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A rectangular solid is changed

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A rectangular solid is changed

by Zach.J.Dragone » Thu Dec 12, 2013 4:59 am
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

Spoiler:E

We are told that we need to solve for the height change i.e. H=4W+9 and H-9=4W....is that simply to get exactly one variable to solve for?

Thanks! :D
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by theCodeToGMAT » Thu Dec 12, 2013 6:08 am
w = l
(h - 9) = 4 (w) ==> h = 4w + 9

TO find: wlh

(w+1)(l+1)(h-9)
(w+1)(w+1)(4w)
(w+1)^2 4w

Old Volume = wlh ==> ww(4w+9)

Old Volume = NEw Volume.
w(4w+9) = (w+1)^2 * 4
4w^2 + 9w = 4(w^2 + 1 + 2w)
w = 4

Volume = (4+1)^2 * 4 * 4= 25 * 16 = 400
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by ganeshrkamath » Thu Dec 12, 2013 6:16 am
Zach.J.Dragone 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

Spoiler:E

We are told that we need to solve for the height change i.e. H=4W+9 and H-9=4W....is that simply to get exactly one variable to solve for?

Thanks! :D
W' = W + 1
L' = L + 1
H' = H - 9
V' = W'*L'*H' = V = W*L*H

W = L
H' = 4W

Write everything in terms of W
W' = W + 1
L' = L + 1 = W + 1
H' = 4W
V' = (W+1)^2 * 4W

L = W
H = H' + 9 = 4W + 9
V = (W)^2 * (4W + 9)

V' = V
(W+1)^2 * 4W = (W)^2 * (4W + 9)
(W^2 + 2W + 1) * 4W = W^2*(4W+9)
4W^2 + 8W + 4 = 4W^2 + 9W
W = 4

Volume = V' = 25*16 = 400

Choose E

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by Mathsbuddy » Thu Dec 12, 2013 7:39 am
a^2*(4a + 9) = (a+1)^2 * 4a
4a^3 + 9a^2 = 4a^3 + 8a^2 + 4a
a(a - 4) = 0
a = 4 (as we ignore a = 0 which it gives a volume of zero, i.e. a non-solid)

Substituting a = 4 into V = 4a^3 - 4a = 400
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by GMATGuruNY » Thu Dec 12, 2013 2:27 pm
Zach.J.Dragone 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

Spoiler:E
The answer choices -- which represent the volume of each solid -- are integers.
Implication:
The dimensions of the two solids also are integers.

When the height of the original solid is DECREASED BY 9, the resulting height is 4 TIMES the length and width of the original solid.
Since the resulting height is a MULTIPLE OF 4, the original height must be 9 MORE THAN A MULTIPLE OF 4.
Options:
4+9 = 13.
8+9 = 17.
12+9 = 21.
16+9 = 25.
Only the option in red yields an original height that is a factor of a majority of the answer choices.

Option in red: Original height = 25, decreased height = 25-9 = 16
Original solid:
Since the decreased height is 4 times the length and width of the original solid, L=4 and W=4.
Resulting volume = LWH = 4*4*25 = 400.

New solid:
Since the original length and width are each increased by 1 inch, L=5 and W=5.
Resulting volume = LWH = 5*5*16 = 400.

Success!
The volume in each case is the same: 400.

The correct answer is E.
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