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Senior Manager  Joined: 13 May 2013
Posts: 399
A rectangular solid is changed such that the width and lengt  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 69% (03:25) correct 31% (03:37) wrong based on 402 sessions

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A rectangular solid is changed such that the width and length are increased by 1 inch apiece and the height is decreased by 9 inches. Despite these changes, the new rectangular solid has the same volume as the original rectangular solid. If the width and length of the original rectangular solid are equal and the height of the new rectangular solid is 4 times the width of the original rectangular solid, what is the volume of the rectangular solid?

(A) 18
(B) 50
(C) 100
(D) 200
(E) 400

Ok, so here is what I know:

~Old Volume = New Volume

~(L+1)(W+1)(H-9) = (L*W*H)

~(H-9) = 4w

~width, length of original rectangular are equal

So, from that I get:

(L+1)(W+1)(4w)=(W*W*H)

But in the book, the equation differs from mine in two ways.

For starters, mine is L+1)(W+1)(4w)=(W*W*H) while theirs is (W+1)(W+1)(4w)=(W*W*H)

Also, because (H-9)=4w, they derive H=4w+9 then plug it in so (W+1)(W+1)(4w)=(W*W*4w+9)

But here is my (apparently incorrect) reasoning.

L=W in the old rectangle, so why plug "W" into the new rectangle volume?

if (H-9)=4w, then why do I plug 4w into the new rectangle volume and h=4w-9 into the old rectangle formula? It seems unnecessary to have to plug in 4w for (h-9) then derive h=4w-9 and plug in on the other side.

What's the reasoning!

Thanks!

Originally posted by WholeLottaLove on 26 May 2013, 12:21.
Last edited by Bunuel on 27 May 2013, 01:49, edited 1 time in total.
Edited the question and moved to PS forum.
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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WholeLottaLove wrote:
A rectangular solid is changed such that the width and length are increased by 1 inch apiece and the height is decreased by 9 inches. Despite these changes, the new rectangular solid has the same volume as the original rectangular solid. If the width and length of the original rectangular solid are equal and the height of the new rectangular solid is 4 times the width of the original rectangular solid, what is the volume of the rectangular solid?

(A) 18
(B) 50
(C) 100
(D) 200
(E) 400

Let's try this question using only the change in volume. Since original volume = final volume, Decrease in volume = Increase in volume

You decrease volume by chopping off 9 inches of the height. Decrease = 9*w*l
You increase the volume now by adding an inch to the width and length. The height remains h-9.
Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) (make a rectangle and increase its width and length by 1 to see how area changes. This tells you how volume changes by just considering the height as well)

So 9*w*l = (h-9)*(l+w+1)
Given that (h-9) = 4w and l = w, substitute both in the equation to get
9*w*w = 4w*(2w+1)
Cancel w from both sides and get w = 4 = l
h-9 = 4w so h = 16+9 = 25

Volume = wlh = 4*4*25 = 400
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Re: A rectangular solid is changed such that  [#permalink]

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Ok, so here is what I know:
~Old Volume = New Volume
~(L+1)(W+1)(H-9) = (L*W*H)
~(H-9) = 4w
~width, length of original rectangular are equal
So, from that I get:
(L+1)(W+1)(4w)=(W*W*H)
But in the book, the equation differs from mine in two ways.
For starters, mine is L+1)(W+1)(4w)=(W*W*H) while theirs is (W+1)(W+1)(4w)=(W*W*H)

If L=W then substitute L with W in the equation and obtain
$$(W+1)(W+1)(4W)=(W*W*H)$$- the one in the book

But here is my (apparently incorrect) reasoning.
L=W in the old rectangle, so why plug "W" into the new rectangle volume?
if (H-9)=4w, then why do I plug 4w into the new rectangle volume and h=4w-9 into the old rectangle formula? It seems unnecessary to have to plug in 4w for (h-9) then derive h=4w-9 and plug in on the other side.

