Home / Digital SAT Math Practice Questions – Medium : Equivalent expressions

# Digital SAT Math Practice Questions – Medium : Equivalent expressions

## SAT MAth Practice questions – all topics

• Advanced Math Weightage: 35% Questions: 13-15
• Equivalent expressions
• Nonlinear equations in one variable and systems of equations in two variables
• Nonlinear functions

## SAT MAth and English  – full syllabus practice tests

[calc]  Question Medium

The expression $$\frac{4 x^{11}}{12 x^5}$$ is equivalent to $$\frac{1}{3} x^b$$, where $$b$$ is a constant and $$x>0$$. What is the value of $$b$$ ?

Ans: 6

To simplify the expression $$\frac{4x^{11}}{12x^5}$$ and find the value of $$b$$, follow these steps:

1. Simplify the coefficients:
$\frac{4}{12} = \frac{1}{3}$

2. Simplify the exponents using the property $$a^m / a^n = a^{m-n}$$:
$\frac{x^{11}}{x^5} = x^{11-5} = x^6$

So the expression becomes:
$\frac{4x^{11}}{12x^5} = \frac{1}{3}x^6$

Therefore, $$b = 6$$, and the value of $$b$$ is:
$\boxed{6}$

[calc]  Question Medium

Which expression is equivalent to $$\sqrt[4]{a^9}$$ ?
A) $$a^{\frac{9}{4}}$$
B) $$a^{\frac{4}{9}}$$
C) $$a^{13}$$
D) $$a^{36}$$

Ans: A

We need to simplify the expression $$\sqrt[4]{a^9}$$.

Using the property of exponents that $$\sqrt[n]{a^m} = a^{\frac{m}{n}}$$, we can write:

$\sqrt[4]{a^9} = a^{\frac{9}{4}}$

So, the correct answer is:
$\boxed{A}$

[Calc]  Question  medium

Which expression is equivalent to $$\left(x^2+4\right)^2+(x-2)(x+2)$$ ?

A. $$x^4+x^2+20$$
B. $$x^4+5 x^2+16$$
C. $$x^4+9 x^2$$
D. $$x^4+9 x^2+12$$

Ans:D

To simplify the expression $$(x^2+4)^2+(x-2)(x+2)$$, let’s first expand $$(x^2+4)^2$$ using the binomial theorem:

$(x^2+4)^2 = (x^2+4)(x^2+4) = x^4 + 4x^2 + 4x^2 + 16 = x^4 + 8x^2 + 16$

Now, let’s distribute $$(x-2)(x+2)$$:

$(x-2)(x+2) = x^2 + 2x – 2x – 4 = x^2 – 4$

Now, add the two expressions together:

$(x^2 + 8x^2 + 16) + (x^2 – 4) = x^4 + 9x^2 + 12$

So, the expression $$\left(x^2+4\right)^2+(x-2)(x+2)$$ is equivalent to $$x^4 + 9x^2 + 12$$. Therefore, the correct answer is option D.

[No calc]  Question  medium

The expression $$\frac{32 x^6}{4 x^3}$$ is equivalent to $$c x^d$$, where $$c$$ and $$d$$ are constants and $$x>0$$. What is the value of $$c+d$$ ?

Ans:11

$\frac{32 x^6}{4 x^3} = \frac{32}{4} \cdot \frac{x^6}{x^3} = 8 \cdot x^{6-3} = 8x^3$

So, $$c = 8$$ and $$d = 3$$, and $$c + d = 8 + 3 = 11$$.

[No calc]  Question  medium

$q=s(r-1)^2$

The given equation relates the positive numbers $$q, r$$, and $$s$$. Which equation gives $$r$$ in terms of $$q$$ and $$s$$, when $$r>1$$ ?

A. $$r=1+\sqrt{\frac{q}{s}}$$
B. $$r=1+\frac{\sqrt{q}}{s}$$
c. $$r=-1-\sqrt{\frac{q}{s}}$$
D. $$r=-1-\frac{\sqrt{q}}{s}$$

Ans:A

To solve for $$r$$ in terms of $$q$$ and $$s$$ in the equation $$q = s(r-1)^2$$, we’ll first isolate $$r$$.

