Full Length THEA Mathematics Practice Test

The correct answer is $20$
Add $8$ both sides of the equation $8\ x \ - \ 8=24$ gives $8 \ x =24 \ + \ 8=32$.
Dividing each side of the equation $8 \ x=32$ by $8$ gives $x=4$.
Substituting $4$ for $x$ in the expression $6\ x \ - \ 4$ gives $6\ (4)\ - \ 4=20$.

The correct answer is $( 4 , -\ 4 )$
Method 1: Plug in the values of $x$ and $y$ provided in the options into both equation
A.$(4,3)$ $x \ + \ y=0→4 \ + \ 3≠0$
B.$(5,4)$ $x \ + \ y=0→5 \ + 4 ≠0$
C.$(4,- \ 4)$ $x \ + \ y=0→4 \ + \ ( -\ 4)=0$
D.$(4,-\ 6)$ $x \ + \ y=0→4 \ + \ (-\ 6)=0$
Only option C is correct.
Method 2: Multiplying each side of $x \ + \ y=0$ by $2$ gives $2 \ x \ + \ 2 \ y=0$.
Then, adding the corresponding side of $2 \ x \ + \ 2 \ y=0$ and $4 \ x \ - \ 2 \ y=24$ gives $6 \ x=24$ .
Dividing each side of $6 \ x=24$ by $6$ gives $x=4$ .
Finally, substituting $4$ for $x$ in $x \ + \ =0$ , or $y= - \ 4$.
Therefore, the solution to the given system of equations is $( 4,-\ 4)$

The correct answer is  $28 \ x \ + \ 6$
If $f(x)=  3 \ x \ + 4\  (\ x \ +\ 1 ) \ +\ 2$ , then find $f(4 \ x)$  by substituting   $4 \ x$ for  every $x$ in the function.
This gives:
$f(4\ x) = 3 \ (4\ x \ ) \ + \ 4 \ (4 \ x \ + \ 1 \ ) \ + \ 2$,
It simplifies to: $f(4 \ x \ )=3 \ (4 \ x \ ) \ + \ 4  \ (4 \ x \ + \ 1 ) \ + \ 2=12 \ x \ + \ 16 \ x \ + \ 4 \ + \ 2=28 \ x \ + \ 6$

The correct answer is $(9,3)$
First, find the equation of the line.
All lines through the origin are of the form $y=m$
so the equation is $y=\frac {1}{3}\ x$ .
Of the given choices, only choice C $(9 \ , \ 3)$ , satisfies this equation:
$y=\frac{1}{3} \ x→\ 3 =\frac {1}{3} \ (9)=3$

The correct answer is $n^2 \ + \ 2 \ n \ + \ 10$
$(3 \ n ^2 \ + \ 2 \ n \ + \ 6 ) \ - \ (2 \ n^2 \ - \ 4  )$
Add like terms together:
$3 \ n^2 \ - \ 2 \ n^2=n^2$
$2 \ n$ doesn’t have like terms.
$6 \ - \ ( - \ 4)=10$
Combine these terms into one expression to find the answer:
$n^2 \ + \ 2 \ n \ + \ 10$

The correct answer is $23$ , $26$
You can find the possible values of $a$ and $b$ in $(a \ x \ + \ 4) \ (b \ x \ + \ 3)$ by using the given equation $a \ + \ b=7$
and finding another equation that relates the variables $a$ and $b$.
Since $(a \ x \ + \ 4)\ (b \ x \ + \ 3)=10 \ x^2 \ + \ c \ x \ + \ 12$,
expand the left side of the equation to obtain
$a \ b \ x^2 \ + \ 4 \ b \ x \ + \ 3 \ a \ x \ + \ 12=10 \ x^2 \ + \ c \ x \ + \ 12$
Since $a \ b$ is the coefficient of $x^2$ on the left side of the equation and $10$ is the coefficient of $x^2$ on the right side of the equation,
it must be true that $a \ b=10$
The coefficient of $x$ on the left side is $4 \ b \ + \ 3 \ a$ and the coefficient of $x$ in the right side is $c$.
Then: $4 \ b \ + \ 3 \ a=c$
$a \ + \ b=7$, then: $a=7 \ - \ b$
Now, plug in the value of a in the equation $a \ b=10$.
Then:
$a \ b=10→(7 \ - \ b) \ b=10→7 \ b \ - \ b^2=10$
Add $\ - \ 7 \ b \ + \ b^2$ both sides.
Then: $b^2 \ - \ 7 \ b \ + \ 10=0$
Solve for $b$ using the factoring method.
$b^2 \ - \ 7 \ b \ + \ 10=0→(b \ - \ 5)\ (b \ - \ 2)=0$
Thus, either $b=2$ and $a = 5$, or $b = 5$ and $a = 2$.
If $b = 2$ and $a = 5$,
then
$4 \ b \ + \ 3 \ a=c→4\ ( \ 2 \ ) \ + \ 3 \ ( \ 5\ )=c→c=23$

