Mathematics Past Question and Answer for Pre-University Students

CREATIVE SOLUTIONS TO OAUPOST–UTME MATHEMATICS

(2006− 2014)

CHRIST OPHER O.O.

i

Creative Solutions:©2014 by Christopher O.O

Published in Nigeria by:

Contact the author on:Mobile:+2347066578808.Email: [email protected]

All rights reserved.No part of this publication may be reproduced, stored in a retrieval system, or transmittedin any form or by any means, mechanical, electronic, photocopying or otherwise withoutthe prior written consent of the copyright owner.

ii

Dedication

This book is dedicated to men and women, young and old in all the nations of the earth,who are living purposefully and adding values to the lives of others.

iii

1

Appreciation

Tremendous thanks to my brilliant friend, colleague, editor and typist JP Okeke. Youare the best I have ever worked with · · · by a million miles!My deep appreciation and thanks to my wonderful Senior partner, from whom I drawcreativity and wisdom for all aspect of life.I also thank all my teachers, from Nursery to University level. I love you all and I say aBig thank you to you all.

Contents

DEDICATION iii

Appreciation 1

Preface 4

1 51.1 Questions, 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Solution, 2014 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 162.1 Question, 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2 Solution, 2013 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18






2

3 CONTENTS



4 CONTENTS

Preface

”Private study is the best form of learning and it should not be replaced byany other form of learning”.

Dr. O.P LayeniDept of Mathematics, OAU

Over the years, millions and billions of people dreamt and desired to get the best educa-tion and to gain admission into a higher Institution of Learning but unfortunately everyyear all over the world many could not achieve this aim.

The few ones who end up attaining their aims on their own merits, have just oneadvantage over the others. They have sought and found access to the best preparatorymaterials and requisite books, in their subject areas and they have devoted their time andenergy to studying them. Our motive is to ensure you are among these few ones.

This book, Creative Solution to OAU POST–UTME Mathematics, gives you all youneed to excel in your Mathematics examinations as the solutions are so detailed and nicelybroken down, that the knowledge you acquire having studied them, will be very usefulin solving other similar questions under the same topic.

The questions are properly typeset to ease understanding. They are arranged in thepattern you will find them in your examination. It is recommended that you first attemptthe questions on your own, set a time limit and mark your attempts. Then go over thesolutions given here to familiarize yourself with the areas you could not deal with.

Wishing you the best in your preparations and I pray that GOD Ease your understand-ing and grant you success. He will bless all your good efforts in this examination and youshall be rewarded with admission into the Institution of your dream. Your happiness andsuccess is our joy!

1

Questions, 2014

1. Simplify30√

2+√

50.

A. 4√

5 B. 20√

2 C. 5√

5 D. 10√

2

2. If m is the gradient of the line

pq − px− qy = 0 and q 6= 0, find1

mA.

q

pB.

p

qC. −q

pD. −p

q

3. Evaluate(

1

25

)−12

+

(1

8

)−23

A. 8 B. 10 C. 9 D. 6

4. Find the remainder when x4− 11x+ 2is divided by xA. 2 B. 6 C. -2 D. 5

5. If cosA =12

13and A is an acute angle,

find(1 + tan2A

)A.

144

25B.

25

144C.

169

25D.

169

144

6. Integrate the function 1 − 2x with re-spect to x.A. x− x2 +K B. x+ x2 +KC. −x− x−2 +K D. x− x−2 +K

7. If equation 6− kx+ 2x2 = 0 has equalroots, find k2 + 4A. 48 B. 52 C. 44 D. 96

8. Simplify log100

√10−1

A. −1

8B. −1

4C.

1

4D.

1

8

9. Obtain the center of the circle7(y2 + 10y) + 7x2 = 1

A. (0, 5) B. (−5, 0)C. (0,−5) D. (5, 0)

10. If α and β are the roots of equationcx2 + ax+ b = 0, find αβ.

A. − ba

B. −ac

C.b

cD.

c

a

11. The binary operation ⊗ is defined bya⊗ b = 2a− 1. Find 3⊗ (2⊗ 1)A. 3 B. 4 C. 5 D. 6

12. Two coins are tossed. Find the proba-bility of having at least two heads.A. 1

2B. 3

4C. 1

4D. 1

13. Solve the equation1

1 +1

x3

= 0

A. 0 B. −1 C. 1 D. ∞

14. If P is directly proportional to√Q;

P = 20 when Q = 4. Find Q whenP = 100A. 200 B. 300 C. 100 D. 400

15. Find the angle in degree which theline x−

√3y = 0 makes with the posi-

tive y−axisA. 30 B. 90 C. 60 D. 180

16. Given∫ a

−a15x2dx = 3430, find the

value of the constant aA. 8 B. 6 C. 7 D. 9

17. Evaluated

dx(ln sin 3x)

A. 3 cot 3x B. 3 tan 3x

C.1

sin 3xD. 3 sin 3x

5

6 Questions, 2014

18. Find the equation of a line whichpasses through a point (−2, 3) andmakes an angle of 45◦ with positivex− axis.A. y − x − 5 = 0 B. y + x + 5 = 0C. x− y − 5 = 0 D. y − x+ 5 = 0

19. Find the sum to infinity of the se-quence 1,−1, 1,−1, 1,−1, · · ·A. 2 B. −1

2C. -2 D. 1

2

20. Differentiate 2 − sin(2 − ax) with re-spect to xA. a cos(2− ax) B. −a sin(2− ax)C. −a cos(2− ax) D. − sin(2− ax)

21. Simplify

(8√n

m32

)2(4−1m2

2n−2

)A. 128n3m−1 B. 8n3m−1

C. 8n3m D. 8n4m

22. If x is a real number and x + 11 < 0,

evaluate|x|x

.A. 0 B. −1 C. 1 D. 2

7 Solution, 2014

Solution, 2014

1. In Surds, we rationalize a fractionwhen its denominator has the rootsign in it. Example

2

1−√

2. We ratio-

nalize by multiplying both the numer-ator and denominator of the fractionwith the conjugate of the denomina-tor. The conjugate of 1 −

√2 is what

you multiply with it that will makethe root sign disappear which is 1+

√2

since (1−√

2)(1 +√

2) = 1− 2 = −1.

Now, to solve30√

2+√

50, we first ra-

tionalize30√

2to have

30√2

=30√

2×√

2√2

=30√

2√4

=30√

2

2= 15

√2

We also simplify√

50 as follows:√

50 =√

2× 25 =√

2×√

25

=√

2× 5√50 = 5

√2

Now,

30√2

+√

50 = 15√

2 + 5√

2

= (15 + 5)√

2

= 20√

2

Hence,

30√2

+√

50 = 20√

2 B.

2. The equation of a straight line whengiven the gradient or slope m and they− intercept c is

y = mx+ c (1.1)

The equation of a straight line whengiven the gradient or slope m and apoint (x1, y1) on the line is

y − y1 = m(x− x1) (1.2)

Now, for the equation;

pq − px− qy = 0

We make y the subject of formula tomake it look like equation (1.1), sothat we can compare the two equa-tions and bring out m, the gradient

−qy = px− pq

y =px− pq−q

, q 6= 0

y = −pqx+

pq

q

= −pqx+ p

soy = −p

qx+ p (1.3)

On comparing (1.1) and (1.3) we ob-serve that

m = −pq

and1

m=

1

−pq

= −qp

Hence,1

m= −q

pC.

3. Note: In Indices;(ab

)− 1n

=

(b

a

) 1n

=n

√b

a

Also,(ab

)−mn

=

(b

a

)mn

=

(n

√b

a

)m

8 Solution, 2014

Now,(1

25

)−12

+

(1

8

)−23

= (25)12 + (8)

23

=2√

25 +(

3√

8)2

= 5 + (2)2

= 5 + 4

= 9

Hence,(1

25

)−12

+

(1

8

)−23

= 9. C.

Kindly see solutions to Q2 of 2012 formore on indices.

4. Note: Polynomial, p(x) = D(x)Q(x) +R(x), where D(x) is the divisor, Q(x)is the Quotient andR(x) is the remain-der.

Now,

P (x) = x4 − 11x+ 2 and D(x) = x

x3 − 11

x)

x4 − 11x + 2− x4

− 11x11x

Hence, the remainder is 2. A.

step 1 :x4

x= x3

(Q(x)

)step 2 : x(x3) = x4

(D(x)Q(x)

)P (x)− x4 = −11x+ 2

step 3 : −11x

x= −11

(Q(x)

)step 4 : x(−11) = −11x

(D(x)×Q(x)

)(−11x+ 2)− (−11x) = 2

(R(x)

)

Kindly see the solution to Q18 of 2009for more explicit explanation on poly-nomials.

5. From trigonometry,

sin2A+ cos2A = 1

On dividing each term of the aboveequation by cos2A, we have

cos2A

cos2A+

sin2A

cos2A=

1

cos2A

which gives

1 + tan2A =1

cos2A= sec2A

(Note that tanA = sinAcosA

, 1cosA

= secA

and sin2 Acos2 A

= tan2A). Now,

cosA =12

13, cos2A =

(12

13

)2

=144

169

1 + tan2A =1

cos2A=

169

144

Hence,

1 + tan2A =169

144, D.

Kindly see solution to Q13,Q14 andQ15 of 2010 for more on trigonometry.

6. Generally,∫axndx =

axn+1

n+ 1+ k,

where k is the constant of integra-tion.

9 Solution, 2014

Now,∫(1− 2x)dx =

∫1dx−

∫2xdx

=

∫1x0dx−

∫2x1dx

=1x0+1

0 + 1− 2x1+1

1 + 1+ k

=1x1

1− 2x2

2+ k

= x− x2 + k

Hence,∫(1− 2x)dx = x− x2 + k A.

(Kindly see the solutions to Q19 of2013, Q17 of 2012 and Q5 of 2011 formore explicit explanations on integra-tion.)

7. For the quadratic equation ax2 + bx +c = 0 to have equal roots then, b2−4acmust be equal to zero, that is, when

b2 − 4ac = 0, equal and real rootsb2 − 4ac < 0, complex /imaginary rootsb2 − 4ac > 0, unequal and real roots

Now, if 6 − kx + 2x2 = 0 has equalroots,

2x2 − kx+ 6 = 0,

a = 2, b = −k, c = 6

then

b2 − 4ac = (−k)2 − 4(2)(6) = 0

which implies that

k2 − 4(12) = 0

k2 − 48 = 0

k2 = 48

From this, k2+4 = 48+4, on adding 4,to both sides of equation above. Thisimplies that

k2 + 4 = 52, B.

8. We note that

i. loga x = b implies that x = ab

ii. m√an = a

nm

iii.√a−n = a−

n2

Now,

log100

√10−1 = log100 10−

12

Let log100 10−12 = x, then using (i.)

10−12 = 100x

10−12 =

(102)x

10−12 = 102x

On equating the powers of 10, wehave

−1

2= 2x

2x = −1

2which implies x = −1

2· 1

2= −1

4

x = −1

4

which implies that

log100

√10−1 = −1

4, B.

Kindly see the solutions to Q5 andQ18 of 2013, Q2, Q3 and Q4 of 2012for more explicit explanations on log-arithms and indices.

10 Solution, 2014

9. The general equation for a circle is

x2 + y2 + 2gx+ 2fy + c = 0 (1.4)

where (−g,−f) is the center of the cir-cle and

√g2 + f 2 − c is the radius of

the circle.

Now, for the given circle equation

7(y2 + 10y) + 7x2 = 1

We expand this as follows

7y2 + 70y + 7x2 = 1

7x2 + 7y2 + 70y = 1

Divide through by 7, to have

x2 + y2 + 10y =1

7

which implies

x2 + y2 + 10y − 1

7= 0 (1.5)

On comparing equations (1.4) and(1.5), we have

2gx = 0, 2fy = 10y and , c = −1

7,

This implies that g = 0, f = 5 andc = −1

7Now,

centre = (−g,−f) = (−0,−5) = (0,−5)

radius =√g2 + f 2 − c =

√02 + 52 − (−1

7)

radius =√

0 + 25 + 17

=

√176

7

Hence, the center of the circle is(0,−5), C.

See the solutions to Q2, Q3 and Q4 of2013, Q18 of 2012 for more explana-tions on equations of a circle.

10. If α and β are the roots of a quadraticequation

ax2 + bx+ c = 0,

then

x2 − (α + β)x+ αβ = 0

is the quadratic equation in terms ofits roots α and β

x2 − (α + β)x+ αβ = 0 (1.6)

Now,cx2 + ax+ b = 0,

Divide through by c

cx2

c+ax

c+b

c= 0

x2 +a

cx+

b

c= 0 (1.7)

Comparing equations (1.6) and (1.7),

−(α + β) =a

c, and αβ =

b

c

Hence,

αβ =b

c, C.

See the solutions to Q9 of 2013 andQ9 of 2011 for more explanations onquadratic equations and its roots.

11. By the definition given, a⊗ b = 2a− 1.So

2⊗ 1 = 2(2)− 1, since a = 2

2⊗ 1 = 4− 1 = 3

Now,

3⊗ (2⊗ 1) = 3⊗ 3

3⊗ 3 = 2(3)− 1, since a = 3

3⊗ 3 = 6− 1 = 5

11 Solution, 2014

so that

3⊗ (2⊗ 1) = 3⊗ 3 = 5

Hence,

3⊗ (2⊗ 1) = 5, C.

See the solution to Q15 of 2011 formore explanation on binary opera-tions.

12. We note that a coin has two sides. Onone side is the head(H) and on theother side is the tail(T ).

When two coins are tossed, the tablebelow shows the outcome. From the

H TH HH HTT TH TT

table, observe that we have only one(HH) that is, two heads (HH) appeartogether, so

Probability(having at least two heads)

=Number of Events

Total number of outcomes

P(at least two heads)

=numbers of times HH appears

Total number of possible outcomes

From the table, we have only four pos-sible outcome and only one HH ap-pearing, so that

P(at least two heads) =1

4, C.

See solution to Q21, Q22,Q23 and Q24of 2008 for more on probability andcoins.

13.1

1 + 1x3

= 0

We simplify this as follows:

1÷(

1 +1

x3

)= 0

1÷(x3 + 1

x3

)= 0, (taking L.C.M)

1×(

x3

x3 + 1

)= 0

so,x3

x3 + 1= 0

On cross multiplying, we have

x3 = 0(x3 + 1)

x3 = 0

which implies that x = 3√

0 = 0

Hence,x = 0, A.

See also Q9 of 2007 for more on this.

14. We note that in variation, if P is di-rectly proportional to

√Q, it means

P ∝√Q

and P = k√Q where k is the constant

of proprotionality.

Next, we find k using the parametersgiven as follows

P = 20, when Q = 4

P = k√Q (1.8)

20 = k√

4

20 = k · 22k = 20

k =20

2= 10

12 Solution, 2014

Substitute, k = 10 in (1.8)

Now, P = 10√Q is the equation con-

necting P and Q. To find Q whenP = 100, we have

P = 10√Q,

100 = 10√Q

100

10=√Q; 10 =

√Q√

Q = 10

Take the square of both sides to findQ, (√

Q)2

= (10)2

Q = 102 = 100

Hence,

Q = 100, C.

See also the solution to Q9 of 2006 andQ11 of 2009 for more explanation onvariation.