You have
$$(W+1)(W+1)(H-9)=(W*W*H)$$ and in order to solve it you need to express H in terms of W, so from $$H-9=4w$$ you get $$H=4W+9$$ and $$H-9=4W$$
$$(W+1)(W+1)(4W)=(W*W*(4W+9))$$

The above equation is in W and H, so have to express all the variables in one term (W) in order to solve it. You have to plug in those values at both sides in order to solve the equation.

Hope it's clear, let me know
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Zarrolou, I followed your approach and I got the final quadratic equation as $$w^2-4w=0$$---->w=4. Then we get h=4w+9=25. Also l=w=4. Volume is 4*4*25=400. Your approach is correct and I don't think there is any other possibility.
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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VeritasPrepKarishma wrote:
WholeLottaLove wrote:
A rectangular solid is changed such that the width and length are increased by 1 inch apiece and the height is decreased by 9 inches. Despite these changes, the new rectangular solid has the same volume as the original rectangular solid. If the width and length of the original rectangular solid are equal and the height of the new rectangular solid is 4 times the width of the original rectangular solid, what is the volume of the rectangular solid?

(A) 18
(B) 50
(C) 100
(D) 200
(E) 400

Let's try this question using only the change in volume. Since original volume = final volume, Decrease in volume = Increase in volume

You decrease volume by chopping off 9 inches of the height. Decrease = 9*w*l
You increase the volume now by adding an inch to the width and length. The height remains h-9.
Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) (make a rectangle and increase its width and length by 1 to see how area changes. This tells you how volume changes by just considering the height as well)

So 9*w*l = (h-9)*(l+w+1)
Given that (h-9) = 4w and l = w, substitute both in the equation to get
9*w*w = 4w*(2w+1)
Cancel w from both sides and get w = 4 = l
h-9 = 4w so h = 16+9 = 25

Volume = wlh = 4*4*25 = 400

Hi Karishma/Bunnel,

Any more such questions?
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Joined: 02 Sep 2009
Posts: 58434
Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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cumulonimbus wrote:
VeritasPrepKarishma wrote:
WholeLottaLove wrote:
A rectangular solid is changed such that the width and length are increased by 1 inch apiece and the height is decreased by 9 inches. Despite these changes, the new rectangular solid has the same volume as the original rectangular solid. If the width and length of the original rectangular solid are equal and the height of the new rectangular solid is 4 times the width of the original rectangular solid, what is the volume of the rectangular solid?

(A) 18
(B) 50
(C) 100
(D) 200
(E) 400

Let's try this question using only the change in volume. Since original volume = final volume, Decrease in volume = Increase in volume

You decrease volume by chopping off 9 inches of the height. Decrease = 9*w*l
You increase the volume now by adding an inch to the width and length. The height remains h-9.
Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) (make a rectangle and increase its width and length by 1 to see how area changes. This tells you how volume changes by just considering the height as well)

So 9*w*l = (h-9)*(l+w+1)
Given that (h-9) = 4w and l = w, substitute both in the equation to get
9*w*w = 4w*(2w+1)
Cancel w from both sides and get w = 4 = l
h-9 = 4w so h = 16+9 = 25

Volume = wlh = 4*4*25 = 400

Hi Karishma/Bunnel,

Any more such questions?

Somewhat similar questions:
a-closed-aluminum-rectangular-box-has-inner-dimensions-x-141049.html
the-measurements-obtained-for-the-interior-dimensions-of-a-160295.html
m01-70731.html

Hope this helps.
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Its hard to understand the increase & decrease thing in provided explanation. Please help me ..
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Here's how i solved it

Rectangle 1: length = l, width = w and height = h. Now l=w.
so Volume V1 = l(squared)h

Rectangle 2: length = l+1, width = w+1 or l+1 and height = h-9.
So, volume V2 = (l+1)squared(h-9)

Now new height = 4 times previous width => h-9 = 4w or (h-9) = 4l => h = 4l + 9

Substituting the value of h:-
Volume V1 = l(squared)(4l+9) = 4l(cube) + 9 l(squared)

and Volume V2 = (l+1)squared(4l) = 4lcube + 8lsquared = 4l

Equating V1 and V2 we get: l = 0, 4. since length isnt 0, l=4 => w=4 and h = 16+9 = 25

therefore volume V = lwh = 4x4x25 = 400
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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luckyme17187 wrote:
Its hard to understand the increase & decrease thing in provided explanation. Please help me ..

