Free Full Length ACT Mathematics Practice Test

The correct answer is $36^\circ$ 
The sum of all angles in a quadrilateral is $360$ degrees.
Let $x$ be the smallest angle in the quadrilateral.
Then the angles are: $x,\ 2\ x\ ,\ 3\ x\ ,\ 4\ x$
$x\ +\ 2\ x\ +\ 3\ x\ +\ 4 \ x=360→10\ x=360→x=36$ 
The angles in the quadrilateral are: $36^\circ, \ 72^\circ,\ 108^\circ$, and $144^\circ$

The correct answer is $0$ 
Solving Systems of Equations by Elimination
Multiply the first equation by $- \ 2$, then add it to the second equation.
$\cfrac{\begin{align} - \ 2 \ (2 \ x \ + \ 3 \ y \ = \ 12) \\ 4 \ x \ - \ 4 \ y \ = - \ 16 \end{align}}{}$
$\cfrac{ \begin{align} - \ 4 \ x \ - \ 6 \ y \ = \ - \ 24 \\ 4 \ x \ - \ 4 \ y \ = - \ 16 \end{align} }{\begin{align} - \ 10\ y \ = - \ 40 \\ ⇒ y \ = \ 4 \end{align}}$
Plug in the value of $y$ into one of the equations and solve for $x$.
$2 \ x \ + \ 3 \ (4)= 12 ⇒$
$2 \ x \ + \ 12= 12 ⇒$
$2 \ x=0⇒ x= 0$

The correct answer is $\frac{2 \ \sqrt{6}}{5}$
sin $A=\frac{1}{5}⇒$
Since sin$θ=\frac{opposite}{hypotenuse}$, we have the following right triangle. Then:
$c=\sqrt{5^2\ -\ 1^2 }= \sqrt{25\ -\ 1}=\sqrt{24}=2 \ \sqrt{6}$
cos $A=\frac{2 \ \sqrt{6}}{5}$

The correct answer is $420$
The ratio of boy to girls is $3:4$.
Therefore, there are $3$ boys out of $4$ students.
To find the answer, first divide the total number of students by $4$, then multiply the result by $3$. 
$560 \ ÷\ 4 = 140 ⇒ 140 \ ×\ 3 = 420$

The correct answer is $\frac{2 \ x \ − \ 1}{x^2 \ + \ 2 \ x}$
$(\frac{f}{g})(x) = \frac{f(x)}{g(x)} = \frac{2 \ x \ − \ 1}{x^2 \ + \ 2 \ x}$

The correct answer is$8$
Plug in the value of each option in the inequality.
A. $1$    $(1\ -\ 2)^2\ +\ 1>3\ (1) \ -\ 1→2>2$        No!
B. $6$    $(6\ -\ 2)^2\ +\ 1>3\ (6)\ -\ 1→17>17$    No!
C. $8$    $(8\ -\ 2)^2\ +\ 1>3\ (8)\ -\ 1→37>23$    Bingo!
D. $3$    $(3\ -\ 2)^2\ +\ 1>3\ (3)\ -\ 1→2>8$        No!
E $4$     $(4\ -\ 2)^2\ +\ 1>3\ (4)\ -\ 1→5>11$      No!

The correct answer is $25\ π$
The equation of a circle in standard form is: 
$(x\ -\ h)^2\ +\ (y\ -\ k)^2=r^2$, where $r$ is the radius of the circle. 
In this circle the radius is $5$.
$r^2=25→r=5$
$(x\ +\ 2)^2\ +\ (y\ -\ 4)^2=25$
Area of a circle: $A=π\ r^2=π\ (5)^2=25\ π$

The correct answer is $y=4 \ x \ − \ 17$
The equation of a line is: $y=m \ x \ + \ b$, where m is the slope and b is the $y-$intercept.
First find the slope: 
$m=\frac{y_{2} \ - \ y_{1}}{x_{2} \ - \ x_{1} }=\frac{15 \ -\ (- \ 5)}{8 \ - \ 3}=\frac{20}{5}=4$
Then, we have: $y=4 \ x \ + \ b$
Choose one point and plug in the values of $x$ and $y$ in the equation to solve for $b$.
Let’s choose the point $(3, - \ 5)$
$y=4 \ x \ + \ b→ \ - \ 5=4 \ (3) \ + \ b→ \ - \ 5=12 \ + \ b→b= \ - \ 17$
The equation of the line is: $y=4 \ x \ - \ 17$

