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If the length of a rectangle is increased by 20 percent and the width

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If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 16 Oct 2017, 09:53
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If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%

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If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 16 Oct 2017, 12:34
Bunuel wrote:
If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%

Assign values
Let L = 10
Let W = 10 (a square is a rectangle)

Original area = (LW) = 10*10 = 100

L increases by 20 percent:
(1.2)(10) = 12

W decreases by 20 percent:
(0.8)(10) = 8

New area: (12)*(8) = 96

Percent change:\(\frac{New-Old}{Old} * 100\)

\(\frac{96-100}{100}=\)
\((-.04) * 100\) = 4 percent decrease

Algebra
L = length
W = width
Area = (1)LW

Length increases by 20 percent, width decreases by 20 percent:

1.2L * 0.8W = (.96) LW

.96 is 4 percent less than 1

The area decreases by 4 percent

Answer B
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If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 23 Oct 2017, 07:02
One thing I try to remind myself is that squares are rectangles.
-So I started with a square with sides of 5, area would equal 25. I chose 5 as the size of the sides because it's easy to increase and decrease by 20%.
-Length is increased by 20% 5+1=6.
-Width is decreased by 20%, 5-1=4
-New area is 4*6=24.
-(24-25)/25=-1/25=-4%.


-My biggest problem is dumb mistakes, if positive 4% was an answer, I might have chosen it. If anyone has any advice on dumb mistakes, please share.
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If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 23 Oct 2017, 09:41
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Plunkster82 wrote:
One thing I try to remind myself is that squares are rectangles.
-So I started with a square with sides of 5, area would equal 25. I chose 5 as the size of the sides because it's easy to increase and decrease by 20%.
-Length is increased by 20% 5+1=6.
-Width is decreased by 20%, 5-1=4
-New area is 4*6=24.
-(24-25)/25=-1/25=-4%.


-My biggest problem is dumb mistakes, if positive 4% was an answer, I might have chosen it. If anyone has any advice on dumb mistakes, please share.

Plunkster82 , there are threads here (and info elsewhere) that answer your question.

My two cents':

1) check your answers. On this kind of question, try different numbers. The square as rectangle is a great idea. But with percentages, try to use values such as 10 and 100 that give you a base of 100. Try 10 here as original L and W. You'll get 96 vs 100: Four percent smaller.

2) write down all but the most simple things, neatly. One of my favorite experts at gmatclub is a fan of mental math. I am not.

3) Check out these threads:

Tip of the Week, posts from Brian Galvin at Veritas Prep and great links from Bunuel: HERE

Another, HERE

Another, HERE

After those, I would run two Google searches with these terms:
1) GMAT dumb silly mistakes gmatclub; and
2) same, but omit "gmatclub"

Hope it helps. :-)
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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 26 Oct 2017, 10:44
Bunuel wrote:
If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%


We can let L = the original length and W = the original width; thus, the original area of the rectangle is LW and the new area is 1.2L x 0.8W = 0.96LW, so the area is decreased by 4%.

Answer: B
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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 07 Mar 2018, 07:38
JeffTargetTestPrep wrote:
Bunuel wrote:
If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%


We can let L = the original length and W = the original width; thus, the original area of the rectangle is LW and the new area is 1.2L x 0.8W = 0.96LW, so the area is decreased by 4%.

Answer: B

New area how come 1.2l?

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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 07 Mar 2018, 07:56
Akshusaya wrote:
New area how come 1.2l?

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Hey Akshusaya ,

1.2 L is the the new length when the original length was increased by 20%.

New area = (1.2 L) * (0.8W) = 0.96LW

Does that make sense?
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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 07 Mar 2018, 08:10
Bunuel wrote:
If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%


\(20 - 20 +\frac{(20*-20)}{100}\)

= \(-4\)

There will be a net decrease of 4% , answer must be (B)
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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 07 Mar 2018, 08:21
abhimahna wrote:
Akshusaya wrote:
New area how come 1.2l?

Sent from my MI 5 using GMAT Club Forum mobile app


Hey Akshusaya ,

1.2 L is the the new length when the original length was increased by 20%.

New area = (1.2 L) * (0.8W) = 0.96LW

Does that make sense?

Yes.Perfect
Thanks Abhimahna

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Re: If the length of a rectangle is increased by 20 percent and the width  [#permalink]

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New post 07 Mar 2018, 08:27
Abhishek009 wrote:
Bunuel wrote:
If the length of a rectangle is increased by 20 percent and the width is decreased by 20 percent, then the area

(A) decreases by 20%
(B) decreases by 4%
(C) stays the same
(D) increases by 10%
(E) increases by 20%


\(20 - 20 +\frac{(20*-20)}{100}\)

= \(-4\)

There will be a net decrease of 4% , answer must be (B)

Perfect solution..
Thanks Abhishek

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Re: If the length of a rectangle is increased by 20 percent and the width &nbs [#permalink] 07 Mar 2018, 08:27
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