2000 - WAEC Mathematics Past Questions and Answers - page 3
The probabilities that Kodjo and Adoga pass an examination are \(\frac{3}{4}\) and \(\frac{3}{5}\) respectively. Find the probability of both boys failing the examination
P(Kodjo passing) = \(\frac{3}{4}\); P(Adoga passing) = \(\frac{3}{5}\)
P(Kodjo failing) = \(\frac{1}{4}\); P(Adoga failing) = \(\frac{2}{5}\)
P(both fail) = \(\frac{1}{4} \times \frac{2}{5}\)
= \(\frac{1}{10}\)
Which of the following statement is not true about a rectangle? I.Each diagonal cuts the rectangle into two congruent triangles. II. A rectangle has four lines of symmetry III. The diagonals intersect at right angles
In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40°. Calculate ∠QSR.
< Q = < R (OQ = OR = radii)
< QOR = 180° - 2(40°) = 100°
< QSR = < RPQ = \(\frac{1}{2}\) < QOR
= \(\frac{100}{2} = 50°\)
Each side of a regular convex polygon subtends an angle of 30° at its center. Calculate each interior angle
If the interior angles of hexagon are 107°, 2x°, 150°, 95°, (2x-15)° and 123°, find x.
Sum of interior angle in a hexagon = (6 - 2) x 180°
= 720°
\(\therefore\) 107° + 2x° + 150° + 95° + (2x - 15)° + 123° = 720°
460 + 4x = 720 \(\implies\) 4x = 720 - 460
4x = 260° \(\implies\) x = 65°
In the diagram, POS and ROT are straight lines, OPQR is a parallelogram. |OS| = |OT| and ∠OST = 50°. Calculate ∠OPQ.
< T = < S = 50° (OS = OT)
< SOT = 180° - 2(50°) = 80°
< ROP = 80° (vertically opposite angle)
\(\therefore\) < OPQ = 180° - 80° = 100° (adjacent angles)
Given that \(x = -\frac{1}{2}and \hspace{1mm} y = 4 \hspace{1mm} evaluate \hspace{1mm} 3x^2y+xy^2\)
\(x = -\frac{1}{2}, y = 4\
3x^2y + xy^2\
3\left[-\frac{1}{2}\right]^2 \times 4 \times + \left(\frac{-1}{2}\right)(4)^2\
3\times \frac{1}{4} \times 4 -\frac{1}{2} \times 16\
3-8 = -5\)
3^{3(1+x)}=3^2\
3(1+x)=2\
3+3x = 2\
3x = -1
x = \frac{-1}{3}\)
If x varies inversely as y and \(x = \frac{2}{3}\) when y = 9, find the value of y when \(x=\frac{3}{4}\)
\(x \propto \frac{1}{y}\)
\(x = \frac{k}{y}\)
\(\frac{2}{3} = \frac{k}{9}\)
\(3k = 18 \implies k = 6\)
\(x = \frac{6}{y}\)
When y = \(\frac{3}{4}\),
x = \(\frac{6}{\frac{3}{4}}\)
= \(\frac{6 \times 4}{3}\)
= 8
Given that (2x + 7) is a factor of \(2x^2 + 3x - 14\), find the other factor
\(2x^2 + 3x - 14\)
\(2x^2 + 7x - 4x - 14\)
\(x(2x + 7) - 2(2x + 7)\)
= \((x - 2)(2x + 7)\)
The other factor = (x - 2).