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

This is the third part (questions 21-30) of the solutions of the Grade 8 MTAP 2015 Elimination Questions. You can read the solutions for questions 1-10 and 2-10.

Although much care was given in solving these problems, the solver is only human. If you found any errors in the solution, please comment in the box below.

Note: If you have old MTAP questions, you can send them to mtapreviewers@gmail.com and I will solve them for you and post it here.

21.) When $P (x)$ is divided by $x^2 + 2$, the quotient and remainder are both $x$. What is $P(-1)$?

Solution
If we divide 5 by 3, then we get a quotient of 1 and a remainder of 2. If we generalize latex $a$ the dividend, $b$ as the divisor, $q$ as the quotient and $r$ as the remainder, we can form the equation

$a = bq + r$ (*).

In the given, we can see that $a = P(x)$, $b = x^2 + 2$, $q = x$ and $r = x$.

Substituting in *, $P(x) = (x^2 + 2) (x) + x$
$= x^3 + 2x + x$
$P(x) = x^3 + 3x$

Now $P(-1) = (-1)^3 + 3(-1) = -4$.

22.) If two more than twice p is four less than twice q, express q in terms of p.

Solution
$2 + 2p = 2q - 4$
$2q = 2p + 2 + 4$
$2q = 2p + 6$
$q = \frac {2p + 6}{2}$
$q = p + 3$

Answer: $q = p + 3$

23.) If x is nonnegative and 3x – 4 ≤ x, what is the least value of x?

Solution
3x – 4 ≤ x
3xx ≤ 4
2x ≤ 4
x ≤ 2

So the lowest non-negative value is x = 0.

24.) In Item 23, what is the maximum value of x?

25.) Solve for $x$ in the equation $x + 2( x+ 1) + 3 (x + 1) + \cdots + 10 (x + 1) = 110$.

Solution

$x + 2(x + 1) + 3(x + 1) + \cdots + 10(x + 1) = 110$
$x + (x + 1) (2 + 3 + \cdots + 10) = 110$
$x + (x + 1) (54) = 110$
$x + 54x + 54 = 110$
$55x + 54 = 110$
$55x = 110 - 54 = 56$
$x = \frac {56}{55}$

26.) Solve for $x$ in the equation $|2x - 3| = 5$.

There are two values for $x$:

$2x - 3 = 5$
$2x = 8$
$x = 4$

Also,

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

27.) If ∠B is the complement of ∠A, and the supplement of ∠A is 138°, what is ∠B – ∠A?

Solution
Complementary angles add up to 90° and supplementary angles add up to 180°. So,
∠A + 138 = 180
∠A = 42.

Now since ∠A and ∠B are complementary,
∠A + ∠B = 90
42 + ∠B = 90
∠B = 90 – 42
∠B = 48.

Now, ∠B – ∠A = 48 – 42 = 6

28.) If the diagonals of rectangle ABCD meet E and ∠AEB = 140°, what is ∠EBC?

Solution

∠A + ∠B + ∠E = 180°
∠A + ∠B + 140 = 180°
∠A + ∠B = 40°

Since ABE is an isosceles triangle, ∠A = ∠B. This means that ∠A = ∠B = 20°.
Clearly, ∠ABC = 90°. So,
∠ABE + ∠EBC = 90°
20° + ∠EBC = 90°
∠EBC = 90 – 20°
∠EBC = 70°

29.) If two exterior angles of a triangle measures 80° and 130°, what is its smallest interior angle?

Solution

The adjacent interior and exterior angles of a triangle are supplementary. Therefore, the largest exterior angle has the smallest adjacent interior angle.

The sum of the exterior angle of any polygon is 360°so,
$80 + 130 + x = 360$
$210 + x = 360$
$x = 360 - 210$
$x = 150$

Among the three angles, 150° is the largest and its adjacent interior angle is 30°.