What is the ratio of the surface area of a cube to the surface area of a rectangular solid identical to the cube in all ways except that its length has been doubled?
A)1/4
B)3/8
C)1/2
D)3/5
E)2
OAD
Surface area
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Hi J-shreyans,
With these types of Geometry questions, you might find it helpful to draw the shapes and label the side lengths accordingly.
Here, we're asked to compare the SURFACE AREAS of a cube and a rectangular-solid that has twice the length of the cube.
Let's start with the dimensions of both shapes:
Cube = (length)(width)(height) = (X)(X)(X)
Rect. = (2length)(width)(height) = (2X)(X)(X)
To find total surface area, we need to find the areas of each of the surfaces and then add them up.
Cube = 6 identical surfaces = 6 squares = 6(X^2)
Rect. Solid = 2 identical square "ends" and 4 identical rectangular "body pieces"
Rect. Solid = 2(X^2) + 4(2X^2) = 2(X^2) + 8(X^2) = 10(X^2)
We're asked for the ratio of the surface area of the cube to the surface area of the rectangular solid:
6(X^2)/10(X^2) = 6/10 = 3/5
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
With these types of Geometry questions, you might find it helpful to draw the shapes and label the side lengths accordingly.
Here, we're asked to compare the SURFACE AREAS of a cube and a rectangular-solid that has twice the length of the cube.
Let's start with the dimensions of both shapes:
Cube = (length)(width)(height) = (X)(X)(X)
Rect. = (2length)(width)(height) = (2X)(X)(X)
To find total surface area, we need to find the areas of each of the surfaces and then add them up.
Cube = 6 identical surfaces = 6 squares = 6(X^2)
Rect. Solid = 2 identical square "ends" and 4 identical rectangular "body pieces"
Rect. Solid = 2(X^2) + 4(2X^2) = 2(X^2) + 8(X^2) = 10(X^2)
We're asked for the ratio of the surface area of the cube to the surface area of the rectangular solid:
6(X^2)/10(X^2) = 6/10 = 3/5
Final Answer: D
GMAT assassins aren't born, they're made,
Rich
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Notice what happens when we double the length of a cube:
The red cube on top has six faces, each of which has an area of x². Its surface area is thus 6x²
The DOUBLED rectangular solid on the bottom is essentially two cubes welded together. This now has TEN faces (as you can see, the two that are welded together both no longer count toward the surface area, so have 6 faces + 6 faces - 2 faces), each with an area of x². Its surface area is thus 10x².
This gives us a ratio of 6/10, or 3/5.
The red cube on top has six faces, each of which has an area of x². Its surface area is thus 6x²
The DOUBLED rectangular solid on the bottom is essentially two cubes welded together. This now has TEN faces (as you can see, the two that are welded together both no longer count toward the surface area, so have 6 faces + 6 faces - 2 faces), each with an area of x². Its surface area is thus 10x².
This gives us a ratio of 6/10, or 3/5.
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A cube has 6 square faces.
A cuboid (as described) could be built from 10 such squares.
6/10 = 3/5
A cuboid (as described) could be built from 10 such squares.
6/10 = 3/5