First: Choose the correct answer (20 Questions):
What is the fourth proportional for the quantities $2$, $4$ and $5$?
Answer: B $10$What is the S.S. for the equation: $\sqrt{5x - 1} = 4$ in $R$?
Answer: A $\{5\}$If $a$ is the first proportional for the two numbers $8$ and $16$.
What is the value of $a$?
The ordered pair $(1, 3)$ belongs to the function $f$ where:
Answer: D $f(x) = -3x + 6$If $\dfrac{a}{b}$ $=$ $\dfrac{c}{d} = \dfrac{2}{3}$, then $\dfrac{a+c}{b+d} = \;...$
Answer: B $\dfrac{2}{3}$The first proportional for the quantities $2$, $3$ and $6$ is ......
Answer: A $1$If $(a + 3, b - 2) = (5, 5)$, what is the value of $a + b$?
Answer: C $9$If $\dfrac{a}{4} = \dfrac{b}{3}$, what is the value of $\dfrac{a+b}{a-b}$?
Answer: D $7$If $X = \{2\}$ and $Y = \{3\}$, then $X \times Y = ....$
Answer: D $\{(2, 3)\}$What is the third proportional for $x^4$ and $2x^2$?
Answer: D $4$If $n(X^2) = 9$ and $n(X \times Y) = 12$, then $n(Y^2) = ....$
Answer: D $16$If $X \times Y = \{(4, 5), (4, 7), (9, 7), (9, 5)\}$, then $X = ...$
Answer: B $\{4, 9\}$If $X = \{5, 7, 8\}$, $Y = \{9, 10\}$ and $R$ is a function from $X$ to $Y$ where: $R = \{(5, 9), (8, 10), (k, 9)\}$. What is the value of $k$?
Answer: B $7$What is the S.S. for the inequality: $5x \leq 25$ in $R$?
Answer: D $]-\infty, 5]$Which of the following Cartesian products does the point $(\frac{1}{5}, \sqrt{5})$ belong to?
Answer: D $R \times R$If $f: R \to R$ and $f(x) = 5$, what is the value of $f(10) - f(7)$?
Answer: A $0$If $5x - 2y = 0$, what is the value of $\dfrac{y}{x}$?
Answer: D $\frac{5}{2}$If $3a = 5b = 4c$, then $a : b : c$ equals ......
Answer: B $20 : 12 : 15$What is the S.S. for the inequality: $3 \geq x > -3$ in $R$?
Answer: B $]-3, 3]$If $X = \{5, 3\}$ and $Y = \{4, 8\}$, then $(4, 3) \in ....$
Answer: B $Y \times X$Second: Answer the following (13 Questions):
Find the value of $x$ that makes the following numbers proportional: $x$, $20$, $30$ and $60$.
Answer: $\dfrac{x}{20} = \dfrac{30}{60}$ $\implies$ $\dfrac{x}{20} = \dfrac{1}{2}$ $\implies$ $x = 10$Find the values of $x$ and $y$ to make the quantities $x$, $6$, $12$, $y$ are in continued proportion.
Answer: $\dfrac{x}{6} = \dfrac{6}{12} = \dfrac{12}{y}$ $\implies$ $\dfrac{x}{6} = \dfrac{1}{2}$ $\implies$ $x = 3$. Also, $\dfrac{1}{2} = \dfrac{12}{y}$ $\implies$ $y = 24$.If $x : y = 3 : 4$, find the value of $(6x + 4y) : (3x + 2y)$.
