2024 - JAMB Mathematics Past Questions and Answers - page 6

We are tasked with solving the inequality:

\( \frac{x+2}{4} - \frac{2x-3}{3} < 4 \)

First, we multiply both sides of the inequality by 12 (the least common multiple of 4 and 3) to eliminate the fractions:

\( 12 \times \left( \frac{x+2}{4} - \frac{2x-3}{3} \right) < 12 \times 4 \)

Which simplifies to:

\( 3(x+2) - 4(2x-3) < 48 \)

Now, distribute and simplify:

\( 3x + 6 - 8x + 12 < 48 \)

\( -5x + 18 < 48 \)

Subtract 18 from both sides:

\( -5x < 30 \)

Divide both sides by -5 (and reverse the inequality sign):

\( x > -6 \)

w = \(\frac{v(2 + cn)}{1 - cn}\)

2v + cnv = w(1 - cn)

2v + cnv = w - cnw

2v - w = -cnv - cnw

Multiply through by negative sign:

-2v + w = cnv + cnw

-2v + w = n(cv + cw)

n = \(\frac{-2v + w}{cv + cw}\)

n = \(\frac{1}{c} \frac{-2v + w}{v + w}\)

Re-arrange:

n = \(\frac{1}{c} \frac{w - 2v}{v + w}\)

We are given a point \(B\) such that its distance from Y is always twice its distance from X. This means the ratio of the distances from B to X and from B to Y is 1:2.

Let d_X represent the distance from B to X, and d_Y represent the distance from B to Y. According to the given condition, we have:

\( d_Y = 2 \cdot d_X \)

This suggests that point B lies on a locus where the distance to Y is always twice the distance to X. This condition is satisfied by a specific geometric construction, which is the perpendicular bisector of the line segment XY. The perpendicular bisector divides the segment XY into two equal parts and places any point on it equidistant from X and Y, satisfying the given ratio of distances.

We are given the equation of a line: \(7x + 5y = 12\), and we need to find the equation of a line passing through the point (5, 7) and parallel to this line. Parallel lines have the same slope. To find the slope of the given line, we rewrite it in slope-intercept form:

\(5y = -7x + 12\)

\(y = -\frac{7}{5}x + \frac{12}{5}\)

The slope is -\(7/5\). Using the point-slope form:

\(y - y_1 = m(x - x_1)\)

Substitute the point (5,7) and slope -7/5:

\(y - 7 = -\frac{7}{5}(x - 5)\)

Expand and simplify:

\(y - 7 = -\frac{7}{5}x + 7\)

\(y = -\frac{7}{5}x + 14\)

Multiply through by 5:

5y = -7x + 70

Simplified: 7x + 5y = 70

We are given the equation:

\[ \frac{x}{a + 1} + \frac{y}{b} = 1 \]

Subtract \(\frac{x}{a + 1}\) from both sides:

\[ \frac{y}{b} = 1 - \frac{x}{a + 1} \]

Multiply both sides by b:

\[ y = b \left(1 - \frac{x}{a + 1}\right) \]

Simplify:

\[ y = \frac{b(a + 1 - x)}{a + 1} \]

\(\sin 45 = \frac{1}{\sqrt{2}}\)

\(\sin 30 = \frac{1}{2}\)

Therefore:

\(\sin 45 + \sin 30 = \frac{1}{\sqrt{2}} + \frac{1}{2} = \frac{\sqrt{2} + 1}{2} = \frac{1 + \sqrt{2}}{2}\)

The problem states that y varies directly as √x. Thus:

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

Given y = 3 when x = 16:

3 = k \sqrt{16} = 4k

k = \frac{3}{4}

Now, when x = 64:

y = \frac{3}{4} \sqrt{64} = \frac{3}{4} \times 8 = 6

μ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}

A = {3, 6, 9, 12, 15, 18}

B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

A ∩ B = {3, 9, 15}

For a geometric progression: S_∞ = a / (1 - r)

-1/10 = -1/8 / (1 - r)

Multiply both sides: -1/10 (1 - r) = -1/8

-(1 - r)/10 = -1/8

1 - r = 10/8 = 5/4

r = 1 - 5/4 = -1/4

(x₁, y₁) = (0,5)

(x₂, y₂) = (5,0)

Using the two-point formula:

\(\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}\)

\(\frac{y - 5}{0 - 5} = \frac{x - 0}{5 - 0}\)

\(\frac{y - 5}{-5} = \frac{x}{5}\)

Multiply both sides by 5:

y - 5 = -x

Thus: x + y = 5

or y = -x + 5