Full Length ACCUPLACER Mathematics Practice Test

Full-Length ACCUPLACER Mathematics Practice Test

If you want to prepare for the ACCUPLACER Mathematics Practice Test? It’s time to take a Full-length ACCUPLACER Mathematics Practice Test. It is the best way to simulate test day. To challenge your skills and simulate the full-length ACCUPLACER Mathematics Practice Test day experience, score your tests using the answer keys.

The Most Comprehensive Accuplacer Math Preparation Bundle

~~$76.99~~ $33.99

56% Off*

Bundle Includes Accuplacer Math Prep Books, Workbooks, and Practice Tests

eBook (PDF Format)

Download

ACCUPLACER Mathematics Practice Test 4 (Non–Calculator) 2 Sections – 40 questions Total time for two sections: No Time Limit You may not use a calculator on this section.
Arithmetic and Elementary Algebra
1-	Last Friday Jacob had $$ 45.12 $ . Over the weekend he received some money for cleaning the attic. He now has $$ 62$ . How much money did he receive?
(A)	$ $ 16.68$
(B)	$ $ 16.78$
(C)	$ $ 16.88$
(D)	$ $ 16.98$
2-	$ x^2 \ - \ 8 \ x \ + \ 15 \ = $ ?
(A)	$(x \ – \ 3) \ (x \ - \ 5)$
(B)	$(x \ + \ 3) \ (x \ - \ 5)$
(C)	$(x \ + \ 3) \ (x \ + \ 5)$
(D)	$(x \ - \ 3) \ (x \ + \ 5)$
3-	What is $6254.74164$ rounded to the nearest tenth?
(A)	$ 6254.7$
(B)	$ 6254.8$
(C)	$ 6254.9$
(D)	$ 6254.0$
4-	In the following diagram, what is the value of $ x$ in the following triangle?
(A)	$40^\circ$
(B)	$46^\circ$
(C)	$60^\circ$
(D)	$45^\circ$
5-	$45$ is what percent of $120$ ?
(A)	$ 37\%$
(B)	$ 37.5\%$
(C)	$ 38.5\%$
(D)	$ 38\%$
6-	Which of the following equations has a graph that is a straight line?
(A)	$3 \ y \ + \ 6 = 4 \ x$
(B)	$ y \ = \ 3 \ x^2 \ + \ 9 $
(C)	$3 \ y^2 \ + \ 6 = x$
(D)	$3 \ y \ + \ 6 = x^2$
7-	Find all values for which $2 \ x^2 \ - \ 5 \ x \ - \ 3 \ = \ 0$
(A)	$- \ 12 , \frac{1}{2}$
(B)	$\frac{1}{2} , 12$
(C)	$- \ \frac{1}{2} , 3$
(D)	$- \ \frac{3}{2} , 12$
8-	What is the distance between the points $ (2, 4)$ and $ (- \ 4, - \ 4) $ ?
(A)	$100$
(B)	$10$
(C)	$5$
(D)	$7$
9-	Simplify $ \frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2}{3} \ - \ \frac{7}{3}} $
(A)	$ \frac{- \ 2 \ x \ - \ 15}{3 \ x^2 \ - \ 21} $
(B)	$ \frac{- \ 2 \ x \ - \ 15}{3 \ x^2 \ +\ 21} $
(C)	$ \frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ + \ 21} $
(D)	$ \frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ - \ 21} $
10-	A man owed $$ 3254 $ on his car. After making $42$ payment of $$ 65 $ each, how much did he have left to pay?
(A)	$$ 234$
(B)	$$ 236$
(C)	$$ 524$
(D)	$$ 253$
11-	Write the $ \frac{7}{120} $ as a decimal.
(A)	$0.059$
(B)	$0.058$
(C)	$0.60$
(D)	$0.62$
12-	$ \sqrt{75} $ is between which two whole numbers?
(A)	$7$ and $8$
(B)	$8$ and $9$
(C)	$6$ and $7$
(D)	$5$ and $6$
13-	Which of the following is one solution of this equation? $3 \ x^2 \ + \ 10 \ x \ - \ 2 \ = \ 0$
(A)	$ \frac{- \ 5 \ ± \ \sqrt{31}}{3} $
(B)	$ \frac{- \ 5 \ - \ \sqrt{31}}{3} , 0.12$
(C)	$ \frac{- \ 5 \ + \ \sqrt{31}}{3} , 0.358$
(D)	$ \frac{- \ 5 \ + \ \sqrt{31}}{3} , 0.438$
14-	$ (x \ + \ 7 ) \ (x^{2} \ - \ 4 \ x \ + \ 3) = $ ?
(A)	$ x^3 \ - \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$
(B)	$ x^3 \ - \ 3 \ x^2 \ + \ 25 \ x \ + \ 21$
(C)	$ x^3 \ - \ 3 \ x^2 \ + \ 25 \ x \ - \ 21$
(D)	$ x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$
15-	$ (x^{5})^{\frac{2}{7}} $
(A)	$ x^{ \frac{11}{7}}$
(B)	$ x^{ \frac{10}{7}}$
(C)	$ x^{ \frac{9}{7}}$
(D)	$ x^{ \frac{8}{7}}$
16-	How many $ 6 \ × \ 6$ squares can fit inside a rectangle with a height of $42$ and width of $18$?
(A)	$21$
