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The smaller rectangle in the figure above represents the original size

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The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 15 Oct 2015, 22:03
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Question Stats:

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The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


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Attachment:
2015-10-16_0901.png
2015-10-16_0901.png [ 6.68 KiB | Viewed 23340 times ]

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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 17 Oct 2015, 17:44
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Hi Nylal,

The 'shortcut' that you're looking for is to actually AVOID doing the Quadratic altogether and instead TEST THE ANSWERS. Since you know that you're looking for a total area that equals 30,000, and you have 5 answers that are numbers, you just have to do enough basic multiplication to find the one answer that gives you that exact area.

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The smaller rectangle in the figure above represents the original size  [#permalink]

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New post Updated on: 16 Oct 2015, 01:43
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


Area of big Rec= 2*Area of smaller rec=30000

(100+w)(150+w)=30000. By scanning and Testing the answers, so 50 will fit.
Answer B

Originally posted by Mo2men on 16 Oct 2015, 01:35.
Last edited by Mo2men on 16 Oct 2015, 01:43, edited 2 times in total.
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 15 Oct 2015, 23:34
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


Area of the small rectangle: 150*100=15000
Area of the big one: 30000

(100+w)*(150+w)=30000 -> \(15000+100w+150w+w^2=30000 --> w^2+250w-15000=0\)
w=-300,50 Answer (B) it cannot be negative.
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 16 Oct 2015, 01:40
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


(100+w)(150+w) = 2*100*150
=> (w-50)(w+300)=0
=> w=50

Ans: B
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 16 Oct 2015, 02:58
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png



My Solution:

Area of smaller rectangle : 100 x 150 = 15000 sqft

Area of larger rectangle : (100+w) X (150+w) = 15000 x 2 (Given,Twice the area of smaller rectangle)

Solving this, w^2+250w-15000=0

Solutions of this equation are w=-300 and 50.

As dimension can't be negative, value of w is 50. Option B

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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 16 Oct 2015, 18:45
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We can write from given equation that area of bigger one is twice the smaller one; so (100+w)(150+w) = 2*100*150
there fore solving above equation we get: w^2 +250w-100*150=0; solving this quadratic equation we get w = 50 or negative value. As negative value is not possible.
Hence answer is 50; B

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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 16 Oct 2015, 21:09
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The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200

Area of smaller rectangle = 100 * 150 = 15000
Area of larger = 2 * smaller = 30,000

Now given that both the length were increased by w .
Hence new length =(100 + W ) and (150 + w)

Area => 30,000 = (100 + W ) and (150 + w)

Now use the option to check the value .
substitute on the place of W .

Option B perfectly fits the bill . hence the answer is B .

Always try to use number and option . It is faster and easy than solving equation .
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 17 Oct 2015, 08:01
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


Area of original Plot = 100*150

Area of Enlarged Plot = (100+w)*(150+w)

Given (100+w)*(150+w) = 2*(100*150)

i.e. (100+w)*(150+w) = 30000

Checking Options:

(A) 25 i.e. (100+w)*(150+w) = (100+25)*(150+25) (unit digit non-zero) is NOT equal to 30000 hence INCORRECT
(B) 50 i.e. (100+w)*(150+w) = (100+50)*(150+50) = 150*200 = 30,000 hence CORRECT
(C) 75
(D) 100
(E) 200

Answer: Option B
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 17 Oct 2015, 13:54
One question (not to the problem itself):

Is there any "shortcut" to find the solutions of the quation 15000+100w+150w+w^2=30000 --> w^2+250w-15000=0?
Easy problem in general but took me (in total) almost 2min to find numbers that sum up to 250 and multiply to -15000?

Maybe I'm missing one of the "basic" concepts?
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 13 Feb 2016, 20:50
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GMATinsight wrote:
Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


Area of original Plot = 100*150

Area of Enlarged Plot = (100+w)*(150+w)

Given (100+w)*(150+w) = 2*(100*150)

i.e. (100+w)*(150+w) = 30000

Checking Options:

(A) 25 i.e. (100+w)*(150+w) = (100+25)*(150+25) (unit digit non-zero) is NOT equal to 30000 hence INCORRECT
(B) 50 i.e. (100+w)*(150+w) = (100+50)*(150+50) = 150*200 = 30,000 hence CORRECT
(C) 75
(D) 100
(E) 200

Answer: Option B


This looks to be a very good approach when compared to traditional Methods....

