2013 - WAEC Mathematics Past Questions and Answers - page 2
11
If x = 64 and y = 27, evaluate: \(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)
A
2\(\frac{1}{5}\)
B
1
C
\(\frac{5}{11}\)
D
\(\frac{11}{43}\)
correct option: c
\(\frac{x^{\frac{1}{2}} - y^{\frac{1}{3}}}{y - x^{\frac{2}{3}}}\)
substitute x = 64 and y = 27
\(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\)
= \(\frac{8 - 3}{27 - 16}\)
= \(\frac{5}{11}\)
Users' Answers & Commentssubstitute x = 64 and y = 27
\(\frac{64^{\frac{1}{2}} - 27^{\frac{1}{3}}}{27 - 64^{\frac{2}{3}}} = \frac{\sqrt{64} - 3\sqrt{27}}{27 - (3\sqrt{64})^2}\)
= \(\frac{8 - 3}{27 - 16}\)
= \(\frac{5}{11}\)
12
If \(\frac{1}{2}\)x + 2y = 3 and \(\frac{3}{2}\)x and \(\frac{3}{2}\)x - 2y = 1, find (x + y)
A
3
B
2
C
1
D
5
correct option: a
\(\frac{1}{2}\)x + 2y = 3......(i)(multiply by 2)
\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2)
x + 4y = 6......(iii)
3x - 4y = 2.....(iv) add (iii) and (iv)
4x = 8, x = \(\frac{8}{4}\) = 2
substitute x = 2 into equation (iii)
x + 4y = 6
2 + 4y = 6
4y = 6 - 2
4y = 4
y = \(\frac{4}{4}\)
= 1(x + y)
2 + 1 = 3
Users' Answers & Comments\(\frac{3}{2}\)x - 2y = 1......(ii)(multiply by 2)
x + 4y = 6......(iii)
3x - 4y = 2.....(iv) add (iii) and (iv)
4x = 8, x = \(\frac{8}{4}\) = 2
substitute x = 2 into equation (iii)
x + 4y = 6
2 + 4y = 6
4y = 6 - 2
4y = 4
y = \(\frac{4}{4}\)
= 1(x + y)
2 + 1 = 3
13
Given that p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\), make q the subject of the equation
A
q = p\(\sqrt{r}\)
B
q = p3r
C
q = pr3
D
q = pr\(\frac{1}{3}\)
correct option: d
p\(\frac{1}{3}\) = \(\frac{3\sqrt{q}}{r}\)(cross multiply)
3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply)
3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side
q = 3\(\sqrt{r}\) x p
q = r\(\frac{1}{3}\)p = pr\(\frac{1}{3}\)
Users' Answers & Comments3\(\sqrt{q}\) = r x 3\(\frac{\sqrt{q}}{r}\)(cross multiply)
3\(\sqrt{q}\) = r x 3\(\sqrt{p}\) cube root both side
q = 3\(\sqrt{r}\) x p
q = r\(\frac{1}{3}\)p = pr\(\frac{1}{3}\)
14
A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord
A
2\(\sqrt{21}\)cm
B
\(\sqrt{42}\)cm
C
2\(\sqrt{19}\)cm
D
\(\sqrt{21}\)cm
correct option: a
From \(\bigtriangleup\) OMQ find /MQ/ by Pythagoras OQ2 = OM2 + MQ2
52 = 22 + MQ2
25 = 4 + MQ2
25 - 4 = MQ2
21 - MQ2
MQ2 = 21
MQ2 = \(\sqrt{21}\)
Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm
Users' Answers & Comments52 = 22 + MQ2
25 = 4 + MQ2
25 - 4 = MQ2
21 - MQ2
MQ2 = 21
MQ2 = \(\sqrt{21}\)
Length of chord = 2 x \(\sqrt{21}\) = 2\(\sqrt{21}\)cm
15
A cube and cuboid have the same base area. The volume of the cube is 64cm3 while that of the cuboid is 80cm3. Find the height of the cuboid
