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What is the area of a rectangle whose length is twice its width and wh

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What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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New post 31 Jan 2019, 22:04
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A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

78% (01:36) correct 22% (01:48) wrong based on 29 sessions

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New post 31 Jan 2019, 23:53
Bunuel wrote:
What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?

A) \(1\)

B) \(6\)

C) \(\frac{2}{3}\)

D) \(\frac{4}{3}\)

E) \(\frac{8}{9}\)


l = 2*w

Perimeter of the rectangle \(= 2*(l+w) = 4*1\) (Side of square with area 1 will be 1)

i.e. \((l+w) = 2\)

i.e. \(2w+w = 3w = 2\)

i.e. \(w = 2/3\) and \(l = 4/3\)

Area of rectangle \(= l*w = (2/3)*(4/3) = 8/9\)

Answer: Option E
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Re: What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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New post 01 Feb 2019, 10:57
Bunuel wrote:
What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?

A) \(1\)

B) \(6\)

C) \(\frac{2}{3}\)

D) \(\frac{4}{3}\)

E) \(\frac{8}{9}\)


l=2w
2*( l+w)= 4
6w=4
w=2/3
l=4/3
l*w= 8/9
IMO E
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What is the area of a rectangle whose length is twice its width and wh  [#permalink]

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New post 03 Feb 2019, 18:30
Bunuel wrote:
What is the area of a rectangle whose length is twice its width and whose perimeter is equal to that of a square whose area is 1?

A) \(1\)

B) \(6\)

C) \(\frac{2}{3}\)

D) \(\frac{4}{3}\)

E) \(\frac{8}{9}\)

This problem just requires meticulousness.

Start with the square. Find side length from area.

(1) area of square = 1
gives side length, \(s\)

Area of square = \(1=s^2\)
\(\sqrt{1}=\sqrt{s^2}\)
\(s=1\)

(2) Perimeter of square =
\(4s=(4*1)=4=\)
perimeter of rectangle

Use rectangle perimeter to find side lengths of rectangle.
From side lengths, find area.

(3) Perimeter of rectangle: side lengths?

Perimeter, \(P=4\)
\(2l+2w=P\)
\(2l+2w=4\)
\(2(l+w)=4\) - (divide by 2)
\((l+w)=2\)

The length of the rectangle is twice its width

\(l=2w\)
Substitute \(2w\) for \(l\)

\((l+w)=2\)
\(2w+w=2\)
\(3w=2\)
\(w=\frac{2}{3}\)
\(l=2w=(2*\frac{2}{3})=\frac{4}{3}\)

(4) Area, A, of rectangle?
\(A=l*w\)

\(A=(\frac{4}{3}*\frac{2}{3})=\frac{8}{9}\)

Answer E
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What is the area of a rectangle whose length is twice its width and wh   [#permalink] 03 Feb 2019, 18:30
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