15. The general equation of a straight linewith slope or gradientm and y− inter-cept c is

y = mx+ c (1.9)

The slope or gradient m is the sameas the tangent of the angle the straightline makes with the x−axis, that is,

m = tan θ,

where θ is the angle the straight linemakes with the x− axis;

Now, from the equation x −√

3y = 0,we make y the subject of formula as

follows:

x−√

3y = 0

−√

3y = −xDivide through by−

√3

y =x√3

y =1√3x (1.10)

On comparing equation (1.9) and(1.10), we have that

m =1√3, c = 0

Since m = tan θ, then

tan θ =1√3

θ = tan−11√3

θ = 30◦

This means the straight line makes anangle of 30◦ with the x− axis. For thepositive y−axis the straight line willmake an angle of (90◦ − 30◦) = 60◦

with the y−axis. C.

Kindly see the solution to Q19 of 2011for more on straight lines and theirslopes.

16. There are two types of integral – Def-inite and Indefinite. The definite in-tegrals have boundaries (which will

13 Solution, 2014

exclude the constant of integration)while the indefinite integrals have noboundary (which will include the con-stant of integration). Example of defi-

nite integral,∫ 2

1

x2dx. Example of in-

definite integral,∫ √

2x+ 1dx.

In this case,∫ a

−a15x2dx = 3430

To solve this definite integral, we willproceed as follows: Integrate the lefthand side of the equation[

15x2+1

2 + 1

]a−a

= 3430[15x3

3

]a−a

= 3430[5x3]a−a

= 3430

Now we substitute the lower and up-per bounds, but the upper bound (a)comes first, before the lower bound(−a) and separated by a negative sign

as follows:

5a3 − 5(−a)3 = 3430

5a3 − 5(−a3) = 3430

5a3 + 5a3 = 3430

10a3 = 3430

a3 =3430

10= 343

a3 = 343

a =3√

343

a = 7

Hence, a = 7 C.

17. We note that

d

dxlnu =

1

u· dudx,

where u = u(x) is a function of x. ddx

means to find the derivative (i.e. todifferentiate) with respect to x.

Also,

d

dxsinu = cosu · du

dx

where u = u(x) is a functions of x.

Now,

d

dx

(ln sin(3x)

)=

1

sin 3x· ddx

sin(3x)

=1

sin 3x· cos 3x · d(3x)

dx

=cos 3x

sin 3x· 3

= 3cos 3x

sin 3x

but

cot θ =1

tan θ,

tan θ =sin θ

cos θand cot θ =

cos θ

sin θ

14 Solution, 2014

so that

d

dx

(ln sin 3x

)= 3 cot 3x, A.

See the solutions to Q21 of 2013, Q15of 2012 and Q4 of 2011 for more ondifferentiation.

18. The equation of a straight line withgradient m and passing through thepoint (x1, y1) is

y − y1 = m(x− x1) (1.11)

Also m = tanθ, where θ is the anglethe straight line makes with the posi-tive x− axis.

Now,

θ = 45◦, (x1, y1) = (−2, 3)

m = tan θ = tan 45◦

m = 1.

Then using (1.11) where m = 1,(x1, y1) = (−2, 3), we have

(y − 3) = 1(x− (−2)

)y − 3 = x+ 2

y − 3− x− 2 = 0

y − x− 5 = 0

Hence, the equation of the line is

y − x− 5 = 0, A.

See also Q4 of 2010

19. For a Geometric Progression, (GP),T2T1

= T3T2

= r, the common ratio. Thegeneral nth term of a Geometric pro-gression is

Tn = arn−1

where a is the first term of the progres-sion.

The nth sum of the G.P is defined as;

Sn =a(rn − 1)

r − 1, when |r| > 1

or

Sn =a(1− rn)

r − 1, when |r| < 1

The sum to infinity of a G.P. is definedas:

S∞ =a

1− r.

Now, for the sequence,

1,−1, 1,−1, · · ·T1 = 1, T2 = −1, T3 = 1,

T2T1

=−1

1=T3T2

=1

−1= −1 = r.

So the sequence is a Geometric se-quence with common ratio r = −1and a = 1(the first term). The sum toinfinity then becomes,

S∞ =a

1− r=

1

1− (−1)

S∞ =1

1 + 1=

1

2

Hence, the sum to infinity is 12, D.

See also Q10 of 2012, Q23 of 2010 andQ12 of 2009 for more on GeometricProgression.

20. To differentiate 2 sin(2 − ax) with re-

15 Solution, 2014

spect to x, we mean to find

d

dx

[2− sin(2− ax)

]=d(2)

dx−d(

sin(2− ax))

dx

= 0− cos(2− ax) · d(2− ax)

dx= − cos(2− ax) · (0− a)

= −(−a) cos(2− ax)

= a cos(2− ax)

(Note: ddx

(constant) = 0, constants likenumbers and parameters.) Hence,

d

dx

[2−sin(2−ax)

]= a cos(2−ax), A.

Kindly see Q17 of 2014, Q21 of 2013,Q15 of 2012 and Q4 of 2011 for moreon differentiation.

21. (8√n

m32

)2(4−1m2

2n−2

)=

(8√n)2

(m32 )2·( 1

4m2

2 · 1n2

)=

(82(√n)2

m3

)·(m2

4÷ 2

n2

)

=

64(n

12 )2

m3

(m2

4× n2

2

)

=64n

m3· m

2n2

8

=8n3m2

m2

=8n3

m= 8n3m−1, B.

See similar solution to the same ques-tion in Q2 of 2012.

22. We note that |x| is a symbol enclosingx and it is defined as:

|x| ={x; when x ≥ 0−x; when x < 0

(1.12)

Now, if x ∈ R, that is, if x is a realnumber, and x + 11 < 0, x < −11which means that x < 0, (as −11 < 0).Since, x < 0, then

|x| = −x

and

|x|x

=−xx

= −1 (since x < 0 as −11 < 0)

Hence,|x|x

= −1

Note that (1.12) means that when x ≥0, |x| = x and when x < 0, |x| = −x.B.

Kindly see Q16 of 2012 for more onabsolute value.

2

Question, 2013

1. If the probability of success in anevent is

y

x. What is the probability of

failure?A.

x− yx

B.y − xx

C.x− yy

D.y − xy

2. What is the circumference of the circle

x2 + y2 =

(7

π

)2

?

A. 16 units B. 14 unitsC. 15 units D. 18 units

3. Find the diameter of the circle2x2 + 2y2 = 50A. −10 units B. 10 unitsC. 25 units D. −25 units

4. What is the coordinate of center of thecircle x2 + y2 + 2x− 4y = 10?A. (−1,−2) B. (1, 2) C. (−1, 2)D. (1,−2)

5. Simplify logx x4 + log4 4x

A. 4x B. 4x

C. 4 + x D. 4x log4x 4x

6. Solve the equation 3x+1 = 271−x

A. 12

B. −12

C. 34

D. −34

7. Given f(x) = 3 + x and g(x) = 3 − x,find g

(f(x)

)A. 6 B. x C. −x D. 0

8. What is the remainder whenx3 + 5x2 − 6x+ 1 is divided by x− 1A. −1 B. 2 C. −2 D. 1

9. If α, β are the roots of equation6 + 5x− x2 = 0, find αβ + α + βA. 11 B. −11 C. 1 D. −1

10. What is the value of y for which the

functiony − 1

y + 1is undefined?

A. −1 B. 1 C. 0 D. 2

11. Resolve1

x(1 + x)into partial frac-

tions.A.

1

x+

1

1 + xB.

1

1 + x− 1

x

C. −1

x− 1

1 + xD.

1

x− 1

1 + x

12. Solve the equation 5x2

= 25x+4

A. −4, 2 B. −4,−2 C. 4,−2D. 4, 2

13. Evaluate4∑

n=2

(2n + 1)

A. 28 B. 31 C. 29 D. 32

14. Find point of intersection of the lines3x− 2y = 5, 2x+ 5y = −7.A. x = 11

19, y = −31

19

B. x = −1119, y = 31

19

C. x = −1119, y = −31

19

D. x = 1119, y = 31

19

15. Solve 4x2 + 20x− 24 = 0.A. 1, 6 B. −1,−6 C. 6,−1D. −6, 1

16. What is the 15th term of the sequence−3, 2, 7, · · ·?A. 65 B. 66 C. 68 D. 67

16

17 Question, 2013

17. What is the distance between thepoints (−1, 5) and (−7,−3)?A. 9 B. 10 C. 11 D. 12

18. Evaluatelog√

27− log√

8

log 3− log 2A. 2

3B. −2

3C. 3

2D. −3

2

19. Integrate 4x3 +1

xwith respect to x

A. lnx+ x4 +K B. x−1 + x4 +KC. 12x2−x−2 +K D. 1

5x5 +x−2 +K

20. If X = {2, 3, 6, 7, 8} andY = {6, 7, 10, 3, 17}, find X ∩ YA. {} B. 3, 6, 7C. {2, 3, 6, 7, 8, 10, 17} D. {6, 3, 7}

21. Differentiate sin(2x − 5) with respectto xA. cos(2x− 5) B. − cos(2x− 5)C. 2 cos(2x− 5) D. −2 cos(2x− 5)

22. If δ, λ are the roots of equation x2−5x+7 = 0, find the value of δ2 + λ2

A. 25 B. −25 C. −11 D. 11

18 Solution, 2013

Solution, 2013

1. The probability of any event (say x),P (x) is between 0 and 1, that is, 0 ≤p(x) ≤ 1. P (x) = 1 when the eventis certain to occur and P (x) = 0 whenthe event will not occur at all. Proba-bilty means chance or likelihood of anevent occurring.

Probability of success+ Probability of failure = 1,

that is,

P(success) + P(failure) = 1

Since P(success) of the event is yx, then

y

x+ P(failure) = 1

P(failure) = 1− y

x

=x− yx

Hence,

P(failure) =x− yx

, A

2. The general equation of a circle withcenter at the origin (0, 0) and radius ris

(x− 0)2 + (y − 0)2 = r2

which implies

x2 + y2 = r2 (2.1)

Our case;

x2 + y2 =

(7

π

)2

(2.2)

Comparing (2.1) and (2.2) implies thatr = 7

πunits (take note r2 =

(7π

)2,which implies r = 7

π), that is, the ra-

dius, r = 7π

units.

But circumference of a curve = 2πr,where r is the radius of the circle andπ = 3.142.

C = 2πr = 2π ×(

7

π

)units

= 2× 7

= 14 units B.

3. Remember, the general equation of acircle with center at the origin (0, 0)and radius r is

x2 + y2 = r2 (2.3)

In this case,

2x2 + 2y2 = 50

Divide through by 2, to have

x2 + y2 = 25

x2 + y2 = 52 (2.4)

Comparing equation (2.3) and (2.4)gives

r2 = 52

which implies r = 5 units. So, our ra-dius, r = 5 units. But

diameter = 2× radius = 2r

= 2× 5 units= 10 units

Hence, the diameter = 10 units, B.

19 Solution, 2013

4. The most general equation of a circleis

x2 + y2 + 2gx+ 2fy + c = 0 (2.5)

where the circle’s center coordinateis (−g,−f) and the radius, r =√f 2 + g2 − c. Note that this is the ap-

proach when the center is not at theorigin.

Our case:

x2 + y2 + 2x− 4y = 10

implies

x2 + y2 + 2x− 4y − 10 = 0 (2.6)

We now proceed to find f, g and c.So comparing (2.5) and (2.6), we havethat

2x = 2gx, −4y = 2fy and −10 = c2 = 2g, −4 = 2f and c = −10g = 1, f = −2

We have g = 1, f = −2 and c = −10

so center = (−g,−f)

=(− 1,−(−2)

)= (−1, 2) C.

5. Laws of Logarithm:

i. loga ax = x loga a

ii. loga a = 1

iii. loga b = x implies b = ax

iv. loga x+ loga y = loga xy

v. loga x− loga y = loga

(xy

)In our case;

logx x4 + log4 4x

Applying laws (i) and (ii) gives

logx x4 = 4 logx x = 4× 1 = 4

log4 4x = x log4 4 = x× 1 = x

Hence,

logx x4 + log4 4x = 4 + x, C.

6.3x+1 = 271−x

But 3 = 31 and 27 = 33, so that wehave

31(x+1) = 33(1−x)

implies3x+1 = 33−3x

Equating the powers of 3 gives

x+ 1 = 3− 3x

Collecting like terms

x+ 3x = 3− 1

4x = 2

x =1

2, A.

7. We note that

g(f(x)

)≡ g ◦ f

is a composite function that is, it is afunction of a function and

g ◦ f 6= f ◦ g

In our case;

f(x) = 3 + x and g(x) = 3− xg ◦ f = g

(f(x)

)= 3− f(x) = 3− (3 + x)

So

g(f(x)

)= 3− (3 + x)

= 3− 3− x= −x

Hence,

g(f(x)

)= −x C.

20 Solution, 2013

8. We note that in the Remainder’s theo-rem, if x− a is a divisor of the polyno-mial P (x) and leaves a remainder, R,then put x − a = 0 implies x = a andP (a) = R, the remainder.

In our case;

P (x) = x3 + 5x2 − 6x+ 1

and the divisor is x−1. We put x−1 =0 which implies x = 1 and

Remainder,R = P (1) = P (x = 1)

R = P (x = 1) = P (1) = 13 + 5(1)2 − 6(1)

+1

= 1 + 5(1)− 6 + 1

= 1 + 5− 6 + 1

= 1

Hence, Remainder R = 1, D.

9. The general form of a quadratic equa-tion in terms of its roots α and β is

x2 − (α + β)x+ αβ = 0 (2.7)

that is,

x2−(sum of roots)x+products of roots = 0.

We note the use of negative sign afterthe first term in the equation (2.7) andNEVER a positive sign.

In our case;

6 + 5x− x2 = 0

We rearrange in decreasing powers ofx, that is,

−x2 + 5x− x2 = 0

We ensure the coefficient of x2 is posi-tive, so we multiply through by nega-tive, to have

x2 − 5x− 6 = 0 (2.8)

Comparing equation (2.7) and (2.8)gives

−5x = −(α + β)x and − 6 = αβ

Divide through by x, to have

−5 = −(α + β)

Divide through by negative, to have

5 = (α + β)

which implies

(α + β) = 5 and αβ = −6

Now

αβ + (α + β) = −6 + 5

= −1

Hence,

αβ + (α + β) = −1, D.

10. A rational function, P (x)Q(x)

, is undefinedor not define when the denominatorQ(x) = 0.

In our case;y − 1

y + 1

is undefined when y + 1 = 0 whichimplies that y = −1. Hence,

y − 1

y + 1

is undefined when y = −1, A.

21 Solution, 2013

11. Partial fractions are splits (two ormore) rational functions of a singlerational function. We note the rulewhich states that when the denomina-tor of a split rational function is a lin-ear function, say (x+ 1), it’ s numera-tor must be a constant sayA and whenthe denominator is a quadratic func-tion, say x2 + 1, it’s numerator mustbe a linear function (Ax+B).

In our case, 1x(1+x)

, since we have twolinear functions in the denominator xand 1 +x, we shall introduce two con-stants A and B such that

1

x(1 + x)=

A

x+

B

1 + x

=A(1 + x) +Bx

x(1 + x)

Multiply through by x(1 + x) gives

1 = A(1 + x) +Bx (2.9)

Now we find A and B. To find A, weeliminate B by letting it’s coefficientx = 0 in equation (2.9). So,

1 = A(1 + 0) +B(0)

1 = A(1) + 0

A = 1

To find B, we eliminate A by lettingit’s coefficient 1 +x = 0 which impliesthat we put x = −1 in equation (2.9).So

1 = A(1 + (−1)) +B(−1)

1 = A(1− 1)−B1 = A(0)−B

1 = 0−B1 = −BB = −1

Now, A = 1 and B = −1 replacethis in the splits rational functions, wehave

1

x(1 + x)=

1

x+

(−1)

1 + x

=1

x− 1

1 + x, D.

