2010 - WAEC Mathematics Past Questions and Answers - page 2
11
Solve the inequality: 3(x + 1) \(\leq\) 5(x + 2) + 15
A
x \(\geq\) -14
B
x \(\leq\) - 14
C
x \(\geq\) -11
D
x \(\leq\) - 11
correct option: d
3(x + 1) \(\leq\) 5(x + 2) + 15
3x + 3 \(\leq\) 5x + 10 + 15
3x - 5x \(\geq\) 10 + 15 - 3
-2x \(\geq\) 22
x \(\leq\) \(\frac{-22}{2}\)
x \(\leq\) -11
12
An empty rectangular tank is 250cm long and 120cm wide. If 180 litres of water is poured into the tank. Calculate the height of the water
A
6.0cm
B
5.5cm
C
5.0cm
D
4.5cm
correct option: a
Volume of water in tank = l x b x h = 180 litres
but 1 litre = 1000cm2
250 x 120 x hw = 180 x 1000
hw = \(\frac{180 \times 1000}{250 \times 120}\)
= 6cm
Users' Answers & Commentsbut 1 litre = 1000cm2
250 x 120 x hw = 180 x 1000
hw = \(\frac{180 \times 1000}{250 \times 120}\)
= 6cm
13
Given that \(\frac{5^{n +3}}{25^{2n -2}}\) = 5o, find n
A
n = 1
B
n = 2
C
n = 3
D
n = 5
correct option: c
\(\frac{5^{n +3}}{25^{2n -2}}\) = 5o
\(\frac{5^{n + 3}}{5^{2(2n - 3)}}\) = 5o
n + 3 - 4n + 6 = 0
-3n + 9 = 0
-3n = -9
n = \(\frac{-9}{-3}\)
n = 3
Users' Answers & Comments\(\frac{5^{n + 3}}{5^{2(2n - 3)}}\) = 5o
n + 3 - 4n + 6 = 0
-3n + 9 = 0
-3n = -9
n = \(\frac{-9}{-3}\)
n = 3
14
\(\begin{array}{c|c} \text{No. of pets} & 0 & 1 & 2 & 3 & 4 \ \hline \text{No. of students} & 8 & 4 & 5 & 10 & 3\end{array}\) The table shows the number of pets kept by 30 students in a class. If a student is picked at random ftom the class. What is the probability that he/she kept more than one pet?
A
\(\frac{1}{5}\)
B
\(\frac{2}{5}\)
C
\(\frac{3}{5}\)
D
\(\frac{4}{5}\)
correct option: c
pr. (More than one pet)
= \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\)
= \(\frac{5 + 10 + 3}{30}\)
= \(\frac{18}{30}\)
= \(\frac{3}{5}\)
Users' Answers & Comments= \(\frac{\text{No. of students with > 1 pet}}{\text{total no. of students}}\)
= \(\frac{5 + 10 + 3}{30}\)
= \(\frac{18}{30}\)
= \(\frac{3}{5}\)
15
Simplify 2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)
A
1
B
\(\frac{1}{3}\sqrt{3}\)
C
2\(\sqrt{3} - 5\frac{2}{3}\)
D
6\(\sqrt{3}\) - 17
correct option: b
2\(\sqrt{3}\) - \(\frac{6}{\sqrt{3}} + \frac{3}{\sqrt{27}}\)
= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)
= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)
= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)
= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)
Users' Answers & Comments= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} + \frac{3}{\sqrt{9 \times 3}}\)
= 2\(\sqrt{3} - \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} + \frac{3}{3\sqrt{3}}\)
= 2\(\sqrt{3} = 6 \frac{\sqrt{3}}{3} + \frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\)
