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

Question

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.

Diwakar003 · Answer

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!

MathRevolution · Answer

==> 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

Therefore, the answer is E.
Answer: E

jy295 · Answer

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

sleepynut · Answer

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 :-)

debtanuB · Answer

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.

dabhishek87 · Answer

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

TheMechanic · Answer

I am getting C as well. :shock:

The OA needs to be corrected. :roll:

rekhabishop · Answer

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

Sash143 · Answer

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   = LB - lb
and wkt L = l + 2d ---- i ; B = b + 2d ----- ii

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

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

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