The correct answer is $\frac{( \ x \ - \ 5 \ ) \ ( \ x \ + \ 4 \ )}{( \ x \ - \ 5 \ ) \ + \ ( \ x \ + \ 4 \ )}$ 
To rewrite $\frac{1}{{\frac{1}{x\ -\ 5}}\ +\ {\frac{1}{x\ +\ 4}}}$,
first simplify ${\frac{1}{x\ - \ 5}}\ + \ {\frac{1}{x\ +\ 4}}$.
${\frac{1}{(x \ - \ 5)}+\frac{1}{( x \ + \ 4)}=\frac{1 \ (x \ + \ 4)}{(x \ - \ 5)(x \ + \ 4)}+\frac{1\ (x \ - \ 5 )}{(x \ + \ 4)\ (x \ - \ 5)}}=\frac {(x \ + \ 4) \ + \ (x \ - \ 5)}{(x \ + \ 4)\ (x \ - \ 5)}$
Then
$\frac{1}{{\frac{1}{x\ -\ 5}}\ +\ {\frac{1}{x\ +\ 4}}}$=$\frac {1}{\frac{( \ x \ + \ 4 \ ) \ + \ ( \ x \ - \ 5 \ ) }{( \ x \ + \ 4 \ )\ ( \ x \ - \ 5 \ )}}$=$\frac{( \ x \ - \ 5 \ ) \ ( \ x \ + \ 4 \ )}{( \ x \ - \ 5 \ ) \ + \ ( \ x \ + \ 4 \ )}$ . (Remember,$\frac{1}{\frac{1}{x }}=x$)
This result is equivalent to the expression in choice A.

The correct answer is $a \ > \ b$
Since $(0 \ , \  0)$ is a solution to the system of inequalities, substituting $0$ for $x$ and $0$ for $y$  in the given system must result in two true inequalities.
After this substitution, $y \ < \ a \ - \ x$ becomes $0 \ < \ a$ ,
and  $y \ > \ x \ + \ b$ becomes  $0 \  > \ b$.
Hence, $a$  is positive and $b$ is negative.
Therefore, $a \ > \ b$.

The correct answer is $6$ , $6$
First find the slope of the line using the slope formula.
$m=\frac {y_2 \ - \ y_1}{x_2 \ - \ x_1}$
Substituting in the known information.
$(\ x_1 \ , \ y_1 )=(2 \ ,4 )$
$( \ x_2 \ ,\ y_2 \ )=( \ 4 \ , \ 5 \ )$
$m=\frac {5 \ - \ 4}{4 \ - \ 2}=\frac{1}{2}$
Now the slope to find the equation of the line passing through these points.
$y=m \ x \ + \ b$
Choose one of the points and plug in the values of $x$ and $y$ in the equation to solve for $b$.
Let’s choose point $(\ 4 \ , \ 5)$. Then:
$\ y=m \ x \ + \ b→5=\frac{1}{2} \ (4) \ + \ b→5=2 \ + \ b→b=5 \ - \ 2=3$
The equation of the line is: $y=\frac{1}{2} \  x \ + \ 3$
Now, plug in the points provided in the choices into the equation of the line.
$(\ 9\ ,\ 9\ )$ $y=\frac {1}{2} \ + \ 3→9=\frac{1}{2}(\ 9 ) \ + \ 3→9=7.5$ This is NOT true.
$(\ 9\ ,\ 6\ )$ $y=\frac{1}{2} x \ + \ 3→6=\frac{1}{2}\ (\ 9) \ + \ 3→6=7.5$ This is NOT true.
$(\ 6\ , \ 9\ )$ $y=\frac{1}{2} x \ + \ 3→9=\frac{1}{2}\ ( 6) \ + \ 3→9=6$ This is NOT true.
$(\ 6\ , \ 6\ )$ $y=\frac{1}{2} x \ + \ 3→6=\frac{1}{2}( \ 6) \ + \ 3→6=6$ This is true!
Therefore, the only point from the choices that lies on the line is $(6 , 6)$ .