$q = s(r-1)^2$

Divide both sides by $$s$$:

$\frac{q}{s} = (r-1)^2$

Now, take the square root of both sides:

$\sqrt{\frac{q}{s}} = |r-1|$

Since $$r > 1$$, we can drop the absolute value:

$r – 1 = \sqrt{\frac{q}{s}}$

Add 1 to both sides:

$r = 1 + \sqrt{\frac{q}{s}}$

So, the equation that gives $$r$$ in terms of $$q$$ and $$s$$ when $$r > 1$$ is:

A. $$r = 1 + \sqrt{\frac{q}{s}}$$

[Calc]  Question   Medium

$$r^q= t^s$$

The given equation relates the distinct positive real numbers q, r, sand t. Which equation correctly expresses tin terms of q, r, and s ?

A) $$t=r^{\frac{q}{s}}$$

B) $$t=r^{\frac{s}{q}}$$

C) $$t=r^{q-s}$$

D) $$t=\frac{r^q}{s}$$

A) $$t=r^{\frac{q}{s}}$$

Given the equation $$r^q = t^s$$, we need to express $$t$$ in terms of $$q$$, $$r$$, and $$s$$.

1. Rewrite the equation:
$r^q = t^s$

2. Take the $$s$$th root on both sides to solve for $$t$$:
$t = \left( r^q \right)^{\frac{1}{s}}$

3. Simplify the exponent:
$t = r^{\frac{q}{s}}$

Thus, the correct expression for $$t$$ in terms of $$q$$, $$r$$, and $$s$$ is:
$\boxed{t = r^{\frac{q}{s}}}$

[No calc]  Question   medium

Which expression is equivalent to  $$16^{\frac{1}{2}x}$$ ?
A. $$4^{x}$$
B. $$8^{x}$$
C. 8𝑥
D. $$16\sqrt{x}$$

Ans: A

We’ll use the properties of exponents to simplify the expression.

$$16^{\left(\frac{1}{2} x\right)} = (2^4)^{\left(\frac{1}{2} x\right)} = 2^{4 \cdot \frac{1}{2} x} = 2^{2x} = (2^2)^x = 4^x$$

So, the answer is A. $$4^x$$.

[No calc]  Question   medium

ax-y = bxy

The given equation relates the positive numbers a, b, x, and y. Which equation correctly expresses a in terms of b, x, and y ?

A. $$a=\frac{bxy+y}{x}$$

B. $$a=\frac{bxy-y}{x}$$

C.  $$a=\frac{x}{bxy+y}$$

D. $$a=\frac{x}{bxy-y}$$

Ans: A

To express $$a$$ in terms of $$b$$, $$x$$, and $$y$$ in the given equation $$a x – y = bxy$$, we can rearrange the equation to solve for $$a$$:

$a x = y + bxy$

Now, divide both sides by $$x$$:
$a = \frac{y + bxy}{x}$

So, the correct expression for $$a$$ in terms of $$b$$, $$x$$, and $$y$$ is:
$a = \frac{y + bxy}{x}$

[No calc]  Question   medium

Which expression is equivalent to $$\sqrt[5]{32a^{3}b^{4}}$$ where a >0 and b>0?

A. $$2a^{\frac{3}{5}}b^{\frac{4}{5}}$$

B. $$2a^{\frac{5}{3}}b^{\frac{5}{4}}$$

C. $$32a^{\frac{3}{5}}b^{\frac{4}{5}}$$

D. $$32a^{\frac{5}{3}}b^{\frac{5}{4}}$$

Ans: A

To simplify the expression $$\sqrt[5]{32a^3b^4}$$, we can break down $$32$$, $$a^3$$, and $$b^4$$ into their prime factors:
$32 = 2^5$
$a^3 = a \cdot a \cdot a$
$b^4 = b \cdot b \cdot b \cdot b$

Now, let’s rewrite the expression:
$\sqrt[5]{32a^3b^4} = \sqrt[5]{(2^5)(a \cdot a \cdot a)(b \cdot b \cdot b \cdot b)}$

Taking the fifth root of each factor, we get:
$\sqrt[5]{32a^3b^4} = 2 \cdot a^{\frac{3}{5}} \cdot b^{\frac{4}{5}}$

So, the equivalent expression is $$2a^{\frac{3}{5}}b^{\frac{4}{5}}$$, which corresponds to option A).

