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

Answer: -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}

This is already correct, but if you want your answer in radical form, the previous expression can be converted to

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

Screen Shot 2016-01-22 at 6.41.57 PM

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.

Answer: 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

Answer: 3/5 or 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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5 Responses to Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 2

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  2. Phoebe says:

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  3. Phoebe says:

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  4. ronnie aljibe says:

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