# Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 4

This is the fourth part (questions 31-40) of the solutions of the Grade 8 MTAP 2015 Elimination Questions. You can read the solutions for questions 1-10 11-20, and 21 – 30

Although these solutions were carefully checked, the solver is only human. Kindly comment in the box below if you see any errors.

31.) Factor completely: $a^2c^2 + b^2d^2 - a^2d^2 - b^2c^2$.

Solution

Rearranging the terms, we obtain
$a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2$.

Factoring by grouping, we have
$a^2(c^2 - d^2) - b^2(c^2 - d^2)$
$= (c^2 - d^2)(a^2 - b^2)$
$= (c - d)(c + d)(a - b)(a + b)$.

Answer: (a + b)(a – b)(c + d)(c – d)

32.) Simplify $\dfrac{x^3- 4x}{x^3-x^2-6x}$.

Solution

$\dfrac{x^3- 4x}{x^3-x^2-6x} = \dfrac{x(x + 2)(x - 2)}{x(x - 3)(x + 2)} = \dfrac{x-2}{x-3}$.

Answer: $\frac{x - 2}{x - 3}$

33.) Perform the indicated operations: $\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{x - 1}$

Solution

$\dfrac{1}{x} - \dfrac{1}{x+1} - \dfrac{1}{x - 1} = \dfrac{(x + 1)(x-1) - x(x-1)-x(x+1)}{x(x+1)(x-1)}$
$= \dfrac{x^2 - 1 - x^2 + x - x^2 -x }{x(x + 1)(x - 1)}$
$= \dfrac{x^2 - 1}{x(x+1)(x-1)} = \dfrac{(x + 1)(x - 1)}{x(x + 1)(x - 1)} = \dfrac{1}{x}$

34.) Perform the indicated operations.

$\left ( \dfrac{2x^2 - 2x - 4}{x^2 - 5x + 6} \right ) \left ( \dfrac{x^2- x-6}{x^2+ 4x+3} \right) \div \dfrac{4x + 8}{3x + 9}$.

Solution

We can factor the given completely into the following expressions and change division into multiplication by multiplying the first two expressions with the reciprocal of the third expression. The result of these operations is shown below.

$\left ( \dfrac{2(x-2)(x + 1)}{(x - 2)(x - 3)} \right ) \left ( \dfrac{(x - 3)(x + 2)}{(x + 3)(x + 1)} \right) \left (\dfrac{3(x + 3)}{(4(x + 2)} \right )$.

After cancelling similar terms, we are left with $2(\frac{3}{4}) = \frac{3}{2}$

35.) Simplify $1 - \dfrac{x}{1 - \frac{x}{1 - x}}$.

Solution

Simplifying the rational expressions, we have

$1 - \dfrac{x}{1 - \frac{x}{1 - x}} = 1 - \dfrac{x}{\frac{1-2x}{1 - x}} = 1 - \dfrac{1 - x^2}{1 - 2x}$.

This simplifies further to

$\dfrac{1 - 3x + x^2}{1 - 2x}$.

Answer: $\dfrac{1 - 3x + x^2}{1 - 2x}$

36.) Solve for x in the equation $\dfrac{x+1}{(x-2)} = \dfrac{x-1}{x-2}$

Solution

Getting the least common denominator of the left hand side and combining the terms, we have

$\dfrac{x(x - 2)+1}{(x-2)} = \dfrac{1}{x - 2} = \dfrac{x-1}{x-2}$
$\dfrac{x^2 - 2x + 1}{x - 2} = \dfrac{x-1}{x-2}$.

Now, we can only equation the numerator.

$x^2 - 2x + 1 = x - 1$ which is equivalent to $x^2 – 3x + 2 = 0$

Getting the solution, we have $(x - 1)(x - 2 ) = 0$

Therefore, $x = 1$ or $x = 2$. But x cannot be equal to 2, because the it will make the denominator of the original expressions undefined. So, the solution is $x = 1$.

37.) The points (0, 0), (2, 3), and (4, 0) form a triangle. What is its perimeter?

Solution

To find the perimeter of the triangle, we need to find the distance between these points and add them. Let’s name the points A, B, and C respectively.

Using the distance formula, we can subtract the corresponding coordinates, square them, and get the square root.

Distance between A and B is $\sqrt{2^2 + 3^2} = \sqrt{13}$
Distance between A and C is $\sqrt{4^2 + 0^2} = 4$
Distnace between B and C is $2^2 + (-3)^2 = \sqrt{13}$

So, the perimeter of ABC is $AB + AC + BC = \sqrt{13} + 4 + \sqrt{13} = 4 + 2\sqrt{13}$

Answer: $4 + 2 \sqrt{13}$.

38.) Find the equation of the perpendicular bisector of the segment joining the points (-3, 2) and (5, 2)?

Solution

Let A be the point with coordinates (-3,2) and B be the point with coordinates (5,2). Notice that AB have the same y-coordinates which mean that it is a horizontal segment. This means that the perpendicular bisector is a vertical line.

To get the perpendicular bisector, we get the midpoint M of AB and find the equation of the vertical line passing through M.

$M = ( \frac{-3 + 5}{2}, \frac{2 + 2}{2}) = (1, 2)$.

So, the equation of the perpendicular bisector of AB is $x = 1$

39.) A man agrees to invest part of his 1-million-peso inheritance at an annual interest rate of 5%, while the rest at 6% interest. If, at the end of the year, he needs a total interest of Php 56, 200, how much should he invest at 5%?

Solution

Let x = amount invested at 5% and 1000 000 – x be invested at 6%.

$0.05x + 0.06(1000000 - x) = 56200$
$0.05x + 60000 - 0.06x = 56200$
$-0.01x = 3800$
$x = 380000$

40.) If 2x + 5y = 10 and x = 3y + 1, what is 11x + 11y?

Solution

Substituting the expression on the right hand side of the second equation to x in the left hand side of the first equation, we have

$2(3y + 1) + 5y = 10$
$6y + 2 + 5y = 10$
$11y = 8$ (*)

Multiplying $2x + 5y = 10$ by 3, we get $6x + 15y = 30$ (#).

Transposing the second equation and multiplying it by 5, we have

$5x - 15y = 5$ (##)

Adding # and ##, we have

$11x = 35$ (**)

By * and **

We have 11x + 11y = 35 + 8 = 43\$.

We will discuss the solution of numbers 41-50 in the next post.

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### 3 Responses to Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 4

1. Tyeng says:

i dont know if im correct but, i think number 35 has an error…that’s in multiplying x with the quantity 1-x..

2. cecillepantonial says:

Thnx. This is a great help to our contestants.

3. Galen Buhain says:

Please double-check the number 33. Thanks!