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

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Intern
Joined: 25 Jan 2013
Posts: 25
In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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20 Feb 2013, 22:02
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Difficulty:

15% (low)

Question Stats:

77% (01:41) correct 23% (02:01) wrong based on 200 sessions

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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%
Manager
Joined: 25 Sep 2012
Posts: 238
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31
GMAT 2: 680 Q48 V34
Re: In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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09 Jun 2014, 11:39
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5
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 % = $$8 - 5 -\frac{8*5}{100} = 2.6$$
##### General Discussion
Intern
Joined: 18 Jan 2013
Posts: 2
Re: In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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20 Feb 2013, 22:43
1
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.
Cheers
Math Expert
Joined: 02 Sep 2009
Posts: 54543
Re: In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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21 Feb 2013, 03:36
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.

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Re: In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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12 Oct 2017, 18:00
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.

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Re: In measuring the sides of a rectangle, one side is taken 8% in excess  [#permalink]

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21 Mar 2019, 17:04
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Re: In measuring the sides of a rectangle, one side is taken 8% in excess   [#permalink] 21 Mar 2019, 17:04
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