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

Paper 1 | Objectives

21 - 30 of 50 Questions

# Question Ans
21.

Find, correct to two decimal places, the mean of 9, 13, 16, 17, 19, 23, 24.

A. 23.00

B. 17.29

C. 16.50

D. 16.33

Detailed Solution

Mean \(= \frac{sum\hspace{1mm}of\hspace{1mm}all\hspace{1mm}the\hspace{1mm}numbers}{Total\hspace{1mm}Number}\\
\frac{9+13+16+17+19+23+24}{7}\\
\frac{121}{7}=17.2857=17.29\)
22.

In the diagram, ∆XYZ is similar to ∆PRQ, |XY| = 5cm, |XZ| =3.5cm and |PR| = 8cm. Find |PQ|

A. 5.6 cm

B. 11.2cm

C. 11.4cm

D. 28.0cm

Detailed Solution

\(\frac{PQ}{PR}=\frac{XZ}{XY}; \frac{PQ}{8}=\frac{3.5}{5}\\
PQ=\frac{8\times 3.5}{5}=\frac{28}{5}=5.6cm\)
23.

Factorise 27p2x2 - 48y2.

A. 9(3px - 4y)2

B. 3(3px - 4y)(3px - 4y)

C. 9(px - 4y)(3p x + 4y)

D. 3(3px -4y)(3px +4y)

Detailed Solution

27p2x2 - 48y2; 3(32 p2x2-42y2)
3[(3px) 2 - (4y) 2]; 3[(3px-4y)(3px+4y)]
24.

What is the volume of a solid cylinder of diameter 7cm and height 7cm? (Take \(\pi = \frac{22}{7}\))

A. 38.5cm3

B. 77cm3

C. 269.5cm3

D. 1078cm3

Detailed Solution

Volume of a solid cylinder = Base area x height
Base area \(=\pi r^2 =\frac{22}{7} \times \frac{7}{2}\times \frac{7}{2} = \frac{77}{2}\)
Volume of a solid cylinder = \(\frac{77}{2}\times \frac{7}{1} = \frac{539}{2}cm^3\\
=269.5cm^3\)
25.

Find the sum of the roots of the equation 2x2 + 3x - 9 = 0

A. -18

B. -6

C. \(-\frac{9}{2}\)

D. \(-\frac{3}{2}\)

Detailed Solution

The general quadratic equation
x2 - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\\
Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)
26.

Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q

A. {2}

B. {1,9}

C. {315.7}

D. {4, 6, 8, 10}

Detailed Solution

ξ = {1, 2, 3, . . . . . . ,10}, P= (2, 3, 5, 7) and Q = {1, 3, 5, 9},
Pl∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}
27.

In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80o. Find ∠QRS.

A. 90o

B. 80o

C. 50o

D. 40o

Detailed Solution

∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 50o
∠QRS = 50o
28.

Find the mean deviation of 6, 7, 8, 9, 10

A. 1.2

B. 1.5

C. 2

D. 8

Detailed Solution

\(\begin{matrix}
x & x-m & |x-m| \\
6 & -2 & 2 \\
7 & -1 & 1 \\
8 & 0 & 0 \\
9 & 1 & 1 \\
10 & 2 & 2 \\
& & 6
\end{matrix}
\)
Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)
29.

A point X is on the bearing 342o from a point Y. What is the bearing of Y from X?

A. 342o

B. 252o

C. 198o

D. 162o

Detailed Solution

If x is the bearing 342ofrom a point y
= 360o - 342o = 18o;
The bearing of y from x = 180o - 18o = 162o
30.

In the diagram, O is the centre of the circle where OS//QR and ∠SOR = 35o

A. 35o

B. 45o

C. 55o

D. 70o

Detailed Solution

∠SOR = ∠ORQ alternate angle.
Also ∠OQR = ∠OQR (Base angle of isosceles)
∠QOR = 108 – 2(35o) = 180 - 70o - 110o ∠QPR = 1/2QOR = 1/2(110o) ∠QPR + 55o (theorem angle at center is twice angle at circumference)
21.

Find, correct to two decimal places, the mean of 9, 13, 16, 17, 19, 23, 24.

A. 23.00

B. 17.29

C. 16.50

D. 16.33

Detailed Solution

Mean \(= \frac{sum\hspace{1mm}of\hspace{1mm}all\hspace{1mm}the\hspace{1mm}numbers}{Total\hspace{1mm}Number}\\
\frac{9+13+16+17+19+23+24}{7}\\
\frac{121}{7}=17.2857=17.29\)
22.

