Year : 
2004
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

1 - 10 of 50 Questions

# Question Ans
1.

1. Evaluate \( 202^2_{three} - 112^2_{three}\)

A. 21120

B. 21121

C. 21112

D. 21011

Detailed Solution

\(202^2_{three}\)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
\(\begin{matrix}
3 & 204\\
3 & 69 &R0\\
3 & 22 & R2\\
3 & 7 & R1\\
2 & 2 & R1\\
& 0& R2 \uparrow\\
\end{matrix} \\
=21120_3\)
2.

If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two

A. 1110

B. 10111

C. 11101

D. 111100

Detailed Solution

\(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13\\
101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\\
2 & 11 &R1\\
2 & 5 & R1\\
2 & 2 & R1\\
2 & 1 & R0\\
& 0& R1 \uparrow\\
\end{matrix} \\
=y=10111_2\)
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

Detailed Solution

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
\(x = \frac{168}{8}=21^{\circ}\)
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. \(9\frac{3}{5}\)

Detailed Solution

\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\
\frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\
9t + 84 – 4t = 132; 5t = 132 – 84\\
5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)
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. \(\frac{1}{6}\)

Detailed Solution

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 = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)
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

Detailed Solution

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (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

Detailed Solution

The sum of angles at point = 360o
(x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o
10x + 80o = 360o
10x = 360o - 80o = 280o
\(\frac{280}{10}=28^{\circ}\)
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. \(\frac{12}{m-12}\)

Detailed Solution

Prob. (a girl) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\\
= \frac{12}{m+12}\)
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

Detailed Solution

\(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}\\

\(\frac{11}{2}\) ÷ \(\frac{33}{2}\) = \(\frac{11}{2}\) * \(\frac{2}{33}\)
= \(\frac{11}{33}\) or \(\frac{1}{3}\) (when simplified)
\(\frac{15}{2}\) - \(\frac{1}{3}\) = \(\frac{43}{6}\)
7.166
The nearest whole number is 7
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

Detailed Solution

Where H is the height of the tower H = ?
\(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\\
H = 36 \times 0.5774 = 20.79\)
1.

1. Evaluate \( 202^2_{three} - 112^2_{three}\)

A. 21120

B. 21121

C. 21112

D. 21011

Detailed Solution

\(202^2_{three}\)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
\(\begin{matrix}
3 & 204\\
3 & 69 &R0\\
3 & 22 & R2\\
3 & 7 & R1\\
2 & 2 & R1\\
& 0& R2 \uparrow\\
\end{matrix} \\
=21120_3\)
2.

If \(y = 23_{five} + 101_{three}\), find y, leaving your answer in base two

A. 1110

B. 10111

C. 11101

D. 111100

Detailed Solution

\(23_{five} = X_{ten}; X_{ten} = 2\times 5^1 + 3\times 5^0 = 10 + 3 = 13\\
101_{three}=P_{ten}; P_{ten} = 1\times 3^2 + 0\times 3^1 + 1\times 3^0=9+0+1=10_{ten}\\
Y = 13+10=23_{ten}\);
Converting to base two
\(\begin{matrix}
2 & 23\\
2 & 11 &R1\\
2 & 5 & R1\\
2 & 2 & R1\\
2 & 1 & R0\\
& 0& R1 \uparrow\\
\end{matrix} \\
=y=10111_2\)
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

Detailed Solution

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
\(x = \frac{168}{8}=21^{\circ}\)
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. \(9\frac{3}{5}\)

Detailed Solution

\(\frac{3}{4}t+\frac{1}{3}(21-t) = 11; \frac{3t}{4} + \frac{7}{1} - \frac{t}{3} = \frac{11}{1}\\
\frac{3 \times 3t + 7\times 12 – 4 \times t = 11 \times 12}{12}\\
9t + 84 – 4t = 132; 5t = 132 – 84\\
5t = 48; t = \frac{48}{5} = 9\frac{3}{5}\)
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. \(\frac{1}{6}\)

Detailed Solution

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 = \(\frac{x}{1}-\left(\frac{x}{4}+\frac{x}{3}\right)\\
\frac{3x+4x}{12}; = \frac{12x-7x}{12} = \frac{5x}{12} = \frac{5}{12}\)
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

Detailed Solution

\(\frac{PR}{QR} = \frac{QR}{QY} = \frac{QP}{YR}\) (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

Detailed Solution

The sum of angles at point = 360o
(x+10)o + (4x+50)o + 20o + 3xo + 2xo = 360o
10x + 80o = 360o
10x = 360o - 80o = 280o
\(\frac{280}{10}=28^{\circ}\)
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. \(\frac{12}{m-12}\)

Detailed Solution

Prob. (a girl) \(=\frac{Number\hspace{1mm} of\hspace{1mm} girls}{Number \hspace{1mm} of \hspace{1mm} boys \hspace{1mm} and \hspace{1mm} girls\hspace{1mm} in \hspace{1mm} class}\\
= \frac{12}{m+12}\)
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

Detailed Solution

\(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}\\

\(\frac{11}{2}\) ÷ \(\frac{33}{2}\) = \(\frac{11}{2}\) * \(\frac{2}{33}\)
= \(\frac{11}{33}\) or \(\frac{1}{3}\) (when simplified)
\(\frac{15}{2}\) - \(\frac{1}{3}\) = \(\frac{43}{6}\)
7.166
The nearest whole number is 7
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

Detailed Solution

Where H is the height of the tower H = ?
\(Tan 30^{\circ} = \frac{H}{36} \Rightarrow H = 36 \times tan30^{\circ}\\
H = 36 \times 0.5774 = 20.79\)