2024 - JAMB Mathematics Past Questions and Answers - page 3

For two lines to be parallel, their slopes must be equal. The slope of the first line is 5 (from the equation p = 5x + 3), and the slope of the second line is w (from the equation p = wx + 5). Therefore, setting the slopes equal gives:

5 = w. Hence, the value of w is 5.

Let the total number of students be 46, those playing football 22, those playing volleyball 26, and 3 students play both games. Using the principle of inclusion and exclusion:

Total students playing either football or volleyball = 22 + 26 - 3 = 45.

Therefore, the number of students who play neither game is 46 - 45 = 1.

The intersection of the lines y = 2x + 4 and y = 7 - x occurs when:

2x + 4 = 7 - x

3x = 3

x = 1

Substitute x = 1 into y = 2x + 4 to find y:

y = 2(1) + 4 = 6.

Now, calculate the distance between the point (4, 3) and (1, 6) using the distance formula:

Distance = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\) = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = \(3\sqrt{2}\).

Given that y varies directly as \(w^2\), the equation becomes:

y = k * w²,

where k is the constant of variation. When y = 8 and w = 2, we can solve for k:

8 = k * 2²,

8 = k * 4,

k = 2.

Now, substitute w = 3 into the equation to find y:

y = 2 * 3² = 2 * 9 = 18.

Since x varies directly as \(\sqrt{y}\), the equation becomes:

x = k * \(\sqrt{y}\),

where k is the constant of variation. When x = 81 and y = 9, we can solve for k:

81 = k * \(\sqrt{9}\),

81 = k * 3,

k = 27.

Now, substitute y = 1\(\frac{7}{9}\) into the equation:

y = 1.777... → \(\sqrt{y}\) ≈ 1.33,

x = 27 * 1.33 ≈ 36.

The curved surface area of a cylinder is given by the formula:

CSA = 2πrh,

where r is the radius and h is the height. Given that CSA = 110 cm² and h = 5 cm, and using \(π = \frac{22}{7}\), we can solve for r:

110 = 2 * \(\frac{22}{7}\) * r * 5

110 = \(\frac{220}{7}\) * r

r = \(\frac{110 * 7}{220}\) = 3.5 cm.

The given expression is \(a^2 - b^2 - 4a + 4\). We can factor it as follows:

\(a^2 - b^2 - 4a + 4 = (a - 2)^2 - b^2\)

Now, apply the difference of squares formula: \((x^2 - y^2) = (x - y)(x + y)\):

\( (a - 2 - b)(a - 2 + b)\).

To find the total marks scored by the children, we multiply each score by the corresponding number of students and sum the results:

• 0 * 4 = 0
• 1 * 3 = 3
• 2 * 7 = 14
• 3 * 8 = 24
• 4 * 6 = 24
• 5 * 2 = 10

Total marks = 0 + 3 + 14 + 24 + 24 + 10 = 75.

To find the value of p in terms of q, substitute x = 2 into both equations and equate the values of y:

For y = px² + q, we get y = 4p + q.

For y = 2x² - 1, we get y = 8 - 1 = 7.

Equating the two expressions for y:

4p + q = 7.

Solving for p:

4p = 7 - q,

p = \(\frac{7 - q}{4}\).