12. We note similar question from Q6

5x2

= 25x+4

Since 5 = 51, 25 = 52 implies

5x2

=(

52)x+4

5x2

= 52(x+4)

Equate the powers of 5 to have,

x2 = 2x+ 8

x2 − 2x− 8 = 0

We now solve this quadratic equation.

Since −8x2 = (−4x)(2x) and−2x = −4x+ 2x, then

x2 − 4x+ 2x− 8 = 0

x(x− 4) + 2(x− 4) = 0

(x− 4)(x+ 2) = 0

(x− 4) = 0 or (x+ 2) = 0

which implies x = 4 or x = −2

Hence,

x = 4,−2, C.

13. We note8∑

n=3

2n2 = 2(3)2 + 2(4)2 + 2(5)2 + 2(6)2

+2(7)2 + 2(8)2

= 2(9) + 2(16) + 2(25) + 2(36)

+2(49) + 2(64)

22 Solution, 2013

This means you add up 2n2 with astarting value of 3 and an end valueof 8 with an increment of 1.

In our case;

4∑n=2

2n + 1

We start adding up 2n + 1 from n = 2and end at n = 4 with an increment of1

4∑n=2

2n + 1 = (22 + 1) + (23 + 1) + (24 + 1)

= (4 + 1) + (8 + 1) + (16 + 1)

= 5 + 9 + 17

= 31, B.

14. The point of intersection of the twolines is the point where the two linesmeet and it is obtained by solving thetwo lines (now equations) simultane-ously

3x− 2y = 5 (i)

2x+ 5y = −7 (ii)

Eqn(i)× 5; 15x− 10y = 25 (iii)

+Eqn(ii)× 2; +4x+ 10y = −14 (iv)

19x = 11

x =11

19

To find y we substitute x = 1119

into oneof the equations, say (i)

3x− 2y = 5

3(11

19

)− 2y = 5

33

19− 2y = 5

33

19− 5 = 2y

2y = −62

19

y = −31

19

Hence,

x =11

19, y = −31

19, A.

15.4x2 + 20x− 24 = 0

Divide through by 4

x2 + 5x− 6 = 0

Since −6x2 = (6x)(−x) and 5x =(6x) + (−x), then

x2 + 6x− x− 6 = 0

x(x+ 6)− (x+ 6) = 0

(x+ 6)(x− 1) = 0

x+ 6 = 0 or x− 1 = 0

x = −6 or x = 1

Hence,

x = −6, 1, D.

16. A sequence is an Arithmetic sequencewhen T2− T1 = T3− T2 = · · · = d, thatis, when the sequence have a commondifference. The general nth term of anArithmetic sequence or progression is

Tn = a+ (n− 1)d,

23 Solution, 2013

while the nth sum is

Sn =n

2

(2a+ (n− 1)d

).

In this case,

−3, 2, 7, · · ·

T1 = first term = −3

T2 = 2, , T3 = 7

T2 − T1 = 2− (−3) = 2 + 3 = 5

T3 − T2 = 7− 2 = 5

The sequence is an A.P. with commondifference d = 5, a = −3 and n = 15

T15 = a+ (15− 1)d

= a+ 14d

= −3 + 14(5)

= −3 + 70

= 67

Hence

T15, the 15th term = 67, D.

17. The general formula for the distancebetween two points P (x1, y1) andQ(x2, y2) is

|PQ| =√

(x2 − x1)2 + (y2 − y1)2

In this case

(x1, y1) = (−1, 5) =⇒ x1 = −1, y1 = 5

and

(x2, y2) = (−7,−3) =⇒ x2 = −7, y2 = −3

The distance is =

√(− 7− (−1)

)2+ (−3− 5)2

=√

(−7 + 1)2 + (−8)2

=√

(−6)2 + (−8)2

=√

36 + 64

=√

100

= 10 units, B.

18. We note the use of laws of logarithmand indices in solution to Q5, (vi) and(vii.)

i. loga bx = x loga b

v. loga x− loga y = loga

(xy

)vi. Indices:

√a = a

12 ; (√a)n = a

n2 .

Also√

27 =√

33 = (33)12 = 3

32 .

vii.√

8 =√

23 = (23)12 = 2

32

log√

27− log√

8

log 3− log 2=

log 332 − log 2

32

log 3− log 2,

=

log

(332

232

)log(32

)=

log(32

)32

log(32

)Since

(xa

ya

)=

(x

y

)a∴ On using (i.), we obtain

=32

log(32

)log(32

)=

3

2, C.

19. The general formula to integrate anymonomials axn is:∫

axndx =axn+1

n+ 1+ C,

where C is the constant of integra-tion.

∫dx is the symbol of integration.

There are also some special integralssuch as ∫

1

xdx = lnx+K,

24 Solution, 2013

whereK is the constant of integration.Or more generally∫

1

u(x)dx =

lnududx

+K,

where u(x) is a linear function of x.

In this case;∫ (4x3 +

1

x

)dx =

∫4x3dx+

∫1

xdx

=4x3+1

3 + 1+ lnx+K

=4x4

4+ lnx+K

= x4 + lnx+K∫ (4x3 +

1

x

)dx = ln x+ x4 +K, A.

20. In set theory, the symbol ∩ called in-tersection represent the set of com-mon elements in two sets. In this case;X = {2, 3, 6, 7, 8}, Y = {6, 7, 10, 3, 17}then X ∩ Y represent the set of com-mon elements in both sets X and Y .

X ∩ Y = {2, 3, 6, 7, 8} ∩ {6, 7, 10, 3, 17}= {3, 6, 7}

∴ X ∩ Y = {3, 6, 7} or {6, 3, 7}, D

Note that the answer cannot be3, 6, 7 because this is not a set but asequence. You only have a set whenit is enclosed in braces, {}.

21. The general formula to differentiateany monomial axn with respect to x is

d(axn)

dx= anxn−1.

There are some special trigonometricderivative such as

d

dxsinu(x) = cosu(x) · du(x)

dx,

where u(x) means u is a function of x.

In our case;

u(x) = 2x− 5

du(x)

dx=

d

dx(2x− 5) =

d(2x)

dx− d(5)

dx= 2

Now,

d

dxsin(2x− 5) = cos(2x− 5)× d

dx(2x− 5)

= 2 cos(2x− 5), C

Note that the derivative of a constant(example 5) is zero. Also derivativeand differentiation means the samething.

22. We know that from expansion;

(α + β)2 = (α + β)(α + β)

= α(α + β) + β(α + β)

= α2 + αβ + βα + β2

= α2 + 2αβ + β2

= α2 + β2 + 2αβ

This implies that

α2 + β2 = (α + β)2 − 2αβ.

With the knowledge that α2 + β2 =(α+β)2−2αβ, we proceed as follows.We need the values of (α+ β) and αβ,recall that if α and β are the roots of aquadratic equation, then

x2 − (α + β)x+ αβ = 0 (2.10)

In this case,

x2 − 5x+ 7 = 0 (2.11)

On comparing (2.10) and (2.11) , wehave that

α + β = 5, and αβ = 7

25 Solution, 2013

Now,

α2 + β2 = (α + β)2 − 2αβ

= 52 − 2(7)

= 25− 14

= 11

Hence,

α2 + β2 = 11, D.

3

Question, 2012

1. Obtain the product of 11002 and 1012

A. 1111002 B. 1101002 C. 22205

D. 11447

2. Simplify

(8√n

m32

)(4−1m2

2n−2

)A. 128n3m−1 B. 8n3m−1 C. 8n4mD. 8n3m

3. Evaluate log8 128 + log3 9A. 19 B. 48 C. 13

3D. 6

4. Find the value of y if 12

log3 y = 2A. 9 B. 18 C. 9

2D. 81

The universal set U consists of allintegers. Subsets of U are defined as:

A = {Y : Y ≤ 3}B = {Y : −5 < Y < 12}C = {Y : −2 ≤ Y < 5}

Use the information above to answerquestion 5.

5. A ∩ (B ∪ C)′ isA. {y < −4} B. ∅ C. {y < 0}D. {−4 ≤ y ≤ 3}

6. Make k the subject of the formula

m =2nk

p+

k

2p

A. k =2mp

2n+ 1B. k =

2mp

4n+ 1

C. k =mp

2n+ 1D. k =

2n+ 1

2mp

7. Which of the following is a perfectsquare?A. x2 + 8

3x−1 = 0 B. x2 +9x+9 = 0

C. x2 − 83− 1 = 0 D. x2 − 9 = 0

8. The quadratic equation whose rootsare (x− 3) and (x+ 1

3) is

A. x2 + 83x−1 = 0 B. x2−2x−3 = 0

C. x2 − 83x− 1 = 0 D. x2 − 9 = 0

9. What is the highest possible value of8

1 + x2if 0 ≤ x ≤ 3?

A. 8 B. 4 C. 2 D. 16

10. The fifth term in the progression9, 27, 81, · · · isA. 243 B. 37 C. 729 D. 38

11. The interior angles of an hexagon are120, 100, 80, 150, x and 130. The valueof x isA. 170◦ B. 20◦ C. 120◦ D. 140◦

12. If the bearing of a town B from A is145◦, the bearing of A from B isA. 305◦ B. 325◦ C. 35◦ D. 145◦

The scores of the students in class testare shown in the table below:Use the information to answer ques-tion 13

Score 1 2 3 4 5 6 7 8No of student 0 1 3 5 3 4 2 0

13. The modal score isA. 5 B. 4 C. 6 D. 8

26

27 Question, 2012

14. A number is selected randomly fromthe set of integers 1 to 30 inclusive.The probability that the number isprime isA. 4

15B. 1

3C. 3

15D. 7

30

15. Differentiate cos ax with respect to x.A. a sin ax B. 1

asin ax

C. −a sin ax D. − 1a

sin ax

16. Obtain the value of x in |x− 9| = 16A. 25 B. −7 C. 25,−7D. −25, 7

17. Integrate√

2x+ 1

A. 13(2x+ 1)

32 + k B. 2

3(2x+ 1)

32 + k

C. −13(2x+1)

32 +k D. −2

3(2x+1)

32 +

k

18. Obtain the center of the circle 3y2 +3(x+ 5)2 = 17A. (0, 5) B. (−5, 0) C. (0,−5)D. (5, 0)

19. If f(x+ 1) =x2 + 1

x3. Find f(2)

A. 58

B. 2 C. 14

D. 1

20. Which of these numbers is an irra-tional number?A. sin 0◦ B. sin 30◦ C. sin 60◦

D. sin 90◦

21. Given∫ a

−a15x2dx = 3430, find the

value of the constant a.A. 8 B. 6 C. 7 D. 9

22. Which of these lines is at right anglewith the line x = −7?A. 2x+ y = −1 B. 2x− y = 1C. y = 0 D. x = 49

28 Solution, 2012

Solution, 2012

1. In Number theory, binary system(i.e base two) has digits 0, 1; Deci-mal system (i.e base ten) has digits0, 1, 2, 3, 4, 5, 6, 7, 8, 9; Base three sys-tems has digits 0, 1, 2; Base four sys-tem has digits 0, 1, 2, 3 etc.

In this case,

11002 × 1012

This multiplication is done normallybut with restriction on the digits (i.e0, 1 only), any number outside of 1 isreduced by subtracting 2 or multiplesof 2 from it, writing down the remain-der.

11002

× 1012

11000000

+11001111002 A.

This is similar to normal base 10 mul-tiplication where maximum digit is 9,but in base 2 the maximum is 1.

2. Note from the laws of indices

i. a−1 = 1a

ii. a−n = 1an

ii. a−1n = 1

n√a

iv. amn = ( n

√a)m

v. a−mn = 1(

n√a)m

vi.(an)m

= anm

vii.(an) 1m = a

nm

viii. (ab)m = ambm

ix.(ab

)m= am

bm

x. an

am= an−m

xi. an × am = an+m

Now,(8√n

m32

)2(4−1m2

2n−2

)=

(8√n)2(

m32 )2·(4−1m2

)(2n−2

)=

82(√n)2

m32×2·(14m2)(

2 · 1n2

)=

64n

m3·(m2

4÷ 2

n2

)=

64n

m3·(m2

4× n2

2

)=

64n

n3· m

2n2

8

=8n3

m

= 8n3 · 1

m= 8n3m−1. B.

3. Using the laws of logarithms in solu-tion Q5 and Q8 of 2013, we have,

log8 128 + log3 9 = X + Y.

We would not apply the law:loga x + loga y = loga xy directly be-cause the logarithm we are consider-ing here do not have the same base(base 8 and base 3).

So we let

log8 128 = X,

which implies 128 = 8x.

Since 128 = 27 and 8 = 23, we have27 = (23)x,

which implies 27 = 23x

By equating the powers of 2, we obtain,7 = 3x

x =7

3.

29 Solution, 2012

Also let log3 9 = y,

which implies 9 = 3y.

Since 9 = 32 and 3 = 31 we have32 = (31)y,

32 = 3y

y = 2

Therefore,

log8 128 + log3 9 =7

3+ 2

=13

2, C.

4. Using the laws in solution Q3,

1

2log3 y = 2

which implies log3 y1/2 = 2

y1/2 = 32

√y = 9

On squaring both sides(√y)2 = 92

then, we have y = 81, D.

5.

U = {set of integers}= {· · · ,−2,−1, 0, 1, 2, 3, · · · }

A = {y : y ≤ 3}= {· · · ,−1, 0, 1, 2, 3}

B = {y : −5 < y < 12}= {−4,−3,−2,−1, 0, 1, 2, · · · , 11}

C = {y : −2 ≤ y < 5}= {−2,−1, 0, 1, · · · , 4}.

Now, to obtain A ∩ (B ∪ C)′, we solveaccordingly

B ∪ C = {−4,−3, · · · , 11},

∪ is the union of sets B and C that is,the collection of all the members of Band C.

(B ∪ C)′ = {· · · ,−5, 12, 13, · · · },

(B ∪ C)′ means the members of theuniversal set U that are not containedin B ∪ C.

A ∩ (B ∪ C)′ = {· · · ,−1, 0, 1, 2, 3}∩ {· · · ,−5, 12, 13, · · · }

= {· · · ,−5}= {y : y < −4}

∩ is the intersection of sets, that is, thecommon members of the two seets.Hence,

A ∩ (B ∪ C)′ = {y : y < −4} A.

6.m =

2nk

p+

k

2p

We do every thing possible with theequation to make k stand alone;

m =2nk

p+

k

2p

m =4nk + k

2p.

Cross multiplying givesm(2p) = 4nk + k.

2mp = k(4n+ 1)

On dividing through by 4n+ 1, we obtain

k =2mp

4n+ 1, B.

7. Consider the general quadratic equa-tion

ax2 + bx+ c = 0.

Divide through by a gives

x2 +b

ax+

c

a= 0. a 6= 0

30 Solution, 2012

Let P = ba

and Q = ca, then

x2 + Px+Q = 0

This equation is a perfect squarewhenever (

P

2

)2

= Q,

that is, whenever the square of half ofthe coefficient of x (P in this case) isequal to the constant (Q in this case).

