Year : 
2012
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

41 - 49 of 49 Questions

# Question Ans
41.

The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.

A. 6.0

B. 3.0

C. 2.4

D. 1.8

Detailed Solution

\(\begin{array}{c|c} x & f & fx\\ \hline 1 & 6 & 6 \\
2 & 8 & 16\\ 3 & 8 & 18\\ 4 & 5 & 20\end{array}\)

mean x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{60}{25}\)

x = 2.4
42.

The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?

A. 2

B. 4

C. 6

D. 8

Detailed Solution

median = (\(\frac{N + 1}{2}\))th = (\(\frac{25 + 1}{2}\))th

= \(\frac{26th}{2}\)

= 13th

the 13th is 2
43.

The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?

A. 11.1

B. 10.5

C. 10.3

D. 9.9

A
44.

In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y

A. 51o

B. 54o

C. 56o

D. 68o

Detailed Solution

In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180o(sum of interior angles of a \(\bigtriangleup\) )

i.e. NPO + 129 + 37 = 180

< NOP = 180 - (129 + 37) = 14o

< MNP = < NOP = 14o (alt. < s)

In \(\bigtriangleup\) MPN

< PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))

i.e. 112 + 14 + y = 180o

y = 180 - (112 + 14) = 180 - 126 = 54o
45.

The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS

A. 144o

B. 72o

C. 40o

D. 36o

Detailed Solution

Reflex < POS = 2x (angle at centre is twice that at circumference)

reflex < POS + < POS = 350o(angles at a point)

i.e. 2x + 8x = 360o

10x = 360o

x = \(\frac{360}{10}\)

= 36o

< PRS = \(\frac{1}{2}\)

< POS(< at centre twice that circumference)

= \(\frac{1}{2}\) x 8x = 4x

4 x 36o

< PRS = 144
46.

The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.

A. 2.32cm

B. 1.84cm

C. 0.62cm

D. 0.26cm

C
47.

In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ

A. 65o

B. 45o

C. 25o

D. 19o

Detailed Solution

< RSQ = < RPQ = 48o (angle in the same segment)

< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))

< SQR = 480

< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))

i.e. < QRS + 48o + 48o = 180

< QRS = 180 - (48 + 48) = 180 - 96 = 84o

but < PRQ + < PRS = < QRS

< PRQ = < QRS - < PRS - 84 - 65

= 19o
48.

The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?

A. -\(\frac{1}{2} \leq x\) < 3

B. -\(\frac{1}{2} < x \leq 3\)

C. -\(\frac{1}{2} < x < 3\)

D. -\(\frac{1}{2} \leq x \leq 3\)

Detailed Solution

2x2 - 5x - 3 = 0

2x2 - 6x + x - 3 = 0

2x(x - 3) + 1(x - 3) = 0

(2x + 1)(x - 3) = 0

2x + 1 = 0

2x = -1

x = -\(\frac{1}{2}\)

x - 3 = 0

-\(\frac{1}{2}\) < x < 3
49.

The diagram is a polygon. Find the largest of its interior angles

A. 300o

B. 100o

C. 120o

D. 150o

Detailed Solution

Sum of interior angles = 3602

i.e 3x + 2x + 2x + x + 4x = 360o

12x = 360

x = \(\frac{360}{12}\)

x = 30o

x + c = 180

30o + c = 180

c = 180 - 30

c = 150o
41.

The bar chart shows the frequency distribution of marks scored by students in a class test. Calculate the mean of the distribution.

A. 6.0

B. 3.0

C. 2.4

D. 1.8

Detailed Solution

\(\begin{array}{c|c} x & f & fx\\ \hline 1 & 6 & 6 \\
2 & 8 & 16\\ 3 & 8 & 18\\ 4 & 5 & 20\end{array}\)

mean x = \(\frac{\sum fx}{\sum f}\)

= \(\frac{60}{25}\)

x = 2.4
42.

