Mathematics 2014 Jamb Examination

First, we determine the total number of integers from 10 to 30 inclusive:\[ \text{Total} = (30 - 10) + 1 = 21 \]Next, we list the numbers between 10 and 30 that are divisible by 3:
12, 15, 18, 21, 24, 27, 30.
Counting these successes, we find there are 7 numbers.
The probability (P) is the number of successes divided by the total count:\[ P = \dfrac{7}{21} \]Simplifying the fraction by dividing both the numerator and the denominator by 7:\[ P = \dfrac{1}{3} \]

The equation of a straight line is generally expressed in the slope-intercept form:\[ y = mx + c \]where \(m\) is the gradient (slope) and \(c\) is the y-intercept. From the question, the line crosses the y-axis at (0, 5), which directly tells us that the y-intercept \(c = 5\).
Next, we calculate the gradient \(m\) using the two points provided, \((x_1, y_1) = (0, 5)\) and \((x_2, y_2) = (5, 0)\):\[ m = \dfrac{y_2 - y_1}{x_2 - x_1} \]\[ m = \dfrac{0 - 5}{5 - 0} = \dfrac{-5}{5} = -1 \]Substituting \(m = -1\) and \(c = 5\) into the slope-intercept equation:\[ y = -1(x) + 5 \]\[ y = -x + 5 \]Alternatively, we can use the intercept form of a linear equation \(\dfrac{x}{a} + \dfrac{y}{b} = 1\), where \(a\) and \(b\) are the x and y intercepts respectively:\[ \frac{x}{5} + \frac{y}{5} = 1 \]Multiplying through by 5 gives \(x + y = 5\), which simplifies to \(y = -x + 5\).

To find the value of x in the right-angled triangle, we use trigonometric ratios. In this triangle, we are given an angle of 60°, the length of the adjacent side (10 cm), and we need to find the length of the opposite side (x).
The trigonometric ratio that relates the opposite side and the adjacent side is the tangent (tan):\[ \tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}} \] Substituting the known values into the formula:\[ \tan 60^\circ = \frac{x}{10} \] From trigonometric tables or the unit circle, we know that \(\tan 60^\circ = \sqrt{3}\). Substituting this into the equation gives:\[ \sqrt{3} = \frac{x}{10} \] To solve for x, multiply both sides by 10:\[ x = 10\sqrt{3} \] Thus, the length of the side x is exactly \(10\sqrt{3}\) cm, which corresponds to option B.

Opposite angles are equal
q = 30
2(p + 2q) + 2q = 360
p + 2q + q = 180
p + 3q = 180
p + 3(30) = 180
p = 180 - 90
p = 90

To solve for ∠KMN, we use the properties of parallel lines and triangles.
Step 1: Identify alternate angles. Since KL is parallel to NM (KL//NM) and LN is a transversal, ∠KLN and ∠LNM are alternate angles. Therefore:\[ \angle LNM = \angle KLN = 54^\circ \] Step 2: Use the bisection property. We are told LN bisects ∠KNM, which means it divides the angle into two equal parts. Thus:\[ \angle KNL = \angle LNM = 54^\circ \] \[ \angle KNM = \angle KNL + \angle LNM = 54^\circ + 54^\circ = 108^\circ \] Step 3: Consider the triangle KNM. The sum of angles in a triangle is 180°. In ΔKNM, we have ∠KNM = 108° and ∠MKN = 35°. Let ∠KMN be x:\[ 108^\circ + 35^\circ + x = 180^\circ \] \[ 143^\circ + x = 180^\circ \] \[ x = 180^\circ - 143^\circ \] \[ x = 37^\circ \] Therefore, the size of ∠KMN is 37°.

By observing the shaded regions in the provided Venn diagram, we can identify which set operations describe them. The diagram shows two distinct shaded areas within circle P.
The first shaded area is the region where circle P and circle Q overlap. In set theory, the overlap or common elements between two sets is known as the intersection, denoted by the symbol ∩. Therefore, this region is represented as \(P \cap Q\).
The second shaded area is the region where circle P and circle R overlap. Similarly, this region is represented as the intersection of P and R, denoted as \(P \cap R\).
Since both of these regions are shaded together, we use the union operation (denoted by ∪) to combine them. The union represents the collection of all elements found in either of the specified regions. Combining these two intersection regions gives us the final expression:\[ (P \cap Q) \cup (P \cap R) \]This expression can also be simplified using the distributive law of set theory to \(P \cap (Q \cup R)\), which represents the part of P that overlaps with the combined area of Q and R. However, based on the options provided, the expanded form is the correct match.

To determine the amount spent on yam, we first need to find the angle of the sector subtended by yam in the pie chart. The total angle in a circle is \(360^\circ\). Based on the distribution provided in the chart, the angles for the other food items (excluding yam) sum up to \(200^\circ\) (specifically \(70^\circ + 80^\circ + 50^\circ = 200^\circ\)).
The angle for yam is calculated as:\[ \text{Angle of Yam} = 360^\circ - 200^\circ = 160^\circ \] Thus twice the what was spent on rice was spent on yam. We are given that the man spent ₦8,000 on rice. So for yam we have 2 x 8,000 = ₦16,000

The perimeter of a sector consists of the lengths of the two radii and the length of the arc. The formula is given by:\[ \text{Perimeter} = 2r + \text{Arc length} \]Where the arc length is calculated as:\[ \text{Arc length} = \frac{\theta}{360} \times 2\pi r \]Given \(r = 10.5\, \text{cm}\), \(\theta = 120^\circ\), and \(\pi = \dfrac{22}{7}\), we first calculate the arc length:\[ \text{Arc length} = \frac{120}{360} \times 2 \times \frac{22}{7} \times 10.5 \]\[ = \frac{1}{3} \times \frac{44}{7} \times 10.5 \]\[ = \frac{1}{3} \times 44 \times 1.5 \]\[ = 44 \times 0.5 = 22\, \text{cm} \]Now, add the lengths of the two radii to the arc length to find the total perimeter:\[ \text{Perimeter} = 2(10.5) + 22 \]\[ = 21 + 22 \]\[ = 43\, \text{cm} \]The sector boundary thus measures 43 cm in total length.