In our case,

For, x2 +8

3x− 1 = 0

P =8

3, Q = −1(

P

2

)2

=

(8

3÷ 2

)2

=

(8

6

)2

=

(4

3

)2

=16

96= −1 6= Q

For, x2 − 9 = 0

P = 0, Q = −9(P

2

)2

= 02 = 0 6= −9 6= Q

For, x2 + 9x+ 9 = 0

P = 9, Q = 9(P

2

)2

=

(9

2

)2

=81

46= 9 6= Q

For, x2 − 8

3x− 1 = 0

P = −8

3, Q = −1(

P

2

)2

=

(−4

3

)2

=16

96= −1 6= Q

None is a perfect square since(P

2

)2

6= Q, in all

8. Suppose x = α and x = β are the rootsof a quadratic equation then we ob-tain the quadratic equation as follow:

x− α = 0 and x− β = 0

then

(x− α)(x− β) = 0

which implies x2 − (α + β)x+ αβ = 0

which is the quadratic equation.

In our case;

(x− 3) = 0 and(x+

1

3

)= 0

has been given . So multiply bothgives

(x− 3)(x+

1

3

)= 0× 0 = 0

x2 +1

3x− 3x− 1 = 0

which implies that x2 − 8

3x− 1 = 0, C.

which is the required equation.

9. Note: A rational function P (x)Q(x)

willbe maximum or highest wheneverthe denominator Q(x) is minimum orlowest, that is, P (x)

Q(x)= maximum

when Q(x) is minimum.

In our case, 81+x2

will be highest when1 + x2 is lowest. But since 0 ≤ x ≤3, so we choose the smallest or lowestvalue of x (which is x = 0) in order to

31 Solution, 2012

make 1 +x2 lowest. So 1 +x2 is lowestwhen x = 0.

8

1 + x2=

8

1 + 02=

8

1= 8

Hence, the highest value of

8

1 + x2= 8, A.

10. A sequence or progression is a Geo-metric progression when

T2T1

=T3T2

= · · · = r

where T1 = first term, r =common ratio, T2 = second term.The general nth term for a geometricprogression is

Tn = arn−1

See more on this in Q19 of 2014.

In our case,

9, 27, 81, · · ·

a = 9, r =27

9=

81

27= 3, n = 5

Using Tn = arn−1,

T5 = 9(3)5−1 = 9(3)4 = 9(81) = 729

Hence, the fifth term T5 is 729, C.

11. Polygon: These are plane shapes witha specific number of sides usuallythree or more. The sum of the interiorangles in any polygon is

Sn = (n− 2)× 180◦

where n is the number of sides of thepolygon. Note that; Hexagon, n = 6

Polygon Number of sides, nTriangle 3Quadrilateral 4Pentagon 5Hexagon 6Heptagon 7Octagon 8Nonagon 9Decagon 10

sides. The sum of the interior anglesin an Hexagon is

S6 = (6− 2)× 180◦

= 4× 180◦

= 720◦

Now, the interiors are120◦, 100◦, 80◦, 150◦, x◦, 130◦. Thisimplies that

120◦ + 100◦ + 80◦ + 150◦ + x◦ + 130◦ = 720◦

450◦ + x◦ + 130◦ = 720◦

x◦ + 580◦ = 720◦

x◦ = 720◦ − 580◦

x◦ = 140◦.

Hence,

x = 140◦, D.

12. Note: The bearing of a point P fromanother point Q is the sum of all theangles from the north of Q in clock-wise direction, to the line connectingQ to P .

In our case,

From the fact that alternating anglesare equal α = 55◦. Now th bearing ofA from B is the sum of all the angles

32 Solution, 2012

from the north of B, in clockwise di-rection, to the line connecting B to A.

So the bearing is = 90◦ + 90◦ + 90◦ + α

= 270◦ + α

= 270◦ + 55◦

= 325◦

Hence, the bearing of A from B is325◦, B.

13. The modal score is the score withhighest frequency, that is, the scoreobtained by most of the students.From the table the scores are from1 to 8 and score 4 has the highestfrequency of 5 students obtaining it.Hence, modal score is 4, B.

14.

Probability of an event occurring

=The number of that eventTotal numbers considered

In our case, the event we are consider-ing are prime numbers and the num-bers of prime numbers from 1 to 30inclusive is the number of that eventwhich is 10 and the total numbers con-sidered are the numbers from 1 to 30inclusive which is 30. We note that 1is not a prime number.

The prime numbers from 1 to 30 in-clusive are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29

which on counting is 10 and the num-bers from 1 to 30, inclusive is 30 so that

Probability that the number is prime

=10

30=

1

3, B.

15. Note the following. This is the general

Functions (y) Derivative(dydx

)sinu(x) cosu(x) · du

dx

cosu(x) − sinu(x) · dudx

tanu(x) sec2 u(x) · dudx

secu(x) secu(x) tanu(x) · dudx

eu(x) eu(x) · dudx

way of differentiating trigonometricfunctions and exponential functions.

Now in our case;

Let y = cos ax

u(x) = ax,du(x)

dx=d(ax)

dx= a

dy

dx=

d

dxcos(ax) = − sin(ax) · d(ax)

dx= − sin(ax) · a= −a sin ax

Hence, the derivative of cos ax withrespect to x is −a sin ax, C.

16. Note: |p| means the absolute value ofp which implies you will neglect anynegative sign attached to p. Example|−4| = 4, |−5| = 5. Also,|| is a symbolwhich is defined for x as

|x| =

{x, if x ≥ 0

−x, if x < 0

In our case

|x− 9| ={x− 9,−(x− 9),

33 Solution, 2012

So we are going to have two answersfor |x− 9| = 16 which are

x− 9 = 16 or − (x− 9) = 16

x = 16 + 9 = 25 or − x+ 9 = 16

x = 25 or x = −7, C.

17. Please check solution Q19 of 2013 onhow to integrate monomials. Now tointegrate

√2x+ 1, we proceed as fol-

lows. We use substitution method,where we let u = 2x + 1. Please notethat you represent the inner functionwith u as done in this case. Then,

du

dx= 2 (differentiation)

du = 2dx

dx =du

2

Now we substitute 2x + 1 = u anddx = du

2in the integral (you can only

work with one variable x or u andnever both), so that we have∫ √

2x+ 1dx =

∫ √u·du

2=

1

2

∫ √udu

Constants with no variable (here u)can leave the integral. From indices√u = u1/2

1

2

∫u1/2du =

1

2· u

1/2+1

12

+ 1+ k

=1

2

u32

32

+ k

=

(1

2÷ 3

2

)· u

32 + k

=

(1

2× 2

3

)u32 + k

=1

3u32 + k

Now, we return back our variable x byreversing u = 2x+ 1. Hence,∫ √

2x+ 1dx =1

3(2x+1)3/2+k, B.

18. Please see solution to Q2,Q3,Q4 of2013 before proceeding. You recallthat the general expression for theequation of a circle is

x2 + y2 + 2gx+ 2fy + c = 0 (3.1)

where the center is (−g,−f) and ra-dius r =

√g2 + f 2 − c. In our case,

3y2 + 3(x+ 5)2 = 17

We open the brackets and follow thesteps below to ensure the equationlooks like or is similar to equation(3.1), the general equation.

3y2 + 3(x+ 5)(x+ 5) = 17

3y2 + 3(x2 + 5x+ 5x+ 25) = 17

3y2 + 3(x2 + 10x+ 25) = 17

Divide through by 3 gives

y2 + x2 + 10x+ 25 =17

3

which implies y2 + x2 + 10x+ 25− 17

3= 0

x2 + y2 + 10x+58

3= 0 (3.2)

On comparing (3.1) and (3.2) we have

2gx = 10x and 2fy = 0

g = 5 and f = 0

Hence, the center = (−g,−f) =(−5, 0), B

19. Please check solution Q7 of 2013 be-fore proceeding.

34 Solution, 2012

Here,

f(x+ 1) =x2 + 1

x3,

to find f(2) you will not substitutex = 2 directly into the function butyou will proceed as follows;

Since f(x+ 1) =x2 + 1

x3and not f(x) =

x2 + 1

x3

To find f(2) we setx+ 1 = 2

which implies x = 2− 1 = 1

Now we set x = 1 into x2+1x3

. This isbecause

f(2) = f(1 + 1) =(1)2 + 1

13= 2

Hence, f(2) = 2, B.

20. Irrational numbers are the numbersthat cannot be expressed in fraction orwhole numbers, e.g e,

√2, π,√

3, theirmultiples and submultiples. Now,consider the table; From the table we

θ 0 30◦ 45◦ 60◦ 90◦

sin θ 0 12

√22

√32

1

cos θ 1√32

√22

12

0

tan θ 0√33

1√

3 ∞

see that the only irrational numbersfor sine are sin 45◦ =

√22

and sin 60◦ =√32

which are sub–multiples of√

2

and√

3 respectively.√

2 and√

3 areknown to be irrational by the exam-ples given above. So the answer issin 60◦ =

√32

, C.

21. Please see solution Q19 of 2013 be-fore proceeding. There are two typesof integral – Definite and Indefinite.

The definite integrals have bound-aries ( which will exclude the constantof integration) while the indefiniteintegrals have no boundary (whichwill include the constant of integra-tion). Example of definite integral,∫ 2

1

x2dx. Example of indefinite inte-

gral,∫ √

2x+ 1dx.

In this case,∫ a

−a15x2dx = 3430

To solve this definite integral, we willproceed as follows: Integrate the lefthand side of the equation[

15x2+1

2 + 1

]a−a

= 3430[15x3

3

]a−a

= 3430[5x3]a−a

= 3430

Now we substitute the lower and up-per bounds, but the upper bound (a)comes first, before the lower bound(−a) and separated by a negative signas follows:

5a3 − 5(−a)3 = 3430

5a3 − 5(−a3) = 3430

5a3 + 5a3 = 3430

10a3 = 3430

a3 =3430

10= 343

a3 = 343

a =3√

343

a = 7

Hence, a = 7 C.

35 Solution, 2012

22. Consider the following lines,

The graph of x = −7 is,

The graph of x = 49 is,

The graph of y = 7 is,

We say two lines are at right angleif they are perpendicular with eachother.

Consider the graph of both x = −7and y = 0

The graph of y = 0 is directly alongthe x−axis and that of x = −7 is per-pendicular to it. Hence, y = 0 is atright angle with the line x = −7. C.

4

Question, 2011

1. If the universal set U =(1, 2, 3, 4, 5, 6, 7, 8, 9, 10),M = (1, 3, 5, 7, 9) and N =(2, 4, 6, 8, 10). Which one of thefollowing is equal to (M ∪N)?A. (M ∩N)′ B. M ′ ∪N ′C. M ′ ∩N ′ D. M ∩N

2. cos(180◦ − θ) is equivalent toA. cos(θ − 180◦) B. cos θC. − cos θ D. − cos(θ + 180◦)

3. Find the equation of the circle withcenter (−1, 3) and radius 4.A. x2 + y2 − 6x+ 2y = 6B. x2 + y2 + 2x− 6y = 16C. x2 + y2 − 6x+ 2y = 16D. x2 + y2 + 2x− 6y = 6

4. Finddy

dx, if y =

3

x.

A. −3

2x−

32 B. 3x−

32

C.3

2x−

32 D.

3

4x−

32

5. Integrate1

2x.

A. not defined B. 0C. 1

2lnx+ C D. 1

4x2 + C

6. A die is tossed twice. What is theprobability of obtaining a total of 6 ifboth numbers are odd?A. 1

12B. 1

18C. 5

36D. 1

6

7. If the mean of the numbers a, b, c, d, eis x. Find the mean of numbers a + k,

b+ 2k, c− k, d− 2k, e.A. x B. x+ k C. x− k D. 2x

8. Factorize a2 − b2 + (a+ b)2.A. 2a2 B. 2a(a− b)C. 2a(a+ b) D. 2b(b− a)

9. Let α and β be roots of quadraticequation x2 + 2x− 3 = 0, then αβ isA. −3 B. −2 C. 2 D. 6

10. Convert 6910 to a number in base two.A. 10011012 B. 10100012

C. 10001012 D. 1001012

11. The reciprocal to

34

14

+ 13

isA. 12

7B. 7

9C. −12

7D. 7

9

12. The speed of 30 kilometers perminute, expressed in centimeters persecond isA. 5 B. 50 C. 500 D. 5000

13. Evaluate x if, log4(x+ 3)(x− 3) = 2A. 3 or −3 B. 5 or −5C. 5 or −3 D. 3 or −5

14. Given that a = 12−√3, b = 1

2+√3. Find

the value of a2 + b2

A. 1437

B. 7 C. 14 + 23

D. 14

15. If the binary operation ∗ is defined asx ∗ y = z, find 2 ∗ (4 ∗ 5)A. 4 B. 5 C. −5 D. 2

36

37 Question, 2011

16. Find the value of√√√√6 +

√6 +

√6 +

√6 +√

6 + · · ·

A. −2 B. 2 C. 6 D. 3

17. Find the value of∫ π2

0

(2π + 2 cos 2x)dx.

A. π2 + 1 B. π2

C. π2 − 4 D. π2 + 3

18. The circle 2x2+2(y− 32)2 = 2 has center

and radius respectively asA. (0, 3

2) and 2 B. (0,−3

2) and 1

C. (32, 0) and 2 D. (0, 3

2) and 1

19. The line perpendicular to the straightline y + 3

2x− 1 = 0 has the gradient

A. −23

B. 32

C. 3 D. 23

20. Find x if 2x2

= 4(x+4).A. −2 or 4 B. −2 or 2C. −4 or 4 D. −4 or 2

21. Express in partial fraction

3x

x2 − 1=

A

x− 1+

B

x+ 1,

then A and B respectively areA. −3, 3 B. 2

3, 23

C. −32, −3

2D. 3

2, 32

22. A square has a perimeter of 40cm.What is its area in cm square?A. 80 B. 1600 C. 100 D. 160

38 Solution, 2011

Solution, 2011

1. Please see solution Q20 of 2013 andsolution Q5 of 2012 before proceed-ing. Note the following

i. ∅c = Uii. U c = ∅

iii. (A ∩B)c = Ac ∪Bc

iv. (A ∪B)c = Ac ∩Bc

(iii) and (iv.) are called De Morgan’slaw in set theory.

Now,

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}M = {1, 3, 5, 7, 9}N = {2, 4, 6, 8, 10}

M ∪N = {1, 3, 5, 7, 9} ∪ {2, 4, 6, 8, 10}= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}= U , the universal set

M ∩N = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8, 10}= {} or ∅

(M ∩N)′ = ∅′ = UFrom the De Morgan’s law above

M ′ ∪N ′ = (M ∩N)′ = U

Hence, (M ∪ N) = U = (M ∩ N)′ =M ′ ∪N ′. We have two answers A andB.

2. Consider the four quadrants intrigonometry;

We see that (180◦ − θ) corresponds tothe second quadrant where the cosineof angles is negative, this implies that

cos(180◦ − θ) = − cos θ, C.

3. Please see solution to Q2 of 2013 be-fore proceeding. The general equationof a circle with center (a, b) and radiusr is

(x− a)2 + (y − b)2 = r2.

In this case, center = (−1, 3) andradius, r = 4. This implies thata = −1, b = 3, r = 4. Substitutingthese values into the general equationabove gives ;(

x− (−1))2

+ (y − 3)2 = 42

(x+ 1)2 + (y − 3)3 = 16

(x+ 1)(x+ 1) + (y − 3)(y − 3) = 16

x2 + 2x+ 1 + y2 − 6y + 9 = 16

x2 + y2 + 2x− 6y + 10− 16 = 0

x2 + y2 + 2x− 6y − 6 = 0

x2 + y2 + 2x− 6y = 6, D.

4. Please see solution Q21 of 2013 andsolution Q2 of 2012 before proceed-ing. To differentiate y = 3√

x. Note that

from indices,

3√x

= 3(√x)−1

= 3(x12)−1

= 3x−12

This implies that y = 3√x

= 3x−12 .