(B)	$16$
(C)	$24$
(D)	$36$
17-	If a vehicle is driven $42$ miles on Monday, $47$ miles on Tuesday, and $31$ miles on Wednesday, what is the average number of miles driven each day?
(A)	$42$ miles
(B)	$45$ miles
(C)	$46$ miles
(D)	$40$ miles
18-	Alex’s average (arithmetic mean) on two mathematics tests is $10$ . What should Liam’s score be on the next test to have an overall of $12$ for all the tests?
(A)	$12$
(B)	$18$
(C)	$16$
(D)	$10$
19-	If $ 6\ - \ 3 \ x \ ≤ \ 18 $ , what is the value of $ x \geq $ ?
(A)	$4$
(B)	$3$
(C)	$-4$
(D)	$-3$
20-	$ 9^{5 } \ × \ 9^{ - \ 8} \ = $?
(A)	$9^{ 3}$
(B)	$9^{ - \ 3}$
(C)	$9^{ 14}$
(D)	$9^{ 10}$
Accuplacer Math for Beginners ~~$24.99~~ $13.99 44% Off* The Ultimate Step by Step Guide to Preparing for the Accuplacer Math Test eBook (PDF Format) All Available Formats & Editions Download
College–Level Mathematics
21-	cos $2 \ \theta = $ ?
(A)	$ 1\ + \ 2 \ sin^2\ \theta$
(B)	$ 1\ - \ 2 \ sin^2\ \theta$
(C)	$ - \ 2 \ sin^2\ \theta$
(D)	$ 2 \ sin^2\ \theta$
22-	If $ \theta $ is an acute angle and sin $ \theta =\frac{4}{5} $ , cos $ \theta = $ ?
(A)	$\frac{4}{5}$
(B)	$\frac{3}{5}$
(C)	$\frac{2}{5}$
(D)	$\frac{1}{5}$
23-	If the center of a circle is at the point $ (4, - \ 1) $ and its circumference equals to $4 \ π$ , what is the standard form equation of the circle?
(A)	$(x \ +\ 4)^2 \ + \ (y \ + \ 1)^2 = 2 $
(B)	$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 4 $
(C)	$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 2 $
(D)	$(x \ - \ 4)^2 \ + \ (y \ - \ 1)^2 \ = \ 4 $
24-	What is the solution of the following system of equations? $ \begin{cases} 4 \ x \ + \ y = 8 \\ - \ 8 \ x \ - \ 4 \ y = 16\end{cases} $
(A)	$(6, 16)$
(B)	$(- \ 6, 16)$
(C)	$(6, - \ 16)$
(D)	$(6, 16)$
25-	What is the center and radius of a circle with the following equation? $ (x \ – \ 6)^2 \ + \ (y \ + \ 4)^2 \ = \ 5 $
(A)	$(6, 4), \ \sqrt{5} $
(B)	$(6 , - \ 4), \ \sqrt{5} $
(C)	$(- \ 6 , - \ 4), \ \sqrt{5} $
(D)	$(- \ 6 , 4), \ \sqrt{5} $
26-	If sin A $=\ \frac{2}{5} $ in a right triangle and the angle A is an acute angle, then what is cos A ?
(A)	$\frac{\sqrt{22}}{5}$
(B)	$\frac{\sqrt{20}}{5}$
(C)	$\frac{\sqrt{21}}{5}$
(D)	$\frac{\sqrt{3}}{5}$
27-	If $ \log_{3}{x \ = \ 6}$ , then $ x \ = $ ?
(A)	$729$
(B)	$243$
(C)	$719$
(D)	$216$
28-	Simplify: $ \frac{2 \sqrt{12}}{9 \sqrt{48}} $
(A)	$\frac{2}{9}$
(B)	$\frac{4}{9}$
(C)	$\frac{1}{9}$
(D)	$\frac{5}{9}$
29-	What’s the reciprocal of $ \frac{25}{x^3} $ ?
(A)	$\frac{x^3}{25 }$
(B)	$\frac{25}{x^3 }$
(C)	$\frac{5}{x }$
(D)	$\frac{5}{x ^3 }$
30-	If $ \log_{4}{x \ = \ 3}$ , then $x \ =$ ?
(A)	$81$
(B)	$16$
(C)	$64$
(D)	$36$
31-	What is sin $60^\circ $?
(A)	$\frac{\sqrt{3}}{2}$
(B)	$\sqrt{3}$
(C)	$- \ \sqrt{3}$
(D)	$- \ \frac{\sqrt{3}}{2}$
32-	Find the inverse function for ln $ (4 \ x \ - \ 3) $
(A)	$\frac{1}{4} (e^{\ x} \ - \ 3)$
(B)	$\frac{1}{2} (e^{\ x} \ + \ 3)$
(C)	$\frac{1}{4} (e^{\ x} \ + \ 3)$
(D)	$\frac{1}{2} (e^{\ x} \ - \ 3)$
33-	Solve. $ \| \ 15 \ – \ (18 \ ÷ \ \| \ 3 \ + \ 6 \ \|)\| = $ ?
(A)	$- \ 13$
(B)	$- \ 12$
(C)	$12$
(D)	$13$
34-	Simplify $( – \ 7 \ + \ 2 \ i) \ (9 \ + \ 4 \ i)$ .
(A)	$ - \ 10 \ i \ - \ 71$
(B)	$ - \ 10 \ i \ + \ 71$
(C)	$ 10 \ i \ + \ 71$
(D)	$ 10 \ i \ - \ 71$
35-	If $ f(x) \ = \ x \ – \frac{4}{5}$ and $f ^{ \ – \ 1} $ is the inverse of $ f(x) $ , what is the value of $ f^{ \ - \ 1}(2) $
(A)	$\frac{7}{5} $
(B)	$\frac{16}{5} $
(C)	$\frac{6}{5} $
(D)	$\frac{14}{5} $
36-	Find tan $ \frac{4 \ π}{3} $
(A)	$ - \ \sqrt{3} $
(B)	$ \sqrt{3} $
(C)	$2 \ \sqrt{3} $
(D)	$ - \ 2 \ \sqrt{3} $

