Digital SAT Math Practice Questions – Advanced : 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

[No calc]  Question  Hard

$$\frac{1}{C}=\frac{1}{d}+\frac{1}{e}+\frac{1}{f}$$

An electric circuit contains three capacitors in a particular arrangement. The given equation relates the equivalent capacitance C of the arrangement to d, e, and f, the capacitances of the individual capacitors. Which equation correctly gives C in terms of d, e, and f ?

A. 𝐶 = 𝑑 + 𝑒 + 𝑓
B. $$C=\frac{def}{d+e+f}$$
C. $$C=\frac{d+e+f}{def}$$
D. $$C=\frac{def}{de+df+ef}$$

Ans: D

Given the equation $$\frac{1}{c}=\frac{1}{d}+\frac{1}{e}+\frac{1}{f}$$, the equivalent capacitance $$C$$ is the reciprocal of the sum of the reciprocals of $$d$$, $$e$$, and $$f$$.

$\frac{1}{C} = \frac{1}{d} + \frac{1}{e} + \frac{1}{f}$

To find $$C$$, we take the reciprocal of both sides:

$C = \frac{1}{\frac{1}{d} + \frac{1}{e} + \frac{1}{f}}$

To simplify the expression in the denominator, we’ll first find a common denominator:

$C = \frac{1}{\frac{ef + df + ed}{def}}$
$C = \frac{def}{ef + df + ed}$

[No calc]  Question  Hard

Which of the following expressions is equivalent to $$(\frac{a}{4}+\frac{b}{3})^{2}$$ ?

A. $$(\frac{a^{2}}{4}+\frac{b^{2}}{3})$$

B. $$(\frac{a^{2}}{16}+\frac{b^{2}}{9})$$

C. $$(\frac{a^{2}}{4}+\frac{ab}{12}+\frac{b^{2}}{3})$$

D. $$(\frac{a^{2}}{16}+\frac{ab}{6}+\frac{b^{2}}{9})$$

Ans: D

To find the expression equivalent to $$\left(\frac{a}{4}+\frac{b}{3}\right)^2$$, we can expand the expression using the distributive property and then simplify:

$\left(\frac{a}{4}+\frac{b}{3}\right)^2=\left(\frac{a}{4}+\frac{b}{3}\right) \cdot\left(\frac{a}{4}+\frac{b}{3}\right)$

Using FOIL (First, Outer, Inner, Last) method:
\begin{aligned} & =\frac{a^2}{16}+\frac{a b}{12}+\frac{a b}{12}+\frac{b^2}{9} \\ & =\frac{a^2}{16}+\frac{2 a b}{12}+\frac{b^2}{9} \\ & =\frac{a^2}{16}+\frac{a b}{6}+\frac{b^2}{9} \end{aligned}

So, the expression equivalent to $$\left(\frac{a}{4}+\frac{b}{3}\right)^2$$ is $$\frac{a^2}{16}+\frac{a b}{6}+\frac{b^2}{9}$$, which corresponds to option D).

[Calc]  Question  Hard

Two numbers, $$a$$ and $$b$$, are each greater than zero, and 4 times the square root of $$a$$ is equal to 9 times the cube root of $$b$$. If $$a=\frac{2}{3}$$, for what value of $$x$$ is $$a^x$$ equal to $$b$$ ?

7.5,15/2

Given:
$$4\sqrt{a} = 9\sqrt[3]{b}$$
$$a = \frac{2}{3}$$

find $$x$$ such that $$a^x = b$$.

solving for $$b$$ using the given values:

$4\sqrt{\frac{2}{3}} = 9\sqrt[3]{b}$

$\sqrt{\frac{2}{3}} = \frac{9}{4}\sqrt[3]{b}$

Now, we can isolate $$b$$ by squarring both sides:

$\frac{2}{3} = \left(\frac{9}{4}\right)^2 b^{\frac{2}{3}}$

$\frac{2}{3} = \frac{81}{16} b^{\frac{2}{3}}$

$b^{\frac{2}{3}} = \frac{2}{3} \times \frac{16}{81}$

$b = \frac{32}{243}^{\frac{3}{2}}$

$b = (\frac{32}{243})^{\frac{3}{2}}$
$b = (\frac{2^5}{3^5})^{\frac{3}{2}}$
$b = ([\frac{2}{3}])^{5\times \frac{3}{2}}$
$b = (\frac{2}{3})^{\frac{15}{2}}$

