An open box is formed from a square piece of cardboard, by removing squares of side 5 in. from each corner and folding up the sides. If the volume of the carton is then 40 in^3, what was the length of a side of the original square of cardboard?
A. 5+2sqrt2 in.
B. 5+4sqrt5 in.
C. 10+2sqrt2 in.
D. 10+4sqrt5 in.
E. 2
Cardboard Box
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 44
- Joined: Sun Jun 28, 2009 8:58 am
- Location: Online
- Thanked: 13 times
- GMAT Score:51
Assume the length of the original card board is x.
Then the volume of the expressed as 5 * (x -10) * (x - 10).
Hence 5 * (x -10) * (x - 10) = 40
(x-10)^2 = 8
x-10 = sqrt(8) or x-10 = -sqrt(8)
x = 10 + 2*sqrt(2) or x = 10 - 2*sqrt(2)
C is your answer.
Then the volume of the expressed as 5 * (x -10) * (x - 10).
Hence 5 * (x -10) * (x - 10) = 40
(x-10)^2 = 8
x-10 = sqrt(8) or x-10 = -sqrt(8)
x = 10 + 2*sqrt(2) or x = 10 - 2*sqrt(2)
C is your answer.
FREE Weekly Online Office Hours
Upcoming FREE Online Class: Effective Calculations - 11/15/09 Sunday
Instructor | GMATPrepMath.com | GMAT Prep Courses Starting at $60 with FREE GMAT Math Formula Sheet
Choose from Topics:
Number Properties - Advanced Counting (Combinatorics) - Probability - Advanced Algebra in DS - Geometry - Word Problems - Fundamental Algebra in PS - Effective Calculations
Upcoming FREE Online Class: Effective Calculations - 11/15/09 Sunday
Instructor | GMATPrepMath.com | GMAT Prep Courses Starting at $60 with FREE GMAT Math Formula Sheet
Choose from Topics:
Number Properties - Advanced Counting (Combinatorics) - Probability - Advanced Algebra in DS - Geometry - Word Problems - Fundamental Algebra in PS - Effective Calculations
- Morgoth
- Master | Next Rank: 500 Posts
- Posts: 316
- Joined: Mon Sep 22, 2008 12:04 am
- Thanked: 36 times
- Followed by:1 members
Two squares of each 5 inches are taken out from the corner.
Therefore minimum length of the original cardboard is 5+5 = 10
You can eliminate all the options except C and D.
The cube or cuboid formed has a volume of 40. We no idea weather it is cuboid or cube. I think this info is missing. However, if we assume it to be a cube:
cube root (40) = 2cuberoot5
10+2cuberoot5
I dont understand why are the options in sqroot.
Therefore minimum length of the original cardboard is 5+5 = 10
You can eliminate all the options except C and D.
The cube or cuboid formed has a volume of 40. We no idea weather it is cuboid or cube. I think this info is missing. However, if we assume it to be a cube:
cube root (40) = 2cuberoot5
10+2cuberoot5
I dont understand why are the options in sqroot.
- Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Great start, but remember that B is also > 10. We can narrow it down to C and D because we know that the length is 5 + 5 + length of the base of the box, so the answer should be in the form 10 + something.Morgoth wrote:Two squares of each 5 inches are taken out from the corner.
Therefore minimum length of the original cardboard is 5+5 = 10
You can eliminate all the options except C and D.
Now we go 1 step further.
We know that the height of our open box is 5, since we lifted a flap of length 5 to form the outside of the box.
We also know that the base of the box is a square, since we started with a square and cut away the same amount from each dimension.
So, we know that our box has height 5 and length and width x.
Accordingly:
5(x^2) = 40
x^2 = 8
x = root8 = 2root2
So, the length of a side of the original square is:
5 (that we lifted up) + 2root2 (what's left on the base) + 5 (that we lifted up)
= 10 + 2root2
choose (C).
Note that letting x be the base of the box instead of the original length of the cardboard makes the math simpler.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course