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

Answer: – 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 AC 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.

Answer: 2

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

Solution: To be posted later.

Answer: 3

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.

Screen Shot 2016-02-06 at 11.38.11 AM

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

Answer: 41

50.) At least how many numbers should be selected from the set {1, 5, 9, 13, … , 125} to be assured that two of the numbers selected have a sum of 146?

Solution: To be posted later.

Answer: 20

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