The correct answer is $10$
The input value is $5$. Then: $x = 5$
$f(x) = x^2\ - \ 3 \ x → f(5) = 5^2 \ - \ 3 \ (5) = 25 \ - \ 15 = 10$

The correct answer is  The $y-$intercept represents the starting height of $6$ inches
To solve this problem, first recall the equation of $a$ line: $y=m \ x \ + \ b$
Where $m=$ slope
$y=y-$intercept
Remember that slope is the rate of change that occurs in a function and that the $y-$intercept is the $y$ value corresponding to $x=0$.
Since the height of John’s plant is $6$ inches tall when he gets it. Time (or $x$) is zero.
The plant grows $4$ inches per year.
Therefore, the rate of change of the plant’s height is $4$.
The $y-$intercept represents the starting height of the plant which is $6$ inches.

The correct answer is $- \ 2$
Multiplying each side of $\frac{4}{x}=\frac{12}{x \ - \ 8}$ by $x \ (x \ - \ 8)$ gives $4 \ ( \ x \ - \ 8 \ )=12 \ ( \ x \ )$,
distributing the $4$ over the values within the parentheses yields $x \ - \ 8=3 \ x$ or $x=- \ 4$.
Therefore, the value of  $\frac{x}{2}$=$\frac{(- \ 4 )}{2}=- \ 2$.

The correct answer is $7$
Adding $6$ to each side of the inequality $4 \ n \ - \ 3 \ ≥ \ 1$ yields the inequality $4 \ n \ + \ 3 \ ≥ \ 7$.
Therefore, the least possible value of $4 \ n \ + \ 3$ is $7$.

The correct answer is $-\frac{8}{7}$
Since$f(x)$ is linear function with a negative slop, then when $x=-\ 2$ , $f(x)$ is maximum and when $x=3$ , $f(x)$ is minimum.
Then the ratio of the minimum value to the maximum value of the function is:$\frac{f(3)}{f(-2)}$=$\frac{- \ 3 \ ( \ 3 \ ) \ + \ 1 \ }{- \ 3 \ ( \ - \ 2) \ + \ 1}$=$\frac  {- \ 8}{7}$=$-\frac {8}{7}$

The correct answer is $2$
 Method 1: There can be $0$, $1$, or $2$ solutions to a quadratic equation.
In standard form, a quadratic equation is written as: $a \ x^2 \ + \ b \ x \ + \ c=0$
For the quadratic equation, the expression $b^2 \ - \ 4 \ a \ c$ is called discriminant.
If discriminant is positive, there are $2$ distinct solutions for the quadratic equation.
If discriminant is $0$, there is one solution for the quadratic equation and if it is negative the equation does not have any solutions.
To find number of solutions for $x^2=4 \ x \ - \ 3$, first, rewrite it $a \ s \ x^2 \ - \ 4 \ x \ + \ 3=0$.
Find the value of the discriminant. $b^2\ -\ 4 \ a \ c=( - \ 4)^2 \ - \ 4 \ (1) \ (3)=16 \ - \ 12=4$
Since the discriminant is positive, the quadratic equation has two distinct solutions.

The correct answer is  $\frac{11}{30}$
Of the $30$ employees, there are $5$ females under age $45$ and $6$ males age $45$ or older.
Therefore, the probability that the person selected will be either a female under age $45$ or a male age $45$ or older is:$\frac{5}{30} \ + \frac {6}{30}=\frac{11}{30}$

The correct answer is $86$
In the figure angle A is labeled $(3 \ x \ - \ 2)$ and it measures $37$.
Thus, $3 \ x \ - \ 2=37$ and $3 \ x=39$ or $x=13$.
That means that angle B, which is labeled $(5 \ x)$, must measure $5 \ × \ 13=65$.
Since the three angles of a triangle must add up to $180$ , $37 \ + \ 65 \ + \ y \ - \ 8=180$ , then:
$y \ + \ 94=108→y=180 \ - \ 94=86$

The correct answer is $22$
Substituting $6$ for $x$ and $14$ for $y$ in $y = n \ + \ 2$ gives $14=( \ n \ )\ ( \ 6 \ )+2$ ,
which gives $n=2$. Hence, $y=2 \ x \ + \ 2$.
Therefore, when$= 10$ , the value of $y$ is
$y=( \ 2 \ ) \ ( \ 10 \ ) \ + \ 2 = 22$.