Answer: Let $x = 3k$ and $y = 4k$. Then $(6(3k) + 4(4k)) : (3(3k) + 2(4k))$ $= (18k + 16k) : (9k + 8k)$ $= 34k : 17k = 2 : 1$.If $X = \{1, 5, 6\}$, $Y = \{5, 7\}$ and $Z = \{2, 3, 5\}$, find:
(a) $(X \times Z)$
(b) $(X \cup Y) \times (X \cap Y)$
(c) $n(X^2)$
(d) $(X - Y) \times (Y - Z)$
(a) $(X \times Z)$
Answer: $X \times Z = \{(1,2), (1,3),$ $(1,5), (5,2), (5,3), (5,5), (6,2),$ $(6,3), (6,5)\}$(b) $(X \cup Y) \times (X \cap Y)$
Answer: $X \cup Y = \{1,5,6,7\}$, $X \cap Y = \{5\}$. So $(X \cup Y) \times (X \cap Y) =$ $\{(1,5), (5,5), (6,5), (7,5)\}$.c) $n(X^2)$
Answer: $n(X) = 3$, so $n(X^2) = 3 \times 3 = 9$.d) $(X - Y) \times (Y - Z)$
Answer: $X - Y = \{1,6\}$ and $Y - Z = \{7\}$.So $(X-Y) \times (Y-Z) =$ $\{(1,7), (6,7)\}$.
Find the solution set in $R$ for the equation: $6x + 3 = 2x + 11$.
Answer: $6x - 2x = 11 - 3$ $\implies$ $4x = 8$ $\implies$ $x = 2$.If $X = \{0, 1, 2, 3\}$, $Y = \{1, 2, 4, 5, 6\}$ and $R$ is a relation from $X$ to $Y$ where $R$ means: "$x + y = 6$" for each $x \in X$ and $y \in Y$.
a) Represent $R$ as a set of ordered pairs.
Answer: $R = \{(0,6), (1,5), (2,4)\}$.b) Write the relation in a table.
Answer:| x | 0 | 1 | 2 |
| y | 6 | 5 | 4 |
c) Represent the relation using the arrow diagram.
Answer: (Diagram description: Arrows from $0 \rightarrow 6$, $1 \rightarrow 5$, $2 \rightarrow 4$).d) Represent the relation using a Cartesian diagram.
Answer: (Points on a graph: $(0,6)$, $(1,5)$, $(2,4)$).e) State the domain and the range of the relation.
Answer: Domain $= \{0,1,2\}$, Range $= \{4,5,6\}$.Find the number which, if added to the numbers $3$, $5$, $8$ and $12$, they will be proportional.
Answer: Let the number be $x$. Then $(3+x):(5+x) = (8+x):(12+x)$. $\implies$ $(3+x)(12+x) = (5+x)(8+x)$ $\implies$ $36 + 15x + x^2 = 40 + 13x + x^2$ $\implies$ $15x - 13x = 40 - 36$ $\implies$ $2x = 4$ $\implies$ $x = 2$.If $b$ is a proportional mean between $a$ and $c$. Prove that: $\dfrac{4a^2 - 9b^2}{4b^2 - 9c^2}$ $= \dfrac{a}{c}$.
Answer: Since $b$ is a proportional mean between $a$ and $c$, then $b^2 = ac$.$\Rightarrow 4a^2 - 9b^2$ $= 4a^2 - 9ac$ $= a(4a - 9c)$.
Also $4b^2 - 9c^2$ $= 4ac - 9c^2$ $= c(4a - 9c)$.
$\therefore \dfrac{4a^2 - 9b^2}{4b^2 - 9c^2}$ $= \dfrac{a(4a - 9c)}{c(4a - 9c)}$ $= \dfrac{a}{c}$.
If $\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=\dfrac{4a+mb+5c}{25}$, find the value of $m$.
Answer: Let $\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k$.$\Rightarrow a=3k,\; b=4k,\; c=5k$.
$\dfrac{4a+mb+5c}{25}=k$.
$\Rightarrow \dfrac{4(3k)+m(4k)+5(5k)}{25}=k$.
$\Rightarrow \dfrac{12k+4mk+25k}{25}=k$.
$\Rightarrow 37k+4mk=25k$.
$\Rightarrow 4mk=-12k$.
$\Rightarrow m=-3$.
Using the opposite Venn diagram, find:
a) $(A \cap C) \times B = \{2\} \times \{7,5,1\}$.
b) $(C \cup B) \times A$.
c) $n(A \times B) = n(A) \times n(B)$.
