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20 Feb 2013, 22:02
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E
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Difficulty:
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76% (01:42) correct
24%
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In measuring the sides of a rectangle, one side is taken 8% in excess and other 5% in deficit. Find the error percentage in the area calculated from these measurements.
A. 3.5%
B. 2.4%
C. 3%
D. 5%
E. 2.6%
A. 3.5%
B. 2.4%
C. 3%
D. 5%
E. 2.6%
09 Jun 2014, 11:39
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Percentage Change Formula
if there are two succesive changes to a quantity ( as in this case ) you can solve it by a simple formula
\(net change = a + b + \frac{ab}{100}\)
e.g. net change of to increase of 50% = \(50 + 50 + \frac{2500}{100} = 125\)
net change of increase in 8% and decrease of 5 % = \(8 - 5 -\frac{8*5}{100} = 2.6\)
if there are two succesive changes to a quantity ( as in this case ) you can solve it by a simple formula
\(net change = a + b + \frac{ab}{100}\)
e.g. net change of to increase of 50% = \(50 + 50 + \frac{2500}{100} = 125\)
net change of increase in 8% and decrease of 5 % = \(8 - 5 -\frac{8*5}{100} = 2.6\)
General Discussion
20 Feb 2013, 22:43
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antoxavier
Let the sides of the rectangle be a and b.
actual area of the rectangle without error = ab
New side a ( 8% in excess) = a+0.08a = 1.08a
New side b ( 5% in deficit) = b-0.05b = 0.95b
Area of the new rectangle = (1.08a)(0.95b)= (1.0260)ab
say a=1 and b=1
area of the original rectangle = 1
area of the new rectangle = 1.026
change in the area calculation = .026
error percentage in the area calculated = 2.6%
Let me know otherwise.
Cheers