37-	Solve the equation: $ \log_{5}{(x \ - \ 4)} \ – \ \log_{5}{(x \ + \ 3)}⁡ \ = \ 1 $
(A)	$\frac{9}{4}$
(B)	$\frac{18}{4}$
(C)	$\frac{14}{4}$
(D)	$-\frac{19}{4}$
38-	If $ f(x) \ = 2 \ x \ + \ 6 $ and $ g(x) \ = x^2 \ + \ 3 \ x$ , then find $ (\frac{f}{g}) (x) $
(A)	$\frac{2 \ x \ - \ 6}{x^2 \ +\ 3 \ x} $
(B)	$\frac{2 \ x \ + \ 6}{x^2 \ +\ 3 \ x} $
(C)	$\frac{2 \ x \ + \ 6}{x^2 \ - \ 3 \ x} $
(D)	$\frac{2 \ x \ - \ 6}{x^2 \ - \ 3 \ x} $
39-	The slop of a line with the equation $ y \ = \ 6 \ x \ + \ 12 $ is …
(A)	$4$
(B)	$\frac{5}{3}$
(C)	$\frac{5}{6}$
(D)	$6$
40-	Solve $ e^{(7 \ x \ + \ 1 )} \ = \ 8 $
(A)	$\frac{ln⁡(8) \ +\ 1}{7}$
(B)	$\frac{ln⁡(8) \ +\ 1}{5}$
(C)	$\frac{ln⁡(8) \ -\ 1}{5}$
(D)	$\frac{ln⁡(8) \ -\ 1}{7}$

