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 the slope-intercept form:

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

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

The slope of this line is \( -\frac{7}{5} \). Now, using the point-slope form of a line equation:

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

Substitute the point (5, 7) and the slope \( -\frac{7}{5} \):

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

Expand and simplify:

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

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

Multiply through by 5 to eliminate fractions:

\( 5y = -7x + 70 \)

Which simplifies to: \( 7x + 5y = 70 \)

Therefore, the correct answer is option b) 7x + 5y = 70.

We are given the equation:

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

To make \(y\) the subject, we need to isolate \(y\) on one side of the equation.

First, subtract \(\frac{x}{a + 1}\) from both sides:

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

Now, multiply both sides of the equation by \(b\) to get rid of the denominator on the left-hand side:

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

Next, simplify the expression inside the parentheses:

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

Thus, the equation for \(y\) becomes:

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

hypotenuse

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

(\sin45 = \frac{1}{\sqrt{2}})

= (\frac{2}{2})

∴ (sin45 + sin30)

= (\frac{1}{\sqrt{2}} + \frac{1}{2})

= (\frac{\sqrt{2}}{2}) + (\frac{1}{2})

= (\frac{\sqrt{2} + 1}{2})

= (\frac{1 + \sqrt{2}}{2})

The problem states that y varies directly as the square root of x and y = 3 when x = 16. This means:

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

where k is a constant. We substitute the given values into the equation to find k:

\( 3 = k\sqrt{16} \)

\( 3 = 4k \)

\( k = \frac{3}{4} \)

Now, we are asked to find y when x = 64. Using the equation:

\( y = \frac{3}{4}\sqrt{64} \)

\( y = \frac{3}{4} \times 8 \)

\( y = 6 \)

Therefore, the correct answer is option b) 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}

S_r = (\frac{a}{1 - r})

(-\frac{1}{10}) = (\frac{1}{8} \times \frac{1}{1 - r})

(-\frac{1}{10}) = (\frac{1}{8(1 - r)})

(-\frac{1}{10}) = (\frac{1}{8 - 8r})

cross multiply...

-1(8 - 8r) = -10

-8 + 8r = -10

8r = -2

r = -1/4

(x₁, y₁) = (0,5)

(x₂, y₂) = (5, 0)

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

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

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

5(y - 5) = -5x

y - 5 = -x

x + y = 5

y = -x + 5