= 2\(\sqrt{3} - 2\sqrt{3} + \frac{\sqrt{3}}{3}\)
= \(\frac{\sqrt{3}}{3} = \frac{1}{3} \sqrt{3}\)
16
If sin 3y = cos 2y and 0o \(\leq\) 90o, find the value of y
A
18o
B
36o
C
54o
D
90o
correct option: a
sin 3y = cos 2y, but sin \(\theta\) = cos(90 - \(\theta\))
sin 3y = cos(90 - 3y)
cos(90 - 3y) = cos 2y
90 - 3y = 2y
5y = 90
y = \(\frac{90}{5}\)
y = 18o
Users' Answers & Commentssin 3y = cos(90 - 3y)
cos(90 - 3y) = cos 2y
90 - 3y = 2y
5y = 90
y = \(\frac{90}{5}\)
y = 18o
17
A rectangle has length xcm and width (x - 1)cm. If the perimeter is 16cm. Find the value of x
A
3\(\frac{1}{2}\)cm
B
4cm
C
4\(\frac{1}{2}\)cm
D
5cm
correct option: c
l = x; b = x - 1
perimeter = 2(l + b) = 16
l + b = \(\frac{16}{2}\) = 8
l + b = 8
x + x - 1 = 8
2x = 8 + 1
2x = 9
x = \(\frac{9}{2}\)cm
x = 4\(\frac{1}{2}\)
Users' Answers & Commentsperimeter = 2(l + b) = 16
l + b = \(\frac{16}{2}\) = 8
l + b = 8
x + x - 1 = 8
2x = 8 + 1
2x = 9
x = \(\frac{9}{2}\)cm
x = 4\(\frac{1}{2}\)
18
Given that tan x = 1, where 0o \(\geq\) x 90o, evaluate \(\frac{1 - \sin^2 x}{\cos x}\)
A
2\(\sqrt{2}\)
B
\(\sqrt{2}\)
C
\(\frac{\sqrt{2}}{2}\)
D
\(\frac{1}{2}\)
correct option: c
Given tan x = 1
x = tan-1(1)
x = 45o
Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
Users' Answers & Commentsx = tan-1(1)
x = 45o
Now, \(\frac{1 - ( \frac{1}{\sqrt{2} )^2}}{\frac{1}{\sqrt{2}}}\)
= \(\frac{1 - \frac{1}{2}}{\frac{1}{2}}\)
= \(\frac{1}{2} + \frac{1}{\sqrt{2}}\)
= \(\frac{1}{2} \times \frac{1}{\sqrt{2}}\)
= \(\frac{\sqrt{2}}{2}\)
19
What is the length of a rectangular garden whose perimeter is 32cm and area 39cm2?
A
25cm
B
18cm
C
13cm
D
9cm
correct option: c
perimeter = 2(l + b) = 32
l + b = \(\frac{32}{2}\)
l + b = 16
b = 16 - 1.......(1)
Area = l + b = 39
lb = 39 .....(2)
put (1) into (2)
l(16 - 1) = 39
16l - l2 = 39
l2 = 13l - 3l + 39 = 0
l(l - 13) - 3(1 - 13) = 0
(l - 3)(l + 130 = 0
l - 3 = 0 or l - 13 = 0
l = 3cm or l = 13cm; The length in 13cm
Users' Answers & Commentsl + b = \(\frac{32}{2}\)
l + b = 16
b = 16 - 1.......(1)
Area = l + b = 39
lb = 39 .....(2)
put (1) into (2)
l(16 - 1) = 39
16l - l2 = 39
l2 = 13l - 3l + 39 = 0
l(l - 13) - 3(1 - 13) = 0
(l - 3)(l + 130 = 0
l - 3 = 0 or l - 13 = 0
l = 3cm or l = 13cm; The length in 13cm
20
Simplify \(\frac{2}{2 + x} + \frac{2}{2 - x}\)
A
\(\frac{4}{4 - x^3}\)
B
\(\frac{8}{4 - x^2}\)
C
\(\frac{4x}{4 - x^2}\)
D
\(\frac{8 - 4x}{4 - x^2}\)
correct option: b
\(\frac{2}{2 + x} + \frac{2}{2 - x}\)
\(\frac{2(2 - x) + 2(2 + x)}{(2 + x)(2 - x)} = \frac{4 - 2x + 4 + 2x}{4 - 2x + 2x - x^2}\)
= \(\frac{8}{4 - x62}\)
Users' Answers & Comments\(\frac{2(2 - x) + 2(2 + x)}{(2 + x)(2 - x)} = \frac{4 - 2x + 4 + 2x}{4 - 2x + 2x - x^2}\)
= \(\frac{8}{4 - x62}\)