2003 - WAEC Mathematics Past Questions and Answers - page 3
(x + 20)o + (x+ 10)o + (2x - 45)o + (x - 25)o = 360o
5xo - 40o = 360o
x = 400/5 = 80o
180o - 5yo + 136o + 180o + 180o - 3yo = 360o
-8yo + 136o = 0
-8yo = -136; y = 17
Probability (Not doctors ad not lawyers) \(=\frac{36}{60}\
=\frac{6}{10}=\frac{3}{5}\)
\(2x^2 - 13x + 15 = 0\
2x^2 - 3x - 10x + 15\
x(2x-3)-5(2x-3) = 0\
(2x-3)(x-5)=0\
x = \frac{3}{2} or x = 5\)
In the diagram, \(P\hat{Q}S = 65^o, R\hat{P}S = 40^2\hspace{1mm}and\hspace{1mm}Q\hat{S}R=20^o\hspace{1mm}P\hat{S}Q\)
< QPS = < PRS = 65° (angles in the same segment)
< PSR + 40° + 65° = 180°
< PSR + 105° = 180°
< PSR = 75°
< PSR = < PSQ + < QSR
75° = < PSQ + 20° \(\implies\) < PSQ = 75° - 20° = 55°
Evaluate \((111_{two})^2 - (101_{two})^2\)
\((111_{2})^2 - (101_{2})^2\)
Difference of two squares
\((111 - 101)(111 + 101)\)
= \((10)(1100)\)
= \(11000_{2}\)
Given that x ≅ 0.0102 correct to 3 significant figures, which of the following cannot be the actual value of x?
3^{1-n}\times 3^{-2(-2n)} = 3^{-2}\
1-n-2(-2n)= -2\
1-n+4n=-2\
n=-1\)
Which of the following is/are not the interior angle(s) of a regular polygon? I.108° II. 116° III. 120°
Using the formula, \((n - 2) \times 180°\) to get the sum of the interior angles. Then we can have
\((n - 2) \times 180° = 108n\) ... (1)
\((n - 2) \times 180° = 116n\) ... (2)
\((n - 2) \times 180° = 120n\) ... (3)
Solving the above given equations, where n is not a positive integer then that angle is not the interior for a regular polygon.
(1): \(180n - 360 = 108n \implies 72n = 360\)
\(n = 5\) (regular pentagon)
(2): \(180n - 360 = 116n \implies 64n = 360\)
\(n = 5.625\)
(3): \(180n - 360 = 120n \implies 60n = 360\)
\(n = 6\) (regular hexagon)
Hence, 116° is not an angle of a regular polygon.
sin 60^o = \frac{|YZ|}{|XZ|}=\frac{6}{P}\
P sin 60^o = 6\
P = \frac{6}{sin60^o}\
=\frac{6}{\sqrt{\frac{3}{2}}}=4\sqrt{3}\)