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# A piece of wire used to fence a rectangular land can be used to fence

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A piece of wire used to fence a rectangular land can be used to fence  [#permalink]

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01 Aug 2017, 02:43
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Difficulty:

35% (medium)

Question Stats:

77% (02:10) correct 23% (03:53) wrong based on 69 sessions

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A piece of wire used to fence a rectangular land can be used to fence a square land that has twice the width of the rectangular land.What is the area of the square land if the area of the rectangular land is 100 sq units less than the area of the square land?

A. 100 sq units
B. 300 sq units
C. 400 sq units
D. 900 sq units
E. 1000 sq units
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Posts: 6559
Re: A piece of wire used to fence a rectangular land can be used to fence  [#permalink]

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01 Aug 2017, 02:56
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DH99 wrote:
A piece of wire used to fence a rectangular land can be used to fence a square land that has twice the width of the rectangular land.What is the area of the square land if the area of the rectangular land is 100 sq units less than the area of the square land?

A. 100 sq units
B. 300 sq units
C. 400 sq units
D. 900 sq units
E. 1000 sq units

Hi..
rectangular field -$$2*( l+b)$$
square= $$(2b)*4=8b$$....
so $$2l+2b=8b.....l=3b$$....

also $$(2b)^2-l*b=100.....4b^2-3b*b=100.....b^2=100....b=10$$

so area of rectangle = $$(2*10)^2=400$$

C
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A piece of wire used to fence a rectangular land can be used to fence  [#permalink]

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01 Aug 2017, 03:39
1
Given data:
Length or wire needed to fence both the rectangle and the square is the same.
The side of the square is twice the width of the rectangle.
Area(Square) - Area(Rectangle) = 100

We have been asked to find the area of the square

Let the length and width of the rectangle be y and x.
From the question stem, side of square is 2x
$$2(y+x) = 4(2x)$$
Therefore, $$y = 4x-x =3x$$

We know that Area(Square) - Area(Rectangle) = 100
Therefore $$4x^2 - x(3x) = 100 => x^2 = 100 => x = 10$$(because side cannot be negative)
Hence, area of square is $$(2x)^2 = 20^2 = 400$$(Option C)
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Re: A piece of wire used to fence a rectangular land can be used to fence  [#permalink]

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11 Apr 2018, 22:40
+1 for option C. Let the length and width of the rectangle be l and w. The side of the square becomes 2w.
Reln b/w l and w is - l=3w. Solving we get the area of square as 400 i.e. option C.
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A piece of wire used to fence a rectangular land can be used to fence  [#permalink]

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13 Apr 2018, 18:04
DHAR wrote:
A piece of wire used to fence a rectangular land can be used to fence a square land that has twice the width of the rectangular land.What is the area of the square land if the area of the rectangular land is 100 sq units less than the area of the square land?

A. 100 sq units
B. 300 sq units
C. 400 sq units
D. 900 sq units
E. 1000 sq units

The square's area is a square number: Eliminate answers B and E

Eliminate Answer A. The difference between the square and rectangular areas is 100 sq units. Rectangular area cannot be 0. Only two answers remain.

Answer C) 400 = square's area
Square side, $$s=\sqrt{400}$$
$$s = 20$$

$$s$$ is twice the width of the rectangle:
$$s = 20 = 2W$$
$$W = 10$$

Rectangle's area is 100 sq units less than square:
$$(400 - 100) = 300$$

Rectangle's area $$300 = L*W$$
$$L =(\frac{300}{W})=(\frac{300}{10})=30$$

That works. KEEP

Square side, $$s = \sqrt{900}=30$$

Square side, $$s = 2W = 30$$
$$W = 15$$

Rectangle's area is 100 sq units less than square:
$$(900 - 100) = 800$$
Rectangle's area, $$A = L * W$$
$$800 = L * 15$$
$$L = ?$$
STOP. 800 cannot be divided evenly by 15. (800/15 = 53 R5). REJECT

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