Year : 
2016
Title : 
Mathematics (Core)
Exam : 
WASSCE/WAEC MAY/JUNE

Paper 1 | Objectives

31 - 40 of 49 Questions

# Question Ans
31.

The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range

A. 16

B. 14

C. 13

D. 11

Detailed Solution

First, arrange the marks in order of magnitude; 3, 4, 5, 7, 10, 13, 14, 16

Hence range = 16 - 3 = 13
32.

If log2(3x - 1) = 5, find x.

A. 2.00

B. 3.67

C. 8.67

D. 11

Detailed Solution

Log2(3x - 1) = 5

Log2(3x - 1) = Log225

Log2(3x - 1) = Log232

3x - 1 = 32

3x = 32 + 1 = 33

x = \(\frac{33}{3}\)

= 11
33.

A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r

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

B. 2

C. 3

D. 6

Detailed Solution

Volume of sphere = Volume of cylinder

i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\)

\(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\)

r3 = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\)

r = 3\(\sqrt{27}\)

= 3
34.

Express 1975 correct to 2 significant figures

A. 20

B. 1,900

C. 1,980

D. 2,000

D
35.

The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of \(\Delta\) QNP = 6cm, calculate the area of the trapezium.

A. 96cm2

B. 90cm2

C. 81cm2

D. 27cm2

Detailed Solution

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2

Area of \(\Delta\) QMN = Area of \(\Delta\) QNP

= Area of \(\Delta\) PNO (triangles between the same parallels)

Hence, area of the trapezium

3 x area of \(\Delta\) QNP

= 3 x 27

= 81cm2
36.

The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector [Take \(\pi = \frac{22}{7}\)]

A. 70o

B. 60o

C. 55o

D. 42o

Detailed Solution

Perimeter of a sector

= 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3

64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3

22 = \(\frac{33\theta}{90}\)

\(\theta = \frac{22 \times 30}{11}\)

= 60o
37.

Examine M' \(\cap\) N from the venn diagram

A. {f, g}

B. {e}

C. {e, f, g}

D. {e, f, g}

Detailed Solution

From the venn diagram given,

M = (a, b, c), N = (c, f, g)

U = (a, b, c, d, e, f, g)

Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g)

= (f, g)
38.

If 20(mod 9) is equivalent to y(mod 6), find y.

A. 1

B. 2

C. 3

D. 4

Detailed Solution

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9)

If 2(mod 9) y(mod 6), then y = 2 by comparism
39.

Simplify \(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

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

B. p - 2r

C. \(\frac{1}{p - 2r}\)

D. \(\frac{2p}{p - 2r}\)

Detailed Solution

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

= \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\

= \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\)

= \(\frac{p^2 - 2pr}{2p(p - 2r)}\)

= \(\frac{p(p - 2r)}{2p(p - 2r)}\)

= \(\frac{1}{2}\)
40.

In the diagram, O is the centre of the circle, < QPS = 100o, < PSQ = 60o and < QSR. Calculate < SQR

A. 20o

B. 40o

C. 60o

D. 80o

Detailed Solution

In the diagram, < RPQ = 80o(angles in same segment)

< SPR = 100o - < RPQ

= 100 - 80

= 20o

< SQR = < SPR = 20o (same reason as above)

< SQR = 20o
31.

The marks of eight students in a test are: 3, 10, 4, 5, 14, 13, 16 and 7. Find the range

A. 16

B. 14

C. 13

D. 11

Detailed Solution

First, arrange the marks in order of magnitude; 3, 4, 5, 7, 10, 13, 14, 16

Hence range = 16 - 3 = 13
32.

If log2(3x - 1) = 5, find x.

A. 2.00

B. 3.67

C. 8.67

D. 11

Detailed Solution

Log2(3x - 1) = 5

Log2(3x - 1) = Log225

Log2(3x - 1) = Log232

3x - 1 = 32

3x = 32 + 1 = 33

x = \(\frac{33}{3}\)

= 11
33.

A sphere of radius rcm has the same volume as cylinder of radius 3cm and height 4cm. Find the value of r

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

B. 2

C. 3

D. 6

Detailed Solution

Volume of sphere = Volume of cylinder

i.e. \(\frac{4}{3} \pi r^3 = \pi r^2 h\)

\(\frac{4}{3} \pi r^3 = \pi \times 3^2 \times 4\)

r3 = \(\frac{\pi \times 9 \times 4 \times 3}{4 \pi}\)

r = 3\(\sqrt{27}\)

= 3
34.

Express 1975 correct to 2 significant figures

A. 20

B. 1,900

C. 1,980

D. 2,000

D
35.

The diagram, MOPQ is a trapezium with QP||MO, MQ||NP, NQ||OP, |QP| = 9cm and the height of \(\Delta\) QNP = 6cm, calculate the area of the trapezium.

A. 96cm2

B. 90cm2

C. 81cm2

D. 27cm2

Detailed Solution

Area of \(\Delta\) QNP = \(\frac{1}{2} \times 9 \times 6 \) = 27cm2

Area of \(\Delta\) QMN = Area of \(\Delta\) QNP

= Area of \(\Delta\) PNO (triangles between the same parallels)

Hence, area of the trapezium

3 x area of \(\Delta\) QNP

= 3 x 27

= 81cm2
36.

The perimeter of a sector of a circle of radius 21cm is 64cm. Find the angle of the sector [Take \(\pi = \frac{22}{7}\)]

A. 70o

B. 60o

C. 55o

D. 42o

Detailed Solution

Perimeter of a sector

= 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21

64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3

64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3

22 = \(\frac{33\theta}{90}\)

\(\theta = \frac{22 \times 30}{11}\)

= 60o
37.

Examine M' \(\cap\) N from the venn diagram

A. {f, g}

B. {e}

C. {e, f, g}

D. {e, f, g}

Detailed Solution

From the venn diagram given,

M = (a, b, c), N = (c, f, g)

U = (a, b, c, d, e, f, g)

Thus M' \(\cap\) N = (e, f, g) \(\cap\) (c, f, g)

= (f, g)
38.

If 20(mod 9) is equivalent to y(mod 6), find y.

A. 1

B. 2

C. 3

D. 4

Detailed Solution

First, reduce 20(mod 9) to its simplest form in mod 9; 9 x 2 + 2 = 2(mod 9)

If 2(mod 9) y(mod 6), then y = 2 by comparism
39.

Simplify \(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

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

B. p - 2r

C. \(\frac{1}{p - 2r}\)

D. \(\frac{2p}{p - 2r}\)

Detailed Solution

\(\frac{(p - r)^2 - r^2}{2p^2 - 4pr}\)

= \(\frac{(p - r)(p - r) - r^2}{2p^2 - 4pr}\)\

= \(\frac{p^2 - 2pr + r^2 - r^2}{2p(p - 2r}\)

= \(\frac{p^2 - 2pr}{2p(p - 2r)}\)

= \(\frac{p(p - 2r)}{2p(p - 2r)}\)

= \(\frac{1}{2}\)
40.

In the diagram, O is the centre of the circle, < QPS = 100o, < PSQ = 60o and < QSR. Calculate < SQR

A. 20o

B. 40o

C. 60o

D. 80o

Detailed Solution

In the diagram, < RPQ = 80o(angles in same segment)

< SPR = 100o - < RPQ

= 100 - 80

= 20o

< SQR = < SPR = 20o (same reason as above)

< SQR = 20o