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# Two rectangles have the same breadth and the difference

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Intern
Joined: 27 Sep 2011
Posts: 13

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01 May 2012, 01:36
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Difficulty:

15% (low)

Question Stats:

89% (02:18) correct 11% (03:34) wrong based on 31 sessions

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Two rectangles have the same breadth and the difference between their perimeters is 56m. If the difference between their areas is 336m^2, find the breadth of the rectangles.

A. 18
B. 10
C. 24
D. 12

Source: Mathematics by L.Harwood Clarke ( page: 168, No.11)

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All I want to be A Self-made Man

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Joined: 25 Apr 2012
Posts: 11

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01 May 2012, 01:44
Let the common breadth be 'b'
and lengths be x and y
Diff. of perimeters = 2(x+b) - 2(y+b) = 2(x-y) = 56
=> x-y = 28
Diff. of areas = bx - by = b(x-y) = 336
=> b = 336/28 = 12
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Ravi Sankar Vemuri

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Joined: 11 Dec 2013
Posts: 114
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)

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02 Feb 2019, 23:18
1
Length of first rectangle = x
Length of first rectangle = y

Difference in perimeters $$= 2(x+b) - 2(y+b) = 2(x-y) = 56 \rightarrow x-y = 28$$
Difference in areas $$= bx - by = b(x-y) = 336 \rightarrow b = 336/28 = 12$$

IMO D
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Joined: 09 Mar 2016
Posts: 1284

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03 Feb 2019, 02:36
4d wrote:
Length of first rectangle = x
Length of first rectangle = y

Difference in perimeters $$= 2(x+b) - 2(y+b) = 2(x-y) = 56 \rightarrow x-y = 28$$
Difference in areas $$= bx - by = b(x-y) = 336 \rightarrow b = 336/28 = 12$$

IMO D

4d
how did you get 28 ?
Manager
Joined: 11 Dec 2013
Posts: 114
Location: India
GMAT Date: 03-15-2015
WE: Education (Education)

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03 Feb 2019, 02:48
1
dave13 wrote:
4d wrote:
Length of first rectangle = x
Length of first rectangle = y

Difference in perimeters $$= 2(x+b) - 2(y+b) = 2(x-y) = 56 \rightarrow x-y = 28$$
Difference in areas $$= bx - by = b(x-y) = 336 \rightarrow b = 336/28 = 12$$

IMO D

4d
how did you get 28 ?

dave13

$$2(x+b)−2(y+b)=2x+2b-2y-2b =2x-2y=2(x−y)=56$$

$$2(x−y)=56$$ [divide by 2]
$$x-y=28$$
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Re: Two rectangles have the same breadth and the difference   [#permalink] 03 Feb 2019, 02:48
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