Year : 
2003
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

11 - 20 of 46 Questions

# Question Ans
11.

Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)

A. \(2\frac{1}{3}\)

B. \(1\frac{1}{3}\)

C. 1

D. \(\frac{3}{7}\)

Detailed Solution

\(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2=\left(\frac{5}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)
Difference of two squares
\(\left(\frac{5}{3}-\frac{2}{3}\right)\left(\frac{5}{3}+\frac{2}{3}\right)=\left(\frac{3}{3}\right)\left(\frac{7}{3}\right)\\
\frac{7}{3}=2\frac{1}{3}\)
12.

Find the value of x which satisfies the equation
5(x-7)=7-2x

A. x =2

B. x=4

C. x=6

D. x = 14

Detailed Solution

5(x - 7) = 7 - 2x
5x - 35 = 7 - 2x
5x + 2x = 7 + 35
7x = 42
x = \(\frac{42}{7}\)
= 6
13.

Evaluate \(\frac{log8}{log\left(\frac{1}{4}\right)}\)

A. -2

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

C. \(\frac{1}{2}\)

D. 4

Detailed Solution

\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\)
14.

A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid

A. 5cm

B. 9cm

C. 12cm

D. 15cm

Detailed Solution

\(V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\\
120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm\)
15.

PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?

A. 180o

B. 150o

C. 105o

D. 75o

Detailed Solution


< QPS = 75°
< QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral)
16.

In the diagram, ZM is a straight line. Calculate the value of x.

A. 27o

B. 30o

C. 35o

D. 37o

Detailed Solution

The sum of angles in a triangle is 180o
2xo + (2x -21)o + ∠ZYP = 180o
Also ∠ZYP = 180o - (3x+14)o
∴2xo + (2x-21)o + 180o - (3x+14)o = 180
x = 35o
17.

Find the value of x in the equation 3x\(^2\) - 8x - 3 = 0

A. \(\frac{1}{3},-3\)

B. \(-\frac{1}{3},-3\)

C. \(-\frac{1}{3},3\)

D. \(\frac{1}{3},3\)

Detailed Solution

3x\(^2\) - 8x - 3 = 0
3x\(^2\) - 9x + x - 3 = 0
3x(x - 3) + 1(x - 3) = 0
(3x + 1)(x - 3) = 0
3x = -1 \(\implies\) x = \(-\frac{1}{3}\)
or x = 3.
18.

In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place.

A. 3.4

B. 4.3

C. 5.9

D. 6.2

Detailed Solution

Using the sine rule
\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\
\frac{x}{sin43^o}=\frac{5}{sin53^o}\\
x = \frac{5\times sin43^o}{sin53^o}\\
=\frac{5\times 0.6820}{0.7986}\\
=4.3\)
19.

Given, that \(4P4_5 = 119_{10}\), find the value of P

A. 1

B. 2

C. 3

D. 4

Detailed Solution

\(4P4_5 = 119_{10}\)
\(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\)
\(100 + 5P + 4 = 119\)
\(5P = 119 - 104 = 15\)
\(P = 3\)
20.

A ladder, 6m long, leans against a vertical wall at an angle 53o to the horizontal. How high up the wall does the ladder reach?

A. 3.611m

B. 4.521m

C. 4.792m

D. 7.962m

Detailed Solution

Sin 53 = \(\frac {h}{6} \)
h = 6 sin 53 = 6 x 0.7986
= 4.792m
11.

Simplify \(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)

A. \(2\frac{1}{3}\)

B. \(1\frac{1}{3}\)

C. 1

D. \(\frac{3}{7}\)

Detailed Solution

\(\left(1\frac{2}{3}\right)^2 - \left(\frac{2}{3}\right)^2=\left(\frac{5}{3}\right)^2 - \left(\frac{2}{3}\right)^2\)
Difference of two squares
\(\left(\frac{5}{3}-\frac{2}{3}\right)\left(\frac{5}{3}+\frac{2}{3}\right)=\left(\frac{3}{3}\right)\left(\frac{7}{3}\right)\\
\frac{7}{3}=2\frac{1}{3}\)
12.

