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!
A rectangular solid is changed
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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
(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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W' = W + 1Zach.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!
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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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
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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The answer choices -- which represent the volume of each solid -- are integers.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
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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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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