The correct answer is $\frac{7\sqrt{π}}{π}$
Formula for the area of a circle is: $A=π\ r^2$ 
Using $49$ for the area of the circle we have:$49=π\ r^2$ Let’s solve for the radius $(r)$. 
$\frac{49}{π}=r^2→r=\sqrt{\frac{49}{π}}=\frac{7}{\sqrt{π}}=\frac{7}{\sqrt{π}}\ ×\ \frac{\sqrt{π}}{\sqrt{π}}=\frac{7\sqrt{π}}{π}$

The correct answer is $$750$
Use simple interest formula:
$I=prt$
($I =$ interest, $p =$ principal, $r =$ rate, $t =$ time)
$I=(15000) \ (0.025) \ (2)=750$

The correct answer is $x\ ≥9 \ ∪ \ x≤ 1$
$x\ -\ 5\ ≥\ 4→x\ ≥4\ +\ 5→x≥9$
Or
$x\ -\ 5≤-\ 4→x≤-\ 4\ +\ 5→x≤ 1$
Then, solution is: $x\ ≥9 \ ∪ \ x≤ 1$

The correct answer is $48$ 
Length of the rectangle is: $\frac{3}{5}\ ×\ 15=9$ 
perimeter of rectangle is: $2\ ×\ (9\ +\ 15)=48$

The correct answer is $60$
First, find the number.
Let $x$ be the number.
Write the equation and solve for $x$. 
$120 \%$ of a number is $90$, then:
$1.2\ ×\ x=90 ⇒ x=90 \ ÷\ 1.2=75$
$80 \%$ of $75$ is: $0.7 \ ×\ 75 = 60$

The correct answer is $40\%$ 
The population is increased by $12\%$ and $25\%$.
$12\%$ increase changes the population to $112\%$ of original population.
For the second increase, multiply the result by $125\%$.
$(1.12) \ ×\ (1.25) = 1.40= 140\%$
$40$ percent of the population is increased after two years.

The correct answer is $40\%$ 
The population is increased by $12\%$ and $25\%$.
$12\%$ increase changes the population to $112\%$ of original population.
For the second increase, multiply the result by $125\%$.
$(1.12) \ ×\ (1.25) = 1.40= 140\%$
$40$ percent of the population is increased after two years.

The correct answer is $\frac{4}{19}$
The probability of choosing a Hearts is $\frac{12}{57} = \frac{4}{19}$

The correct answer is $9\ x^3\ y^5\ -\ 5 \ x^2\ y^3$
$4\ x^2\ \ y^3\ + \ 5\ x^3\ y^5\ –\ 9\ x^2\ y^3\ +\ 4\ x^3\ y^5\ =$
$4\ x^2\ y^3\ -\ 9\ x^2\ y^3\ +\ 5\ x^3\ y^5\ + \ 4 \ x^3 \ y^5\ =$
$9\ x^3\ y^5\ -\ 5 \ x^2\ y^3$

The correct answer is $142$ degree 
The angle $x$ and $35$ are complementary angles.
Therefore: $x\ +\ 38=180$ 
$180^\circ\ - \ 38^\circ=142^\circ$

The correct answer is $\frac{3}{5}$
tan ⁡$=\frac{opposite}{djacent}$, and tan⁡ $x=\frac{6}{8}$, therefore, the opposite side of the angle $x$ is $6$ and the adjacent side is $8$.
Let’s draw the triangle.
Using Pythagorean theorem, we have:
$a^2\ +\ b^2=c^2→6^2\ +\ 8^2=c^2→36\ +\ 64=c^2→c=10$
sin ⁡$=\frac{opposite}{hypotnuse}=\frac{6}{10} = \frac{3}{5}$

The correct answer is $36$
The sum of supplement angles is $180$. Let $x$ be that angle.
Therefore, $x \ +\ 4\ x\ = 180$
$5\ x\ = 180$, divide both sides by $5$:
$x = 36$

