2004 - WAEC Mathematics Past Questions and Answers - page 1
1
1. Evaluate \( 202^2_{three} - 112^2_{three}\)
A
21120
B
21121
C
21112
D
21011
correct option: a
\(202^2_{three}\)when converted to base ten \(=(202_3)^2\
202_3 = 2 \times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\
=20_{ten}; (202_3)^2 = (20)^2_{ten} = 400\
112^2_{three}\)when converted to base ten \(= (112_3)^2\
112_3 = 1 \times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\
(112_3)^2 = (14)^2_{ten} = 196_{ten}\
Evaluate \Longrightarrow 400-196 = 204\)
Reconvert to base three
\(\begin{matrix}
3 & 204\
3 & 69 &R0\
3 & 22 & R2\
3 & 7 & R1\
2 & 2 & R1\
& 0& R2 \uparrow\
\end{matrix} \
=21120_3\)
Users' Answers & Comments202_3 = 2 \times 3^2 + 0 \times 3^1 + 2\times 3^0 = 18 + 0 + 2\
=20_{ten}; (202_3)^2 = (20)^2_{ten} = 400\
112^2_{three}\)when converted to base ten \(= (112_3)^2\
112_3 = 1 \times 3^2 + 1 \times 3^1 + 2\times 3^0 = 9+3+2=14_{ten}\
(112_3)^2 = (14)^2_{ten} = 196_{ten}\
Evaluate \Longrightarrow 400-196 = 204\)
Reconvert to base three
\(\begin{matrix}
3 & 204\
3 & 69 &R0\
3 & 22 & R2\
3 & 7 & R1\
2 & 2 & R1\
& 0& R2 \uparrow\
\end{matrix} \
=21120_3\)
2
If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two
A
1110
B
10111
C
11101
D
111100
correct option: b
\(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13\
101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\
2 & 11 &R1\
2 & 5 & R1\
2 & 2 & R1\
2 & 1 & R0\
& 0& R1 \uparrow\
\end{matrix} \
=y=10111_2\)
Users' Answers & Comments101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\
2 & 11 &R1\
2 & 5 & R1\
2 & 2 & R1\
2 & 1 & R0\
& 0& R1 \uparrow\
\end{matrix} \
=y=10111_2\)
3
Given that sin (5x - 28)o = cos (3x - 50)o,0 < x < 90o, find the value of x
A
14o
B
21o
C
32o
D
39o
correct option: b
Sin (5x – 28)o = cos (3x - 50)o
Since by the trigonometry relation
Sin(5x – 28)o = cos[90 – (5x – 28)]o
Hence cos(3x – 50)o = cos[90 – (5x – 28)]o
3x – 50 = 90 - (5x-28)
3x – 50 = 90 – 5x + 28
3x + 5x = 90 + 28 + 50
8x = 168
\(x = \frac{168}{8}=21^{\circ}\)
Users' Answers & CommentsSince by the trigonometry relation
Sin(5x – 28)o = cos[90 – (5x – 28)]o
Hence cos(3x – 50)o = cos[90 – (5x – 28)]o
3x – 50 = 90 - (5x-28)
3x – 50 = 90 – 5x + 28
3x + 5x = 90 + 28 + 50
8x = 168
\(x = \frac{168}{8}=21^{\circ}\)
4
Solve for t in the equation \(\frac{3}{4}t+\frac{1}{3}(21-t)\) = 11,
A
\(\frac{9}{13}\)
B
\(\frac{9}{5}\)
C
5
D
\(9\frac{3}{5}\)
correct option: d
\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\
\frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\
9t + 84 – 4t = 132; 5t = 132 – 84\
5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)
Users' Answers & Comments\frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\
9t + 84 – 4t = 132; 5t = 132 – 84\
5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)
5
A school girl spends \(\frac{1}{4}\) of her pocket money on books and \(\frac{1}{3}\) on dress. What fraction remains?
A
\(\frac{5}{6}\)
B
\(\frac{7}{12}\)
C
\(\frac{5}{12}\)
D
\(\frac{1}{6}\)
correct option: c
Let the girls pocket money be rep. by x. The amount spent on books = \(\frac{1}{4}of\hspace{1mm}x = \frac{x}{4}\)
The amount spent on dress \(=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}\)
∴The fraction that remains = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)
Users' Answers & CommentsThe amount spent on dress \(=\frac{1}{3} of \hspace{1mm}x=\frac{x}{3}\)
∴The fraction that remains = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)
6
In the diagram, \(R\hat{P}Q = Q\hat{R}Y, \hspace{1mm} P\hat{Q}R = R\hat{Y}Q, \ \hspace{1mm}|QP| = 3cm \hspace{1mm}|QY| = 4cm \hspace{1mm}and \hspace{1mm}|RY | = 5cm. \hspace{1mm} Find \hspace{1mm}|QR|\)
A
2.0cm
B
2.5cm
C
6.4cm
D
10.0cm
correct option: c
\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (SSS Congruence)
\(\frac{QR}{4} = \frac{8}{5}\)
\(QR = \frac{4 \times 8}{5}\)
= 6.4 cm
7
Find the value of x in the diagram
A
10o
B
28o
C
36o
D
44o
correct option: b
The sum of angles at point = 360o
(x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o
10x + 80o = 360o
10x = 360o - 80o = 280o
\(\frac{280}{10}=28^{\circ}\)
Users' Answers & Comments(x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o
10x + 80o = 360o
10x = 360o - 80o = 280o
\(\frac{280}{10}=28^{\circ}\)
8
There are m boys and 12 girls in a class. What is the probability of selecting at random a girl from the class?
A
\(\frac{m}{12}\)
B
\(\frac{12}{m}\)
C
\(\frac{12}{m+12}\)
D
\(\frac{12}{m-12}\)
correct option: c
Prob. (a girl) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\
= \frac{12}{m+12}\)
Users' Answers & Comments= \frac{12}{m+12}\)
9
Simplify \(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}\)and correct your answer to the nearest whole number
A
33
B
8
C
7
D
0
correct option: d
\(7\frac{1}{2}-\left(2\frac{1}{2}+3\right)\div16\frac{1}{2}; \frac{15}{2} - \left(\frac{5}{2}+\frac{3}{1}\right)\div \frac{33}{2}\
\frac{15}{2} - \left(\frac{5+6}{2}\right)\div \frac{33}{2}; \frac{15}{2} - \frac{11}{2} \div \frac{33}{2}\
\frac{15-11}{2}\div \frac{33}{2}; \frac{4}{2} \times \frac{2}{33} = \frac{4}{33} = 0.1212\)
The nearest whole number is 0
10
The angle of elevation of the top of a tower from a point on the ground which is 36m away from the foot of the tower is 30o. Calculate the height of the tower.
A
62.35m
B
20.78m
C
18.00m
D
10.39m
correct option: b
Where H is the height of the tower H = ?
\(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\
H = 36 \times 0.5774 = 20.79\)
Users' Answers & Comments\(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\
H = 36 \times 0.5774 = 20.79\)