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The perimeter of rectangle A is 200 meters. The length of rectangle B

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The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 25 Jul 2017, 11:22
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The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60

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The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 25 Jul 2017, 12:04
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carcass wrote:
The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60

For the square (rectangle B) the net change in perimeter, from decreased L (-10), twice, and (+10) in W, twice, is 0.

All sides of a square are equal. If we add 10 to two sides, and subtract 10 from two sides,
because we started with equal side lengths, the perimeter does not change.

If rectangle-square B has side = 50.
Perimeter is 4s = 200.
Each L increases by 10 (to make rectangle A): two lengths = 60
Each W decreases by 10 (to make rectangle A): two widths = 40
Perimeter of rectangle A? 60 + 60 + 40 + 40 = 200 (same as rectangle-square B)

So the perimeter of rectangle A equals perimeter of square (rectangle B).

Perimeter of square = 4s

4s = 200
s = 50

If the side of the square (B) is 50, then the width of rectangle A is (50 - 10) = 40.

Answer C
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 25 Jul 2017, 12:17
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carcass wrote:
The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60


For Rectangle A, 2(l+b) = 200 or l+b = 100 -------(1) where l & b are the length and breadth of rectangle A
For Rectangle B, which is a square: l-10 = b+10 ----- (2) (as the sides of the squares are equal)
Using (1) and (2), b= 40
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 25 Jul 2017, 12:29
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2(L+W)=200. Then L+W=100

Now For B,

Lenght of B= L-10. And Width of B= W+10.

We are told that B is square. Then Length of B= Width of B

L-10=W+10. So L= W+20.

W+20+W=100

2W=80

W=40. So the Answer is C.
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 10 Sep 2017, 05:25
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carcass wrote:
The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60





Permiter = 2(length + width)

A = 200 = 2(l +w)
B = 2((l-10) + (w +10)
given B = square, you know that B's length is equal to its width
(l-10) = (10 + w)
l = 20 + w
200 = 2((20 +w) + w)
100 = ((20 + w) + w)
80 = 2w
40 =w

Therefore the answer is (C)
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 02 Oct 2018, 19:18
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carcass wrote:
The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60


We can let the width of rectangle A = W and the length of rectangle A = L and create the equations:

2L + 2W = 200

L + W = 100

We also know that the width of rectangle B is (10 + W) and its length is (L - 10). Since B is a square, we have:

10 + W = L - 10

W - L = -20

Adding the two equations, we have:

2W = 80

W = 40

Alternate solution:

Since the length of rectangle B is 10 less than the length of rectangle A but the width of B is 10 more than the width of A, we know that B and A must have equal perimeters. Therefore, the perimeter of B is 200. Since we are told that B is a square, a side of B must be 50. Since this is 10 more than the width of A, the width of A must be 40.

Answer: C
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B  [#permalink]

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New post 30 Oct 2018, 07:06
carcass wrote:
The perimeter of rectangle A is 200 meters. The length of rectangle B is 10 meters less than the length of rectangle A and the width of rectangle B is 10 meters more than the width of rectangle A. If rectangle B is a square, what is the width, in meters, of rectangle A ?

A. 10

B. 20

C. 40

D. 50

E. 60


\(Perimeter of A = 200 = 2 (L_A+W_A)\)

\(Perimeter of A = 100 = L_A+W_A\)

\(LB = L_A - 10\)

\(WB = W_A + 10\)

We can then substitute back in the equation so that \(100 = L_B + W_B\)

Given that B is a square so \(L_B\) and \(W_B\) are both 50.

\(W_A = W_B - 10 = 50 - 10 = 40.\)

Answer choice C
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Re: The perimeter of rectangle A is 200 meters. The length of rectangle B   [#permalink] 30 Oct 2018, 07:06
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