The correct answer is $-\ 2$
Subtracting $2 \ x$ and adding $5$ to both sides of $2 \ x \  - \  5 \ ≥ \ 3 \ x \ - \ 1$ gives $- \ 4 \ ≥ \ x$.
Therefore, $x$ is a solution to $2 \ x \ - \ 5 \ ≥ \ 3 \ x \ - \ 1$ if and only if $x$ is less than or equal to $- \ 4$ and
$x$ is NOT a solution to $2 \ x \ - \ 5 \ ≥ \ 3 \ x  \ - 1$ if and only if $x$ is greater than $- \ 4$.
Of the choices given, only $- \ 2$ is greater than $- \ 4$ and,
therefore, cannot be a value of $x$.

The correct answer is $12$
Given the two equations,
substitute the numerical value of $a$ into the second equation to solve for $x\ . \ a=\sqrt{3}$,
$4\ a=\sqrt{4 \ x}$
Substituting the numerical value for ainto the equation with $x$ is as follows.
$4 \ (\sqrt{3})=\sqrt{4 \ x}$,
From here, distribute the $4$ . $4\sqrt{3}=\sqrt { \ 4 \ x}$
Now square both side of the equation.
$(4\sqrt{3})^2=(\sqrt{4 \ x})^2$
Remember to square both terms within the parentheses.
Also, recall that squaring a square root sign cancels them out.
$4^2 \sqrt{3}^2=4 \ x$ ,
$16 \ (3)=4 \ x$ ,
$48=4 \ x$ ,
$x=12$

The correct answer is $1 \ , \ 3$
First square both sides of the equation to get $4 \ m-3=m^2$
Subtracting both sides by $4 \ m \ - \ 3$ gives us the equation $\ m^2 \ - \ 4 \ m \ + \ 3=0$
Here you can solve the quadratic equation by factoring to get $(\ m \ - \ 1) \ ( \ m \ - \ 3  )=0$
For the expression $( \ m \ - \ 1) \ ( \ m \ - \ 3  )$ to equal zero, $m=1$ or $m=3$

The correct answer is $\frac{5}{4}$
To solve the equation for $y$ , multiply both sides of the equation by the reciprocal of  $\frac{6}{5}$ , which is  $\frac{5}{6}$ ,
 this gives $\frac{(5)}{(6)}×\frac {6}{5} y= \frac {3}{2} \ × \frac {(5)}{(6)}$, which simplifies to $y=\frac{15}{12}=\frac{5}{4}$.

The correct answer is $720$ m
Because Jack walks $30$ meters in $15$ seconds, and $6$ minutes is equal to $360$ seconds,
use the proportion to solve.
$\frac{30\  meters}{15\ s ec}=\frac{x \ meters}{360 \ sec}$
The proportion can be simplified to $\frac{30}{15}=\frac{x}{360}$ then each side of the equation can be multiplied by $360$ ,
giving $\frac{(360)(30)}{15}=x=720$.
Therefore, $720$ meters is the distance Jack will walk in $6$ minutes.

A zero of a function corresponds to an $x-$intercept of the graph of the function in the $x \ y-$plane.
Therefore, the graph of the function $g ()$, which has three distinct zeros, must have three $x-$intercepts.
Only the graph in choice C has three $x-$intercepts.

The correct answer is $3$
The fastest way to find the answer is to pick numbers.
Pick a number for that has a remainder of $2$ when divided by $8$, such as $10$. 
Increase the number you picked by $9$. In this case $10 \ + \ 9 \ =19$ .
Now divide $19$ by $8$, which gives you remainder $3$. Therefore, the answer is $3$.

