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Math Revolution GMAT Instructor
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A larger playground has a rectangularshaped track shaded shown as abo
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26 Apr 2017, 01:16
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52% (01:58) correct 48% (02:12) wrong based on 137 sessions
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A larger playground has a rectangularshaped 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 rectangularshaped track shaded shown as abo
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26 Apr 2017, 09:25
MathRevolution wrote: Attachment: 1.png A larger playground has a rectangularshaped 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. Cheers!



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Re: A larger playground has a rectangularshaped track shaded shown as abo
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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 rectangularshaped track shaded shown as abo
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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 rectangularshaped track shaded shown as abo
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19 May 2017, 08:53
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 rectangularshaped track shaded shown as abo
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20 May 2017, 23:34
MathRevolution wrote: Attachment: 1.png A larger playground has a rectangularshaped 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^2ab=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 rectangularshaped track shaded shown as abo
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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+4ddlb 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 rectangularshaped track shaded shown as abo
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17 Jun 2017, 12:10
I am getting C as well. The OA needs to be corrected.
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Re: A larger playground has a rectangularshaped track shaded shown as abo
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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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A larger playground has a rectangularshaped track shaded shown as abo
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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 rectangularshaped track shaded shown as abo
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