2012 - WAEC Mathematics Past Questions and Answers - page 5
41
The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.
A
6.0
B
3.0
C
2.4
D
1.8
correct option: c
\(\begin{array}{c|c} x & f & fx\ \hline 1 & 6 & 6 \
2 & 8 & 16\ 3 & 8 & 18\ 4 & 5 & 20\end{array}\)
mean x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{60}{25}\)
x = 2.4
Users' Answers & Comments2 & 8 & 16\ 3 & 8 & 18\ 4 & 5 & 20\end{array}\)
mean x = \(\frac{\sum fx}{\sum f}\)
= \(\frac{60}{25}\)
x = 2.4
42
The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?
A
2
B
4
C
6
D
8
correct option: a
median = (\(\frac{N + 1}{2}\))th = (\(\frac{25 + 1}{2}\))th
= \(\frac{26th}{2}\)
= 13th
the 13th is 2
Users' Answers & Comments= \(\frac{26th}{2}\)
= 13th
the 13th is 2
43
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?
A
11.1
B
10.5
C
10.3
D
9.9
correct option: a
Users' Answers & Comments44
In the diagram, MN//PO, < PMN = 112o, < PNO = 129o
A
51o
B
54o
C
56o
D
68o
correct option: b
Users' Answers & Comments45
The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS
A
144o
B
72o
C
40o
D
36o
correct option: a
Reflex < POS = 2x (angle at centre is twice that at circumference)
reflex < POS + < POS = 350o(angles at a point)
i.e. 2x + 8x = 360o
10x = 360o
x = \(\frac{360}{10}\)
= 36o
< PRS = \(\frac{1}{2}\)
< POS(< at centre twice that circumference)
= \(\frac{1}{2}\) x 8x = 4x
4 x 36o
< PRS = 144
Users' Answers & Commentsreflex < POS + < POS = 350o(angles at a point)
i.e. 2x + 8x = 360o
10x = 360o
x = \(\frac{360}{10}\)
= 36o
< PRS = \(\frac{1}{2}\)
< POS(< at centre twice that circumference)
= \(\frac{1}{2}\) x 8x = 4x
4 x 36o
< PRS = 144
46
The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.
A
2.32cm
B
1.84cm
C
0.62cm
D
0.26cm
correct option: c
Users' Answers & Comments47
In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ
A
65o
B
45o
C
25o
D
19o
correct option: d
< RSQ = < RPQ = 48o (angle in the same segment)
< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
< SQR = 480
< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))
i.e. < QRS + 48o + 48o = 180
< QRS = 180 - (48 + 48) = 180 - 96 = 84o
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19o
Users' Answers & Comments< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))
< SQR = 480
< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))
i.e. < QRS + 48o + 48o = 180
< QRS = 180 - (48 + 48) = 180 - 96 = 84o
but < PRQ + < PRS = < QRS
< PRQ = < QRS - < PRS - 84 - 65
= 19o
48
The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?
A
-\(\frac{1}{2} \leq x\) < 3
B
-\(\frac{1}{2} < x \leq 3\)
C
-\(\frac{1}{2} < x < 3\)
D
-\(\frac{1}{2} \leq x \leq 3\)
correct option: c
2x2 - 5x - 3 = 0
2x2 - 6x + x - 3 = 0
2x(x - 3) + 1(x - 3) = 0
(2x + 1)(x - 3) = 0
2x + 1 = 0
2x = -1
x = -\(\frac{1}{2}\)
x - 3 = 0
-\(\frac{1}{2}\) < x < 3
Users' Answers & Comments2x2 - 6x + x - 3 = 0
2x(x - 3) + 1(x - 3) = 0
(2x + 1)(x - 3) = 0
2x + 1 = 0
2x = -1
x = -\(\frac{1}{2}\)
x - 3 = 0
-\(\frac{1}{2}\) < x < 3
49
The diagram is a polygon. Find the largest of its interior angles
A
300o
B
100o
C
120o
D
150o
correct option: d
Sum of interior angles = 3602
i.e 3x + 2x + 2x + x + 4x = 360o
12x = 360
x = \(\frac{360}{12}\)
x = 30o
x + c = 180
30o + c = 180
c = 180 - 30
c = 150o
Users' Answers & Commentsi.e 3x + 2x + 2x + x + 4x = 360o
12x = 360
x = \(\frac{360}{12}\)
x = 30o
x + c = 180
30o + c = 180
c = 180 - 30
c = 150o