ACCUPLACER Mathematics Practice Test Answers and Explanations

1- Choice C is correct

The correct answer is $ $ 16.88$
$$ 62 \ - \ $ 45.12 \ = \ $ 16.88$

2- Choice A is correct

The correct answer is $(x \ – \ 3) \ (x \ - \ 5)$
$ x^2 \ - \ 8 \ x \ + \ 15 \ = \ (x \ – \ 3) \ (x \ - \ 5) $

3- Choice A is correct

The correct answer is $ 6254.7$
Underline the tenth place:
$6254.\underline{\\7}4164$
Look to the right if it is $7 $ or above, give it a shove.
Then, round up to $ 6254.7$

4- Choice D is correct

The correct answer is $45^\circ$
$90^\circ \ + \ 45^\circ = 135^\circ$
$180^\circ \ - \ 135^\circ = 45^\circ$

5- Choice B is correct

The correct answer is $ 37.5\%$
$ 120\ × \ \frac{x}{100} \ = \ 45 \ ⇒ \ 120 \ × \ x \ = \ 4500 \ ⇒ \ x \ = \ \frac{4500}{120} =37.5\% $

6- Choice A is correct

The correct answer is $3 \ y \ + \ 6 = 4 \ x$
$ 3 \ x \ + \ 6 = 4 \ x $ has a graph that is a straight line

7- Choice C is correct

The correct answer is $-\frac{1}{2} , 3 $
$ x_{1,2} \ = \ \frac{- \ b \ ± \ \sqrt{b^2 \ - \ 4 \ a \ c}}{2 \ a} $
$ a \ x^2 \ + \ b \ x \ + \ c \ = \ 0 $
$ 2 \ x^2 \ - \ 5 \ x \ – \ 3 \ = \ 0 \ ⇒$ then: $a \ = \ 2, b \ = - \ 5 \ $ and $ \ c \ = \ – \ 3 $
$x \ = \ \frac{- \ (- \ 5) \ + \ \sqrt{(5^2) \ - \ (4) \ (2) \ ( - \ 3)}}{2 .2} = 3$
$ x \ = \ \frac{- \ (- \ 5 )\ - \ \sqrt{(5^2) \ - \ (4)\ (2) \ (- \ 3)}}{2.2} \ = -\ \frac{1}{2}$

8- Choice B is correct

The correct answer is $10$
$C \ = \ \sqrt{(x_A \ - \ x_B)^2 \ + \ (y_A \ - \ y_B)^2 }$
$C \ = \ \sqrt{(2 \ - \ (- \ 4) )^2 \ + \ ( 4 \ - \ (- \ 4))^2 }$
$C \ = \ \sqrt{(6)^2 \ + \ ( 8)^2 } $
$C \ = \ \sqrt{36 \ + \ 64}$
$C \ = \ \sqrt{100} = 10 $

9- Choice D is correct

The correct answer is $ \frac{- \ 2 \ x \ + \ 15}{3 \ x^2 \ - \ 21} $
Simplify:
$ \frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2}{3} \ - \ \frac{7}{3}} = \frac{\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9}}{\frac{x^2 \ - \ 7}{3}} = \frac{ 3 \ (\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9 })}{x^2 \ - \ 7} $
⇒Simplify: $\frac{4}{3} \ - \ \frac{2 \ x \ - \ 3}{9 } = \frac{ -\ 2\ x \ + \ 15 }{9}$
then: $\frac{3(\frac{ - \ 2 \ x \ + \ 15 }{9})}{x^2 \ - \ 7} = \frac{\frac{ -\ 2 \ x \ + \ 15}{3 }}{x^2 \ - \ 7} = \frac{ - \ 2 \ x \ + \ 15}{3\ (x^2 \ - \ 7)} = \frac{-2 \ x \ +\ 15}{3 \ x^2 \ - \ 21} $