Answers:a) $(A \cap C) \times B$ $= \{2\} \times \{7,5,1\}$ $= \{(2,7),(2,5),(2,1)\}$.
b) $(C \cup B) \times A$ $= \{7,5,1,2,3\} \times \{1,4,2\}$.
c) $n(A \times B) = n(A) \times n(B)$.
$A=\{1,4,2\},\; B=\{7,5,1\}$
$\Rightarrow n(A \times B)=3 \times 3=9$.
If $\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}$ and $x+y-2z=12$. Find the value of $z$.
Answer: Let $\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{5}=k$.$\Rightarrow x=2k,\; y=4k,\; z=5k$.
Substitute in $x+y-2z=12$.
$\Rightarrow 2k+4k-2(5k)=12$.
$\Rightarrow 6k-10k=12$.
$\Rightarrow -4k=12$.
$\Rightarrow k=-3$.
$\Rightarrow z=5k=5(-3)=-15$.
If the quantities $a, b, c, d$ are proportional prove that: $\dfrac{4b+7d}{4a+7c}=\dfrac{5b-8d}{5a-8c}$.
Answer: Since $a,b,c,d$ are proportional $\Rightarrow \dfrac{a}{b}=\dfrac{c}{d}$.$\Rightarrow ad=bc$.
Compare by cross multiplication:
$(4b+7d)(5a-8c)$ and $(5b-8d)(4a+7c)$.
LHS $=20ab-32bc+35ad-56cd$.
RHS $=20ab+35bc-32ad-56cd$.
Since $ad=bc$, substitute:
LHS $=20ab-32bc+35bc-56cd=20ab+3bc-56cd$.
RHS $=20ab+35bc-32bc-56cd=20ab+3bc-56cd$.
$\therefore$ LHS = RHS $\Rightarrow$ $\dfrac{4b+7d}{4a+7c}=\dfrac{5b-8d}{5a-8c}$.
If $X=\{1,3,4\}$ and $R$ a relation on $X$ where $xRy$ means $(x+y)$ is an odd number for each $x \in X,\; y \in X$.
a) Represent $R$ as a set of ordered pairs.
b) Represent the relation using the arrow diagram.
c) Represent the relation using a cartesian diagram.
d) State the domain and the range of the relation.
Since $x+y$ must be odd $\Rightarrow$ one number is odd and the other is even.
$X=\{1,3,4\}$ where $1,3$ are odd and $4$ is even.
a) $R=\{(1,4),(3,4),(4,1),(4,3)\}$.
b) Arrow diagram:
$1 \rightarrow 4$, $3 \rightarrow 4$, $4 \rightarrow 1$, $4 \rightarrow 3$.
c) Cartesian diagram points:
$(1,4),(3,4),(4,1),(4,3)$.
d) Domain $=\{1,3,4\}$
Range $=\{1,3,4\}$.
Find the value of $x$ if the numbers $x, 24, 144$ are proportional.
Answer: Since the numbers $x,24,144$ are proportional.$\Rightarrow \dfrac{x}{24}=\dfrac{24}{144}$.
$\Rightarrow 144x=24 \times 24$.
$\Rightarrow 144x=576$.
$\Rightarrow x=4$.
If $(x-4,9)=(5,x+y)$, find the value of $\sqrt{4x+2y}$.
Answer: Since $(x-4,9)=(5,x+y)$ then the corresponding components are equal.$\Rightarrow x-4=5$ and $9=x+y$.
$\Rightarrow x=9$.
Substitute in $9=x+y$.
$\Rightarrow 9=9+y \Rightarrow y=0$.
$\Rightarrow \sqrt{4x+2y}=\sqrt{4(9)+2(0)}=\sqrt{36}=6$.
If $a,3,9$ and $b$ are in continued proportion, find the value of $a$ and $b$.
Answer: Since $a,3,9,b$ are in continued proportion.$\Rightarrow a:3=3:9=9:b$.
From $a:3=3:9$
$\Rightarrow \dfrac{a}{3}=\dfrac{3}{9}$.