The correct answer is $4$
sin$^2 \ a\ +$ cos$^2 \ a=1$, then: $x\ +\ 1=5, x=4$

The correct answer is $24\%$
The question is this: $474.24$ is what percent of $624$?
Use percent formula: part $= \frac{percent}{100} \ ×$ whole
$474.24 = \frac{percent}{100} \ ×\ 624 ⇒ 474.24= \frac{percent \ ×\ 624}{100} ⇒47424 =$ percent $×\ 624 ⇒$ 
percent $= \frac{47424}{624} = 83.39$
$47424$ is $76 \%$ of $624$. Therefore, the discount is: $100\% \ –\ 76\% = 24\%$

The correct answer is $\frac{\sqrt{5} }{2 \ \sqrt{3}} \ x$
Simplify the expression.
$\sqrt{\frac{x^2}{3}\ +\ \frac{x^2}{12}}=\sqrt{\frac{4 \ x^2}{12}\ +\ \frac{x^2}{12}}=\sqrt{\frac{5x^2}{12}}=\sqrt{\frac{5}{12} \ x^2 }= \sqrt{\frac{5}{12}} \ ×\ \sqrt{x^2} = \frac{\sqrt{5}}{\sqrt{12}}\ ×\ \sqrt{x^2}= \frac{\sqrt{5}}{2 \ \sqrt{3}}\ ×\ \sqrt{x^2}=\frac{\sqrt{5} }{2 \ \sqrt{3}} \ x$

The correct answer is $45,000$ 
Three times of $18,000$ is $54,000$. One sixth of them cancelled their tickets.
One sixth of $54,000$ equals $9,000\ (\frac{1}{6} \ ×\ 54000 = 9000)$. 
$45,000 \ (54000 \ –\ 9000 = 45000)$ fans are attending this week

The correct answer is $27.5$
Let $x$ be the smallest number.
Then, these are the numbers:
$x, x\ + \ 1, x \ + \ 2, x \ + \ 3, x \ + \ 4, x \ + \ 5$
average $= \frac{sum \ of \ terms}{number \ of \ terms} ⇒$
$30 = \frac{x \ + \ (x \ + \ 1) \ + \ (x \ + \ 2) \ + \ (x \ + \ 3) \ + \ (x \ + \ 4)\ + \ (x \ + \ 5)}{6}⇒$
$30=\frac{6 \ x \ + \ 15}{6} ⇒$ 
$180 = 6 \ x \ + \ 15 ⇒$
$165 = 6 \ x ⇒ x=27.5$

The correct answer is $− \ \frac{1}{2}$
The equation of a line in slope intercept form is: $y=m \ x \ + \ b$
Solve for $y$.
$4 \ x \ - \ 2 \ y=16 ⇒$
$- \ 2 \ y=16 \ - \ 4 \ x ⇒$
$y=(16 \ - \ 4 \ x) \ ÷ \ (- \ 2) ⇒$
$y=2 \ x \ - \ 8$
The slope is $2$.
The slope of the line perpendicular to this line is:
$m_{1} \ × \ m{2} = \ - \ 1 ⇒$
$2 \ × \ m_{2} = \ - \ 1 ⇒ m_{2} = \ - \ \frac{1}{2}$

The correct answer is $30$
Plug in the value of $x$ and $y$. 
$x=2$ and $y=-\ 2$
$5\ (x\ -\ 2\ y)\ +\ (2 \ - \ x)^2=5\ (2\ -\ 2(-\ 2))\ +\ (2\ -\ 2)^2=5\ (2\ +\ 4)\ +\ (0)^2 = 30\ +\ 0=30$

The correct answer is $3.2\ ×\ 10^5$
$320000=3.2\ ×\ 10^5$

The correct answer is $\frac{y}{8}$
Solve for $x$.
$\sqrt{8 \ x}=\sqrt{y}$
Square both sides of the equation:
$(\sqrt{8 \ x})^2=(\sqrt{y})^2$  ,  $8\ x=y$  ,   $x=\frac{y}{8}$

The correct answer is Length of AD equal to length BC
In any rectangle opposite sides are equal.

