SAT Area & Volume Question 6 | Free SAT Prep | Aniko

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Question 6·Hard·Area and Volume

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A cube has side length $s$ . On each of its six faces, a smaller cube is attached so that the whole face of the smaller cube coincides with the face of the original cube and no two smaller cubes overlap. The side length of each smaller cube is chosen so that the total surface area of the new solid is $60\%$ greater than the surface area of the original cube.

By approximately what percent, to the nearest whole percent, is the volume of the new solid greater than the volume of the original cube?

For solids with smaller copies attached, first introduce a scale factor (like $k$ ) for the smaller pieces. Write how one attachment changes surface area by carefully counting what area is hidden and what new faces are exposed, then multiply by the number of attachments and set this equal to the given percent change (remember that "60% greater" means a factor of $1.6$ ). After solving for $k$ , switch to volume: the new solid is the original cube plus all the small cubes, so use $s^3$ for the original and $(ks)^3$ for each small cube. Compare the increase to the original volume, multiply by 100, and round at the very end to match the answer choices.

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Introduce a variable for the smaller cube

Let the original cube have side length $s$ and the smaller cubes have side length $ks$ . How can you write the surface area of each and the area of one face of a smaller cube?

Track surface area change for one attached cube

When you glue a smaller cube onto a face, part of the big cube's face is hidden and 5 faces of the smaller cube become exposed. How much surface area is lost and how much is gained, in terms of $k$ and $s$ ?

Use the 60% increase condition

Write an equation saying that the new surface area equals $1.6$ times the original surface area. Substitute your expression for the new surface area involving $k$ and solve for $k^2$ .

Now compare volumes, not areas

Once you know $k$ , remember that volume uses side length cubed. The new solid is the original cube plus 6 smaller cubes. Find the fractional increase in volume and turn it into a percent.

Desmos Guide

Solve for the side-length ratio k using surface area

In Desmos, enter the two equations y = 1 + 4x^2 and y = 1.6. Zoom so you can see their intersection, then tap the positive intersection point. The x-coordinate of this point is the value of $k$ (since the equation $1+4k^2 = 1.6$ determines $k$ ).

Compute the percent volume increase

On a new line, type 6*(a)^3*100, replacing a with the positive x-value you just read for $k$ . The output is the percent increase in volume; round this value to the nearest whole percent and select the matching answer choice.

Step-by-step Explanation

Represent the side length of the smaller cubes

Let the side length of the original cube be $s$ . Let the side length of each smaller cube be $ks$ , where $0<k<1$ .

Surface area of the original cube:

S_{\text{orig}} = 6s^2

Area of one face of a smaller cube:

(ks)^2 = k^2 s^2

Express how one attached cube changes surface area

When you attach one smaller cube to a face of the big cube:

The part of the big cube's face that is covered (area $k^2 s^2$ ) is no longer on the outside, so you lose $k^2 s^2$ of surface area.
The smaller cube contributes 5 new outer faces, each of area $k^2 s^2$ , so you gain $5k^2 s^2$ of surface area.

Net change in surface area from one small cube:

(+5k^2 s^2) - (k^2 s^2) = 4k^2 s^2

With 6 smaller cubes attached, the new total surface area is

S_{\text{new}} = 6s^2 + 6\cdot 4k^2 s^2 = 6s^2(1+4k^2).

Use the 60% increase in surface area to find k

"60% greater" means the new surface area is $1.6$ times the original:

S_{\text{new}} = 1.6\,S_{\text{orig}}.

Substitute the expressions:

6s^2(1+4k^2) = 1.6\cdot 6s^2.

Divide both sides by $6s^2$ :

1 + 4k^2 = 1.6.

4k^2 = 0.6 \quad\Rightarrow\quad k^2 = 0.15, \quad k = \sqrt{0.15}.

Compare the volumes to find the percent increase

Original volume:

V_{\text{orig}} = s^3.

Each smaller cube has volume

V_{\text{small}} = (ks)^3 = k^3 s^3.

There are 6 smaller cubes, so the new total volume is

V_{\text{new}} = s^3 + 6k^3 s^3 = s^3(1+6k^3).

The increase in volume is

\Delta V = V_{\text{new}} - V_{\text{orig}} = 6k^3 s^3.

The percent increase is

\frac{\Delta V}{V_{\text{orig}}}\cdot 100\% = 6k^3\cdot 100\%.

Use $k^2 = 0.15$ so $k^3 = k\cdot k^2 = 0.15k$ and

6k^3 = 6\cdot 0.15k = 0.9k.

Since $k = \sqrt{0.15} \approx 0.387$ , we get

0.9k \approx 0.9\times 0.387 \approx 0.349,

which is about a $35\%$ increase in volume. So the correct choice is 35% (B).

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Step-by-step Explanation

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Step-by-step Explanation