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.

Answer: -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.

Answer: x = 0

24.) In Item 23, what is the maximum value of x?
Answer: x = 2 (Obvious!)

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}

Answer: 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.

Answer: -1 and 4

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

Answer:

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°

rectangle

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°

Answer: 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°.

Answer: 30°

30.) In a grouped data, if 100 – 200 forms one class, what is the class interval of the data?
Answer:

Solution

The class interval is the difference between the upper class limit and the lower class limit of a class. Here, the upper class limit is 200 and the lower class limit is 100. So, the class interval is 200 – 100 = 100.

Answer: 100

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