The correct answer is $2$ cm
Formula for the Surface area of a cylinder is:
S A $=2 \ π \ r^2 \ + \ 2 \ π \ r \ h→$
$48 \ π=2 \ π \ r^2 \ + \ 2 \ π \ r \ (10)→$
$r^2 \ + \ 10 \ r \ - \ 24=0$
Factorize and solve for $r$.
$(r \ + \ 12)(r \ - \ 2)=0→r=2$ or $r= \ - \ 12$ (unacceptable)

The correct answer is $8, 4$
$(x\ +\ 2)\ (x\ +\ p)=x^2\ +\ (2\ +\ p)x\ +\ 2\ p→2\ +\ p=6→p=4$ and
$r=2\ p=8$

The correct answer is $(1,6), (2,5), (− \ 5,8)$
Since the triangle ABC is reflected over the $y-$axis, then all values of $y$’s of the points don’t change and the sign of all $x$’s change. 
(remember that when a point is reflected over the $y-$axis, the value of $y$ does not change and when a point is reflected over the $x-$axis, the value of $x$ does not change).
Therefore:
$(− \ 1,6)$ changes to $(1, 6)$
$(− \ 2, 5)$ changes to $(2, 5)$
$(5, 8)$ changes to $(− \ 5, 8)$

The correct answer is $170$ miles
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $a^2 \ + \ b^2 = c^2$
$80^2 \ + \ 150^2 = c^2 ⇒$
$6400 \ + \ 22500 = c^2 ⇒$
$28900 = c^2 ⇒ c = 170$

The correct answer is $\frac{1}{10}$
Write the ratio of $2 \ a$ to $3 \ b$.
$\frac{2 \ a}{3 \ b}=\frac{1}{15}$
Use cross multiplication and then simplify.
$2 \ a \ × \ 15=3 \ b \ ×\ 1→30 \ a=3 \ b→a=\frac{3 \ b}{30}=\frac{b}{10}$
Now, find the ratio of $a$ to $b$.
$\frac{a}{b}=\frac{\frac{b}{10}}{b}→\frac{b}{10} \ ÷\ b=\frac{b}{10} \ × \ \frac{1}{b}=\frac{b}{10 \ b}=\frac{1}{10}$

The correct answer is $$4.5$
Let $x$ be the cost of one-kilogram orange, then: $4\ x\ + \ (2 \ × \ 4.2)=25.4→$
$4\ x\ +\ 8.4=25.4→4\ x=25.4\ -\ 8.4→4\ x=18→x=\frac{18}{4}=$4.5$

The correct answer is $7$
Let $x$ be the number of adult tickets and $y$ be the number of student tickets. Then:
$x \ + \ y =12, \ 12.50 \ x \ + \ 7.50y=125$
Use elimination method to solve this system of equation.
Multiply the first equation by $- \ 7.5$ and add it to the second equation.
$- \ 7.5 \ (x \ + \ y=12), \ - \ 7.5 \ x \ - \ 7.5 \ y= \ - \ 90, 12.50 \ x \ + \ 7.50y=125$
$5 \ x=35, \ x=7$
There are $7$ adult tickets and $5$ student tickets.

The correct answer is $\begin{bmatrix}-\ 4&5\\ 4&-\ 8\end{bmatrix}$
First, find $2$ A.
A$=\begin{bmatrix}-\ 1 & 3\\ 1 & -\ 3\end{bmatrix}$
$2$A$=2\ ×\ \begin{bmatrix}-\ 1&3\\ 1 & -\ 3\end{bmatrix}=\begin{bmatrix} -\ 2& 6\\ 2 & -\ 6\end {bmatrix}$
Now, solve for $2$ A $-$ B.
$2$ A$-$B$=\begin{bmatrix} -\ 2&6 \\ 2& -\ 6\end {bmatrix}\ -\ \begin{bmatrix} 2&1\\ -\ 2 & 2\end{bmatrix}=\begin{bmatrix}-\ 2 \ -\ 2 & 6 \ -\ 1\\ 2\ -\ (-\ 2) & -\ 6\ -\ 2\end{bmatrix}=\begin{bmatrix}-\ 4&5\\ 4&-\ 8\end{bmatrix}$

The correct answer is $x ^ {\frac{15}{7}}$
$(x^5)^{\frac{3}{7}} = x^{5 \ ×\ \frac{3}{7}} =x ^ {\frac{15}{7}}$

The corrcet answer is $3,375$
If the length of the box is $45$, then the width of the box is one third of it, $15$, and the height of the box is $5$ (one third of the width).
The volume of the box is:
$V = lwh = (45)\ (15)\ (5) = 3,375$

The correct answer is $2$
The amplitude in the graph of the equation $y=a$ cos $b\ x$ is $a$. ($a$ and $b$ are constant)
In the equation $y\ -\ 1=2$ cos $3\ x$, the amplitude is $2$.

