2007 - WAEC Mathematics Past Questions and Answers - page 5
41
Divide the sum of 8, 6, 7, 2, 0, 4, 7, 2, 3, by their mean
A
9
B
8
C
7
D
6
correct option: a
Sum = 8 + 6 + 7 + 2 + 0 + 4 + 7 + 2 + 3 = 39
mean = \(\frac{39}{9}\) = 4.33
then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\)
= 9
Users' Answers & Commentsmean = \(\frac{39}{9}\) = 4.33
then \(\frac{sum}{mean}\) = \(\frac{39}{4.33}\)
= 9
42
Q is 32 km away from P on a bearing 042o and R is 25km from P on a bearing of 132o. Calculate the bearing of R from Q.
A
122o
B
184o
C
190o
D
226o
correct option: b
Users' Answers & Comments43
The bearing of a point P from another point Q is 310o. If |PQ| = 200m, how far west of Q is P?
A
128.6m
B
153.2m
C
167.8m
D
187.9m
correct option: b
cos 40 = \(\frac{x}{200}\)
x - 200 x cosx
= 200 x 0.7660
x = 153.2m
Users' Answers & Commentsx - 200 x cosx
= 200 x 0.7660
x = 153.2m
44
Find the value of a if log10 a + log10a2 = 0.9030
A
4.0
B
2.0
C
1.6
D
0.0
correct option: c
Log a + log a2 = 0.9030
log (a x a2) = 0.9030
log a3 = 0.9030
a3 = 10^.9030(antilogarithm table)
a3 = 7.998
a = 3\(\sqrt{7.998}\)
a = 1.6
45
A messenger was paid N2.50 an hour during the normal working hours and 4.00 n hour during overtime. If he received N31.00 for 10 hours work, how many hours are for overtime?
A
2
B
3
C
4
D
6
correct option: c
Let normal working hour = x
let overtime = y
after 10 hours = N31.00
1 x 2.50x + 4y = 31
4 x x + y = 10
2.50x + 4y = 31
- 4x + 4y = 30
----------------
-1.5x = -9
x = 6; x + y = 10
6 + y = 10
y = 10 - 4
y = 4hrs for over time
Users' Answers & Commentslet overtime = y
after 10 hours = N31.00
1 x 2.50x + 4y = 31
4 x x + y = 10
2.50x + 4y = 31
- 4x + 4y = 30
----------------
-1.5x = -9
x = 6; x + y = 10
6 + y = 10
y = 10 - 4
y = 4hrs for over time
46
At what rate per cent per annum will N520.00 yield a simple interest of N39.00 in three years?
A
4%
B
3\(\frac{1}{2}\)%
C
3%
D
2\(\frac{1}{2}\)%
correct option: d
SI = \(\frac{PRT}{100}\)
39 = \(\frac{520 \times R \times 3}{100}\)
1560R = 3900
R = \(\frac{3900}{1560}\)
= 2.5%
R = 2\(\frac{1}{2}\)%
Users' Answers & Comments39 = \(\frac{520 \times R \times 3}{100}\)
1560R = 3900
R = \(\frac{3900}{1560}\)
= 2.5%
R = 2\(\frac{1}{2}\)%
47
PQR is an equilateral triangle with sides 2\(\sqrt{3cm}\). calculate its height
A
1.7cm
B
3.0cm
C
3.2cm
D
3.9cm
correct option: b
By Pythagoras theorem
h2 = (2\(\sqrt{3})^2 - (\sqrt{3})^2\)
= 22(\(\sqrt{3})^2 - 3\)
4(3) - 3 = 112 - 3
h2 = 9
h = \(\sqrt{9}\)
= 3cm
Users' Answers & Commentsh2 = (2\(\sqrt{3})^2 - (\sqrt{3})^2\)
= 22(\(\sqrt{3})^2 - 3\)
4(3) - 3 = 112 - 3
h2 = 9
h = \(\sqrt{9}\)
= 3cm
48
Which of the following is a factor of 2 - x - x2?
A
1 - x
B
1 + x
C
x - 1
D
2 - x
correct option: a
2 - x - x2; (2 - 2x) + (x - x2)
= 2(1 - x) + x(1 - x)
= (1 - x)(2 + x)
= 1 - x
Users' Answers & Comments= 2(1 - x) + x(1 - x)
= (1 - x)(2 + x)
= 1 - x
49
make w the subject of the relation \(\frac{a + bc}{wd + f}\) = g
A
\(\frac{a + bc - fg}{dg}\)
B
\(\frac{a - bc + fg}{dg}\)
C
\(\frac{a + bc - f}{dg}\)
D
\(\frac{a + bc - dg}{dg}\)
correct option: a
\(\frac{a + bc}{wd + f}\) = g(cross multiply)
a = bc + wdg + fg
wdg = a + bc - fg
w = \(\frac{a + bc - fg}{dg}\)
Users' Answers & Commentsa = bc + wdg + fg
wdg = a + bc - fg
w = \(\frac{a + bc - fg}{dg}\)
50
For what range of values of x is 4x - 3(2x - 1) > 1?
A
x > -1
B
x > 1
C
x < 1
D
x < -1
correct option: c
4x - 3(2x - 1) > 1
4x - 6x + 3 > 1
-2x > 1 - 3; 2x > -2
x < \(\frac{-2}{-2}\)
= x < 1
Users' Answers & Comments4x - 6x + 3 > 1
-2x > 1 - 3; 2x > -2
x < \(\frac{-2}{-2}\)
= x < 1