# Grade 8 MTAP 2015 Elimination Questions with Solutions – Part 5

This is the fifth part (questions 41-50) of the solutions of the Grade 8 MTAP 2015 Elimination Questions. You can read the solutions for questions 1-10 11-20, 21 – 30, and 31 – 40

41.) If $f (2x) = 2 - 3x$, what is $f(10)$?

Solution $2x = 10$, $x = 5$. $f(10) = 3 - 3(5) = -13$

42.) What is the equation of the line that is parallel to $2x + 5y + 6 = 0$ and passes through $(1, 1)$?

Solution

Parallel lines have the same slope, so we get the slope of the given line. That is, $2x + 5y = -6$ $5y = -2x - 6$ $y = \frac{-2}{5}x - \frac{6}{5}$

So, the slope of the given line is $-\frac{2}{5}$

By the slope-intercept form, we get the equation of the line parallel to it and passing through $(1,1)$. $y - 1 = - \frac{2}{5}(x - 1)$

Multiplying both sides by $5$, we obtain $5(y - 1) = - 2(x - 1)$

Simplifying, we have $5y - 5 = -2x + 2$ $2x + 5y - 7 = 0$

Answer: $2x + 5y - 7 = 0$

43.) What is the domain of the function $f(x) = \sqrt{x - 1} - 2$ ?

Solution

We know that we cannot have a negative square root, so $\sqrt{x - 1} \geq 0$. By squaring both sides and simplifying, this means that $x \geq 1$.Since we can substitute any value for $x$ except the mentioned restriction, the domain is therefore the set of real numbers greater than or equal to 1.

Answer: the set of real numbers greater than or equal to 1

44.) Find the range of the function in Item 43.

Solution

The minimum value for $\sqrt{x - 1} = 0$ and $0 - 2 = -2$, therefore, the range of $f$ is the set of real numbers greater than or equal to $-2$.

Answer: set of real numbers greater than or equal to -2.

45.) If $\triangle ABC \cong \triangle DEF$, $AB = x - 2$ cm, and $DE = 5y + 3$ cm, what is $x$ when $y = 10$?

Solution $DE = 5y + 3 = 5(10) + 3 = 53$

Since corresponding sides of congruent triangles are congruent, $AB = DE$. So, $2x - 2 = 53$ $2x = 55$ $x = \frac{55}{2}$.

Answer: $55/2$ or $27.5$.

46.) Let $ABC$ be an isosceles right triangle with $A$C as its hypotenuse, and let $D$ and $E$ be midpoints on $AB$ and $AC$, respectively, such that $DE||BC$. If $AB = BC = x$ and $DE = y$, what is the area of the trapezoid $DECB$ in terms of $x$ and $y$?

Solution

The area $A$ of a trapezoid is $A = \frac{1}{2}h(b_1 + b_2)$ where $b_1$ and $b_2$ are the bases, and $h$ is the height. From the given we can see that $b_1 = \frac{x}{2}$, $b_2 = x$ and $h = \frac{x}{2}$.

Substituting we have $A = \frac{1}{2}(\frac{x}{2})(x + y) = \frac{x}{2}(\frac{x}{2} + \frac{y}{2})$.

This simplifies to $A = \frac{x^2}{4} + \frac{xy}{4}$

Answer: $A = \frac{x^2}{4} + \frac{xy}{4}$

47.) If $|x| + x + y = 8$ and $x + |y| - y = 14$, what is $x + y$?

Solution

Assuming that $y > 0$. From the second equation $x = 14$. This is impossible because $14 + 14 + y \geq 8$. So, y is negative.

This means that the second equation becomes $x - 2y = 14$. Now, suppose $x$ is negative, then the first equation becomes $y = 8$ which is impossible because we have already shown that $y$ is negative. So, we are left with the systems of linear equations. $2x + y = 8$ $x - 2y = 14$

This gives us $x = 6$ and $y = -4$. Therefore, $x + y = 2$.

48.) If $90^a = 2$ and $90^b = 5$, what is $45^{\frac{1 - a - b}{2 - 2a}}$?

Solution: To be posted later.

49.) Let $ABCD$ be a square. Three parallel lines $l_1$, $l_2$, and $l_3$ pass through A, B, C, respectively. The distance between l1 and l2 is 4cm, and the distance between l2 and l3 is 5cm. Find the area of the square.

Solution

Draw line $l_4$ perpendicular to $l_2$ and passing through $B$. Let $P$ be the intersection of $l_1$ and $l_4$ and $Q$ be the intersection of $l_3$ and $l_4$. Then, $\triangle BAP \cong \triangle CBQ$ (Why?). If we let $s$ be the side of the square, $\dfrac{s}{\sqrt{s^2 - 4^2}} = \dfrac{s}{5}$

which means that $s^2 - 4^2 = 5^2$.

So, $s^2 = 4^2 + 5^2 = 41$

1. tracy