The correct answer is $\frac{\sqrt{3}}{2}$
The relationship among all sides of right triangle $30^\circ \ - \ 60^\circ \ - \ 90^\circ$ is provided in the following triangle: 
cos of $30^\circ$ equals to:$\frac{adjacent}{hypotenuse}=\frac{x \ \sqrt{3}}{2\ x}=\frac{\sqrt{3}}{2}$

The correct answer is $43$
The area of rectangle is: $8 \ × \ 4=32$ cm$^2$
The area of circle is: $π \ r^2=π \ × \ ( \frac{10}{2})^2=3 \ × \ 25=75$ cm$^2$
Difference of areas is: $75 \ - \ 32=43$

The correct answer is $62.84$ kg
average$= \frac{sum \ of\ terms }{number \ of\ terms}$
The sum of the weight of all girls is: $18 \ ×\ 60 = 900$ kg
The sum of the weight of all boys is: $37 \ ×\ 64 = 2368$ kg
The sum of the weight of all students is: $900 \ +\ 2368 = 3268$ kg
average $= \frac{3268}{52} = 62.84$

The correct answer is January and February
First find the number of pants sold in each month.
A. January: $110$, February: $88$, March: $90$, April: $70$, May: $85$, June: $65$
Check each option provided.
January and February, 
$(\frac{110 \ - \ 88}{110}) \ × \ 100=\frac{22}{110} \ × \ 100=20\%$
B. February and March, there is an increase from February to March.
C. March and April
$(\frac{90 \ - \ 70}{90}) \ × \ 100=\frac{20}{90} \ × \ 100=22.22\%$
D. April and May: there is an increase from April to May
May and June
$(\frac{85 \ - \ 65}{85}) \ × \ 100=\frac{20}{85} \ × \ 100=23.53\%$

The correct answer is $147.5, 30$
Let’s order number of shirts sold per month:
$130,140,145,150,160,170$
median is: $\frac{145 \ + \ 150}{2}=147.5$
Let’s list the number of shoes sold per month:
$20,25,25,35,35,40$
mean is: $\frac{20 \ + \ 25 \ + \ 25 \ + \ 35 \ + \ 35 \ + \ 40}{6}=\frac{180}{6}=30$

The correct answer is $50$
Let $x$ be the number of shoes need to be added in April. Then:
$\frac{70}{20 \ + \ x}=(\frac{5}{17}) \ (\frac{85}{25}) →\frac{70}{20 \ + \ x}=\frac{425}{425}=1→$
$70=20 \ + \ x→x=50$

The correct answer is $5$
$3125=5^5→5^{ \ x}=5^5→x=5$

The correct answer is $18$
Plug in the value of $x$ in the equation and solve for $y$.
$3 \ y=\frac{2 \ x^3}{5} \ + \ 4→$
$3 \ y =\frac{2(5)^3}{5} \ + \ 4→$
$3 \ y=\frac{2 \ (125)}{5} \ + \ 4→$
$3 \ y= 50 \ + \ 4=54$
$3 \ y = 54→y=18$

The correct answer is $\frac{15 \ \sqrt{3}}{2}$
Based on triangle similarity theorem:
$\frac{a}{a\ +\ b}=\frac{c}{4}→c=\frac{4\ a}{a\ +\ b}=\frac{4\sqrt3}{4\sqrt3}=1→$ 
area of shaded region is:
$(\frac{c\ +\ 4}{2})\ (b)=\frac{5}{2}\ ×\ 3\sqrt{3}=\frac{15 \ \sqrt{3}}{2}$

