any error in my logic?

Let's call length and width of the playground as L and W, respectively. We are trying to find (L + 2d)*(W+2d) - (L*W).

Statement 1: 2(L+W) = 100 ==> L+W = 50. Not Sufficient.
Statement 2: 2(L+2d) + 2(W+2d) = 200 ==> 2L + 4d + 2W + 4d = 200 ==> 2(L+W) + 8d = 200. Not Sufficient

Combining Statement 1 and Statement 2: 2(L+W) + 8d = 200 and we know from Statement 1 that 2(L+W) = 100 ==> 100 + 8d = 200 ==> 8d = 100 ==> d = 12.5
Plugging 12.5 back into the original equation, we get (L+25)*(W+25) - (L*W) = ?. Multiplying out the equation, we get (L*W) + 25L + 25W + 625 - (L*W) = ? ==> 25L + 25W + 625.
Simplifies into 25*(L+W) - 625 ==> From statement 1, we know (L+W) = 50, so (25*50) - 625 = 625. Statement 1 + Statement 2 is sufficient
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 19 May 2017, 08:53
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jy295 wrote:
I'm getting C. Can someone please check my math and see if there's any error in my logic?

Let's call length and width of the playground as L and W, respectively. We are trying to find (L + 2d)*(W+2d) - (L*W).

Statement 1: 2(L+W) = 100 ==> L+W = 50. Not Sufficient.
Statement 2: 2(L+2d) + 2(W+2d) = 200 ==> 2L + 4d + 2W + 4d = 200 ==> 2(L+W) + 8d = 200. Not Sufficient

Combining Statement 1 and Statement 2: 2(L+W) + 8d = 200 and we know from Statement 1 that 2(L+W) = 100 ==> 100 + 8d = 200 ==> 8d = 100 ==> d = 12.5
Plugging 12.5 back into the original equation, we get (L+25)*(W+25) - (L*W) = ?. Multiplying out the equation, we get (L*W) + 25L + 25W + 625 - (L*W) = ? ==> 25L + 25W + 625.
Simplifies into 25*(L+W) - 625 ==> From statement 1, we know (L+W) = 50, so (25*50) - 625 = 625. Statement 1 + Statement 2 is sufficient


Hi Bunuel,
I am with the above poster.Except that his last simplification is wrong.
It should be 25*(L+W) + 625 = 1250+625 = 1875
Thus,together it is SUFFICIENT

Please help clarify
Thanks :-)
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 20 May 2017, 23:34
MathRevolution wrote:
Attachment:
1.png


A larger playground has a rectangular-shaped track shaded shown as above figure such that the track has a uniform d as its width. What is the track’s area?

1) The perimeter of the smaller playground is 100.
2) The perimeter of the larger playground is 200.


I think the correct answer is C.
suppose 'a' be the length & 'b' be the width of the ground. Now as the width of the tract is uniform the the total length of the figure is 'a+2d' & total width is 'b+2d'.
from (1) we have, a+b=100/2=50
from (2) we have, (a+2d+b+2d)=200/2=a+b+4d=100. Hence, we have 4d=50 or, d=50/4.
Now, we have to find net area of the track ,i.e, (a+2d)x(b+2d)-ab=ab+2ad+2bd+4d^2-ab=2d(a+b)+4d^2=2x(50/4)x(50)+4x(50/4)^2=1250+625=1875.
Hence, (1) & (2) are together sufficient.
Bunuel plz correct me if I am wrong.
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 21 May 2017, 07:12
I think answer should be C.

A and B are clearly not sufficient.
Combing 1 and 2
inner rectangle 2*(l+b) = 100 gives l+b = 50....eq1
outer rectangle 2(l+2d+b+2d) = 200 gives l+b+4d = 100....eq2
using eq1 and eq2
50+4d = 100=>4d = 50=>d=25/2..eq3

now area of shaded portion is (l+2d)*(b+2d) - l*b
gives lb+2ld+2bd+4dd-lb
gives 2d*(l+b) +4 d^2
we know d from eq3 and l+b from eq1, we can get area
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 17 Jun 2017, 12:10
I am getting C as well. :shock: The OA needs to be corrected. :roll:
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Re: A larger playground has a rectangular-shaped track shaded shown as abo  [#permalink]

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New post 01 Jul 2017, 23:59
jy295 wrote:
I'm getting C. Can someone please check my math and see if there's any error in my logic?

Let's call length and width of the playground as L and W, respectively. We are trying to find (L + 2d)*(W+2d) - (L*W).

Statement 1: 2(L+W) = 100 ==> L+W = 50. Not Sufficient.
Statement 2: 2(L+2d) + 2(W+2d) = 200 ==> 2L + 4d + 2W + 4d = 200 ==> 2(L+W) + 8d = 200. Not Sufficient

Combining Statement 1 and Statement 2: 2(L+W) + 8d = 200 and we know from Statement 1 that 2(L+W) = 100 ==> 100 + 8d = 200 ==> 8d = 100 ==> d = 12.5
Plugging 12.5 back into the original equation, we get (L+25)*(W+25) - (L*W) = ?. Multiplying out the equation, we get (L*W) + 25L + 25W + 625 - (L*W) = ? ==> 25L + 25W + 625.
Simplifies into 25*(L+W) - 625 ==> From statement 1, we know (L+W) = 50, so (25*50) - 625 = 625. Statement 1 + Statement 2 is sufficient


Absolutely!
I agree. The OA should be changed to C.
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New post 03 Jul 2017, 05:47
MathRevolution wrote:
==> In the original condition, if you set the width and the height of the playground as a and b, there are 3 variables (a,b,d). In order to match the number of variables to the number of equations, there must be 3 equations. Since there is 1 for con 1) and 1 for con 2), E is most likely to be the answer. By solving con 1) and con 2), from 2(a+b)=100 and 2(a+b)+8d=200, you cannot find a and b in a unique way, hence it is not sufficient.

Therefore, the answer is E.
Answer: E


But I believe we can find d with these two equations -

Please correct me if I m wrong - the area of the shaded region is [assuming the length and width of larger rectangle are L & B resp and those of smaller rectangle are l & b] = LB - lb
and wkt L = l + 2d ---- i ; B = b + 2d ----- ii [since d is uniform]

1) l + b = 50 insuff
2) L + B = 100 insuff

1) + 2)

From 2) we can find d i.e. from i and ii l + 2d + b + 2d = 100

Substituting value of d in LB - lb
------> (l + 2d)(b + 2d) - lb

Simplifying,
2d(l + b) + 4\(d^2\). We know value of d and l + b
Hence sufficient. Option C
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A larger playground has a rectangular-shaped track shaded shown as abo   [#permalink] 03 Jul 2017, 05:47
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