10- Choice C is correct

The correct answer is $$ 254$
$42\ × \ $ 65 \ = \ $ 2730$ Payable amount is: $$ 3254 \ - \ $ 2730 \ = \ $ 524$

11- Choice B is correct

The correct answer is $0.058$
$ \frac{7}{120} \ = \ \frac{1}{17.14} \ =0.05833333333 \ \cong \ 0.058 $

12- Choice B is correct

The correct answer is $8$ and $9$
$ \sqrt{75} \ = 8.66025 $ ...
then:
$ \sqrt{75 }$ is between $ 8$ and $9$

13- Choice A is correct

The correct answer is $ \frac{- \ 5 \ ± \ \sqrt{31}}{3} $
$ x_{1,2} \ = \ \frac{- \ b \ ± \ \sqrt{b^2 \ - \ 4 \ a \ c}}{2 \ a} $
$ a \ x^2 \ + \ b \ x \ + \ c \ = \ 0 $
$ 3 \ x^2 \ + \ 10 \ x \ - \ 2 = 0 \ ⇒ $ then: a $= \ 3$, b $ = 10 $ and c $ = -\ 2 $
$ x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3 )(- \ 2 )}}{2.3} \ =\frac{- \ 5 \ - \ \sqrt{31}}{3} = -3.523 $
$ x \ = \frac{- \ 10 \ + \ \sqrt{10^2 \ - \ (4) .(3) .(-\ 2)}}{2.3} \ = \frac{- \ 5 \ + \ \sqrt{31}}{3} = 0.189$

14- Choice D is correct

The correct answer is $ x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21$
Use FOIL (First, Out, In, Last)
$ (x \ + \ 7) (x^2 \ - \ 4 \ x \ + \ 3) = x^3 \ - \ 4 \ x^2 \ + \ 3 \ x \ + \ 7 \ x^2 \ – \ 28 \ x \ + \ 21
= x^3 \ + \ 3 \ x^2 \ – \ 25 \ x \ + \ 21 $

15- Choice B is correct

The correct answer is $ x^{ \frac{10}{7}}$
$ (x^5)^{\frac{2}{7}} = x^{5 \ × \ \frac{2}{7}} \ = \ x^{ \frac{10}{7}} \ = \ x^{ \frac{10}{7}} $

16- Choice A is correct

The correct answer is $21$
Number of squares equal to: $ \frac{42 \ × \ 18}{6 \ × \ 6} \ = \ 7 \ × \ 3 = 21$

17- Choice D is correct

The correct answer is $40$ miles
$ 42 \ + \ 47 \ + \ 31 = 120 $
Average $ = \ \frac{120}{3} = 40 $ miles

18- Choice C is correct

The correct answer is $16$
$ \frac{a \ + \ b}{2} \ = \ 10\ ⇒ \ a \ + \ b \ = \ 20 $
$ \frac{a \ + \ b \ + \ c}{3} \ = 12 \ ⇒ \ a \ + \ b \ + \ c \ = 36 $
$20 \ + \ c \ = 36 \ ⇒ \ c \ = \ 36\ – \ 20 \ = \ 16$

19- Choice C is correct

The correct answer is $-4$
Simplify:
$ 6 \ - \ 3 \ x \ ≤ \ 18 \ ⇒ - \ 3 \ x \ ≤ \ 18 \ – \ 6 \ ⇒ - \ 3 \ x \ ≤ \ 12 ⇒ x \ \geq \ -4 $

20- Choice B is correct

The correct answer is $9^{ - \ 3}$
$ 9^{ 5} \ × \ 9^{ - \ 8} \ = 9^{ 5 \ + (- \ 8)} = 9^{- \ 3} $