A
1cm
B
3cm
C
5cm
D
6cm
correct option: a
Users' Answers & Comments16
If sin x = \(\frac{5}{13}\) and 0o \(\leq\) x \(\leq\) 90o, find the value of (cos x - tan x)
A
\(\frac{7}{13}\)
B
\(\frac{12}{13}\)
C
\(\frac{79}{156}\)
D
\(\frac{209}{156}\)
correct option: c
Sin x = \(\frac{5}{13}\)
0o \(\leq\) x \(\leq\) 90o, (cos x - tan x)
AC2 = AB2 + BC2
132 = 52 + BC2
169 - 25 + BC2
169 - 25 = BC2
144 = BC2
Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)
BC = \(\sqrt{144}\)
BC = 12
tan x = \(\frac{opp}{adj} = \frac{5}{12}\)
BC = 12
cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)
\(\frac{144 - 65}{156} = \frac{79}{156}\)
Users' Answers & Comments0o \(\leq\) x \(\leq\) 90o, (cos x - tan x)
AC2 = AB2 + BC2
132 = 52 + BC2
169 - 25 + BC2
169 - 25 = BC2
144 = BC2
Cos x = \(\frac{Adj}{Hyp}\) = \(\frac{12}{13}\)
BC = \(\sqrt{144}\)
BC = 12
tan x = \(\frac{opp}{adj} = \frac{5}{12}\)
BC = 12
cos x - tan x = \(\frac{12}{13} - \frac{5}{12}\)
\(\frac{144 - 65}{156} = \frac{79}{156}\)
17
An object is 6m away from the base of a mast. The angle of depression of the object from the top pf the mast is 50o, Find, correct to 2 decimal places, the height of the mast
A
8.60m
B
7.51m
C
7.15m
D
1.19m
correct option: c
Tan 50o = \(\frac{h}{6}\)
h = 6 tan 50
= 6 x 1.1917
= 7.1505
= 7.15
Users' Answers & Commentsh = 6 tan 50
= 6 x 1.1917
= 7.1505
= 7.15
18
The bearing of Y from X is 060o and the bearing of Z from Y = 060o. Find the bearing of X from Z
A
300o
B
240o
C
180o
D
120o
correct option: b
Users' Answers & Comments19
Which of the following is not a probability of Mary scoring 85% in a mathematics test?
A
0.15
B
0.57
C
0.94
D
1.01
correct option: d
The probability not scoring 85% is = 1 - pro(scoring 85%)
= 1 - \(\frac{85}{100}\) = 1 - 0.85
= 0.15
not scoring would be less than 1
Users' Answers & Comments= 1 - \(\frac{85}{100}\) = 1 - 0.85
= 0.15
not scoring would be less than 1
20
If 2 log x (3\(\frac{3}{8}\)) = 6, find the value of x
A
\(\frac{3}{2}\)
B
\(\frac{4}{3}\)
C
\(\frac{2}{3}\)
D
\(\frac{1}{2}\)
correct option: a
2 log x (3\(\frac{3}{8}\)) = 6(divide both sides by 2)
\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\)
\(\log_x \frac{27}{8} = 3\)
\(\frac{27}{8} = x^3\)
\(x^3 = \frac{27}{8}\)
x = \(\sqrt{\frac{27}{8}}\)
= (\(\frac{27}{8}\))\(\frac{1}{3}\)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\)
= \(\frac{3}{2}\)
Users' Answers & Comments\(\frac{2 log_x(3 \frac{3}{8})}{2} = \frac{6}{2}\)
\(\log_x \frac{27}{8} = 3\)
\(\frac{27}{8} = x^3\)
\(x^3 = \frac{27}{8}\)
x = \(\sqrt{\frac{27}{8}}\)
= (\(\frac{27}{8}\))\(\frac{1}{3}\)
= \(\frac{(3^3)^{\frac{1}{3}}}{(2^3)^{\frac{1}{3}}}\)
= \(\frac{3}{2}\)