The bar chart shows the frequency distribution of marks scored by students in a class test. What is the median of the distribution?

A. 2

B. 4

C. 6

D. 8

Detailed Solution

median = (\(\frac{N + 1}{2}\))th = (\(\frac{25 + 1}{2}\))th

= \(\frac{26th}{2}\)

= 13th

the 13th is 2
43.

The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? What is the gradient of y = 2x2 - 5x - 3 at the point x = 4?

A. 11.1

B. 10.5

C. 10.3

D. 9.9

A
44.

In the diagram, MN//PO, < PMN = 112o, < PNO = 129oo and < MPN = yo. Find the value of y

A. 51o

B. 54o

C. 56o

D. 68o

Detailed Solution

In \(\bigtriangleup\) NPO + PNO + PNO + < NOP = 180o(sum of interior angles of a \(\bigtriangleup\) )

i.e. NPO + 129 + 37 = 180

< NOP = 180 - (129 + 37) = 14o

< MNP = < NOP = 14o (alt. < s)

In \(\bigtriangleup\) MPN

< PMN + < MNP + y = 180(sum of interior angles of a \(\bigtriangleup\))

i.e. 112 + 14 + y = 180o

y = 180 - (112 + 14) = 180 - 126 = 54o
45.

The diagram is a circle with centre P. PRST are points on the circle. Find the value of < PRS

A. 144o

B. 72o

C. 40o

D. 36o

Detailed Solution

Reflex < POS = 2x (angle at centre is twice that at circumference)

reflex < POS + < POS = 350o(angles at a point)

i.e. 2x + 8x = 360o

10x = 360o

x = \(\frac{360}{10}\)

= 36o

< PRS = \(\frac{1}{2}\)

< POS(< at centre twice that circumference)

= \(\frac{1}{2}\) x 8x = 4x

4 x 36o

< PRS = 144
46.

The diagram is a circle of radius |QR| = 4cm. \(\bar{R}\) is a tangent to the circle at R. If TPO = 120o, find |PQ|.

A. 2.32cm

B. 1.84cm

C. 0.62cm

D. 0.26cm

C
47.

In the diagram, |SR| = |QR|. < SRP = 65o and < RPQ = 48o, find < PRQ

A. 65o

B. 45o

C. 25o

D. 19o

Detailed Solution

< RSQ = < RPQ = 48o (angle in the same segment)

< SQR < RSQ (Base angle of an isosceles \(\bigtriangleup\))

< SQR = 480

< QRS + < RSQ + < RSQ = 180o(sum of interior angles of a \(\bigtriangleup\))

i.e. < QRS + 48o + 48o = 180

< QRS = 180 - (48 + 48) = 180 - 96 = 84o

but < PRQ + < PRS = < QRS

< PRQ = < QRS - < PRS - 84 - 65

= 19o
48.

The graph is that of y = 2x2 - 5x - 3. For what value of x will y be negative? For what value of x will y be negative?

A. -\(\frac{1}{2} \leq x\) < 3

B. -\(\frac{1}{2} < x \leq 3\)

C. -\(\frac{1}{2} < x < 3\)

D. -\(\frac{1}{2} \leq x \leq 3\)

Detailed Solution

2x2 - 5x - 3 = 0

2x2 - 6x + x - 3 = 0

2x(x - 3) + 1(x - 3) = 0

(2x + 1)(x - 3) = 0

2x + 1 = 0

2x = -1

x = -\(\frac{1}{2}\)

x - 3 = 0

-\(\frac{1}{2}\) < x < 3
49.

The diagram is a polygon. Find the largest of its interior angles

A. 300o

B. 100o

C. 120o

D. 150o

Detailed Solution

Sum of interior angles = 3602

i.e 3x + 2x + 2x + x + 4x = 360o

12x = 360

x = \(\frac{360}{12}\)

x = 30o

x + c = 180

30o + c = 180

c = 180 - 30

c = 150o