# Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 2

This is the second part (questions 11-20) of the solutions of the Grade 8 MTAP 2015 Elimination Questions. You can read the solutions for questions 1-10 here. If you found any errors in the solution, please comment in the box below.

11.) If $x = 4$ and $y = -3$, what is $x^2y + xy^2$?

Solution

$x^2y + xy^2 = (4^2) (-3) + (4) (-3)^2 = (16) (-3) + 4(9)$
$= (-48) + 36 = -12$

12.) Simplify $x(1 + y) - 2y(x - 2) + xy$

Solution

$x(1 + y) - 2y (x - 2) + xy$
$= x + xy - 2xy + 4y + xy$
$= x + 2xy - 2xy + 4y$
$= x + 4y$

Answer: $x + 4y$

13.) If $a$ and $b$ are positive constants, simplify $\dfrac{ab \sqrt{ab}}{\sqrt[3]{a^4} \sqrt[4]{b^3}}$.

Solution

Note that $\sqrt {ab} = (ab)^{1/2}$$\sqrt [3]{a^4} = a^{4/3}$ and $\sqrt [4] {b^3} = b^{3/4}$

Now $\dfrac { (ab) (ab)^{1/2}} { a^{4/3} b^{3/4}}$
$= \dfrac{ (ab) (a^{1/2}) (b^{1/2}) } {a^{4/3} b^{3/4}}$
$=\dfrac {a^{3/2} b^{3/2}} {a^{4/3} b^{3/4}}$
$= a^{(3/2 - 4/3)} b^{(3/2 - 3/4)}$
$= a^{(9/6 - 8/6)} b^{(6/4 -3/4)}$
$= a^{1/6}b^{3/4}$

$a^{2/12}b^{9/12} = \sqrt [12] {a^2b^9}$

Answer: $a^{1/6}b^{3/4}$ or $\sqrt [12] {a^2b^9}$

14.) What is the quotient when $6x^4 + x^3 + 4x^2 + x + 2$ is divided by $3x^2 - x + 1$?

Solution

Answer: $2x^2 + x + 1$ remainder $x + 1$.

15.) In Item 14, what is the remainder?

Answer: $x + 1$

16.) If $A + B = x - 2y$, what is $A^2 + 2AB + B^2 + 4xy$?

Solution

$(A + B)^2 = (x-2y)^2$
$A^2 + 2AB + B^2 = x^2 - 4xy + 4y^2$
$A^2 + 2AB + B^2 + 4xy = x^2 - 4xy + 4xy + 4y^2$
$A^2 + 2AB + B^2 + 4xy = x^2 + 4y^2$

Answer: $x^2 + 4y^2$

17.) If $x + y = 7$ and $xy = 5$, what is $x^3 + y^3$?

Solution

$x + y = 7$ and $xy = 5$
$(x + y)^3 = 7^3$
$x^3 + 3x^2y + 3xy^2 + y^3 = 343$
$x^3 + 3xy (x + y) + y^3 = 343$

Substituting the given values above,

$x^3 + 3 (5)(7) + y^3 = 343$
$x^3 + 105 + y^3 = 343$
$x^3 + y^3 = 238$.

18.) If the length, width, and height of an open-top rectangular box are $(x + 3)$ cm, $x$ cm, and $(x - 3)$ cm, what is its surface area?

Solution

The formula for finding the surface area $S$ of a rectangular prism with length $l$, width $w$ and height $h$ is $S = 12lh + 2lw + 2wh$. Since the box is open, we subtract lw, which is the top face. So, the surface area of the open box is $S = 2lh + lw + 2wh$.

Substituting, we have

$S = 2(x + 3) (x - 3) + x(x + 3) + 2x(x - 3)$
$= 2(x^2 - 9) + x^2 + 3x + 2x^2 - 6x$
$= 2x^2 - 18 + x^2 + 3x + 2x^2 - 6x$
$=15x^2 - 3x - 18$

Answer: $15x^2 - 3x - 18$

19.) A man walked $x$ km for 2.5 hrs, then jogged $(2x + 3)$ km for 3.5 hrs, and finally walked again $(5x - 3)$ km for 4 hrs. If his average speed for the entire exercise was 4 kph, what is x?

Solution

$\dfrac{2.5x + 3.5(2x + 3) + 4(5x -3)}{x + 2x + 5x} = 4$
$\dfrac{29.5x - 1.5}{8x} = 4$
$29.5x - 1.5 = 32x$
$2.5x = 1.5$
$x = 3/5 = 0.6$

20.) Simplify $(a - 3) (a + 3) (a^2 + 3a + 9) (a^2 - 3a + 9)$.

Solution

We can group the expressions as sum and difference of two cubes.

$[(a - 3) (a^2 + 3a + 9)] [(a + 3) (a^2 - 3a + 9)]$
$= (a^3 - 3^3) (a^3 + 3^3)$
$= (a^3 - 27) (a^3 + 27)$

Now, this is in the form of the difference of two squares $(x + y) (x - y) = x^2 - y^2$.

So, $(a^3 - 27) (a^3 + 27) = (a^3)^2 - 27^2 = a^6 - 729$

Answer: $a^6 - 729$

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