A solid cube of clay has a volume of cubic centimeters. A sculptor uniformly increases each edge of the cube by centimeters, creating a larger cube whose volume is cubic centimeters greater than the original cube’s volume. What is ?

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SAT Nonlinear Functions Question 216 | Free SAT Prep | Aniko

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Question 216·Hard·Nonlinear Functions

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A solid cube of clay has a volume of $V$ cubic centimeters. A sculptor uniformly increases each edge of the cube by $2$ centimeters, creating a larger cube whose volume is $386$ cubic centimeters greater than the original cube’s volume. What is $V$ ?

Your answer

(Express the answer as an integer)

For word problems about changing dimensions of solids, first assign a variable to the key length (here, the cube’s edge) and express all volumes in terms of that variable. Translate the verbal statement about how much the volume changes into an equation (new volume minus original volume equals the given change), then simplify carefully—often the difference of volumes will reduce to a manageable quadratic. Solve the quadratic, discard any negative length, and finally plug the valid length back into the volume formula to get the requested value. This approach avoids guessing and keeps the algebra organized and quick on the SAT.

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Connect volume to edge length

For a cube, how do you express the volume in terms of its edge length $s$ ? Write the original cube’s volume using $s$ .

Describe the new cube

If each edge of the cube is increased by 2 cm, what is the new edge length in terms of $s$ ? Write the volume of this larger cube using that new edge length.

Use the given volume increase

The larger cube’s volume is 386 cubic centimeters more than the original. How can you express this as an equation involving the new volume, the original volume, and 386?

Solve and interpret

After you expand $(s+2)^3$ and simplify, you will get a quadratic equation in $s$ . Solve it, keep only the positive value for $s$ , and then cube that value to find $V$ .

Desmos Guide

Graph the volume difference equation

In Desmos, enter the expression y = (x + 2)^3 - x^3 on one line and y = 386 on another line. These represent the volume increase as a function of $x$ and the constant difference 386.

Find the original edge length

Look for the point where the two graphs intersect. The $x$ -coordinate of this intersection is the original cube’s edge length (ignore any negative intersection, since edge lengths cannot be negative).

Compute the original volume

In a new expression line, type x^3 and substitute the positive $x$ -value you found from the intersection (for example, you can enter something like (value)^3). The resulting output is the original volume $V$ in cubic centimeters.

Step-by-step Explanation

Define the variable and relate it to volume

Let $s$ be the edge length (in centimeters) of the original cube.

The volume of a cube with edge length $s$ is $s^3$ .
The problem tells us this original volume is $V$ , so

V = s^3.

When each edge is increased by 2 cm, the new edge length becomes $s+2$ , and the new volume is $(s+2)^3$ .

Use the volume increase to write an equation

We are told that the larger cube’s volume is 386 cubic centimeters greater than the original cube’s volume. That means:

(s+2)^3 - s^3 = 386.

This equation relates the original edge length $s$ to the given volume increase.

Expand and simplify to get a quadratic equation

First expand $(s+2)^3$ .

You can think of it as $(s+2)(s+2)(s+2)$ , or use the pattern for a cube of a binomial:

(s+2)^3 = s^3 + 6s^2 + 12s + 8.

Substitute this into the equation:

\begin{aligned} (s+2)^3 - s^3 &= 386 \\ (s^3 + 6s^2 + 12s + 8) - s^3 &= 386. \end{aligned}

The $s^3$ terms cancel:

6s^2 + 12s + 8 = 386.

Move 386 to the left side:

\begin{aligned} 6s^2 + 12s + 8 - 386 &= 0 \\ 6s^2 + 12s - 378 &= 0. \end{aligned}

Divide every term by 6 to simplify:

s^2 + 2s - 63 = 0.

Solve the quadratic for the edge length

Now factor the quadratic equation:

s^2 + 2s - 63 = 0.

We look for two numbers that multiply to $-63$ and add to $2$ . Those numbers are $9$ and $-7$ , so:

s^2 + 2s - 63 = (s+9)(s-7) = 0.

Set each factor equal to zero:

$s+9 = 0 \Rightarrow s = -9$
$s-7 = 0 \Rightarrow s = 7$

Because an edge length cannot be negative, we discard $s=-9$ and keep $s=7$ .

So, the original cube’s edge length is $7$ centimeters.

Find the original volume V

Recall that the original volume is $V = s^3$ .

We found $s = 7$ , so:

V = 7^3 = 7 \cdot 7 \cdot 7 = 343.

Therefore, $V = 343$ cubic centimeters.

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Step-by-step Explanation

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Step-by-step Explanation