The answer is  $c=0.35 \ ( \ 60 \ h)$
$$0.35$ per minute to use car.
This per-minute rate can be converted to the hourly rate using the conversion $1$ hour = $60$ minutes, as shown below.
$\frac{0.35}{minute} \ × \frac {60 minutes}{1 hours}=\frac{$ \ ( \ 0.35 \ × \ 60 \ )}{ hour}$
Thus, the car costs $$( \ 0.35 \ × \ 60 \ )$ per hour.
Therefore, the cost $c$ , in dollars, for h hours of use is $c=( \ 0.35 \ × \ 60) \ h$,
Which is equivalent to $c=0.35 \ (60 \ h)$

The correct answer is $100$
The best way to deal with changing averages is to use the sum.
Use the old average to figure out the total of the first $4$ scores:
Sum of first $4$ scores: $(4) \ (90) = 360$
Use the new average to figure out the total she needs after the $5^{th}$ score:
Sum of score: $(5) \ (92) = 460$ 
To get her sum from $360$ to $460$ ,
Mary needs to score $460 \ - \ 360=100$.

The correct answer is $2$
To solve a quadratic equation, put it in the ${a \ x}^2 \ + \ b \ x \ + \ c=0$ form, factor the left side,
and set each factor equal to $0$ separately to get the two solutions.
To solve ${x}^2=5 \ x \ - \ 4$ , first, rewrite it as ${x}^2 \ - \ 5 \ x \ + \ 4=0$.
Then factor the left side: ${x}^2 \ - \ 5 \ x \ + \ 4=0$ , $(x \ - \ 4) \ (x \ - \ 1)=0$
$x=1$ Or $x=4$ , There are two solutions for the equation.

The correct answer is $31,752$
$30\%$ of the books are Mathematics books and $15\%$ of the books are English books. Thus, number of Mathematics books: $0.3 \ × \ 840=252$
Number of English books: $0.15 \ × \ 840=126$
The product of number of Mathematics and number of English books: $252 \ × \ 126=31,752$

The correct answer is $108^\circ$, $54^\circ$
All central angles in a circle sum up to $360$ degrees.
Thus, the angle $α$  is: $0.3 \ × \ 360=108^\circ$

The correct answer is $120$
According to the graph, $50\%$ of the books are in the Mathematics and Chemistry sections.
Therefore, there are $420$ books in these two sections.
$0.50 \ × \ 840 = 420$
$γ \ + \ α=420$, and $γ=\frac{2}{5} \ α$
Replace$γ$ by $\frac{2}{5} \ α$ in the first equation.
$γ \ + \ α=420 → \frac{2}{5} α \ + \ α=420 → \frac{7}{5} α=420$ → multiply both sides by $\frac{5}{7}$
$\frac{(5)}{(7)} \frac{7}{5} \ α = 420 \ ×  \frac{(5)}{(7)} → α = \frac{420 \ × \ 5}{7}=300$
$α=300 → γ =  \frac{2}{5}\ α  → γ = \frac {2}{5} \ × \ 300=120$
There are $120$ books in the Chemistry section.

The correct answer is $6 \sqrt{2}$
The line passes through the origin, $(6 \ , \ m)$ and $(m \ , \ 12)$.
Any two of these points can be used to find the slope of the line.
Since the line passes through $(0 \ , \ 0)$ and $(6 \ , \ m)$,
the slope of the line is equal to$\frac{m \ - \ 0}{6 \ - \ 0}$=$\frac{m}{6}$.
Similarly, since the line passes through $(0 \ , \ 0)$ and $(m \ , \ 12)$ ,
 the slope of the line is equal to  $\frac{12 \ - \ 0}{m \ - \ 0}$=$\frac{12}{m}$.
since each expression gives the slope of the same line, it must be true that $\frac{m}{6}=\frac{12}{m}$
Using cross multiplication gives
$\frac{m}{6}=\frac{12}{m}→m^2=72→m=± \sqrt{72}=± \sqrt{36 \ × \ 2}=± \sqrt{36} \ × \sqrt{2}=± \ 6 \sqrt{2}$

The correct answer is $6$
It is given that $g(5)=4$. Therefore, to find the value of $f(g(5))$, then $f(g(5))=f(4)=6$

