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In measuring the sides of a rectangle, one side is taken 8% in excess

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antoxavier
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In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  20 Feb 2013, 22:02
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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%
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b2bt
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  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 % = \(8 - 5 -\frac{8*5}{100} = 2.6\)
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sanman
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  20 Feb 2013, 22:43
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antoxavier wrote:
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%


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.
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Bunuel
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  21 Feb 2013, 03:36
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antoxavier wrote:
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%


Say both sides of the rectangle are equal to 100 (so consider that we have a square). In this case the area is 100*100=10,000.

Now, the area obtained with wrong measurements would be 108*95=10,260, which is 2.6% greater than the actual area.

Answer: E.
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  12 Oct 2017, 18:00
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antoxavier wrote:
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%


We can let the original area = LW. Thus, the new area is:

1.08L x 0.95W = 1.026LW

Thus, the error percentage is 0.026 = 2.6 percent.

Answer: E
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  13 Aug 2020, 10:47
antoxavier wrote:
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%


We will take for ease of calculations that length increased and breadth decreased
Excess of 8% in length= (108) l

Deficit in breadth = (095)b

Therefore new area= (1.08) l * (95)b=102.60
Which is difference of 2.6%
E
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink] New post  04 Oct 2023, 12:19
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess [#permalink]
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