Grade 9 MTAP 2015 Elimination Questions with Solutions – Part 1

This is the first part (questions 1-10) of the solutions of the Grade 9 MTAP 2015 Elimination Questions. Although much care has been given to make the solution as accurate as possible, the solver is also human. Please comment below if you see any errors.

Also available: Questions 11-20

1.) Simplify $\sqrt{\frac{3}{2}} - \sqrt{\frac{2}{3}}$

Solution

$\sqrt {\dfrac{3}{2}} - \sqrt {\dfrac{2}{3}} = \dfrac {\sqrt {3}}{\sqrt {2}} - \dfrac {\sqrt {2}}{\sqrt {3}} = \dfrac {(\sqrt {3})^2-(\sqrt {2})^2}{\sqrt {2}\sqrt {3}} = \dfrac {1}{\sqrt {6}}$.

Rationalizing the denominator,

$\dfrac {1}{\sqrt {6}} \times \dfrac {\sqrt {6}}{\sqrt {6}} = \dfrac {\sqrt {6}}{6}$.

Answer: $\sqrt{6}/6$

2.) Evaluate $\dfrac{2^0 + 2{^-1}}{2^{-2} + 2^{-3}}$

Solution

$\dfrac {2^0 + 2^{-1}}{2^{-2} + 2^{-3}} = \dfrac {1 + \frac {1}{2}}{\frac {1}{4} + \frac {1}{8}} = \dfrac {\frac {3}{2}}{\frac {3}{8}} = \dfrac {3}{2} \times \dfrac {8}{3} = 4$

3.) Simplify $\sqrt{ \dfrac{1}{9} + \dfrac{1}{16}}$.

Solution

$\sqrt{\dfrac{1}{9} + \dfrac {1}{16}} = \sqrt {\dfrac{16 + 9}{(9)(16)}} = \sqrt{\dfrac{25}{(9)(16)}} =\dfrac{5}{(3)(4)} = \dfrac{5}{12}$.

4.) Simplify $\frac{x^-1 - x^-1}{x^{1/3} - y^{1/3}}$

Answer: I am not sure if I am on the right track because the simplified form is more complex than the given. I will post the solution once I verified the answer.

5.) If $a$, $b$ and $c$ are real numbers such that $b/a = 5$ and $b/c = 2$, what is the value of $\frac{a + b}{b + c}$?

Solution

$\dfrac{b}{a} = 5 \Leftrightarrow \dfrac{a}{b} = \dfrac{1}{5}$.
$\dfrac{b}{c} = 2 \Leftrightarrow \dfrac{c}{b} = \dfrac{1}{2}$
$\dfrac{a}{b} + \dfrac{c}{b} = \frac{a + c}{b}$
$\dfrac {1}{5} + \dfrac {1}{2} = \dfrac {7}{10}$

So, $a + c = 7$ and $b = 10$.

$\dfrac{b}{a} = \dfrac{10}{a} = 5$, so $a = 2$
$\dfrac{b}{c} = \dfrac{10}{c} = 2$, so $c = 5$

Therefore,

$\dfrac {a + b}{b + c}$ = $\dfrac {2 + 10}{10 + 5}$ = $\dfrac {12}{15}$ = $\dfrac {4}{5}$.

Answer: $\dfrac {4}{5}$

6.) If $x \neq 0$, $x/9 = y^2$ and $x/3 = 3y$, what is $x$?

Solution

$\frac {x}{9} = y^2$, so $x = 9y^2$
$\frac {x}{3} = 3y$, so $x = 9y$

Since $9y^2 = 9y$, $y=1$.

Therefore, $x = 9(1)^2 = 9$

Answer: $x = 9$

7.) If $f(x) = x^2 + 6x + 9$, what is $f(x - 3)$?

$f(x - 3) = (x - 3)^2 + 6(x - 3) + 9$
$f(x - 3) = x^2 - 6x + 9 + 6x - 18 + 9$
$= x^2$

Answer: $x^2$

8.) If $f(x) = x^2 + 1$ and $g(x) = x - 1$, for all real numbers $x$, for what real numbers a does $f(g(-a)) = g(f(-a))$?

Solution
$f(g(-a)) = f(-a - 1) = (-a - 1)^2 + 1 = (a^2 + 29 + 1) + 1 = a^2 + 2a + 2$
$g(f(-a)) = g((-a)^2 + 1) = a^2 + 1 - 1 = a^2$

So, $a^2 = a^2 + 2a + 2$
$2a + 2 = 0$
$a = -1$

9.) Solve for $x$ in $\sqrt{3 + \sqrt{x}} = 3$.

Solution

$\sqrt {3 + \sqrt {x}} = 3$
Squaring both sides of the equation,

$3 + \sqrt {x} = 9$
$\sqrt {x} = 6$.

Squaring both sides,
$x = 36$.

10.) Solve for $x$ in the equation $\frac{2x}{x + 2} + \frac{x + 2}{2x} = 2$.

Solution

Multiplying both sides of the equation by $2x (x + 2)$
$(2x)(2x) + (x + 2) (x + 2) = 2(2x)(x + 2)$
$4x^2 + x^2 + 4x + 4 = 4x(x + 2)$
$5x^2 + 4x + 4 = 4x^2 + 8x$
$x^2 - 4x + 4 = 0$
$(x- 2)(x-2) = 0$
$x = 2, x = 2$