The correct answer is  $16\sqrt{3 }$ cm$^2$
Area of the triangle is: $\frac{1}{2}$ AD$×$BC and AD is perpendicular to BC.
Triangle ADC is a $30^° \ - \ 60^° \ - \ 90^°$ right triangle.
The relationship among all sides of right triangle $30^° \ - \ 60^° \ - \ 90^°$ is provided in the following triangle:
In this triangle, the opposite side of $30^°$ angle is half of the hypotenuse.
And the opposite side of $60^°$ is opposite of  $30^° \ × \sqrt{3}$
CD = $4$ , then AD = $4 \ ×\sqrt{3}$
Area of the triangle ABC is :$\frac {1}{2}$ AD×BC =$\frac {1}{2}  4\sqrt {3} \ × \ 8=16\sqrt{3}$

The correct answer is $32$
It is given that $g(5)=8$.
Therefore, to find the value of $f(g(5) )$, substitute $8$ for $g(5)$.
$f(g(5) )=f(8)=32$.

The correct answer is  $c=- \ 2\ ,\ d=6$
Substituting $5$ for yin $y=c \ x^2 \ + \ d$ gives $5=c \ x^2 \ + \ d$ which can be rewritten as $5 \ - \ d=c \ x^2$ .
Since $y = 5$ is one of the equations in the given system, any solution $x$ of $5 \ - \ d=c \ x^2$ corresponds to the solution $(x \ , \ 5)$ of the given system.
Since the square of a real number is always nonnegative, and a positive number has two square roots,
the equation $5 \ - \ d=c \ x^2$ will have two solutions for $x$ if and only if $(1) \ c \ > \ 0$ and $d \ < \ 5$ or $(2) \ c \ < \ 0$ and $d \ > \ 5$.
Of the values for cand dgiven in the choices, only $c=- \ 2$, $d=6$ satisfy one of these pairs of conditions.
Alternatively, if $c=- \ 2$ and $d=6$ , then the second equation would be
$y=- \ 2 \ x^2 \ + \ 6$
The equation above has two real answer.

The correct answer is  $y \ + \ 4 \ x=z$
$x$ and $z$ are colinear. $y$ and $5 \ x$ are colinear. Therefore,
$x \ + \ z=y \ + \ 5 \ x$ ,subtractxfrombothsides,then,  $z=y \ + \ 4 \ x$

The correct answer is $29$
Here we can substitute $8$ for $x$ in the equation.
Thus, $y \ - \ 3=2 \ (8 \ + \ 5)$ , $- \ 3=26$
Adding $3$ to both side of the equation:
$y=26 \ + \ 3$ , $y=29$

The correct answer is I and III only
Let’s review the options:
I. $|a| \ < \ 1→- \ 1 \ < \ a \ < \ 1$
Multiply all sides by $b$. Since, $b \ > \ 0→- \ b \ < \ b \ a \ < \ b$
II. Since , $- \ 1 \ < \ a \ < \ 1$ , and $a \ < \ 0→- \ a \ > \ a^2 \ > \ a$  (plug in $\frac{- \ 1}{2}$, and check!)
III. $- \ 1 \ < \ a \ < \ 1$ ,multiply ll sides by $2$ , then:
$- \ 2 \ < \ 2 \ a \ < \ 2$ , subtract $3$ from all sides,then:
$- \ 2 \ - \ 3 \ < \ 2 a \ - \ 3 \ < \ 2 \ - \ 3→- \ 5 \ < \ 2 \ a \ - \ 3 \ < \ - \ 1$
I and III are correct.

The correct answer is  $a \ > \ 1$
The equation can be rewritten as
$c \ - \ d=a \ c$ →(divide both sides by $c$ ) $1 \ - \frac {d}{c}=a$ ,
since $c \ < 0$ and $d \ > \ 0$ , the value of $- \frac{d}{c}$ is positive.
Therefore, $1$ plus a positive number is positive. $a$  must be greater than $1$.
$a \ > \ 1$

The correct answer is $8$
Squaring both sides of the equation gives $2 \ m \ + \ 48=m^2$
Subtracting both sides by $2 \ m \ + \ 48$ gives us the equation $m^2 \ - \ 2 \ m \ - \ 48=0$
Here you can solve the quadratic by factoring to get $(m \ - \ 8) \ (m \ + \ 6)=0$
For the expression $(m \ - \ 8) \ (m \ + \ 6)$ to equal zero, $m=8$  or  $m=- \ 6$
Since $m$ is a positive integer, $8$ is the answer.