39 Solution, 2011

From differentiation,

dy

dx=

d

dx

(3x−

12)

= 3 · (−1

2) · x−

12−1

= −3

2x−

32

Hence,

dy

dx= −3

2x−

32 , A.

5. Please see solution Q19 of 2013 beforeproceeding. Recall:∫

1

u(x)dx =

lnu(x)dudx

+ c

where u(x) is a linear function of xand c is the constant of integration.

In this case,∫1

2xdx =

1

2

∫1

xdx

=1

2(lnx+B)

=1

2lnx+

B

2

= ln x12 + C,

Hence,∫1

2xdx = lnx1/2 + C, C.

6. A die has 6 sides with number1, 2, 3, 4, 5 and 6 written on each side.So when the die is tossed twice, theoutcomes are as shown in the table be-low:

From the table, we obtain a total of6 from the outcomes 1, 5; 2, 4; 3, 3;4, 2

1 2 3 4 5 61 1,1 1,2 1,3 1,4 1,5 1,62 2,1 2,2 2,3 2,4 2,5 2,63 3,1 3,2 3,3 3,4 3,5 3,64 4,1 4,2 4,3 4,4 4,5 4,65 5,1 5,2 5,3 5,4 5,5 5,66 6,1 6,2 6,3 6,4 6,5 6,6

and 5, 1. But outcomes 1, 5; 3, 3; and5, 1 have both numbers to be odd. So,

Probability(total 6, numbers odd)

=Number of events

total possible outcomes

=3

36=

1

12, A.

7.

Mean =Sum of the numbers

number count

x =a+ b+ c+ d+ e

55x = a+ b+ c+ d+ e

The mean of numbers a + k, b + 2k,c− k, d− 2k, e becomes

Mean =(a+ k) + (b+ 2k) + (c− k) + (d− 2k) + e

5

Mean =a+ b+ c+ d+ e+ k + 2k − k − 2k

5

But since a + b + c + d + e = 5x, thisimplies

Mean =5x+ 3k − 3k

5

=5x

5= x

Hence, mean = x, A.

40 Solution, 2011

8.

a2 − b2 + (a+ b)2

= a2 − b2 + (a+ b)(a+ b)

= a2 − b2 + a(a+ b) + b(a+ b)

= a2 − b2 + a2 + ab+ ba+ b2

= a2 + a2 − b2 + b2 + 2ab

= 2a2 + 2ab

= 2a(a+ b), C.

9. Please see solution Q9 and Q22 of2013 before proceeding.

Recall: If α and β are the roots of thegeneral quadratic equation ax2+bx+cthen

x2 − (α + β)x+ αβ = 0 (4.1)

is the required equation.

In this case,

x2 + 2x− 3 = 0 (4.2)

On comparing (4.1) and (4.2) we havethat −(α + β) = 2x and αβ = −3.Hence, αβ = −3, A.

10. To convert numbers in base 10 tonumbers in base two, we proceed asfollows:

We divide 69 totally by 2 and also re-tain the remainder. What we need are

2 692 34 rm 12 17 rm 02 8 rm 12 4 rm 02 2 rm 02 1 rm 0

0 rm 1

the remainder arranged upwards re-sulting to

6910 = 10001012, C.

11. The reciprocal of a number or functionis also the same as the inverse of thatnumber or fraction. Example, the re-ciprocal of a is 1

aand the reciprocal of

ab

is ba.

In our case;34

14

+ 13

We first simplify the expression aboveas a single fraction

34

14

+ 13

=3

4÷(

1

4+

1

3

)=

3

4÷(

3 + 4

12

)=

3

4÷ 7

12

=3

4× 12

7

= 3× 3

7

=9

7

Since34

14

+ 13

=9

7

The reciprocal of34

14+13

is 79, D.

12. Note:

100 cm = 1 m and 60 seconds = 1 minutes1000m = 1km and 60 minutes = 1 hour

30 Kilometers per minutes =30km

1 minutes

41 Solution, 2011

Since 1 km = 1000m, then 30km = 30, 000m

and since 1m = 100cm, then 30, 000m = 3, 000, 000cm.

Also 1 mins=60 seconds, so that

30km/min =30km

1 minute

=3, 000, 000cm

60 secs= 50, 000cm/secs

13. Please see solution Q5 of 2013, law(iii)before proceeding.

Recall:loga b = c implies b = ac. In ourcase;

log4(x+ 3)(x− 3) = 2

(x+ 3)(x− 3) = 42

x2 + 3x− 3x− 9 = 16

x2 − 9 = 16

x2 = 25

x =√

25 = ±5

Hence, x = 5 or −5, B.

14.a =

1

2−√

3, b =

1

2 +√

3

We first rationalize a and b.

For a, the conjugate of 2−√

3 is 2+√

3,so that

a =1

2−√

3· (2 +

√3)

(2 +√

3)

=2 +√

3

(2−√

3)(2 +√

3)

=2 +√

3

4−√

9

=2 +√

3

4− 3

= 2 +√

3

∴ a = 2 +√

3.

For b, the conjugate of 2+√

3 is 2−√

3,so that

b =1

2 +√

3· (2−

√3)

(2−√

3)

=2−√

3

(2 +√

3)(2−√

3)

=2−√

3

4−√

9

=2−√

3

4− 3

= 2−√

3

∴ b = 2−√

3

Now;

a2 + b2 = (2 +√

3)2 + (2−√

3)2

= (2 +√

3)(2 +√

3) + (2−√

3)(2−√

3)

= 4 + 4√

3 +√

9 + 4− 4√

3 +√

9

= 4 + 4 + 4√

3− 4√

3 +√

9 +√

9

= 8 + 3 + 3

= 14

Hence,

a2 + b2 = 14, D.

15. Binary operations are like rules thatmust be followed in the same waythey are defined.

x ∗ y = 2.

This rule means that whenever youoperate any two numbers, the resultor outcome will always be 2.

Now, to find 2∗(4∗5), we take it one byone as follows: We get that accordingto the rule, 4∗5 = 2, so that 2∗(4∗5) =2 ∗ 2, (since 4 ∗ 5 = 2). Now 2 ∗ 2 = 2,(according to the rule).

Hence, 2 ∗ (4 ∗ 5) = 2, D.

42 Solution, 2011

16. To solve this we proceed as follows:

Let

x =

√√√√6 +

√6 +

√6 +

√6 +√

6 + ·

On taking the square of both sides, wehave

x2 =

√√√√

6 +

√6 +

√6 +

√6 +√

6 + · · ·

2

. This implies that

x2 = 6 +

√6 +

√6 +

√6 +√

6 + · · ·

Since the roots of 6 contin-ues till infinity, it means that√

6 +

√6 +

√6 +√

6 + · · · is thesame as x in the first equation andthus we write

x2 = 6 + x

x2 − x− 6 = 0

x2 − 3x+ 2x− 6 = 0

x(x− 3) + 2(x− 3) = 0

(x+ 2)(x− 3) = 0

x = −2 or x = 3

The two answers x = −2 or x = 3 areboth correct on testing with the ques-tion. Hence,

x =

√√√√6 +

√6 +

√6 +

√6 +√

6 + · · ·

= 3 or − 2,

A or D.

Function y Integration of y

sinu(x) −cosu(x)dudx

+ C

cosu(x)sinu(x)

dudx

+ C

sec2 u(x)tanu(x)

dudx

+ C

eu(x)eu(x)

dudx

+ C

where u(x) is a linear function of x.

17. Please see solution to Q7 of 2013 andsolution to Q21 of 2012 before pro-ceeding. Note the table above, whereC is the constant of integration.

In this case;∫ π2

0

(2 + 2 cos 2x)dx

=

∫ π2

0

2πdx+

∫ π2

0

2 cos 2xdx

=

∫ π2

0

2πx0dx+

∫ π2

0

2 cos 2xdx

=

[2πx0+1

0 + 1

]π2

0

+

[2 sin 2xd(2x)dx

]π2

0

=

[2πx1

1

]π2

0

+

[2 sin 2x

2

]π2

0

=

[2πx

]π2

0

+

[sin 2x

]π2

0

= 2π[π

2

]− 2π(0) + sin 2

[π2

]− sin 2(0)

= π2 − 0 + sin π − sin 0

= π2 − 0 + 0− 0

= π2, B.

18. Please see solution to Q2, Q3 and Q4of 2013, and Q18 of 2012 before pro-ceeding: Recall the general equation

43 Solution, 2011

of a circle is

x2 + y2 + 2gx+ 2fy + c = 0

where center = (−g,−f) and radius=√g2 + f 2 − c.

In our case;

2x2 + 2(y − 32)2 = 2

2x2 + 2(y − 32)(y − 3

2) = 2

2x2 + 2[y2 − 3y + 94] = 2

We divide through by 2 to havex2 + y2 − 3y + 9

4= 1

x2 + y2 − 3y + 94− 1 = 0

x2 + y2 − 3y + 54

= 0

We proceed to find g, f and C.

On comparing the last equation withthe general circle equation, there is noterm involving x in the last equation,and we have that

2gx = 0, 2fy = −3y, and c = 54

g = 0, 2f = −3, and c = 54

g = 0, f = −32, and c = 5

4

Now,

center = (−g,−f)

=(− 0,−(−3

2))

= (0, 32)

and

radius =√g2 + f 2 − c

=√

02 + (−32)2 − 5

4

=

√0 +

9

4− 5

4

=

√4

4=√

1

= 1

Hence, centre= (0, 32), radius = 1,

D.

19. Two lines y = m1x + c1 and y =m2x+c2 are said to be perpendicular ifthe product of their gradients m1 andm2 is −1, that is, m1 · m2 = −1 andm1 = m2 when the two lines are par-allel. Also note that the general equa-tion for a line is

y = mx+ c (4.3)

wherem is the slope or gradient of theline and c is the intercept of the line onthe y–axis.

In our case; Line 1:

y + 32x− 1 = 0.

We re–write this line equation in thegeneral form to obtain its gradient,that is,

y = −32x+ 1

On comparing this equation with(4.3), we see that the gradient m =−3

2and y–intercept c = 1 for line 1.

Let the slope or gradient of line 2 bem2, since line 2 is perpendicular to thestraight line 1, we have that

m ·m2 = −1,

m2 = − 1

m= −1÷m,

but m = −3

2, we have,

m2 = −1÷−32

m2 = −1×−23,

m2 = 23.

Hence, the gradient of the perpendic-ular line is 2

3. D.

20. Please see solution to Q6 and Q12 of

44 Solution, 2011

2013 before proceeding. In this case;

2x2

= 4(x+4).

Since 2 = 21 and 4 = 22, we have,

(21)x2

= (22)x+4

2x2

= 22(x+4),

2x2

= 22x+8.

On equating the powers of 2, we get,x2 = 2x+ 8,

x2 − 2x− 8 = 0.

On solving this quadratic equation,x2 − 4x+ 2x− 8 = 0,

x(x− 4) + 2(x− 4) = 0,

(x− 4)(x+ 2) = 0,

x = 4, or x = −2.

Hence, x = −2 or x = 4, A.

21. Please see solution to Q11 of 2013 be-fore proceeding, the concept of partialfraction has been explicitly explainedthere.

In this case,

3x

x2 − 1=

A

x− 1+

B

x+ 1

=A(x+ 1) +B(x− 1)

(x− 1)(x+ 1)

Thus

3x

x2 − 1=A(x+ 1) +B(x− 1)

(x− 1)(x+ 1).

On multiplying through by x2 − 1, wehave

3x = A(x+ 1) +B(x− 1) (*)

We eliminate A by setting x + 1 = 0,

that is, x = −1.

We then substitute x = −1 into (*) to findB,3(−1) = A(−1 + 1) +B(−1− 1),

−3 = A(0) +B(−2),

−3 = 0− 2B,

−2B = −3,

B =3

2.

Similarly, we eliminate B by settingx− 1 = 0 which implies x = 1.

We then substitute x = 1 into (*) to find A,

3(1) = A(1 + 1) +B(1− 1),

3 = A(2) +B(0),

3 = 2A,

A =3

2.

Hence, A = 32, B = 3

2, D.

22. A square is a four sided plane shapewith equal length. That is, a squarehas all its four sides equal.

Area of a square = l2

where l is the length of one of the sidesand

Perimeter of a square = 4l.

Now,

Perimeter = 40cm,

4l = 40cm

l =40

4= 10cm.

The length of the square is 10cm. Areaof the square = l2 = (10cm)2 =100cm2. Hence, the area of the squarein cm is 100, C.

5

Question, 2010

1. If(

3

2

)x(2

3

)y=

32

27,

find the value of 3y − 2x.A. −1 B. 7 C. 1 D. −7

2. The integral values of y which satisfythe inequality −1 < 5− 2y ≤ 7 areA. −1, 0, 1, 2 B. 0, 1, 2, 3C. −1, 0, 1, 2, 3 D. −1, 0, 2, 3

3. If x2 − 5x+ 6 = (x− a)2 + b, the valueof b isA. −1

4B. 5

2C. 2 D. 3

4. Find the equation of the line perpen-dicular to the line y = 2x + 1 passingthrough a point (3, 1).A. y = 1

2x+ 5

2B. y = −1

2x+ 5

2

C. y = x+ 5 D. 2y = x+ 5

5. What is the distance between point(1, 2) and (4, 5) on a plane?A. 3√

2 B. 2√

3 C. 3 D. 9

6. Integrate ∫2 tan(2x+ π)dx.

A. 2 cot(2x+ π) + k

B. log[

cos(2x+ π)]

+ k

C. − log[

cos(2x+ π)]

+ k

D. 4 cot(2x+ π) + k

7. Find the value of x for which

5 + 2x− 3x2 = 0.

A. −2 and 65

B. −1 and 53

C. −2 and −1 D. 65

and 53

8. A businessman invested a total ofN200, 000 in two companies whichpay dividends of 5% and 7% re-spectively. If he received a total ofN11, 600, how much did he invest at7%?A. N140, 000 B. N160, 000C. N80, 000 D. N100, 000

9. If a√

5 + b√

2 is the square root of95 − 30

√10, the values of a and b are

respectivelyA. 5, 2 B. 2,−5C. −5, 3 D. 3,−5

10. The scores of 16 studentsin a mathematics test are65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70,60, 65, 65, 70. What is the sum of themedian and modal scores?A. 125 B. 130 C. 140 D. 150

11. Find x if x2 − 2x− 15 = 0A. 3,−5 B. −3, 5C. 1, 15 D. −2, 15

12. A father leaves a legacy of N45 mil-lion for his children, Peter, David andPaul, to be shared in the ratio 7 : 5 : 3.What amount in million naira wouldeach receive respectively?A. N14, N7, N3 B. N15, N5, N3C. N21, N15, N9 D. N20, N16, N10

13. As θ tends to zero, what does cos θtends to ?A. sin θ B. 0 C. 1

2D. 1

14. The expression 2 cos2 θ + 2 sin2 θ hasthe numerical value ofA. 1 B. 2 C. 4 D. 0

45

46 Question, 2010

15. If tanx = sinxcosx

, find tan(90◦ + x), foracute value of x.A. − cotx B. − tanxC. cotx D. tanx

16. Evaluate the length of perpendicularfrom A to BC.A.

√5212cm B. 12√

52cm

C. 24√52cm2 D. 24√

52cm

17. The indefinite integral of xex, for anyreal constant C isA. c B. x+ ex + cC. x2 + ex + c D. ex(x− 1) + c

18. Find the area under the curve y(x) =sinx between x = 10 and x = π.A. 2 B. 1 C. −2 D. π

19. Let the letters P,Q,R and S denoteparallelogram, quadrilateral, rectan-gle and square respectively. Usingsubset notation, which of these inclu-sions is correct?A. Q ⊂ R ⊂ P ⊂ SB. R ⊂ Q ⊂ P ⊂ SC. S ⊂ P ⊂ R ⊂ QD. S ⊂ R ⊂ P ⊂ Q

20. In a convex polygon with n sides, thesum of interior angles isA. (n− 2)π B. 2(n− 1)πC. 4(n− 1)π D. (2n+ 4)π

21. Ifx

y=z

w= c, find the value of

3x2 − xz + z2

3y2 − yw + w2

in terms of cA. 3c2 B. 17c2

4C. 2c− c2 D. c2

22. Express5y − 12

(y − 2)(y − 3)in partial frac-

tions.