Thanks for your solution....
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 07 Feb 2017, 14:24
Nylal wrote:
One question (not to the problem itself):

Is there any "shortcut" to find the solutions of the quation 15000+100w+150w+w^2=30000 --> w^2+250w-15000=0?
Easy problem in general but took me (in total) almost 2min to find numbers that sum up to 250 and multiply to -15000?

Maybe I'm missing one of the "basic" concepts?


Hi Nylal how about :

(150+W) * (100+W)= 2* 100*150
150*100+(100+150)w + W^2= 2*100*150
--> 250w +w^2 = 2*100*150-(150*100)
--> w(250+w) = 150*100

Now from the option we get Option B as 50 *300 or (50*3)*100 = 150*100.
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 01 Jun 2017, 04:09
Solution for Quadratic equation: aX2 + bX+c=0
is, X=[-b+-Sqrt(b2-4ac)]/2
in this case the equation is: w2+250w-150*100=0
hence, w=[-250+-sqrt(250*250+4*1*150*1000)]/2
or, w=[-250+-350]/2=50,-300
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 19 Jun 2017, 23:01
hi..
I have a kind of vauge understanding of the question. Correct me, where i am going wrong..
"If the area of the enlarged lot is twice the area of the original lot"
Original is 15000. ENlarged lot is 30000.
So the total area would be 45000...
I know everyone cannot be wrong... help me understand the way all do...
Thanks in advance
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 20 Jun 2017, 19:14
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Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


We are given a diagram, which represents a parking lot, with a smaller rectangle inside a larger rectangle. The dimensions of the smaller rectangle are 100 ft. by 150 ft., and the dimensions of the larger rectangle are (100 + w) ft. by (150 + w) ft. Since the area of the larger rectangle is twice that of the smaller rectangle, we can create the following equation:

(w + 100)(w + 150) = 2(100 x 150)

w^2 + 250w + 15,000 = 30,000

w^2 + 250w – 15,000 = 0

(w + 300)(w – 50) = 0

w = -300 or w = 50

Since w must be positive, w must be 50.

Answer: B
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 20 Jun 2017, 19:15
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Hi JJSHHShank,

Based on the drawing, we can see that the original lot is INSIDE the enlarged lot, so it's not correct to call the total area 45,000 sq. ft. (since that would 'count' the area of the original lot twice). If you wanted to, you can use the two areas (enlarged lot = 30,000 sq. ft. and the original lot = 15,000 sq. ft.) to determine the area that is NOT a part of the original lot. That 'extra area' is 30,000 - 15,000 = 15,000 additional sq. ft.

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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 20 Jun 2017, 23:12
imo b
(100+w) X (150+w) = 15000 x 2
w^2+250w-15000=0
w=50,-300
so w=50
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 23 Jun 2017, 08:58
Thanks Arvind,Empower and Jeff prep for the reply:
From the discussion above, i understand that the enlarged lot includes the original lot too...
So shudnt add it twice...
Is there something wrong with the english understanding i have, or its just that i hv to start understanding it that way??
Thanks
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 26 Jun 2017, 07:47
30000 = (100+w)*(150+w)

only option B satisfies this equation.

hence answer is B
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The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 26 Jun 2017, 11:32
Bunuel wrote:
Image
The smaller rectangle in the figure above represents the original size of a parking lot before its length and width were each extended by w feet to make the larger rectangular lot shown. If the area of the enlarged lot is twice the area of the original lot, what is the value of w?

(A) 25
(B) 50
(C) 75
(D) 100
(E) 200


Kudos for a correct solution.

Attachment:
2015-10-16_0901.png


As given
let area of the bigger rectangle be a
then,
a=2(150*100)
a=30p
where p=1000

can be solved in two ways.

1.substitute the values back , and we quickly come to know that it is B
(150+50)*(100+50)=30p

2. 2(150*100)=(100+w)*(150+w)
a bit lengthy , but it does give w=50
so B
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The smaller rectangle in the figure above represents the original size   [#permalink] 26 Jun 2017, 11:32

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