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01 May 2012, 02:36
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D
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Difficulty:
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83% (02:17) correct
17%
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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)
A. 18
B. 10
C. 24
D. 12
Source: Mathematics by L.Harwood Clarke ( page: 168, No.11)
01 May 2012, 02:44
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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
Answer "D"
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
Answer "D"
03 Feb 2019, 00:18
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breadth = b
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
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