A.2

y − 2− 3

y − 3

B.2

y − 2+

3

y − 3

C.2

y − 3− 3

y − 2

D.25

y − 3− 4

y − 2

23. The second term of an infinite geo-metric series is −1

2and the third term

is 14. Find the sum of the series

A. 2 B. 1 C. 32

D. 23

24. In the figure below, AB and AD aretangents to the circle. If ∠BCD = 55◦

and ∠BDC = 48◦, find ∠BAD

A. 55◦ B. 70◦ C. 77◦ D. 84◦

25. Find the area of the triangle.A. 24cm B. 24cm2

C. 12cm2 D. 12cm

47 Solution, 2010

Solution, 2010

1. Please see solution to Q20 of 2011 be-fore proceeding. Here,(

3

4

)x(2

3

)y=

32

27

Since 4 = 22, 32 = 25 and 27 = 33, wehave (

3

22

)x(2

3

)y=

25

33

3x

22x· 2y

3y=

25

33

3x

3y· 2y

22x=

25

33

3x−y · 2y−2x =1

33· 25

3x−y · 2y−2x = 3−3 · 25

By equating the powers of 3 and 2, wehave

x− y = −3 (i)

y − 2x = 5 (ii)

This is a simultaneous equation andnow solved accordingly.

From (i)x = −3 + y (iii)

So we substitute x = −3 + y in equa-tion (ii) to have

y − 2(−3 + y) = 5

y + 6− 2y = 5

y − 2y = 5− 6

−y = −1

y = 1

From (iii),

x = −3 + y

x = −3 + 1 = −2

x = −2, y = 1

Now,

3y − 2x = 3(1)− 2(−2)

= 3 + 4

= 7, B.

2. Note: If you divide an inequalitiesthrough by a negative, the inequalitysign will change and if you multiplyan inequality through by a negative,the inequality sign will also change.

Now,−1 < 5− 2y ≤ 7

We find y

Subtract 5 from each side of the inequality−1− 5 < 5− 2y − 5 ≤ 7− 5

−6 < 5− 5− 2y ≤ 2

−6 < −2y ≤ 2

Divide the inequality through by − 2 to have−6

−2>−2y

−2≥ 2

−2

The inequality signs will changewhen divided by a negative number,so we have

3 > y ≥ −1

On re–arranging the inequality, wehave

−1 ≤ y < 3

which is

−1, 0, 1, 2, A.

3.x2 − 5x+ 6 = (x− a)2 + b

48 Solution, 2010

We expand the right hand side

x2 − 5x+ 6 = (x− a)(x− a) + b

= x2 − 2ax+ a2 + b

= x2 − (2a)x+ (a2 + b)

On comparing the two sides, we ob-serve that

−5 = −2a

6 = a2 + b

From

−5 = −2a

a =−5

−2=

5

2

From

a2 + b = 6(5

2

)2

+ b = 6

25

4+ b = 6

b = 6− 25

4

b = −1

4, A.

4. Please see solution to Q19 of 2011 be-fore proceeding.

The equation of a straight line whengiven the gradient or slope m and apoint (x1, y1) on the line is

y − y1 = m(x− x1).

In this case, two lines are perpendicu-lar if the product of their gradients orslopes is −1, that is, m1 ·m2 = −1. Letm1 be the gradient of y = 2x + 1. On

comparing with y = mx + c, we ob-serve that m1 = 2. Let m2 be the gra-dient of the perpendicular line then

m1 ·m2 = −1

2 ·m2 = −1

m2 = −1

2

Now, we have gradient m2 = −12

ofthe perpendicular line and the point(3, 1) on the line, then the equation ofthe perpendicular line is:

(x1, y1) = (3, 1)

x1 = 3, y1 = 1

m2 = −1

2y − y1 = m2(x− x1)y − 1 = −1

2(x− 3)

y − 1 +−x2

+3

2

y = −x2

+3

2+ 1

y = −x2

+5

2, B.


The distance between two pointsA(x1, y1) and B(x2, y2) is

|AB| =√

(x2 − x1)2 + (y2 − y1)2

In our case;

(x1, y1) = (1, 2), implies that x1 = 1, y1 = 2

(x2, y2) = (4, 5) implies that x2 = 4, y2 = 5

49 Solution, 2010

Distance =√

(4− 1)2 + (5− 2)2

=√

32 + 32

=√

9 + 9

=√

18

=√

2× 9

=√

2×√

9

= 3√

2, A.


Here, generally;∫f ′(x)

f(x)dx = loge f(x)+C, or ln f(x)+C

where C is the constant of integrationand f ′(x) is the derivative of f(x).Thismeans that in integration of rationalfunctions, if the derivative of the de-nominator gives the numerator, thenthe result of the integration is the log-arithm of the denominator plus C, theconstant of the integration. Also notethat in trigonometry, tan θ = sin θ

cos θ

In this case,∫2 tan(2x+π)dx = 2

∫tan(2x+π)dx

There is no direct integration for tan-gent. So we integrate as follows usingthe tools above;∫

2 tan(2x+π)dx =

∫2

sin(2x+ π)

cos(2x+ π)dx.

Since ddx

cos(2x + π) = −2 sin(2x + π),

we have that∫2 tan(2x+ π)dx

= −∫−2 sin(2x+ π)

cos(2x+ π)dx

= −(

log[

cos(2x+ π)]

+ C

)= − log

[cos(2x+ π)

]− C

= − log[

cos(2x+ π)]

+K, C.

7.

5 + 2x− 3x2 = 0

−3x2 + 2x+ 5 = 0

−3x2 × 5 = −15x2

−15x2 :: (5x)(−3x)

2x :: (5x) + (−3x)

−3x2 + 5x− 3x+ 5 = 0

−x(3x− 5)− 1(3x− 5) = 0

(3x− 5)(−x− 1) = 0

3x− 5 = 0 or − x− 1 = 0

3x = 5 or − x = 1

x =5

3or x = −1, B.

8. Let X be the amount of money he in-vested at 5% and let Y be the amountof money he invested at 7%. Then

X + Y = N200, 000 (i)

5

100X +

7

100Y = N11, 600 (ii)

Note that the total amount of moneyinvested is X + Y and amount re-ceived from dividends of 5% is 5

100X ,

50 Solution, 2010

amount received from dividends of7% is 7

100Y and the total dividends re-

ceived is 5100X + 7

100Y = N11, 600.

Now multiply (ii) by 100 to have

5X + 7Y = N1, 160, 000 (iii)

We solve this equation simultane-ously with (i) to obtain

(i)× 5 : 5X + 5Y = N1, 000, 000−(iii) : −5X − 7Y = −N1, 160, 000

− 2Y = −N160, 000

Y =−N160, 000

−2Y = N80, 000

Hence, the amount he invested at 7%is N80, 000, C.

9. If the square root of 95− 30√

10 isa√

5 + b√

2, then√(95− 30

√10) = a

√5 + b

√2

Squaring both sides gives,(√(95− 30

√10)

)2

=(a√

5 + b√

2)2

95− 30√

10 = (a√

5 + b√

2)(a√

5 + b√

2)

= a2√

25 + ab√

10 + ba√

10

+b2√

4

= 5a2 + 2√

10ab+ 2b2

= 5a2 + 2b2 + 2√

10ab

On comparing both sides, we havethat

95 = 5a2 + 2b2 (i)

−30√

10 = 2√

10ab (ii)

Equation (ii) implies ab = −30√10

2√10

ab = −15 (iii)

Substitute a = −15b

into equation (i)gives

95 = 5

(−15

b

)2

+ 2b2

= 5

(225

b2

)+ 2b2

let b2 = y, we have

95 = 5

(225

y

)+ 2y

Multiply through by y gives

95y = 5(225) + 2y2

2y2 − 95y + 1125 = 0

We solve this quadratic equation2y2 − 45y − 50y + 1125 = 0

y(2y − 45)− 25(2y − 45) = 0

(y − 25)(2y − 45) = 0

y − 25 = 0 or 2y − 45 = 0

y = 25 or y =45

2We choose y = 25, but b2 = y

This implies b2 = 25, which givesb = ±5, that is, b = 5 or b = −5. Alsoa = −15

bfrom equation (iii). This im-

pliesa = − 15

±5 = ∓3, that is, a = −3 ora = 3. Hence,

a = 3 b = −5 or a = −3, b = 5, D.

10. The median of a set of data is the dataappearing in the middle after the setof data has been arranged in ascend-ing order of magnitude. The modal

51 Solution, 2010

score is the score with the highest fre-quency , that is, the score with thehighest number of students obtainingit.

In our case; 65, 65, 55, 60, 60, 65, 60, 70,75, 70, 65, 70, 60, 60, 65, 65, 70.We first arrange the data or scoresin ascending order as follows55, 60, 60, 60, 60, 65, 65, 65, 65, 65, 65, 70,70, 70, 70, 75. Now, the median is theaverage of the two middle scores65, 65.

Median =65 + 65

2= 65

The modal score is also 65. It occurred6 times in the data.

The sum of the median and modalscores is = 65 + 65 = 130, B.

11.

x2 − 2x− 15 = 0

x2 − 5x+ 3x− 15 = 0

x(x− 5) + 3(x− 5) = 0

(x+ 3)(x− 5) = 0

x+ 3 = 0 or x− 5 = 0

x = −3 or x = 5, B.

12.

Total legacy = N45 million.Peter : David : Paul = 7 : 5 : 3

Total ratio = 7 + 5 + 3 = 15

Peter’s amount =7

15×N45 million

= N21 million

David’s amount =5

15×N45 million

= N15 million

Paul’s amount =3

15×N45 million

= N9 million

Hence, in million naira, they received,N21, N15, N9 respectively, C.

13. Note

As θ tends to zero, sin θ tends to zeroAs θ tends to zero, cos θ tends to one

As θ tends to zero, tan θ tends to zero, D.

14. In trigonometry,

sin2 θ + cos2 θ = 1

tan2 θ + 1 = sec2 θ

1 + cot2 θ = cosec2θ

where sec θ = 1cos θ

, cosec θ = 1sin θ

andcot θ = 1

tan θ

Now;

2 cos2 θ + 2 sin2 θ = 2[cos2 θ + sin2 θ]

= 2 · 1= 2, B.

15. Since tanx = sinxcosx

, then

tan(90◦ + x) =sin(90◦ + x)

cos(90◦ + x)

We note from identities that

sin(A+B) = sinA cosB + cosA sinB

cos(A+B) = cosA cosB − sinA sinB

Now,

sin(90◦ + x) = sin 90◦ cosx+ cos 90◦ sinx

= 1 · cosx+ 0 · sinx= cos x+ 0

= cos x

Also

cos(90◦ + x) = cos 90◦ cosx− sin 90◦ sinx

= 0 · cosx− 1 · sinx= 0− sinx

= − sinx

52 Solution, 2010

(Hint; sin 90◦ = 1, cos 90◦ = 0) Wehave

sin(90◦ + x) = cos x

cos(90◦ + x) = − sinx

tan(90◦ + x) =sin(90◦ + x)

cos(90◦ + x

= −cosx

sinx

= − 1

tanx= − cotx

(Hint; tanx = sinxcosx

, cotx = cosxsinx

)

Hence, tan(90◦ + x) = − cotx, A.

16. Note: The length of perpendicularfrom a point A(x1, y1) to a line ax +by + c = 0 is

ay1 + bx1 + c√a2 + b2

In our case, the line BC and point Aare missing in the question.

17. Please see solution to Q17 of 2011 be-fore proceeding.In differentiation, to find the deriva-tive of products of functions e.g y =u(x)v(x), we use the product rule dy

dx=

u dvdx

+ v dudx

. But in integration, to findthe integral of products of functionse.g y =

∫u(x)dv(x), we use the inte-

gration by parts method which is∫udv = uv −

∫vdu (*)

In our case,

To integrate ∫xexdx,

we set u = x and dv = exdx.du = 1dx and v =

∫dv =

∫exdx = ex.

Now, using (*),∫xexdx = xex −

∫ex1dx

= xex −∫exdx

= xex − ex + C

= ex(x− 1) + C, D.


We note the area under any curve y =f(x) from x = a to x = b is

Area =

∫ b

a

f(x)dx

In our case,

y(x) = sin x, x = 0, x = π

Area =

∫ π

0

sinxdx

=[− cosx

]π0

= (− cosπ)− (− cos 0◦)

= − cosπ + cos 0◦

= − cos 180◦ + cos 0◦

Since we are dealing with degrees, weuse π = 180◦. So,

Area = −(−1) + 1

= 1 + 1

= 2 units square, A.

(Note; cos 180◦ = −1, cos 0◦ = 1).

19. In set theory, the symbol ⊂, meansit is contained in. Example: A ⊂B means A is contained in B. Note

53 Solution, 2010

that quadrilateral is the general namefor all four sided plane shapes. Wecan obtain at least one square froma rectangle which implies S ⊂ R.We can also obtain at least a rectan-gle from a parallelogram, that is, R ⊂P . Since Quadrilateral is the generalname given to all four sided planeshapes, then S ⊂ R ⊂ P ⊂ Q.

Hence,

S ⊂ R ⊂ P ⊂ Q, D.


A convex polygon is any polygonwhose interior angles are less than180◦. The sum of these interior anglesis

S = (n− 2)× 180◦

But since 180◦ = π, we can write

S = (n− 2)× πwhere n is the number of sides of thepolygon, A.

21. If xy

= zw

= c, thenx

y= c and

z

w= c

which implies

x = yc and z = wc.

We now substitute x = yc and z = wcinto 3x2−xz+z2

3y2−yw+w2 , which gives

3x2 − xz + z2

3y2 − yw + w2= 3(yc)2−(yc)(wc)+(wc)2

3y2−yw+w2

=3y2c2 − ywc2 + w2c2

3y2 − yw + w2

=c2[3y2 − yw + w2]

3y2 − yw + w2

= c2 · 1= c2, D.

22. Please see solution to Q11 of 2013 andQ21 of 2011 before proceeding.

Here, let

5y − 12

(y − 2)(y − 3)=

A

y − 2+

B

y − 3,

5y − 12

(y − 2)(y − 3)=A(y − 3) +B(y − 2)

(y − 2)(y − 3)

whereA andB are constants. This im-plies,

5y − 12 = A(y − 3) +B(y − 2) (*)

To find B, we eliminate A, by settingy − 3 = 0 which implies y = 3. Wesubstitute y = 3 into (*). This gives

5(3)− 12 = A(3− 3) +B(3− 2)

15− 12 = A(0) +B(1)

3 = 0 +B

B = 3

To find A, we eliminate B, by settingy−2 = 0 which implies that y = 2. Wesubstitute y = 2 into equation (*). Thisgives

5(2)− 12 = A(2− 3) +B(2− 2)

10− 12 = A(−1) +B(0)

−2 = −AA = 2

Hence,

5y − 12

(y − 2)(y − 3)=

2

y − 2+

3

y − 3, B.