Questions

$\frac{2}{3 x^2}-\frac{1}{6 x^2}$

Which of the following expressions is equivalent to the expression above for $x>0$ ?
A. $-\frac{1}{2 x^2}$
B. $-\frac{1}{3 x^2}$
C. $\frac{1}{3 x^2}$
D. $\frac{1}{2 x^2}$

Ans: D

[No calc]  Question  medium

Which expression is equivalent $$\frac{2+3x}{16-81x^4}$$, where x>1?

A) $$\frac{1}{8-27x^3}$$

B) $$8-27x^3$$

C) $$\frac{1}{(4+9x^2)(2-3x)}$$

D) $$(4+9x^2)(2-3x)$$

C) $$\frac{1}{(4+9x^2)(2-3x)}$$

We need to simplify the expression $$\frac{2 + 3x}{16 – 81x^4}$$.

Factor the denominator $$16 – 81x^4$$:
$16 – 81x^4 = (4^2) – (9x^2)^2 = (4 – 9x^2)(4 + 9x^2)$

Rewrite the expression using this factorization:
$\frac{2 + 3x}{(4 – 9x^2)(4 + 9x^2)}$

$$4 – 9x^2$$ can be factored further as:
$4 – 9x^2 = (2 – 3x)(2 + 3x)$

Substitute this back into the expression:
$\frac{2 + 3x}{(2 – 3x)(2 + 3x)(4 + 9x^2)}$

The $$2 + 3x$$ term cancels out:
$\frac{1}{(2 – 3x)(4 + 9x^2)}$

Thus, the equivalent expression is:
$\frac{1}{(4 + 9x^2)(2 – 3x)}$

[No calc]  Question  medium

Which expression is equivalent to$$k^\frac{5}{16}{(k^\frac{3}{2})}^\frac{5}{8}$$   k > 0 ?

A) $$\sqrt{k}$$

B) $$\sqrt[4]{k^5}$$

C) $$\sqrt[8]{k^5}$$

D) $$\sqrt[15]{k^{16}}$$

B) $$\sqrt[4]{k^5}$$

To simplify the expression $$k^{\frac{5}{16}}\left(k^{\frac{3}{2}}\right)^{\frac{5}{8}}$$, follow these steps:

Simplify the exponent inside the parentheses:
$\left(k^{\frac{3}{2}}\right)^{\frac{5}{8}} = k^{\left(\frac{3}{2} \cdot \frac{5}{8}\right)} = k^{\frac{15}{16}}$

Combine the exponents:
$k^{\frac{5}{16}} \cdot k^{\frac{15}{16}} = k^{\left(\frac{5}{16} + \frac{15}{16}\right)} = k^{\frac{20}{16}} = k^{\frac{5}{4}}$

Express $$k^{\frac{5}{4}}$$ in a simpler form:
$k^{\frac{5}{4}} = \left(k^{5}\right)^{\frac{1}{4}} = \sqrt[4]{k^{5}}$

[Calc]  Question Medium

$$\sqrt{a^{2}b^{6}}$$

If a and b are positive numbers, which of the following is equivalent to the expression above?
A. ab4

B. ab3

c. $$ab\sqrt{ab^{4}}$$

D. $$ab\sqrt{b^{3}}$$

Ans: B

To simplify $$\sqrt{a^2 b^6}$$, we can separate the square root into two parts: $$\sqrt{a^2} \cdot \sqrt{b^6}$$. Since $$a$$ and $$b$$ are positive numbers, we can simplify this expression as follows:

$\sqrt{a^2} \cdot \sqrt{b^6} =(a^2)^\frac{1}{2}.(b^6)^\frac{1}{2} \Rightarrow ab^3$

So, the expression is equivalent to $$ab^3$$. Therefore, the correct answer is:

$$\boxed{\text{B) } ab^3}$$

[No calc]  Question  medium

The equation $$y= \sqrt{\frac{hg}{x}}$$ relates to the positive numbers g, h, x, and y. Which equation correctly expresses h in terms of g, x, and y?