Find the value of x which satisfies the equation
5(x-7)=7-2x

A. x =2

B. x=4

C. x=6

D. x = 14

Detailed Solution

5(x - 7) = 7 - 2x
5x - 35 = 7 - 2x
5x + 2x = 7 + 35
7x = 42
x = \(\frac{42}{7}\)
= 6
13.

Evaluate \(\frac{log8}{log\left(\frac{1}{4}\right)}\)

A. -2

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

C. \(\frac{1}{2}\)

D. 4

Detailed Solution

\(\frac{log8}{log\frac{1}{4}}=\frac{log2^3}{log2^{-2}}=\frac{3log2}{-2log2}=-\frac{3}{2}\)
14.

A pyramid of volume 120cm3 has a rectangular base which measures 5cm by 6cm. Calculate the height of the pyramid

A. 5cm

B. 9cm

C. 12cm

D. 15cm

Detailed Solution

\(V= \frac{1}{2}\hspace{1mm}base\hspace{1mm}area\hspace{1mm}\times\hspace{1mm}height\\
120=\frac{1}{3}\times 5 \times 6 \times h; h = \frac{120}{10}=12cm\)
15.

PQRS is a cyclic quadrilateral. If ∠QPS = 75°, what is the size of ∠QRS?

A. 180o

B. 150o

C. 105o

D. 75o

Detailed Solution


< QPS = 75°
< QRS = 180° - 75° = 105° (opposite angles of a cyclic quadrilateral)
16.

In the diagram, ZM is a straight line. Calculate the value of x.

A. 27o

B. 30o

C. 35o

D. 37o

Detailed Solution

The sum of angles in a triangle is 180o
2xo + (2x -21)o + ∠ZYP = 180o
Also ∠ZYP = 180o - (3x+14)o
∴2xo + (2x-21)o + 180o - (3x+14)o = 180
x = 35o
17.

Find the value of x in the equation 3x\(^2\) - 8x - 3 = 0

A. \(\frac{1}{3},-3\)

B. \(-\frac{1}{3},-3\)

C. \(-\frac{1}{3},3\)

D. \(\frac{1}{3},3\)

Detailed Solution

3x\(^2\) - 8x - 3 = 0
3x\(^2\) - 9x + x - 3 = 0
3x(x - 3) + 1(x - 3) = 0
(3x + 1)(x - 3) = 0
3x = -1 \(\implies\) x = \(-\frac{1}{3}\)
or x = 3.
18.

In\( ∆ PQR, P\hat{Q}P = 84^°, |Q\hat{P}R |= 43^°\) and |PQ| = 5cm. Find /QR/ in cm, correct to 1 decimal place.

A. 3.4

B. 4.3

C. 5.9

D. 6.2

Detailed Solution

Using the sine rule
\(\frac{|QR|}{sinP}=\frac{|PQ|}{sinR}\\
\frac{x}{sin43^o}=\frac{5}{sin53^o}\\
x = \frac{5\times sin43^o}{sin53^o}\\
=\frac{5\times 0.6820}{0.7986}\\
=4.3\)
19.

Given, that \(4P4_5 = 119_{10}\), find the value of P

A. 1

B. 2

C. 3

D. 4

Detailed Solution

\(4P4_5 = 119_{10}\)
\(4 \times 5^2 + P \times 5^1 + 4 \times 5^0 = 119\)
\(100 + 5P + 4 = 119\)
\(5P = 119 - 104 = 15\)
\(P = 3\)
20.

A ladder, 6m long, leans against a vertical wall at an angle 53o to the horizontal. How high up the wall does the ladder reach?

A. 3.611m

B. 4.521m

C. 4.792m

D. 7.962m

Detailed Solution

Sin 53 = \(\frac {h}{6} \)
h = 6 sin 53 = 6 x 0.7986
= 4.792m