2015 - WAEC Mathematics Past Questions and Answers - page 2
Then 4 + x = x2
0 = x2 - x - 42
or x2 - x - 42 = 0
x2 - 7x + 6x - 42 = 0
x(x - 7) + 6(x - 7) = 0
= (x + 6)(x - 7) = 0
x = -6 or x = 7
Hence, x = 7
Using y = y, gives
x2 - x - 2 = 2x - 1
x2 - 3x - 2 + 1 = 0
therefore, x2 - 3x - 1 = 0
so, 38\(\frac{1}{2} = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 3\)
\(\frac{77}{2} = \frac{22}{7} \times r^2\)
r2 = \(\frac{77 \times 7}{2 \times 22}\)
r2 = \(\frac{49}{4}\)
Hence, r = \(\sqrt{\frac{49}{4}}\)
= 3\(\frac{1}{2}\)
= 2 x 7 x 11
= 154cm3
volume of cylindrical tank = \(\pi r^2h\)
154 = \(\frac{22}{7} \times r^2 \times 4\)
r2 = \(\frac{154 \times 7}{22 \times 4}\)
= \(\frac{49}{4}\)
r = \(\sqrt{\frac{49}{4}} = \frac{7}{2}\)
= 3\(\frac{1}{2}\)m
In the diagram, PTR is a tangent to the centre O. If angles TON = 108°, Calculate the size of angle PTN
In the diagram; 108° + x + x = 180° (sum of angle in a triangle)
108° + 2x = 180°
x = 180° - 108°
= 72°
x = \(\frac{72^o}{2}\)
= 36°
(Angle between tangent and a chord through the point of contact)
Hence, angle PTN = 90 + 36
= 126°
x = 180o - b (angles on a str. line)
x = 180o - 48o
= 132o
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and <TOS = 76\(^o\). Using the diagram, calculate the value of angle PTR.
In the diagram given, < PRT = 3\(^o\) (Change in same segment)
< TPR = 90\(^o\) (angle in a semicircle)
Hence, < PTR = 180\(^o\) - (90 + 33)\(^o\)
= 180\(^o\) - 123\(^o\)
= 57\(^o\)
In the diagram, O is the centre, \(\bar{RT}\) is a diameter, < PQT = 33\(^o\) and < TOS = 76\(^o\). Using the diagram, find the size of angle PRS.
In the diagram, < TRS = \(\frac{76^o}{2}\) = 38\(^o\)
Hence, angle PRS = 33\(^o\) + 38\(^o\)
= 71\(^o\)
x & 0 & 1\frac{1}{2} & 2 & 4\
\hline
y & 0 & 5\frac{1}{2} & &
\end{array}\)
The table given shows some values for a linear graph. Find the gradient of the line
(x2, y2) = (\(\frac{5}{4}, \frac{11}{2}\))
Using gradient, m = \(\frac{y_2 - y_2}{x_2 - x_1}\)
= \(\frac{\frac{11}{2} - 3}{\frac{5}{4} - 0}\)
= \(\frac{11 - 6}{2} + \frac{5}{4}\)
= \(\frac{5}{} + \frac{5}{4}\)
= \(\frac{5}{2} \times \frac{4}{5}\)
= 2