20- Choice B is correct

The correct answer is $9^{ - \ 3}$
$ 9^{ 5} \ × \ 9^{ - \ 8} \ = 9^{ 5 \ + (- \ 8)} = 9^{- \ 3} $

21- Choice B is correct

The correct answer is $ 1\ - \ 2 \ sin^2\ \theta$
$ cos\ 2\ \theta = cos^2\ \theta \ - \ sin^2\ \theta = 2 \ cos^2\ \theta \ -\ 1 = 1 \ -\ 2\ sin^2\ \theta$

22- Choice B is correct

The correct answer is $\frac{3}{5}$
sin $θ \ = \ \frac{4}{5} \ ⇒ $ we have following triangle, then:
c $= \ \sqrt{5^2 \ - \ 4 ^2} \ = \ \sqrt{25 \ - \ 16} \ = \ \sqrt{9} \ = \ 3$
cos $θ \ = \ \frac{3}{5}$

23- Choice B is correct

The correct answer is $(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 2 $
$ (x \ – \ h)^2 \ + \ (y \ – \ k)^2 \ = \ r^2 \ ⇒ $ center: $ (h,k)$ and radius: $ r$
center: $ (4 , - \ 1 ) \ ⇒ \ h = 4, k = - \ 1 $
circumference $ = 4 \ π ⇒ $ circumference $ = 2 \ π \ r = 4 \ π \ ⇒ \ r \ = \ 2$
$(x \ - \ 4)^2 \ + \ (y \ + \ 1)^2 \ = \ 4 $

24- Choice C is correct

The correct answer is $(6, - \ 16)$
$ \begin{cases}4 \ x \ + \ y \ = 8 \\ - \ 8 \ x \ - \ 4 \ y \ = 16\end{cases}$⇒ Multiplication $ (4) $ in first equation $⇒ \ \begin{cases}16 \ x \ + \ 4 \ y \ = 32 \ \\ - \ 8 \ x \ - \ 4 \ y = 16\end{cases}$
Add two equations together $ ⇒\ 8 \ x \ = 48 \ ⇒ x = 6 $ then: $ y = - \ 16 $

25- Choice B is correct

The correct answer is $(6, - \ 4), \ \sqrt{5} $
$ (x \ – \ h)^2 \ + (y \ – \ k)^2 = r^2 \ ⇒ $ center: $(h,k)$ and radius: $r$
$ (x \ – \ 6)^2 \ + \ (y \ + \ 4)^2 = 5 ⇒ $ center: $(6, - \ 4)$ and radius:$ \ \sqrt{5} $

26- Choice C is correct

The correct answer is $\frac{\sqrt{21}}{5}$
sin A$= \ \frac{2}{5 } \ ⇒ $ we have following triangle, then
$ c \ = \ \sqrt{5 ^{2} \ - \ 2^{2}} \ = \ \sqrt{25 \ - \ 4} \ = \ \sqrt{21} $
cos⁡ A $= \ \frac{adjacent}{hypotenuse} \ → $ cos A $= \ \frac{\sqrt{21}}{5} $

27- Choice A is correct

The correct answer is $729$
METHOD ONE
$ \log_{3}{x} \ = \ 6 $
Apply logarithm rule: $ a \ = \ \log_{3}{(b^a)} $
$ 6 \ = \ \log_{3}{3^6} \ = \ \log_{3}{729}$
$ \log_{3}{x} \ = \ \log_{3}{729} $
When the logs have the same base: $ \log_{b}{(f (x))} \ = \ \log_{b}{(g (x))} \ ⇒ \ f (x) \ = \ g (x) $
then: $ x \ = \ 729 $
METHOD TWO
We know that: $ \log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c \ \log_{3}{x} \ = 6 \ ⇒ \ x \ = \ 3^6 \ = \ 729 $

28- Choice C is correct

The correct answer is $\frac{1}{9}$
$ \frac{ 2 \ \sqrt{12}}{ 9 \sqrt{48 }}$
$ \sqrt{48} \ = 2 \ \sqrt{12 }$
$ \frac{ 2 \sqrt{12}}{9 .2 \ \sqrt{12}} \ = \ \frac{1}{9} $