The correct answer is $3$
Since we are dealing with an absolute value, $f(a)=20$ means that either $11 \ - \ a^2=20$  or
$11 \ - \ a^2=- \ 20$
Let’s start with the positive value $(20)$ and see what we get.
 If $11 \ - \ a^2=20$ , then $a^2=9$
Taking the square root, we get $a=3$ or $- \ 3$
On the other hand, if $11 \ - \ a^2=- \ 20$,
 then $a=\sqrt {- \ 31}$
Notice that the question states that $a$ is a positive integer, therefore the answer is $3$.

The correct answer is $20\%$ 
Use this formula: Percent of Change
$\frac{New Value \ - \ Old Value}{Old Value}×100\%$
$\frac{16000 \ - \ 20000}{20000} \ × \ 100\%=20\%$ and $\frac{12800 \ - \ 16000}{16000} \ × \ 100 \%=20\%$

The correct answer is $729$ cm$^3$
If the length of the box is $27$ , then the width of the box is one third of it, $9$ , and the height of the box is $3$ (one third of the width). The volume of the box is:
$V = lwh = (27) \ (9) \ (3) = 729$

The correct answer is $12$
Let $x$ represent the number of liters of the $30\%$ solution.
The amount of salt in the $30\%$ solution $(0.30\ x)$ plus the amount of salt in the $75\%$ solution $(0.75) \ × ( \ 5 )$ must be equal to the amount of salt in the $39\%$ mixture $(0.39 \ × \ (x \ + \ 5))$.
Write the equation and solve for $x$.
$0.30 \ x \ + \ 0.75 \ ( 3) = 0.39 \ (x \ + \ 3) →0.30 \ x \ + \ 2.25 = 0.39 \ x \ + \ 1.17 →  0.39 \ x \ - \ 0.30 \ x = 2.25 \ - \ 1.17→0.09 \ x = 1.08 → x = \frac{1.08}{0.09}=12$

The correct answer is $50$ melis
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: $a^2 \ + \ b^2=c^2$
$40^2 \ + \ 30^2$= $c^2$⇒ $1600 \ + \ 900$ = $c^2$ ⇒ $2500$ = $c^2$ ⇒ $c=50$

The correct answer is $360$
One of the four numbers is $x$ ;
let the other three numbers be $y$ , $z$ and $w$.
Since the sum of four numbers is $600$ , the equation $x \ + \ y \ + \ z \ + \ w = 600$ is true.
The statement that $x$ is $50\%$ more than the sum of the other three numbers can be represented as
$x = 1.5 \ (y \ + \ w)$ or $\frac{x}{1.5} =  y \ + \ z \ + \ w → \frac{2x}{3} = y \ + \ z \ + \ w$
Substituting the value $y \ + \ z \ + \ w$ in the equation $x \ + \ y \ + \ z \ + \ w=600$
gives $x \ + \frac {2 \ x}{3} = 600 →  \frac{5 \ x}{3} =  600  →  5 \ x = 1,800 →  x  =  \frac{1,800}{5} = 360$

The correct answer is $570$
This is a simple matter of substituting values for variables.
We are given that the $50$ cars were washed today, therefore we can substitute that for $a$.
Giving us the expression $\frac{40\ (50)\ -\ 500}{50} \ + \ b$
We are also given that the profit was $$600$, which we can substitute for $f(a)$.
Which gives us the equation $600= \frac{40\ (50)\ -\ 500}{50} \ + \ b$
Simplifying the fraction gives us the equation $600=30 \ + \ b$
And subtracting both sides of the equation by $30$ gives us $b=570$, which is the answer.

The correct answer is $7.32$
The weight of $12.2$ meters of this rope is: $12.2 \ × \ 600$ g = $7,320$ g
$1$ kg $= 1,000$ g , therefore , $7,320$ g $÷ \ 1000 = 7.32$ kg

The correct answer is $40$
The area of ∆BED is $16$, then: $\frac{4 × \ AB}{2}$=$16→4 \ × \ AB=32→AB=8$
The area of ∆BDF is $18$, then: $\frac{3×BC } {2}=18→3 \ × \ BC=36→BC=12$
The perimeter of the rectangle is = $2 \ × \ (8 \ + \ 12)=40$