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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, 21:03
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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 37081 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, 16:44
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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, 00:43
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Bunuel wrote:
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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


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, 00:35.
Last edited by Mo2men on 16 Oct 2015, 00:43, 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, 22:34
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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, 00:40
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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, 01:58
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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, 17:45
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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

Thanks,
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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, 20:09
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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, 07:01
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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, 12:54
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 19 Jun 2017, 22:01
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, 18:14
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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, 18:15
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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 14 Sep 2017, 21:18
Instead of taking it as 100 and 150 ft,
take it as 10 ft and 15 ft( multiplication is easier)
so ares becomes 15*10 = 150,
new parking lot area should be 300 sq ft
(15+w)*(10+w) = 300
by scanning the answer(instead of solving quadratic equation scanning the answer is easier, always start with the option C first
C(7.5, divide by 10, the 0 which we left in the first line) answer does not holds good, go up, B,
(10+5)*(15+5) = 300 holds good
Answer should be B
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Re: The smaller rectangle in the figure above represents the original size  [#permalink]

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New post 26 Jan 2019, 14:50
Bunuel wrote:
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






The area of the original triangle is:
100*150 = 15,000 square feet

The enlarged lot is twice the area of the original area:
15,000 * 2 = 30,000 square feet

Now we need to know what value (w) is added to 100 and 150 which will equal 30,000
(100+w) * (150+w) = 30,000

In this case, backsolving using the answer choice is quicker than solving for W.
Start with answer choice B (If B is too small, you can cross out A and B, and then test D, because the answers are in ascending order)

(100+50) * (150+50) =
150 * 200 = 30,000

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

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New post 09 Sep 2019, 03:28
Hey guys,

can you please, tell me how you got from this step:
w^2 + 250w – 15,000 = 0

to this final one:
(w + 300)(w – 50) = 0

?

I just cannot figure out how you factorised and ended up there.

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

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New post 09 Sep 2019, 03:37
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rudywip wrote:
Hey guys,

can you please, tell me how you got from this step:
w^2 + 250w – 15,000 = 0

to this final one:
(w + 300)(w – 50) = 0

?

I just cannot figure out how you factorised and ended up there.

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

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New post 09 Sep 2019, 03:47
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 satisfies the above equation.

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

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New post 09 Sep 2019, 05:39
given
2*(100*150)=(100+w)*(150+w)
plugin answer options ; w=50
IMO B

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

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New post 13 Oct 2019, 06:23
Basic geometry question that triggers your speed of thought and execution:

- Most important piece of information: We know that new area = 2x old area with old area= 150x100=15x10=150+2 zeroes at the end=15000. So 2x15000=30000.

-A=25 is too small, try with B=50 so it will be 200x150=20x15=300 with 2 zeroes at the end-30000.

Answer is B, with a very easy pattern that'll lead us to test answers to save time!
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The smaller rectangle in the figure above represents the original size   [#permalink] 13 Oct 2019, 06:23

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