The correct answer is $\sqrt{6}\ - \ 1$
$x_{1,2} =\frac{ -\ b \ ±\ \sqrt{b^2\ -\ 4\ a\ c }}{2\ a}$ 
$a\ x^2\ +\ b\ x\ +\ c = 0$
$x^2 \ +\ 2\ x\ –\ 5 = 0 ⇒$ then:
$a = 1, b = 2$ and $c = \ – \ 5$ 
$x =\frac{ -\ 2 \ +\ \sqrt{2^2 \ -\ 4 .1 .-5}}{2 .1} = \sqrt{6} \ –\ 1$
$x = \frac{-\ 2 \ -\ \sqrt{2^2 \ -\ 4 .1 .-\ 5 }}{2 .1} = –\ 1 \ –\ \sqrt{6}$

The correct answer is $110 \ x \ + \ 16,000 \ ≤ \ 23,000$
Let $x$ be the number of shoes the team can purchase.
Therefore, the team can purchase $110 \ x$.
The team had $$23,000$ and spent $$16000$.
Now the team can spend on new shoes $$7000$ at most.
Now, write the inequality: $110 \ x \ + \ 16,000 \ ≤ \ 23,000$

The correct answer is $\frac{3}{x^2} \ + \ 5$
$f(g(x))=3 \ × \ (- \ \frac{1}{x})^2 \ + \ 5=\frac{3}{x^2} \ + \ 5$

The correct answer is $64$ m$^2$
Let $L$ be the length of the rectangular and $W$ be the with of the rectangular. Then,
$L=3 \ W \ + \ 4$
The perimeter of the rectangle is $40$ meters.
Therefore: $2 \ L \ + \ 2 \ W=40$
$L \ + \ W=20$
Replace the value of $L$ from the first equation into the second equation and solve for $W$:
$(3 \ W \ + \ 4) \ + \ W=20→4 \ W \ + \ 4=20→4 \ W=16→W=4$
The width of the rectangle is $4$ meters and its length is:
$L=3 \ W \ + \ 4=3 \ (4) \ + \ 4=16$
The area of the rectangle is:
length $×$ width $= 4 \ × \ 16 = 64$

The correct answer is $120$
Here is the list of all prime numbers between $20$ and $40$:
$23, 29, 31, 37$
The sum of all prime numbers between $20$ and $40$ is:
$23 \ + \ 29 \ + \ 31 \ + \ 37 = 120$

The correct answer is $i\ +\frac{ 3}{4}$
To simplify the fraction, multiply both numerator and denominator by $i$. 
$\frac{4\ -\ 3\ i}{-\ 4\ i}\ ×\ \frac{i}{i}=\frac{4\ i\ -\ 3\ i^2} { -\ 4\ i^2}$
$i^2\ -\ 1$, Then: 
$\frac{4\ i\ -\ 3 \ i^2}{-\ 4\ i^2 }=\frac{4\ i\ -\ 3(-\ 1)}{-\ 4(-\ 1)}=\frac{4\ i\ +\ 3}{4}=\frac{4\ i}{4}\ +\ \frac{3}{4}=i\ +\frac{ 3}{4}$

The correct answer is $0, -\ 2 , -\  3$
Frist factor the function:
$f(x)=x^3\ +\ 5\ x^2\ +\ 6\ x\ =x\ (x\ +\ 2)\ (x\ +\ 3)$
To find the zeros, $f(x)$ should be zero.
$f(x)=x \ (x\ +\ 2)\ (x\ +\ 3)=0$
Therefore, the zeros are:
$x=0$
$(x\ +\ 2)=0 ⇒ x= -\ 2$
$(x\ +\ 3)=0 ⇒ x= -\ 3$

The correct answer is $24$
Let $x$ be the length of AB, then:
$12=\frac{x×3}{2}→x=8$
The length of AC $=\sqrt{6^2\ +\ 8^2 }=\sqrt{100}=10$
The perimeter of ∆ABC$=6\ +\ 8\ +\ 10=24$

The correct answer is $12.5$ feet
Use formula of rectangle prism volume.
$V =$ (length) (width) (height) $⇒ 5000 = (20) \ (20)$ (height) $⇒$ height $= 5000 \ ÷ \ 400 = 12.5$

The correct answer is $25 \ x^8$
$y=(-\ 5 \ x^4)^2=(-\ 5^2) \  (x^4)^2= 25 \ x^8$