2016 - WAEC Mathematics Past Questions and Answers - page 2
11
Given that 2x + y = 7 and 3x - 2y = 3, by how much is 7x greater than 10?
A
1
B
3
C
7
D
17
correct option: c
2x + y = 7...(1)
3x - 2y = 3...(2)
From (1), y = 7 - 2x for y in (2)
3x - 2(7 - 2x) = 3
3x - 14 + 4x = 3
7x + 3 + 14 = 17
x = \(\frac{17}{7}\)
Hence, 7 x \(\frac{17}{7}\)
= 17 - 10
= 7
Users' Answers & Comments3x - 2y = 3...(2)
From (1), y = 7 - 2x for y in (2)
3x - 2(7 - 2x) = 3
3x - 14 + 4x = 3
7x + 3 + 14 = 17
x = \(\frac{17}{7}\)
Hence, 7 x \(\frac{17}{7}\)
= 17 - 10
= 7
12
Simplify; \(\frac{2}{1 - x} - \frac{1}{x}\)
A
\(\frac{x + 1}{x(1 - x)}\)
B
\(\frac{3x - 1}{ x(1 - x)}\)
C
\(\frac{3x + 1}{ x(1 - x)}\)
D
\(\frac{x + 1}{ x(1 - x)}\)
correct option: b
\(\frac{2}{1 - x} - \frac{1}{x}\) = \(\frac{2x - 1(1 - x)}{x(1 - x)}\)
= \(\frac{2x - 1(1 + x)}{x(1 - x)}\)
= \(\frac{3x - 1}{x(1 - x)}\)
Users' Answers & Comments= \(\frac{2x - 1(1 + x)}{x(1 - x)}\)
= \(\frac{3x - 1}{x(1 - x)}\)
13
Make s the subject of the relation: P = S + \(\frac{sm^2}{nr}\)
A
s = \(\frac{mrp}{nr + m^2}\)
B
s = \(\frac{nr + m^2}{mrp}\)
C
s = \(\frac{nrp}{mr + m^2}\)
D
s = \(\frac{nrp}{nr + m^2}\)
correct option: d
P = S + \(\frac{sm^2}{nr}\)
P = S(1 + \(\frac{m^2}{nr}\))
P = S(1 + \(\frac{nr + m^2}{nr}\))
nrp = S(nr + m2)
S = \(\frac{nrp}{nr + m^2}\)
Users' Answers & CommentsP = S(1 + \(\frac{m^2}{nr}\))
P = S(1 + \(\frac{nr + m^2}{nr}\))
nrp = S(nr + m2)
S = \(\frac{nrp}{nr + m^2}\)
14
Factorize; (2x + 3y)2 - (x - 4y)2
A
(3x - y)(x + 7y)
B
(3x + y)(2x - 7y)
C
(3x + y)(x - 7y)
D
(3x - y)(2x + 7y)
correct option: a
(2x + 3y)2 - (x - 4y)2
= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)
= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)
= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2
= 3x2 + 20xy - 7y2
= 3x2 + 21xy - xy - 7y2
= 3x(x + 7y) - y(x + 7y)
= (3x - y)(x + 7y)
Users' Answers & Comments= (2x + 3y)(2x + 3y) - (x - 4y)(x - 4y)
= 4x2 + 12xy + 9y2 - (x2 - 8xy+ 16y2)
= 4x2 + 12xy + 19y2 - x2 + 8xy - 16y2
= 3x2 + 20xy - 7y2
= 3x2 + 21xy - xy - 7y2
= 3x(x + 7y) - y(x + 7y)
= (3x - y)(x + 7y)
15
The curve surface area of a cylinder, 5cm high is 110cm 2. Find the radius of its base. [Take \(\pi = \frac{22}{7}\)]
A
2.6cm
B
3.5cm
C
3.6cm
D
7.0cm
correct option: b
Curved surface area of cylinder = 2\(\pi\)rh
110 = 2 x \(\frac{22}{7}\) x r x 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
Users' Answers & Comments110 = 2 x \(\frac{22}{7}\) x r x 5
r = \(\frac{110 \times 7}{44 \times 5}\)
= 3.5cm
16
The volume of a pyramid with height 15cm is 90cm3. If its base is a rectangle with dimension xcm by 6cm, find the value of x
A
3
B
5
C
6
D
8
correct option: a
Volume of pyramid = \(\frac{1}{3}\)lbh
90 = \(\frac{1}{3} \times x \times 6 \times 15\)
x = \(\frac{90 \times 33}{6 \times 15}\)
= 3
Users' Answers & Comments90 = \(\frac{1}{3} \times x \times 6 \times 15\)
x = \(\frac{90 \times 33}{6 \times 15}\)
= 3
17
A straight line passes through the points P(1,2) and Q(5,8). Calculate the gradient of the line PQ
A
\(\frac{3}{5}\)
B
\(\frac{2}{3}\)
C
\(\frac{3}{2}\)
D
\(\frac{5}{3}\)
correct option: c
Let: (x1, y1) = (1, 2)
(x2, y2) = (5, 8)
The gradient m of \(\bar{PQ}\) is given by
m = \(\frac{y_2 y_1}{x_2 - x_1}\)
= \(\frac{8 - 2}{5 - 1}\)
= \(\frac{6}{4}\)
= \(\frac{3}{2}\)
18
A straight line passes through the point P(1,2) and Q
(5,8). Calculate the length PQ
(5,8). Calculate the length PQ
A
\(4\sqrt{11}\)
B
\(4\sqrt{10}\)
C
\(2\sqrt{17}\)
D
\(2\sqrt{13}\)
correct option: d
|PQ| = \(\sqrt{(x_2 - X- 1) + (y_2 - y_1)^2}\)
= \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\)
= \(\sqrt{4^2 + 6^2}\)
= \(\sqrt{16 + 36}\)
= \(\sqrt{52}\)
= 2\(\sqrt{13}\)
Users' Answers & Comments= \(\sqrt{(5 - 1)^2 + (8 - 2)^2}\)
= \(\sqrt{4^2 + 6^2}\)
= \(\sqrt{16 + 36}\)
= \(\sqrt{52}\)
= 2\(\sqrt{13}\)
19
If cos \(\theta\) = x and sin 60o = x + 0.5 0o < \(\theta\) < 90o, find, correct to the nearest degree, the value of \(\theta\)
A
32o
B
40o
C
60o
D
69o
correct option: d
sin 60o = x + 0.5 0o(given)
0.8660 = x + 0.5
0.8660 - 0.5 = x
x = 0.3660
cos\(\theta\) = x(given)
cos\(\theta\) = 0.3660
Hence, \(\theta\) = cos-1(0.3660)
= 68.53o
= 69o (nearest degree)
Users' Answers & Comments0.8660 = x + 0.5
0.8660 - 0.5 = x
x = 0.3660
cos\(\theta\) = x(given)
cos\(\theta\) = 0.3660
Hence, \(\theta\) = cos-1(0.3660)
= 68.53o
= 69o (nearest degree)
20
\(\begin{array}{c|c}
Age(years) & 13 & 14 & 15 & 16 & 17 \
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
The table shows the ages of students in a club. How many students are in the club?
Age(years) & 13 & 14 & 15 & 16 & 17 \
\hline
Frequency & 10 & 24 & 8 & 5 & 3
\end{array}\)
The table shows the ages of students in a club. How many students are in the club?
A
50
B
55
C
60
D
65
correct option: a
Number of students in the club is
10 + 24 + 8 + 5 + 3 = 50
Users' Answers & Comments10 + 24 + 8 + 5 + 3 = 50