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If the length of a rectangle is increased by 20 percent and the width [#permalink]
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Question Stats: Hide Show timer StatisticsIf 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% _________________
Originally posted by Bunuel on 16 Oct 2017, 08:53. Renamed the topic and edited the question.
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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink]
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 valuesLet L = 10Let W = 10 (a square is a rectangle)Original area = (LW) = 10*10 = 100L increases by 20 percent: (1.2)(10) = 12W decreases by 20 percent:(0.8)(10) = 8New area: (12)*(8) = 96Percent change:\(\frac{New-Old}{Old} * 100\)\(\frac{96-100}{100}=\) \((-.04) * 100\) = 4 percent decrease Algebra L = lengthW = widthArea = (1)LWLength increases by 20 percent, width decreases by 20 percent:1.2L * 0.8W = (.96) LW.96 is 4 percent less than 1The area decreases by 4 percentAnswer B _________________
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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink] 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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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink]
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.
—The only thing more dangerous than ignorance is arrogance. ~Einstein —I stand with Ukraine.
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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink]
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]
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?Sent from my MI 5 using GMAT Club Forum mobile app
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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink]
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) = 0.96LWDoes that make sense? _________________
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Re: If the length of a rectangle is increased by 20 percent and the width [#permalink]
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] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: If the length of a rectangle is increased by 20 percent and the width [#permalink] 21 Apr 2022, 23:33 |
What is the percentage change in the area of rectangle if its length decreases by 15% and its width is decreased by 20%?
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