Ninth Grade Math Exam Notes

Question 1

• Part A: Find the slope of the straight line passing through the points (3, 4) and (-3, 2).

  • The formula for the slope (m) between two points $(x<em>1, y</em>1)$ and $(x<em>2, y</em>2)$ is: $m = \frac{y<em>2 - y</em>1}{x<em>2 - x</em>1}$.
  • Substituting the given points: With $(x<em>1, y</em>1) = (-3, 2)$ and $(x<em>2, y</em>2) = (3, 4)$, the slope is $m = \frac{4 - (2)}{3 - (-3)} = \frac{6}{6} = 1$.

• Part B: In the given figure, M is the intersection point of the angle bisectors of the internal angles of triangle ABC. Given that angle B = 70° and angle A = 15°, find (with proof) the measure of angle (BMC).

  • Since M is the intersection of the angle bisectors, it lies at the incenter of the triangle.
  • Given angles: $\angle B = 70^\circ$ and $\angle A = 15^\circ$.
  • Finding angle C: The sum of angles in a triangle is 180°, so
    $\angle C = 180^\circ - (70^\circ + 15^\circ) = 180^\circ - 85^\circ = 95^\circ$.
  • Since M is the incenter, MB and MC bisect angles B and C respectively. Therefore,
    $\angle MBC = \frac{1}{2} \angle B = \frac{1}{2} \times 70^\circ = 35^\circ$
    $\angle MCB = \frac{1}{2} \angle C = \frac{1}{2} \times 95^\circ = 47.5^\circ$.
  • Finding angle BMC: Using the sum of angles in triangle BMC,
    $\angle BMC = 180^\circ - (35^\circ + 47.5^\circ) = 180^\circ - 82.5^\circ = 97.5^\circ$.

• Part C: What is the original price of a watch if it was sold for 120 dinars after a 20% discount?

  • Let the original price be $x$.
  • The selling price after a discount is given by: $Selling Price = Original Price \times (1 - Discount Percentage)$.
  • We have: $120 = x \times (1 - 0.20)$, which simplifies to $120 = 0.8x$.
  • Solving for $x$: $x = \frac{120}{0.8} = 150$ dinars. Therefore, the original price of the watch was 150 dinars.

Question 2

• Part A: If $S = {-2, 0, 2}$ and $T = {1, 5}$, and the mapping $f: S \rightarrow T$ is defined by $f(x) = x^2 + 1$, find the range of the mapping $f$ and determine whether it is surjective (onto), injective (one-to-one), or bijective (one-to-one and onto), providing reasons.

  • Calculating the range of $f$:

    • $f(-2) = (-2)^2 + 1 = 4 + 1 = 5$
    • $f(0) = (0)^2 + 1 = 0 + 1 = 1$
    • $f(2) = (2)^2 + 1 = 4 + 1 = 5$

  • The range of $f$ is therefore ${1, 5}$.

  • Determining the type of mapping:

    • Surjective (onto): A mapping is surjective if its range is equal to its codomain. In this case, the range of $f$ is ${1, 5}$, which is equal to the codomain $T = {1, 5}$. Therefore, $f$ is surjective.
    • Injective (one-to-one): A mapping is injective if every element of the range corresponds to a unique element of the domain. Here, $f(-2) = 5$ and $f(2) = 5$, so two different elements in the domain (-2 and 2) map to the same element in the range (5). Therefore, $f$ is not injective.

    *Bijective (one-to-one and onto): A mapping is bijective if it is both injective and surjective. Since $f$ is surjective but not injective, it is not bijective.

• Part B: Represent the function $y = (x - 3)^2 + 2$ graphically using the graphical representation of the quadratic function $y = x^2$.

  • The function $y = (x - 3)^2 + 2$ represents a transformation of the basic quadratic function $y = x^2$.
  • The term $(x - 3)$ indicates a horizontal shift of the graph by 3 units to the right.
  • The term $+2$ indicates a vertical shift of the graph by 2 units upwards.

• Part C: Find the percentage decrease if the final value is 200 and the initial value is 500.

  • The formula for percentage decrease is: $Percentage Decrease = \frac{Initial Value - Final Value}{Initial Value} \times 100\%$.
  • Substituting the given values: $Percentage Decrease = \frac{500 - 200}{500} \times 100\% = \frac{300}{500} \times 100\% = 0.6 \times 100\% = 60\%$.
  • Therefore, the percentage decrease is 60%.

Question 3

• Part A: For the given right circular cone (assume $\pi = 3.14$), find the surface area of the cone.

  • The formula for the surface area (SA) of a right circular cone is: $SA = \pi r (r + l)$, where $r$ is the radius and $l$ is the slant height.
  • Given that the radius $r = 10$ cm and the slant height $l = 20$ cm.
  • Substituting the values: $SA = 3.14 \times 10 \times (10 + 20) = 3.14 \times 10 \times 30 = 3.14 \times 300 = 942$ cm$\text{}^2$.

• Part B: Represent graphically the common solution region for the inequalities y > x + 1 and $y \leq 3 - x$.

  • To represent the inequalities graphically:
    • y > x + 1: Draw the line $y = x + 1$ as a dashed line (since it is a strict inequality).
    • $y \leq 3 - x$: Draw the line $y = 3 - x$ as a solid line (since it includes equality).

• Part C: In triangle ABC, M is the intersection point of the perpendicular bisectors of the sides of the triangle, AM = 5 cm, BW = 4 cm, and W is the midpoint of BC. Find, with proof:

  1. MB
  2. MW
  • Since M is the intersection point of the perpendicular bisectors, it is the circumcenter of triangle ABC.
    1. MB:
       • Since M is the circumcenter, the distance from M to each vertex is the same. Thus, MB = MA = 5 cm.
    2. MW:
       • Since W is the midpoint of BC, BW = WC = 4 cm. Also, MW is perpendicular to BC. Now, consider the right triangle M WB.
       • By the Pythagorean theorem: $MB^2 = MW^2 + WB^2$
       • Substituting the known values: $5^2 = MW^2 + 4^2$
       • $25 = MW^2 + 16$
       • $MW^2 = 25 - 16 = 9$
       • $MW = \sqrt{9} = 3$ cm

Question 4

• Part A: Triangle $\triangle ABC$ is a right-angled triangle at $\angle B$, where $\angle C = 30^\circ$, $AB = 9$ cm, and M is the intersection point of the medians of the triangle. Find, with proof, the following:

  1. AC

• Part B: From the given figure, write down the elements of each of the

Second: Objective Questions

• Answer Key:
  • Items (1-4): Mark (A) if the statement is correct, and (B) if the statement is incorrect.
  • Items (5-12): For each item, there are four options, only one of which is correct. Shade the symbol corresponding to the correct answer.