$\Rightarrow 9a=9 \Rightarrow a=1$.
From $3:9=9:b$
$\Rightarrow \dfrac{3}{9}=\dfrac{9}{b}$.
$\Rightarrow 3b=81 \Rightarrow b=27$.
Find the S.S. in $R$ for the following inequality and represent it on the number line: $2 \le 3x-1 \le 14$.
Answer: $2 \le 3x-1 \le 14$.Add $1$ to all parts:
$\Rightarrow 3 \le 3x \le 15$.
Divide by $3$:
$\Rightarrow 1 \le x \le 5$.
$\therefore$ S.S. $=\{x \in R : 1 \le x \le 5\}$.
Represent graphically the function: $f(x)=2x-5$.
Answer: Choose two points to draw the line.If $x=0 \Rightarrow f(0)=2(0)-5=-5$.
If $x=3 \Rightarrow f(3)=2(3)-5=1$.
The graph is the straight line passing through the points $(0,-5)$ and $(3,1)$.
The following table is the set of ordered pairs for the function $f$ where $f(x)=3x+4$. Find the value of $(a+b)$.
| x | 2 | a | 7 |
| f(x) | 10 | 16 | b |
$\Rightarrow f(a)=16$
$\Rightarrow 3a+4=16$
$\Rightarrow 3a=12$
$\Rightarrow a=4$
$\Rightarrow b=f(7)$
$\Rightarrow b=3(7)+4$
$\Rightarrow b=21+4=25$
$\therefore a+b=4+25=29$
If $f(x)=2x-b$, $f(3)=10$. Find the value of $3b^2-5b+7$.
Answer: Since $f(x)=2x-b$$\Rightarrow f(3)=2(3)-b$
$\Rightarrow 6-b=10$
$\Rightarrow -b=4$
$\Rightarrow b=-4$
$\therefore 3b^2-5b+7$
$=3(-4)^2-5(-4)+7$
$=3(16)+20+7$
$=48+20+7$
$=75$
If $X=\{1,3,5\}$ and $R$ is a function on $X$ where
$R=\{(a,3),(b,1),(1,5)\}$.
Find $(a+b)$.
$\Rightarrow$ the first elements of the ordered pairs must be $1,3,5$.
Given $R=\{(a,3),(b,1),(1,5)\}$
$\Rightarrow a=3$ and $b=5$
$\therefore a+b=3+5=8$
If $X=\{1,2,3,4,5\}$, $N$ is the set of natural numbers and $f$ is a function from $X$ to $N$ where $f(x)=x^2-1$.
a) Represent $R$ as a set of ordered pairs.
b) Find the range of the function.
$\Rightarrow f(1)=1^2-1=0$
$\Rightarrow f(2)=2^2-1=3$
$\Rightarrow f(3)=3^2-1=8$
$\Rightarrow f(4)=4^2-1=15$
$\Rightarrow f(5)=5^2-1=24$
$\therefore R=\{(1,0),(2,3),(3,8),(4,15),(5,24)\}$
$\therefore$ Range $=\{0,3,8,15,24\}$
In the opposite figure:
If $\angle A$ is an obtuse angle.
Find the possible values of $x$.
and $\angle A$ is an obtuse angle
$\Rightarrow 90^\circ < (5x+10) < 180^\circ$
$\Rightarrow 80 < 5x < 170$
$\Rightarrow 16 < x < 34$
How to study for the March Math Test (Prep 2):
1. Master the concept of Cartesian product of two sets and its properties: $n(X \times Y) = n(X) \times n(Y)$.
2. Understand the definition of a relation vs. a function. A function must assign exactly one output to each input.
3. Practice finding the domain and range of functions represented as sets of ordered pairs or by formulas.
4. Review angle types: acute ($0^\circ < \theta < 90^\circ$), right ($\theta=90^\circ$), obtuse ($90^\circ < \theta < 180^\circ$).
5. Practice solving inequalities involving angle measures, remembering that angles are positive.
6. Review all 20 multiple-choice questions and 23 essay questions in this revision guide.