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JAMB UTME - Mathematics - 2002

Tambaya 1 Rahoto

Find the coordinates of the mid-point of x and y intercepts of the line 2y = 4x - 8

(2, 0)

(1, -2)

(-1, -2)

(1, 2)

Bayanin Amsa

To find the x-intercept of the line 2y = 4x - 8, we set y = 0 and solve for x. So, we have: 2(0) = 4x - 8 4x = 8 x = 2 This means the x-intercept is (2, 0). To find the y-intercept of the line, we set x = 0 and solve for y. So, we have: 2y = 4(0) - 8 2y = -8 y = -4 This means the y-intercept is (0, -4). The midpoint of two points (x1, y1) and (x2, y2) is: ((x1 + x2)/2, (y1 + y2)/2) Using this formula, we can find the midpoint of the x and y intercepts: ((2 + 0)/2, (0 + (-4))/2) = (1, -2) Therefore, the coordinates of the midpoint of the x and y intercepts of the line 2y = 4x - 8 is (1, -2). Answer option number 2.

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Coordinate Geometry

Graphs Of Linear And Quadratic Functions

Tambaya 10 Rahoto

Using the graph find the values of \(p\) and \(q\) if \(px + qy \ge 4\)

p = 2, q = -1

p = -1, q = 2

p = 2, q = 2

p = 1, q = 2

Bayanin Amsa

m =

\frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 - 0}{0 - (4)} = \frac{2}{4} = \frac{1}{2}

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} \geq m$

$\frac{y - 0}{x + 4} \geq \frac{1}{2}$

2y $\geq$ x + 4, -x + 2y $\geq$ 4 = px + qy $\geq$ 4

p = -1, q = 2

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Graphs Of Linear And Quadratic Functions

Tambaya 11 Rahoto

Make \(r\) subject of the formula given that \(\frac{x}{r+a}=\frac{a}{r}\)

\frac{a^{2}}{(x - a)}

\frac{a^{2}}{(x + a)}

\frac{a}{x - a}

\frac{a}{x + a}

Introductory Calculus

Change Of Subject Of A Formula/Relation

Tambaya 16 Rahoto

In the diagram, a cylinder is surmounted by a hemispherical bowl. Calculate the volume of the solid

198

π c m^{3}

180

π c m^{3}

162

π c m^{3}

216

π c m^{3}

Bayanin Amsa

V =

π r^{2} h + \frac{2}{3} π r^{3}

N.B h = r

V = $π x (3)^{2}$ x 20 = $\frac{2}{3} \times π \times (3)^{3}$

180 $π + 18 π = 198 c m^{3}$

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Euclidean Geometry

Tambaya 21 Rahoto

The triangle PQR above is

an obtuse-angled triangle

a scalene triangle

an isosceles triangle

an equilateral triangle

Bayanin Amsa

∠PQR = 180 - 128 (∠s on a straight line)
= 52

^{\circ}

∠QPR + 76 + 52 = 180(extr ∠ = sum of intr. opp ∠s)
∠QPR + 128 = 180
∠QPR = 180 - 128
= 52

^{\circ}

∴ΔPQR is an isosceles triangle

Karanta bayanin darasi akan Circles (WAEC)

Karanta bayanin darasi akan Triangles And Polygons (WAEC)

Triangles And Polygons

Tambaya 44 Rahoto

In the diagram are two concentric circles of radii r and R respectively with centre O. If \(r = \frac{2}{5}R\), express the area of the shaded portion in terms of \(\pi\) and R

\frac{5}{9} π R^{2}

\frac{21}{25} π R^{2}

\frac{21}{23} π R^{2}

\frac{9}{25} π R^{2}

Bayanin Amsa

A

shaded portion

=

π R^{2} - π r^{2}

r =

\frac{2}{5}

R

A $shaded portion$ = $π R^{2} - π (\frac{2}{3} R)^{2} = π R^{2} - \frac{4 π R^{2}}{25}$

= $π R^{2} (1 - \frac{2}{3}) = \frac{21}{25} π R^{2}$

Karanta bayanin darasi akan Circles (WAEC)

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