A) h=gxy

B)h=$$gxy^2$$

C) h=$$\frac{gy^2}{x}$$

D)h=$$\frac{xy^2}{g}$$

D)h=$$\frac{xy^2}{g}$$

The equation $$y=\sqrt{\frac{hg}{x}}$$ relates to the positive numbers $$g, h, x$$, and $$y$$. We need to express $$h$$ in terms of $$g$$, $$x$$, and $$y$$.

Start with the given equation: $$y=\sqrt{\frac{hg}{x}}$$.

Square both sides to eliminate the square root:
$y^2 = \frac{hg}{x}$

Multiply both sides by $$x$$ to isolate $$h$$:
$hx = gy^2$

Divide both sides by $$g$$ to solve for $$h$$:
$h = \frac{gy^2}{x}$

[No- Calc]  Question  Medium

4T-8D=12H
The given equation can be rewritten as T = aD + bH, where a and b are constants. What is the value of a ?

Ans: 2

We’re given the equation $$4T – 8D = 12H$$, and we’re asked to rewrite it in the form $$T = aD + bH$$, where $$a$$ and $$b$$ are constants. To do this, let’s isolate $$T$$ on one side of the equation:

$4T = 8D + 12H$

Dividing both sides by $$4$$:

$T = 2D + 3H$

So, comparing this with the desired form $$T = aD + bH$$, we see that $$a = 2$$.

Therefore, the value of $$a$$ is 2.

[No calc]  Question  medium

Which expression is equivalent to

$$(5x^3-2x+1) – (2x^3+2x+1)$$

A) $$3x^3$$

B) $$3x^3+2$$

C) $$3x^3-4x$$

D) $$3x^3-4x+2$$

C) $$3x^3-4x$$

We need to simplify the expression $$\left(5x^3-2x+1\right)-\left(2x^3+2x+1\right)$$.

Distribute the negative sign in the second expression:
$\left(5x^3-2x+1\right) – \left(2x^3+2x+1\right) = 5x^3 – 2x + 1 – 2x^3 – 2x – 1$

Combine like terms:
$(5x^3 – 2x^3) + (-2x – 2x) + (1 – 1) = 3x^3 – 4x$

So, the expression $$\left(5x^3-2x+1\right)-\left(2x^3+2x+1\right)$$ simplifies to $$3x^3 – 4x$$.

[No- Calc]  Question  Medium

Which expression is equivalent to

$$(2x^{2}+3x-2)-(5x^{2}-x-7)$$

A. $$(7x^{2}+4x+9)$$

B. $$(3x^{2}+4x+5)$$

C. $$(-3x^{2}+2x-9)$$

D. $$(-3x^{2}+4x+5)$$

Ans: D

To find the expression equivalent to $$\left(2x^2+3x-2\right)-\left(5x^2-x-7\right)$$, we distribute the negative sign inside the second parenthesis:

$= 2x^2 + 3x – 2 – 5x^2 + x + 7$

Combining like terms:

$= (2x^2 – 5x^2) + (3x + x) + (-2 + 7)$

$= -3x^2 + 4x + 5$

So, the equivalent expression is D) $$-3x^2 + 4x + 5$$.

[No- Calc]  Question  Medium

Which expression is equivalent to $$\frac{12 \mathrm{x}^3}{8 \mathrm{x}^2}$$, where $$\mathrm{x}$$ is not equal to 0 ?

A) $$\frac{4}{x}$$
B) $$\frac{3 x}{2}$$
C) $$4 x$$
D) $$20 x^5$$

Ans:B

To simplify the expression $$\frac{12x^3}{8x^2}$$, we can cancel out a common factor of $$x^2$$ from both the numerator and the denominator. This leaves us with:

$\frac{12x^3}{8x^2} = \frac{12}{8} \times x^3 \div x^2 = \frac{3}{2}x$

So, the correct answer is B) $$\frac{3x}{2}$$.

