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

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

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.

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

I am getting C as well. The OA needs to be corrected.

Absolutely!
I agree. The OA should be changed to C.

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