Now, we need to find $$x$$ such that $$a^x = b$$:

$\left(\frac{2}{3}\right)^x = b = (\frac{2}{3})^{\frac{15}{2}}$

$x = \frac{15}{2}$

[Calc]  Question   Hard

Which of the following expressions is equivalent to $${(2\sqrt{x}-\sqrt{y})}^{\frac{2}{5}}$$ , where x>y and y>0?

A) $${(4x-y)}^5$$

B) $$\sqrt[5]{4x-y}$$

C) $${(4x-4 \sqrt{xy}+y)}^{\frac{1}{5}}$$

D) $$\sqrt[5]{4x-4xy+y}$$

C) $${(4x-4\sqrt{xy}+y)}^{\frac{1}{5}}$$

We need to determine the expression equivalent to $$(2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}$$.

1. Simplify the expression inside the parentheses:
Given: $$(2 \sqrt{x} – \sqrt{y})^{\frac{2}{5}}$$

2. Examine each option:

Option A: $$(4 x – y)^5$$
$(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}$

Option B: $$\sqrt[5]{4 x – y}$$
$(4x – y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}$

Option C: $$(4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}$$
$(4x – 4\sqrt{xy} + y)^{\frac{1}{5}} = (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}$

Option D: $$\sqrt[5]{4 x – 4xy + y}$$
$(4x – 4xy + y)^{\frac{1}{5}} \neq (2\sqrt{x} – \sqrt{y})^{\frac{2}{5}}$

Thus, the correct expression is:
$\boxed{(4 x – 4 \sqrt{x y} + y)^{\frac{1}{5}}}$

[No calc]  Question  Hard

The function f is defined by $$f(x)=\frac{1}{5}x+\frac{9}{10}$$ For what value of x does f(x)=1?

Ans: 1/2, .5

To find the value of $$x$$ for which $$f(x) = 1$$, we set $$f(x)$$ equal to $$1$$ and solve for $$x$$:

$f(x) = \frac{1}{5}x + \frac{9}{10} = 1$

To solve for $$x$$, we first subtract $$\frac{9}{10}$$ from both sides:

$\frac{1}{5}x = 1 – \frac{9}{10}$
$\frac{1}{5}x = \frac{10}{10} – \frac{9}{10}$
$\frac{1}{5}x = \frac{1}{10}$

Now, multiply both sides by $$5$$ to isolate $$x$$:

$x = \frac{1}{10} \times 5$
$x = \frac{5}{10}$
$x = \frac{1}{2}$

So, $$f(x) = 1$$ when $$x = \frac{1}{2}$$.

[No calc]  Question   Hard

Which expression is equivalent to $$y^\frac{1}{8}{(y^\frac{3}{4})}^\frac{3}{2}$$, where y>0?

A) $$\sqrt[4]{y^5}$$

B) $$\sqrt[3]{y^4}$$

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

D) $$\sqrt[8]{y^7}$$

A) $$\sqrt[4]{y^5}$$

To find an expression equivalent to $$y^{\frac{1}{8}}\left(y^{\frac{3}{4}}\right)^{\frac{3}{2}}$$, let’s simplify the given expression step by step.

$y^{\frac{1}{8}} \left( y^{\frac{3}{4}} \right)^{\frac{3}{2}}$

Apply the power of a power rule $$(a^m)^n = a^{m \cdot n}$$ to the term $$\left( y^{\frac{3}{4}} \right)^{\frac{3}{2}}$$:
$\left( y^{\frac{3}{4}} \right)^{\frac{3}{2}} = y^{\frac{3}{4} \cdot \frac{3}{2}} = y^{\frac{9}{8}}$

$y^{\frac{1}{8}} \cdot y^{\frac{9}{8}} = y^{\frac{1}{8} + \frac{9}{8}} = y^{\frac{10}{8}} = y^{\frac{5}{4}}$

Convert $$y^{\frac{5}{4}}$$ to a root form:
$y^{\frac{5}{4}} = \left( y^5 \right)^{\frac{1}{4}} = \sqrt[4]{y^5}$

[Calc]  Question  Hard

$\left(x^3-6 x+5\right)\left(3 x^2+x\right)$

If the given expression is rewritten in the form $$a x^5+b x^4+c x^3+d x^2+e x$$, where $$a, b, c, d$$, and $$e$$ are constants, what is the value of $$d$$ ?