Second term, T2 = −1

2

Third term, T3 =1

4.

54 Solution, 2010

The general nth term for a GeometricProgression(GP) is Tn = arn−1 wherer is the common ratio.

T2 = ar2−1 = ar1 = −1

2(i)

T3 = ar3−1 = ar2 =1

4(ii)

Divide equation (ii) by (i) we have,

ar2

ar=

1

4÷−1

2

r =1

4×−2

1= −1

2

∴ r = −1

2

So, since the series is an infinite series,the sum to infinity is

S∞ =a

1− rwhere a is the first term.

To find a, we substitute r = −12

into (i)

a

(−1

2

)= −1

2, =⇒ a = 1

S∞ =1

1−(−1

2

)=

1

1 + 12

=132

=2

3

Hence, the sum of the infinite series is23, D.

24. Considering4BCD,

∠BCD + ∠BDC + ∠DBC = 180◦

55◦ + 48◦ + ∠DBC = 180◦

∠DBC = 180◦ − 103◦ = 77◦

We note the following facts in circlegeometry:

i. ∠CDY = ∠DBC (when tangentAD is extended to Y )

ii. ∠CBX = ∠BDC(when tangentAB is extended to X)

So,

using fact (i), ∠CDY = ∠DBC = 77◦,

Using fact (ii), ∠CBX = ∠BDC = 48◦

Now considering straight line ADY

∠ADB + ∠BDC + ∠CDY = 180◦

∠ADB + 48◦ + 77◦ = 180◦

∠ADB = 180◦ − 125◦

∠ADB = 55◦

Also, considering, straight line ABX ,

∠ABD + ∠DBC + ∠CBX = 180◦

∠ABD + 77◦ + 48◦ = 180◦

∠ABD = 180◦ − 125◦ = 55◦

Now, considering triangle4ADB,

∠BAD + ∠ADB + ∠ABD = 180◦

∠BAD + 55◦ + 55◦ = 180◦

∠BAD + 110◦ = 180◦

∠BAD = 180◦ − 110◦

∠BAD = 70◦,

Hence, B.

25. Insufficient question (No diagram)

6

Question, 2009

1. Factorize 6x2 − 14x− 12A. 2(x+3)(3x−2) B. 6(x−2)(x+1)C. 2(x−3)(3x+2) D. 6(x+2)(x−1)E. (3x+ 4)(2x+ 3)

2. What is the product of 275∗ (3)−3 and

(15)−1

A. 5 B. 3 C. 2 D. 1 E. 125

3. If the lengths of the sides of a right-angled triangle are (3x+ 1)cm,(3x− 1)cm and xcm. What is x?A. 2 B. 6 C. 18 D. 12 E. 0

4. Evaluatexy2 − x2yx2 − xy

whenx = 2 and y = 3

. A. −3 B. 45

C. 35

D. 3 E. 4

5. Two fair dice are rolled. What isthe probability that both show up thesame number of point?A. 1

10B. 1

4C. 1

2D. 1

3E. 1

5

6. In 1984, Tolu was 24 years old and hisfather 45 years. In what year was Toluexactly half his father’s age?A. 1982 B. 1981C. 1983 D. 1979 E. 1978

7. Find the probability that a number se-lected at random from 40 to 50 is aprime.A. 2

11B. 5

11C. 5

10D. 4

10E. 1

12

8. A man kept 6 black, 5 brown and 7purple shirts in a drawer. What isthe probability of him picking a pur-ple shirt with his eyes closed?A. 1

7B. 7

18C. 11

18D. 7

11E. 63

10

9. An (n − 2)2 sided figure has n diago-nals. Find the number n of diagonalsfor a 25 sided figure.A. 8 B. 7 C. 6 D. 9 E. 10

10. Find the probability of selecting a fig-ure which is parallelogram from asquare, a rectangle, a rhombus, a kiteand a trapezium.A. 3

5B. 2

5C. 4

5D. 1

5E. 5

11. If P varies inversely as V and V variesdirectly as R2, find the relationshipbetween P and R given that R = 7where P = 2.A. P = 98R2 B. PR2 = 98C. P 2R = 98 D. P = 1

98R

E. P = R2

98

12. If 7 and 189 are the first and fourthterms of a geometric progression re-spectively. Find the sum of the firstthree terms of the progression.A. 182 B. 180 C. 91 D. 63E. 28

13. Find the positive number a such thatthrice its square is equal to twelvetime the number.A. 1 B. 4 C. 2 D. 5 E. 9

14. A sector of a circle of radius 7.2cmwhich subtends an angle of 300◦ at thecenter is used to form a cone. What is

55

56 Question, 2009

the radius of the base of the cone?A. 6cm B. 7cm C. 8cm D. 9cmE. 5cm

15. If pq + 1 = q2 and r = 1p− 1

pq. Express

r in terms of q.A. 1

p+qB. 1

q−p C. 1q+1

D. 1p+1

E. 1pq

16. Given a regular hexagon, calculateeach interior angle of the hexagon.A. 60◦ B. 30◦ C. 120◦ D. 45◦

E. 135◦

17. Find n if log2 4 + log2 2− log2 n = −1.A. 10 B. 14 C. 12 D. 27 E. 16

18. If x = 1 is a root of the equationx3− 2x2− 5x+ 6. Find the other roots.A. −3, 2 B. −2, 2 C. 3,−2D. 1, 3 E. −3, 1

19. The value of (0.303)3 − (0.02)3 isA. 0.019 B. 0.0019 C. 0.00019D. 0.000019 E. 0.000035

20. List all integers satisfying the inequal-ity −2 < 2x− 6 < 4.A. 2, 3, 4, 5 B. 2, 3, 4 C. 2, 5D. 3, 4, 5 E. 4, 5

21. If 32y − 6(3y) = 27. Find y.A. 3 B. 1 C. 2 D. −3 E. 1

22. A number of pencils were sharedamong Peter, Paul and Audu in theratio 2, 3, 5 respectively if Peter got 5,how many were shared?A. 15 B. 25 C. 30 D. 50 E. 55

23. A sum of money was invested at 8%per annum simple interest. If after 4years the money became N330, whatis the amount originally invested?A. N180 B. N165 C. N150D. N200 E. N250

57 Solution, 2009

Solution, 2009

1.6x2 − 14x− 12

Since this is a quadratic expressionand not a quadratic equation (con-taining equality), we will not dividethrough by 2 but, we will factorize out2 to have 2(3x2 − 7x− 6).

From 3x2 × (−6) = −18x2, thenSince − 18x2 = (−9x)(2x)

−7x = (−9x) + (2x)

We have

= 2(3x2 − 9x+ 2x− 6)

= 2(3x(x− 3) + 2(x− 3)

)= 2((3x+ 2)(x− 3)

)= 2(x− 3)(3x+ 2)

Hence,

6x2−14x−12 = 2(x−3)(3x+2), C.

2. The product of(275∗ 3−3

)and (1

5)−1 is

(1

5)−1(

27

5∗ 3−3

)= 5

(27

5∗ 1

27

)= 5

(1

5

)= 1

Hence, answer equals 1 D.

3. For a right– angled triangle;

Hypotenuse side2

= Opposite side2 + Adjacent side2

From the sides given, (3x + 1)cm isthe hypotenuse side since it is the

largest side when x is given any posi-tive value.

By Pythagorean theorem:

(3x+ 1)2 = (3x− 1)2 + x2

(3x+ 1)(3x+ 1) = (3x− 1)(3x− 1) + x2

9x2 + 6x+ 1 = 9x2 − 6x+ 1 + x2

9x2 + 6x+ 1 = 10x2 − 6x+ 1

On collecting like terms, we obtain,9x2 − 10x2 + 6x+ 6x+ 1− 1 = 0

−x2 + 12x = 0

x2 − 12x = 0

x(x− 12) = 0

x = 0 or (x− 12) = 0

Therefore, x = 0 cm or x = 12cm.Since a side cannot be 0cm, our an-swer is 12cm.

Hence, x = 12cm, D.

4.(xy2 − x2y)

x2 − xy.

We factorize the numerator and de-nominator

xy(y − x)

x(x− y).

58 Solution, 2009

Since −(x− y) = −x+ y = y − x, then

(xy2 − x2y)

x2 − xy=

xy(− (x− y)

)x(x− y)

= −�xy��(x− y)

�x��(x− y)= −1 · y · 1= −y

Hence,

(xy2 − x2y)

x2 − xy= −y

Since y = 3, then the answer equals -3.A.


On using the table in solution Q6of 2011, we observe that the out-comes 1, 1, ; 2, 2; 3, 3; 4, 4; 5, 5; and 6, 6showed the same number of point outof a total outcome of 36. Therefore,probability( both have same numberof point)= 6

36= 1

6, E.

6. Let x be the number of years from1984 in which Tolu (24 years) is ex-actly half his father’s (45 years) age.Then

24 + x =1

2(45 + x)

where (24+x) years is Tolu’s age whenhe is half his Father’s age and (45 + x)years is the father’s age at that time.So, on cross–multiplying, we have

2(24 + x) = 45 + x

48 + 2x = 45 + x

2x− x = 45− 48

x = −3

This means 3 years before 1984, Toluwas half his Father’s age.

Hence, the year is 1984 − 3 = 1981,B.


Prime numbers are numbers divisibleby 1 and itself. Note that 1 is nota prime number. Examples of primenumbers are 2, 3, 5, 7, 11, 13, · · ·Now, the primes from 40 to 50 are41, 43 and 47. Numbers from 40 to 50is 11. Probability( prime from 40 to50)= 3

11, A.

8. Total shirts = 6 + 5 + 7 = 18 shirts.

Probability( a purple shirt)= 718

, sincewe have 7 purple shirts. B.

9. If an (n− 2)2 sided figure has n diago-nals then,

25 = (n− 2)2√

25 =√

(n− 2)2

5 = n− 2

n = 7

Hence, a 25 sided figure has 7 diago-nals, B

10. Total plane shapes = 5 (squares,rectangle, rhombus, kite and trapez-ium). Now, plane shapes that areparallelograms= 3, (square, rectangleand rhombus).

Note:A Parallelogram is a quadrilacteralwhich has both pairs of its oppositesides parrallel.

A square is a rectangle with sides ofequal lenght.

59 Solution, 2009

A rectangle is a quadrilacteral whichhas all its angles as right angles.

A rhombus is a parallelogram withsides of equal lenght.

A kite is a quadrilacteral in which thediagonals intersect at right angles andthe two pairs of adjacent sides areequal.

A trapezium is a quadrilacteral withtwo parallel sides.

Also note that rhombus, rectanglesand squares are all special examplesof parrallelograms. Any property aparallelogram has will also be true ofa rhombus, rectangle or square.Probability(a parallelogram)= 3

5, A.

11.

P ∝ 1

V(Inverse Variation) (6.1)

V ∝ R2 (Direct Variation) (6.2)

From (6.1),

P =K

V(*)

whereK is the constant of proportion-ality.

V = LR2 (**)

where L is the constant of proportion-ality.

Substituting V = LR2 into (*), gives

P =K

LR2

P =K

L· 1

R2

Since K and L are constants. Let Q =KL

, we have

P =Q

R2

whereQ is the constant of proportion-ality connecting P and R.

60 Solution, 2009

When R = 7 and P = 2, then

2 =Q

72

2× 72 = Q

Q = 2× 49 = 98

So, the equation connecting or the re-lationship between P and R is,

P =98

R2

PR2 = 98

Hence, the relationship between Pand R is PR2 = 98, B


Here, generally Tn = arn−1

T1 = a = 7 (i)

T4 = ar4−1 = ar3 = 189 (ii)

We proceed to find r, the common ra-tio.

Substitute (i) into (ii) to havear3 = 189

7r3 = 189

r3 =189

7= 27

r3 = 27

r =3√

27 = 3

Since |r| = |3| = 3 > 1, we proceed tofind the sum of the first three (n = 3)terms using,

Sn =a(rn − 1)

r − 1

S3 =7(33 − 1)

3− 1

=7(27− 1)

2

=7

2· 26

= 7× 13

= 91, C.

13. This is a word problem.

The positive number isnIts square is n2

Thrice its square is 3n2

Twelve times the number is 12n

Mathematically,

3n2 = 12n

3n2 − 12n = 0

3n(n− 4) = 0

3n = 0, or n− 4 = 0

n = 0, or n = 4

Th number is either 0 or 4. But sinceit is a positive number then n = 4.Hence, B.

14. The transformation of the sectorabove to the cone is given by the for-mula

θ

360◦× 2πr0 = 2πr1

where θ is the subtended angle , r0 isthe radius of the circle and r1 is the

61 Solution, 2009

base radius of the cone

θ = 300◦, r0 = 7.2cm, r1 =?

θ

360◦× 2πr0 = 2πr1

300◦

360◦× 2π(7.2cm) = 2πr1

r1 =30

36× 7.2

r1 =10

12× 7.2 cm = 6 cm

Hence, the base radius of the cone is6cm, A.

15.pq + 1 = q2

andr =

1

p− 1

pq(6.3)

From

pq + 1 = q2

pq = q2 − 1

p =q2 − 1

q(6.4)

Substitute (6.4) into (6.3), so that

r =1(

q2−1q

) − 1(q2−1q

)× q

r =q

q2 − 1− 1

q2 − 1

r =q − 1

q2 − 1

From difference of two squareA2 − B2 = (A + B)(A − B), so thatq2 − 1 = (q + 1)(q − 1), so

r =(q − 1)

(q + 1)(q − 1).

Hence,

r =1

q + 1, C.


A regular polygon is a polygon whosesides are all equal and whose interiorangles are also all equal. A hexagonhas 6 sides , so n = 6.

Note: For any regular polygon,

Interior angle + Exterior angle = 180◦

Exterior angle =360◦

n

where n is the number of sides.

In our case, hexagon, n = 6.

Exterior angle =360◦

6= 60◦


Interior angle = 180◦ − 60◦

Interior angle = 120◦

Hence, each interior angle of thehexagon is 120◦ C.


Using laws (i), (ii), (iii)

log2 4 + log2 2− log2 n = −1

log2 22 + log2 2− log2 n = −1

2(1) + 1− log2 n = −1

2 + 1− log2 n = −1

2 + 1 + 1 = log2 n

4 = log2 n

log2 n = 4

n = 24 = 16

Hence, n = 16 E.

18. Note: Polynomials

P (x) = Q(x)D(x) +R(x)

Polynomial = Quotient×Divisor + Remainder

62 Solution, 2009

In our case, if x = 1 is a root of thequation x3−2x2−5x+6, then (x−1)is a factor of the polynomial, that is,D(x) = x− 1, P (x) = x3− 2x2− 5x+ 6

x2 − x− 6

x− 1)

x3 − 2x2 − 5x + 6− x3 + x2

− x2 − 5xx2 − x− 6x + 6

6x− 6

0

So, from P (x) = Q(x)D(x) +R(x)

x3 − 2x2 − 5x+ 6 = (x2 − x− 6)(x− 1) + 0

= (x2 − x− 6)(x− 1)

We now factorize the quadratic(x2 − x− 6) normally

x2 − x− 6 = x2 − 3x+ 2x− 6

= x(x− 3) + 2(x− 3)

= (x+ 2)(x− 3)

so that

x3−2x2−5x+6 = (x−1)(x+2)(x−3).