29- Choice A is correct

The correct answer is $\frac{x^3}{25 }$
$ \frac{25}{x^3} ⇒$ reciprocal is : $ \frac{x^3}{25 } $

30- Choice C is correct

The correct answer is $64$
METHOD ONE
$ \log_{4}{x}⁡ = 3 $
Apply logarithm rule: $ a \ = \ log_b⁡(b^a) $
$ 3 = \log_{4}{4^3} = \log_{4}{64} $
$ \log_{4}{64} = \log_{4}{x} $
When the logs have the same base: $ \log_{b}{(f (x))} \ = \ \log_{b}{(g (x))} \ ⇒ \ f (x) \ = \ g (x) $
then: $ x = 64 $
METHOD TWO
We know that: $ \log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c $ $ \log_{4}{x}= 3 ⇒ x = 4^3 = 64 $

31- Choice A is correct

The correct answer is $\frac{\sqrt{3}}{2}$
sin $60^{°} \ = \ \frac{\sqrt{3}}{2} $

32- Choice C is correct

The correct answer is $\frac{1}{4} (e^{\ x} \ + \ 3)$
$ f (x) \ = \ ln \ (4\ x \ - \ 3) $
$ y \ = \ ln \ (4 \ x \ - \ 3 ) $
Change variables $ x $ and $ y: x \ = \ ln \ (4 \ y \ - \ 3) $
solve: $ x \ = \ ln \ (4 \ y \ - \ 3 ) $
$ y \ = \ \frac{\ e^{\ x} \ + \ 3}{4 } \ = \ \frac{1}{4} (e^{\ x} \ + \ 3) $

33- Choice D is correct

The correct answer is $13$
$ | \ 15 \ – \ (18 \ ÷ \ | \ 3 \ + \ 6 \ |)| = | (15 \ - \ (18 \ ÷ \ | 9 |))| \ =$
$| \ 15 \ - \ (18 \ ÷ \ 9) \ | = |15 \ - \ 2| \ = \ | \ 13 \ | \ = \ 13 $

34- Choice A is correct

The correct answer is $ - \ 10 \ i \ + \ 71$
We know that: $ i \ = \ \sqrt{\ - \ 1} \ ⇒ \ i^2 \ = \ - \ 1 $
$ (- \ 7 \ + \ 2\ i) \ (9 \ + \ 4 \ i) = - \ 63 \ - \ 28 \ i \ + \ 18 \ i \ + \ 8 \ i^2 \ =$
$ - \ 63 \ + \ 10 \ i \ - \ 8 \ =- \ 10 \ i \ - \ 71 $

35- Choice D is correct

The correct answer is $\frac{8}{5} $
$ f(x) \ = \ x \ – \frac{4}{5} \ ⇒ \ y \ = \ x \ – \ \frac{4}{5} \ ⇒ \ y \ + \ \frac{4}{5} \ = \ x $
$ f^{ \ - \ 1} \ = \ x \ + \ \frac{4}{5} $
$ f ^{ \ – \ 1}(2) \ = 2 \ + \ \frac{4}{5} \ = \ \frac{14}{5} $

36- Choice B is correct

The correct answer is $ \sqrt{3} $
tan$ \frac{4 \ π}{3} \ = \ \frac{sin \ \frac{4 \ π}{3}}{cos \ \frac{4 \ π}{3}} = \frac{\frac{ - \ \sqrt{3}}{2}}{- \ \frac{1}{2}} \ = \sqrt{3} $