In the diagram, ∆XYZ is similar to ∆PRQ, |XY| = 5cm, |XZ| =3.5cm and |PR| = 8cm. Find |PQ|

A. 5.6 cm

B. 11.2cm

C. 11.4cm

D. 28.0cm

Detailed Solution

\(\frac{PQ}{PR}=\frac{XZ}{XY}; \frac{PQ}{8}=\frac{3.5}{5}\\
PQ=\frac{8\times 3.5}{5}=\frac{28}{5}=5.6cm\)
23.

Factorise 27p2x2 - 48y2.

A. 9(3px - 4y)2

B. 3(3px - 4y)(3px - 4y)

C. 9(px - 4y)(3p x + 4y)

D. 3(3px -4y)(3px +4y)

Detailed Solution

27p2x2 - 48y2; 3(32 p2x2-42y2)
3[(3px) 2 - (4y) 2]; 3[(3px-4y)(3px+4y)]
24.

What is the volume of a solid cylinder of diameter 7cm and height 7cm? (Take \(\pi = \frac{22}{7}\))

A. 38.5cm3

B. 77cm3

C. 269.5cm3

D. 1078cm3

Detailed Solution

Volume of a solid cylinder = Base area x height
Base area \(=\pi r^2 =\frac{22}{7} \times \frac{7}{2}\times \frac{7}{2} = \frac{77}{2}\)
Volume of a solid cylinder = \(\frac{77}{2}\times \frac{7}{1} = \frac{539}{2}cm^3\\
=269.5cm^3\)
25.

Find the sum of the roots of the equation 2x2 + 3x - 9 = 0

A. -18

B. -6

C. \(-\frac{9}{2}\)

D. \(-\frac{3}{2}\)

Detailed Solution

The general quadratic equation
x2 - (sum of roots) x + (product of root) = 0
Comparing with the given equation
2x2 +3x2 - 9 = 0; \(x^2 + \frac{3}{2}x - \frac{9}{2} = 0\\
Sum \hspace{1mm} of \hspace{1mm}roots = -\frac{3}{2}\)
26.

Given that ξ = {1, 2, 3, . . . . . . ,10}, P= (x : x is prime) and Q = {y : y is odd}, find Pl∩Q

A. {2}

B. {1,9}

C. {315.7}

D. {4, 6, 8, 10}

Detailed Solution

ξ = {1, 2, 3, . . . . . . ,10}, P= (2, 3, 5, 7) and Q = {1, 3, 5, 9},
Pl∩Q = {1, 4, 6, 8, 9, 10}∩{1, 3, 5, 7, 9} = {1, 9}
27.

In the diagram, KS is a tangent to the circle centre O at R and ∠ROQ = 80o. Find ∠QRS.

A. 90o

B. 80o

C. 50o

D. 40o

Detailed Solution

∠QRS = ∠RPQ (theorem: angle in the alternate segment are equal)
∠RPQ = 50o
∠QRS = 50o
28.

Find the mean deviation of 6, 7, 8, 9, 10

A. 1.2

B. 1.5

C. 2

D. 8

Detailed Solution

\(\begin{matrix}
x & x-m & |x-m| \\
6 & -2 & 2 \\
7 & -1 & 1 \\
8 & 0 & 0 \\
9 & 1 & 1 \\
10 & 2 & 2 \\
& & 6
\end{matrix}
\)
Mean deviation\(=\frac{\sum|x-m|}{n}=\frac{6}{5} = 1.2\)
29.

A point X is on the bearing 342o from a point Y. What is the bearing of Y from X?

A. 342o

B. 252o

C. 198o

D. 162o

Detailed Solution

If x is the bearing 342ofrom a point y
= 360o - 342o = 18o;
The bearing of y from x = 180o - 18o = 162o
30.

In the diagram, O is the centre of the circle where OS//QR and ∠SOR = 35o

A. 35o

B. 45o

C. 55o

D. 70o

Detailed Solution

∠SOR = ∠ORQ alternate angle.
Also ∠OQR = ∠OQR (Base angle of isosceles)
∠QOR = 108 – 2(35o) = 180 - 70o - 110o ∠QPR = 1/2QOR = 1/2(110o) ∠QPR + 55o (theorem angle at center is twice angle at circumference)