[Calc]  Question  Medium

The equation $y=\frac{x+w}{z}$ relates the positive numbers $w, x, y$, and $z$. Which equation correctly expresses $x$ in terms of $w, y$, and $z$ ?
A) $x=y z-w$
B) $x=y z+w$
C) $x=\frac{z}{w y}$
D) $x=\frac{y}{z w}$

A

[Calc]  Question Medium

$$\left(x^3+x\right)+\left(x^2-x\right)$$

Which of the following is equivalent to the given expression?
A) $x^5-x^2$
B) $x^5-x^4+x^3-x^2$
C) $x^3+x^2$
D) $x^3+x^2+2 x$

C

[Calc]  Question  Medium

$$\frac{x+2}{(x+2)^2}$$

Which of the following expressions is equivalent to the given expression, where $x \neq-2$ ?
A) $x+2$
B) $\frac{1}{x+2}$
C) $x^2+2 x+4$
D) $\frac{1}{x^2+2 x+4}$

B

Question

$$k$$$$a$$+$$n$$$$b$$=10

The given equation relates the positive numbers $$a$$, $$b$$,$$k$$, and $$n$$. Which of the following correctly expresses $$a$$ in terms of $$b$$,$$k$$, and $$n$$?

1. $$a=10-\frac{k}{nb}$$
2. $$a=10-\frac{nb}{k}$$
3. $$a=\frac{10-k}{nb}$$
4. $$a=\frac{10-nb}{k}$$

D

Question

. Which expression is equivalent to $$x$$4-18$$x$$2+81?

1. ($$x$$-3)4
2. ($$x$$-3)($$x$$+3)3
3. ($$x$$-3)3($$x$$+3)
4. ($$x$$-3)2($$x$$+3)2

D

Question

Which of the following is equivalent to the given expression?

B

Question

The equation shown gives in terms of ,, and , where and are not equal to . Which  equation gives in terms of , , and

1. $$t=\frac{A}{P}-\frac{1}{r}$$
2. $$t=\frac{A}{Pr}-\frac{1}{Pr}$$
3. $$t=\frac{A}{Pr}-\frac{1}{r}$$
4. $$t=\frac{A}{r}-\frac{P}{r}$$

C

Question

$$x$$($$x$$+2)2=$$x$$3+b$$x$$2+c$$x$$

In the equation above, b and c are constants. If the equation is true for all values of $$x$$, what is the value of b+c ?

1. 4
2. 6
3. 8
4. 16

C

Questions

. $x^4-8 x^2+16$

Which of the following is equivalent to the expression above?
A. $(x-2)^2(x+2)^2$
B. $\left(x^2+4\right)(x+2)(x-2)$
C. $(x-2)^4$
D. $(x-4)^4$

Ans: A

Questions

$V=\frac{M}{D}$

The formula above relates volume $V$, mass $M$, and density $D$. What is density in terms of volume and mass?
A. $D=\frac{1}{M V}$
B. $D=\frac{M}{V}$
C. $D=\frac{V}{M}$
D. $D=M V$

Ans: B

Questions

$\frac{1}{2} m v^2=m g h$

Torricelli’s law is given by the equation above, where $m$ represents the mass, $h$ represents the height, $v$ represents the velocity, and $g$ is a constant. According to the equation from Torricelli’s law, which of the following is equivalent to the velocity, $v$ ?
A. $2 g h$
B. $\frac{1}{2} g h m^2$
C. $\sqrt{2 g h}$
D. $\sqrt{\frac{1}{2} m g h}$

Ans: C

Questions

Which of the following is equivalent to the sum of $3 x^4+2 x^3$ and $4 x^4+7 x^3$ ?
A. $16 x^4$
B. $7 x^8+9 x^6$
C. $12 x^4+14 x^3$
D. $7 x^4+9 x^3$

Ans: D

Questions

Blood volume, $V_B$, in a human can be determined using the equation $V_B=\frac{V_P}{1-H}$, where $V_P$ is the plasma volume and $H$ is the hematocrit (the fraction of blood volume that is red blood cells). Which of the following correctly expresses the hematocrit in terms of the blood volume and the plasma volume?
A. $H=1-\frac{V_P}{V_B}$
B. $H=\frac{V_B}{V_P}$
C. $H=1+\frac{V_B}{V_P}$
D. $H=V_B-V_P$

Ans: A

Questions

Which of the following is an equivalent form of the expression $(2 x-2)^2-(2 x-2)$ ?
A. $2 x^2-6 x+6$
B. $4 x^2-10 x+2$
C. $(2 x-2)(2 x-2)$
D. $(2 x-3)(2 x-2)$