9

To find the value of $$d$$ in the expression $$\left(x^3-6x+5\right)\left(3x^2+x\right)$$, we need to expand the product and identify the coefficient of the $$x^2$$ term.

First, distribute each term of the first polynomial by each term of the second polynomial:
$\left(x^3 – 6x + 5\right) \left(3x^2 + x\right)$

We’ll multiply term by term:

1. $$x^3 \cdot 3x^2 = 3x^5$$
2. $$x^3 \cdot x = x^4$$
3. $$-6x \cdot 3x^2 = -18x^3$$
4. $$-6x \cdot x = -6x^2$$
5. $$5 \cdot 3x^2 = 15x^2$$
6. $$5 \cdot x = 5x$$

Now, combine all the terms:
$3x^5 + x^4 – 18x^3 – 6x^2 + 15x^2 + 5x$

Combine like terms:
$3x^5 + x^4 – 18x^3 + (15x^2 – 6x^2) + 5x$
$3x^5 + x^4 – 18x^3 + 9x^2 + 5x$

From this expanded form, we can identify the coefficients:
$$a = 3$$
$$b = 1$$
$$c = -18$$
$$d = 9$$
$$e = 5$$

Thus, the value of $$d$$ is:
$\boxed{9}$

[Calc]  Question  Hard

Which of the following is equivalent to the expression $$x^{4}-8x^{2}+16$$ ?
I. $$(x+2)^{2}(x-2)^{2}$$
II. $$(x^{2}-4)^{2}$$
A) I only
B) II only
C) I and II
D) Neither I nor II

Ans: C

I. $$(x+2)^2(x-2)^2$$

Expanding this expression, we get:

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

$= x^4 – 4x^3 + 4x^2 + 4x^3 – 16x^2 + 16x + 4x^2 – 16x + 16$

$= x^4 – 16x^2 + 16$

This does match the original expression $$x^4 – 8x^2 + 16$$.

Now, let’s check option II:

II. $$(x^{2}-4)^{2}$$

$(x^2 – 4)^2 = (x^2 – 4)(x^2 – 4)$

$= x^4 – 4x^2 – 4x^2 + 16$

$= x^4 – 8x^2 + 16$

This also matches the original expression $$x^4 – 8x^2 + 16$$.

So, both options I and II are equivalent to the given expression.

The correct answer is $$\mathbf{C}$$ – I and II.

[Calc]  Question  Hard

The expressions $x^2+b x+10$ and $(x-3)^2+c$, where $b$ and $c$ are constants, are equivalent. What is the value of $b+c$ ?
A) 7
B) 4
C) 3
D) -5

D

Question

The expression $$3xy(2y + 3)$$ is equivalent to $$bxy^{2} + 9xy$$, where $$b$$ is a constant. What is the value of $$b$$? 1.6

6

Question

Which of the following is equivalent to $r^{\frac{2}{5}} \cdot \sqrt{r}$, where $r>0$ ?
A. $r^{\frac{1}{5}}$
B. $r^{\frac{3}{10}}$
C. $r^{\frac{3}{7}}$
D. $r^{\frac{9}{10}}$

Ans: D

Question

$$x(x + 1) + 2(x + 1) = ax^2 +bx+ c$$

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

1. 6
2. 5
3. 4
4. 3

Ans: A

Questions

Which of the following expressions is equivalent to $\left(-4 x^3\right)^{\frac{2}{3}}$ ?
A. $-2 x^3 \cdot \sqrt[3]{2}$
B. $-x^3 \cdot \sqrt[3]{16}$
C. $2 x^2 \cdot \sqrt[3]{2}$
D. $2 x^2 \cdot \sqrt[3]{16}$