37- Choice D is correct

The correct answer is $-\frac{19}{4}$
METHOD ONE
$ \log_{5}{(x \ - \ 4)} – \log_{5}{(x \ + \ 3)} \ = \ 1 $
Add $ \log_{5}{(x \ + \ 3)} $ to both sides
$\log_{5}{(x \ - \ 4)} \ - \ \log_{5}{(x \ + \ 3)} \ + \ \log_{5}{(x \ + \ 3)} \ = \ 1 \ + \ \log_{5}{(x \ + \ 3)}$
Apply logarithm rule: $ a \ = \ \log_{b}{b^a} \ ⇒ \ 1 \ = \ \log_{5}{5^1} \ = \ \log_{5}{5} $
then: $ \log_{5}{(x \ - \ 4)} \ = \ \log_{5}{5} \ + \ \log_{5}{(x \ + \ 3)} $
Logarithm rule: $ \log_{c}{a} \ + \ \log_{c}{b} \ = \ \log_{c}{ab} $
then: $ \log_{5}{5} \ + \ \log_{5}{(x \ + \ 3)} = \log_{5}{(5(x \ + \ 3))} $
$ \log_{5}{(x \ - \ 4)} \ = \ \log_{5}{(5 \ (x \ + \ 3))} $
When the logs have the same base: $ \log_{b}{(f(x))} \ = \ \log_{b}{(g(x))} \ ⇒ \ f(x) \ = \ g(x) $
$ (x \ - \ 4) \ = 5 \ (x \ + \ 3)$
$ x \ =- \ \frac{19}{4} $
METHOD TWO
We know that: $ \log_{a}{b} \ - \ \log_{a}{c} \ = \ \log_{a}{\frac{b}{c}} $ and $ \log_{a}{b} \ = \ c \ ⇒ \ b \ = \ a^c $
Then: $ \log_{5}{x \ - \ 4} \ - \ \log_{5}{x \ +\ 3} \ = \ \log_{5}{\frac{x \ - \ 4} {x \ + \ 3}} \ = \ 1 \ ⇒$
$\frac{x \ - \ 4}{x \ + \ 3} \ = \ 5^1 \ = \ 5 \ ⇒ \ x \ - \ 4 \ = \ 5 \ (x \ + \ 3)$
$⇒ \ x \ - \ 4\ = \ 5 \ x \ + \ 15 \ ⇒ \ 5 \ x \ - \ x \ =\ -15 \ - \ 4 \ → \ 4 \ x \ = \ -19 \ ⇒ \ x \ =- \ \frac{19}{4} $

38- Choice B is correct

The correct answer is $\frac{2 \ x \ + \ 6}{x^2 \ +\ 3 \ x} $
$ (\frac{f}{g}) (x) \ = \ \frac{f(x)}{g(x)} \ = \ \frac{2 \ x \ + \ 6}{x^2 \ +\ 3\ x} $

39- Choice D is correct

The correct answer is $6$
$ y \ = \ m \ x \ + \ b $
$ m \ =$ slop
$ y \ = \ 6 \ x \ + \ 12$
$ m \ = \ 6 $

40- Choice D is correct

The correct answer is $\frac{ln⁡(8) \ - 1}{7}$
$ e^{7 \ x \ + \ 1 } \ = \ 8$
If $ f(x) \ = \ g(x)$ , then $ ln(f(x)) \ = \ ln(g(x)) $
$ ln⁡(e^{7 \ x \ + \ 1 }) \ = \ ln(8) $
Apply logarithm rule: $ \log_{a}{x^b} \ = \ b \log_{a}{x} $
$ ln⁡(e^{7 \ x \ + \ 1 }) = (7 \ x \ + \ 1) \ ln(e) $
$ (7 \ x \ + \ 1) \ ln \ (e) \ = \ ln \ (8) $
$ (7 \ x \ + \ 1) \ ln \ (e) \ = \ (7 \ x \ + \ 1) $
$ (7 \ x \ + \ 1) \ = \ ln \ (8) \ ⇒ \ x \ = \ \frac{ln⁡(8) \ - 1}{7} $

The Most Comprehensive Accuplacer Math Preparation Bundle

~~$76.99~~ $33.99

56% Off*

Bundle Includes Accuplacer Math Prep Books, Workbooks, and Practice Tests

eBook (PDF Format)

Download

More Free ACCUPLACER Mathematics Practice Test

Practice Test 1

Simulate test day with an official practice test. Then, score your test. The answers come with explanations so you can learn from your mistakes.

Start Practice Test

Practice Test 2

Simulate test day with an official practice test. Then, score your test. The answers come with explanations so you can learn from your mistakes.

Start Practice Test

Full Length ACCUPLACER Mathematics Practice Test