The increase/decrease method has fewer calculations and steps but its certainly a little trickier...
Try to make diagrams at each step to understand what is going on.
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Hi VeritasPrepKarishma,
I tried to solve the question by increase=decrease method. I took increase first and equated the same to decrease keeping the length and breadth same. Equation was:
1*b*h + l*1*h = (l+1)(b+1)*9 -> Keeping the length and breadth same. As l=b,
-> 2lh = (l+1)² *9 ; h-9= 4l -> h = 9+4l
-> 2l*(4l+9) = (l²+2l+1)9
-> 8l²+18l = 9l²+18l+9
As you can see , there is not definite solution with this equation. Can you please point out the flaw in the method?
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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VeritasPrepKarishma wrote:
WholeLottaLove wrote:
A rectangular solid is changed such that the width and length are increased by 1 inch apiece and the height is decreased by 9 inches. Despite these changes, the new rectangular solid has the same volume as the original rectangular solid. If the width and length of the original rectangular solid are equal and the height of the new rectangular solid is 4 times the width of the original rectangular solid, what is the volume of the rectangular solid?

(A) 18
(B) 50
(C) 100
(D) 200
(E) 400

Let's try this question using only the change in volume. Since original volume = final volume, Decrease in volume = Increase in volume

You decrease volume by chopping off 9 inches of the height. Decrease = 9*w*l
You increase the volume now by adding an inch to the width and length. The height remains h-9.
Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) (make a rectangle and increase its width and length by 1 to see how area changes. This tells you how volume changes by just considering the height as well)

So 9*w*l = (h-9)*(l+w+1)
Given that (h-9) = 4w and l = w, substitute both in the equation to get
9*w*w = 4w*(2w+1)
Cancel w from both sides and get w = 4 = l
h-9 = 4w so h = 16+9 = 25

Volume = wlh = 4*4*25 = 400

Shouldn't Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) be Increase = 1*(l+1)*(h-9) + 1*(w+1)*(h-9) = (h-9)*(l+w+2) since the width increased by 1 too?
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Width .............. Length .................. Height

a ........................ a .......................... b (Original dimensions; Length = Width)

a+1 ................... a+1 ....................... b-9 (Post changes)

Given that (b-9) = 4a

b = 4a+9 .............. (1)

Original Volume = Post change Volume

$$a^2* b = (a+1)^2 * (b-9)$$

Placing value of b from (1) in the above equation

$$a^2* (4a+9) = (a+1)^2 * (4a+9-9)$$

a = 4

Volume = 4 * 4 * 25 = 400

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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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Hello,

I really liked the increase/decrease method.

Could you please suggest how you derived this equation:

Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1)

thank you.
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Re: A rectangular solid is changed such that the width and lengt  [#permalink]

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vikasbansal227 wrote:
Hello,

I really liked the increase/decrease method.

Could you please suggest how you derived this equation:

Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1)

thank you.

Attachment: InceaseDecrease.jpg [ 470.15 KiB | Viewed 7976 times ]

So in step 1, you chop off a block to decrease the area. The area of that block is l*w*9

Then in step 2, you add a block to increase the length. The area of this block is 1*w*(h-9)

Then you add another block to increase the width. The area of this block is 1*(l + 1)*(h - 9)

The decrease = Both increases
l*w*9 = 1*w*(h-9) + 1*(l+1)*(h - 9)
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