The roots are:

P (x) = 0

(x− 1)(x+ 2)(x− 3) = 0

x− 1 = 0, x+ 2 = 0, x− 3 = 0

x = 1, x = −2, x = 3

Hence, the other roots are x = −2 orx = 3, C.

19. Note that:

a3 − b3 = (a− b)(a2 + ab+ b2)

So,

(0.303)3 − (0.02)3

= (0.303− 0.02)(0.3032 + (0.303)(0.02) + 0.022)

= (0.283)(0.091809 + 0.00606 + 0.0004)

= (0.283)(0.092209 + 0.00606)

= (0.283)(0.098269)

= 0.0278

a2 − b2 = (a− b)(a+ b), so

(0.303)2 − (0.02)2

= (0.303− 0.02)(0.303 + 0.02)

= (0.283)(0.323)

= 0.091409

Hence, 0.3033 − 0.023 = 0.0278.


Here, −2 < 2x − 6 < 4. We then findx.

Add 6 to all the sides−2 + 6 < 2x− 6 + 6 < 4 + 6

4 < 2x < 10

Divide through by 2 to have,4

2<

2x

2<

10

22 < x < 5

Hence, x = 3, 4.

21. Please see solution to Q6 and Q12 of2013 before proceeding.

63 Solution, 2009

Here,

32y − 6(3y) = 27

(3y)2 − 6(3y) = 27

Let A = 3y, thenA2 − 6A = 27

A2 − 6A− 27 = 0

A2 − 9A+ 3A− 27 = 0

A(A− 9) + 3(A− 9) = 0

(A+ 3)(A− 9) = 0

A+ 3 = 0, or A− 9 = 0

A = −3, A = 9

When A = −3, from A = 3y,−3 = 3y(impossible).When A = 9, fromA = 3y, 9 = 3y, 32 = 3y, y = 2,C.


Ratio

Peter : Paul : Audu = 2 : 3 : 5

Total ratio = 2 + 3 + 5 = 10

Peter’s portion =2

10× T = 5

where T is the total number of pencil shared.2T

10= 5

2T = 50

T = 25

Hence, 25 pencils were shared B

23.

Simple Interest, I

=Principal,P × Rate, R× Time, T

100

I =PRT

100(i)

Total amount = Interest, I+Principal, P

A = I + P (ii)

Here, R = 8%, T = 4, A =N330.00, P =?

From (i),

100I = PRT

100I = P (8)(4) = 32P

25I = 8P

25I − 8P = 0 (iii)

From (ii),

A = I + P

330 = I + P (iv)

From (iii)

25I = 8P

I =8

25P

Substitute I =8

25P into (iv), we have

330 = I + P

330 =8

25P + P

330 =8P + 25P

25

330 =33P

2533P = 330× 25

P = N250.00

(Note: Principal= Original amount in-vested).

Hence, the amount originally in-vested is N250.00 E.

7

Question, 2008

1. The expression a3 + b3 is equal toA. (a2 + b)(a− ab+ b2)B. (a+ b)(a2 − ab+ b2)C. (a− b2)(a2 − ab+ b)D. (a− b)(a2 + ab+ b2)

2. Factorize 16(3x+ 2y)2 − 25(a+ 2b)2

A. (12x+8y−5a−10b)(12x+8y−5a−10b)B. 20(3x+2y−a−2b)(3x+2y+a+2b)C. (12x+8y+5a+10b)(12x+8y−5a−10b)D. 20(3x+2y+a+2b)(3x+2y+a+2b)

3. A cone has base radius 4cm andheight 3cm. The area of its curved sur-face isA. 12πcm2 B. 20πcm2

C. 24πcm2 D. 15πcm2

4. A cylinder has height 4cm and baseradius 5cm. Its volume to 3 significantfigure isA. 314.2cm2 B. 31.42cm2

C. 251.4cm2 D. 251cm2

5. Let log y + log 3 = 3, then y isA.(x10

)3 B.(x10

)−3C.(10x

)−13 D.

(10x

)36. If

x2

a2− y2

b2= 1

then y isA. ± b

a

√a2 − x2 B. ± b

a

√x2 − a2

C. ±ab

√a2 − x2 D. ±a

b

√x2 − a2

7. A Cyclist rode for 30 minutes atx km/hr and due to a breakdown hehad to push the bike for 2hrs at (x −5)km/hr. If the total distance coveredis less than 60 km, what is the range ofvalue of x?A. x < 14 B. x < 29C. x < 28 D. x < 20

8. The expression ax2+bx takes the value6 when x = 1 and 10 when x = 2. Findits value when x = 5A. 10 B. 12 C. 6 D. −10

9. Dividing 2x3−x2−5x+1 by x+3 givesthe remainderA. −3 B. 47 C. 61 D. −47

10. Let f(x) = 2x3 − 3x2 − 5x + 6, if x −1 divides f(x). Find the zeros of thefunctionA. 1, 2, 3

2B. 1, 2,−3

2

C. −1, 2, 3 D. 1,−2,−32

11. The difference of two numbers is 10,while their product is 39. Find thesenumbersA. −3 and 10 or 13 and 10B. 3 and −10 or 3 and 13C. 3 and −3 or 3 and 13D. −3 and−13 or 13 and 3

12. The average age of x pupils in a classis 14 years 2 months. A pupil of 15years 2 months joins the class andthe average age is increased by onemonth. Find x.A. 12 B. 6 C. 11 D. 13

64

65 Question, 2008

All the 120 pupils in a school learnYoruba or Igbo or both. Given that 75learn Yoruba and 60 learn Igbo

13. How many learn both languages?A. 60 B. 45 C. 15 D. 120

14. How many learn Igbo only?A. 45 B. 30 C. 15 D. 60

Suppose we have matrices

A =

(1 −12 3

)and B =

(0 24 3

)15. Find A2 + AB − 2A

A.(−5 −912 14

)B.(−1 −48 7

)C.(−4 −412 13

)D.(

0 −4−8 −6

)16. Evaluate the integral∫ 2

1

(x2 +1

x)dx

A. 83

+ ln 2 B. 73

+ ln 2C. 7

3− ln 3 D. 8

3+ ln 3

17. If the distance covered by a body intime t seconds is s = t3−6t2+5t, whatis its initial velocity?A. 0ms−1 B. −4ms−1

C. (3t2 − 12t+ 5)ms−1 D. 5ms−1

Suppose D,E and P are subsets ofa universal set U . Let U be the setof natural numbers not greater than10, while D,E and P are respectivelythe set of odd numbers, even numbersand prime numbers. For any set X , itscomplementary X ′ and ∅ denote theempty set.

18. Display the set D′ ∩ PA. {3, 5, 7} B. {2}C. {4, 6, 8, 10} D. {2, 3, 5, 7}

19. Find D ∩ EA. {2} B. {2, 3}C. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} D. {∅}

20. The Trigonometric expressioncos 2A+ sin 2A can be written asA. cosA(cosA− sinA)B. cos2A+ sin2A− 2 sinA cosAC. 2 sinA cosA + cos2A + sin2AD. cos2A+ sin2A− 2 sinA cosA

A bag contains 10 balls of which 3 arered and 7 are white. Two balls aredrawn at random. Find the probabil-ity that none of the balls is red if thedraw is

21. With replacementA. 0.9 B. 1 C. 0.4 D. 0.49

22. Without replacementA. 0.1 B. 0.47 C. 0.42 D. 0.21

23. In a throw of a fair die the probabilityof obtaining an even number isA. 1 B. 2

3C. 1

6D. 2

3

24. Two fair coins are tossed simultane-ously. What is the probability of ob-taining at least 1 tail turns up?A. 1

4B. 3

4C. 1

2D. 1

Let α and β be the root of the equation

x2 − 5x+ 4 = 0

25. Find the value of1

α− 1

βA. ±4

3B. 3

4C. ±3

4D. 1

5

26. A regular polygon has each of its an-gles as 160◦. What is the number ofsides of the polygon?A. 18 B. 36 C. 9 D. 20

66 Question, 2008

27. One angle of an octagon is 100◦ whilethe other sides are equal. Find each ofthese exterior anglesA. 80◦ B. 60◦ C. 140◦ D. 0◦

67 Solution, 2008

Solution, 2008

1.

a3 − b3 = (a− b)(a2 + ab+ b2)

a3 + b3 = (a+ b)(a2 − ab+ b2)

a4 − b4 = (a− b)(a3 + ab2 + a2b+ b3), B

2.

16(3x+ 2y)2 − 25(a+ 2b)2

= 42(3x+ 2y)2 − 52(a+ 2b)2

=[4(3x+ 2y)

]2 − [5(a+ 2b)]2

= [12x+ 8y]2 − [5a+ 10b]2

From the difference of two squares,

A2 −B2 = (A+B)(A−B),

so that we have

=[(12x+ 8y) + (5a+ 10b)

]×[(12x+ 8y)− (5a+ 10b)

]= (12x+ 8y + 5a+ 10b)

×(12x+ 8y − 5a− 10b)

Hence, C.

3. We note that the curve surface area ofa cone is πrl, where r is the base ra-dius and l is the slant height. The totalsurface area of a cone is

πrl + πr2 = πr(l + r).

The volume of a cone is

V =1

3πr2h

where h is the height of the cone andπ = 22

7. We remark that slant height

l, is different from the height h of thecone.

Now, r = 4cm, h = 3cm, l =?

To calculate the curve surface area ofa cone, we must first find the slantheight l.

Considering right angle 4COB, us-ing Pythagoras’s theorem

l2 = h2 + r2

l2 = 32 + 42

l2 = 9 + 16

l2 = 25

l =√

25

l = 5

Curve surface area = πrl

= π · 4 · 5= 20π, B.

4. We note that curve surface area of acylinder is 2πrh, where r is the radius,h is the height of the cylinder. Totalsurface area of a cylinder is

2πrh+ 2πr2 = 2πr(h+ r).

Volume of a cylinder is

V = πr2h.

Now, h = 4cm, r = 5cm, π = 3.142

Volume = πr2h

= (3.142)(5)2(4)

= (3.142)(25)(4)

= (3.142)(100)

= 314.2 cm3

68 Solution, 2008

Hence, the volume is 314.2 cm3

5. Please see solution to Q17 of 2009 be-fore proceeding

log y + log x3 = 3

log x3y = 3

log10 x3y = 3

x3y = 103

y =103

x3=

(10

x

)3

Hence, y =(10x

)3, D

6.x2

a2− y2

b2= 1

We make y the subject of the formula.Taking the LCM, we have

b2x2 − a2y2

a2b2= 1

b2x2 − a2y2 = a2b2

b2x2 − a2b2 = a2y2

b2(x2 − a2) = a2y2

a2y2 = b2(x2 − a2)

y2 =b2

a2(x2 − a2)

Taking the square root of both sides, give√y2 =

√b2

a2(x2 − a2)

y =

√b2

a2·√

(x2 − a2)

y = ± ba

√(x2 − a2), B.

7. NoteSpeed =

Distancetime

.

Distance = Speed× time

Let the distance the cyclist rode thebike be P , then

x km/hr =P

30mins.

We shall need to convert the time at 30mins to hours

60mins = 1hr

30mins =1

2hr, so that

x km/hr =P12hr

x× 1

2= P

P =x

2km

Let the distance the cyclist push the bike be Q, then

(x− 5) km/hr =Q

2hrsQ = 2(x− 5) km

The total distance the cyclist covered = P +Q

Total distance, T = P +Q =x

2+ 2(x− 5) km

T =x

2+

2x− 10

1km

T =x+ 2(2x− 10)

2km

T =x+ 4x− 20

2km

T =5x− 20

2km

But total distance, T < 60 km

T =5x− 20

2< 60

5x− 20 < 120

5x < 120 + 20

5x < 140

x <140

5= 28

Hence, x < 28, C.

69 Solution, 2008

8. The function f(x) = ax2 + bx. Butf(1) = 6 and f(x = 2) = 10. We areto find f(x = 5).

Since,

f(x) = ax2 + bx,

f(1) = a(12) + b(1) = 6

a · 1 + b(1) = 6

a+ b = 6

f(2) = a(22) + b(2) = 10

a · 4 + b · 2 = 10

4a+ 2b = 10

Divide through by 2, gives2a+ b = 5

This is a pair of simultaneous equa-tions in a and b, we solve as followsusing elimination method;

a+ b = 6−2a+ b = 5−a+ 0 = 1

a = −1

Substitute a = −1 in a+ b = 6, thena+ b = 6

−1 + b = 6

b = 6 + 1 = 7

So a = −1 and b = 7. This implies that

f(x) = (−1)x2 + (7)x.

This implies that f(x) = −x2 + 7x.

Now,

f(x = 5) = f(5) = −(5)2 + 7(5)

= −25 + 35

= 10

Hence, f(x = 5) = 10, A


By referring to the remainder’s theo-rem in solution to Q8 of 2013, we havethat

f(a) = Remainder, R.

When x− a is a divisor of the polyno-mial f(x) leaving a remainder R.

Here,

f(x) = 2x3 − x2 − 5x+ 1. D(x) = x+ 3

Let x+ 3 = 0, which implies x = −3

f(x = −3) = 2(−3)3 − (−3)2 − 5(−3) + 1

= 2(−27)− (9) + 15 + 1

= −54− 9 + 16

= −54 + 7

= −47

Hence, the remainder is −47, D.


P (x) = 2x3−3x2−5x+6, D(x) = x−1

2x2 − x− 6

x− 1)

2x3 − 3x2 − 5x + 6− 2x3 + 2x2

− x2 − 5xx2 − x− 6x + 6

6x− 6

0

From P (x) = Q(x)D(x) + R(x), wehave

P (x) = 2x3−3x2−5x+6 = (2x2−x−6)(x−1).

70 Solution, 2008

We proceed to factorize the quadraticquotient 2x2 − x− 6 as follows:

2x2 − x− 6

−12x2 = (−4x)(3x)

−x = (−4x) + (3x)

This implies 2x2 − 4x+ 3x− 6

= 2x(x− 2) + 3(x− 2)

= (2x+ 3)(x− 2)

Therefore;

P (x) = (2x+ 3)(x− 2)(x− 1)

The zeros of the function is at P (x) =0 which is

P (x) = (2x+ 3)(x− 2)(x− 1) = 0

2x+ 3 = 0 or x− 2 = 0 or x− 1 = 0

2x = −3 or x = 2 or x = 1

x = −32

or 2 or 1

Hence, x = 1, 2,−32, B.

11. Let the two numbers be A and B.Their difference is A − B and theirproduct is AB.

A−B = 10 (7.1)AB = 39 (7.2)

This is a simultaneous equationand we solve using the substitutionmethod as follows:

From (7.1), A−B = 10

A = 10 +B

Substituting this equation into (7.2), that is(10 +B)B = 39

10B +B2 = 39

B2 + 10B − 39 = 0

Solving this yields,

B2 + 13B − 3B − 39 = 0

B(B + 13)− 3(B + 13) = 0

(B − 3)(B + 13) = 0

B − 3 = 0 or B + 13 = 0

B = 3 or B = −13

Now;

When B = 3, from A = 10 +B

A = 10 + 3 = 13

A = 13, B = 3

When B = −13, from A = 10 +B

A = 10 +B = 10 + (−13) = 10− 13 = −3

Hence, A = 13, B = 3 or A =−3, B = −13, D.


Please note 12 months makes a year.

Average age of n pupils =Total age of npupils

n.

Now,

Let T be the total age of the x pupils, then

14 years 2 months =T

x

We convert 14 years 2months intomonths:

14 years = 12× 14 months

= 168months

14 years 2months = 168months+ 2months

= 170months

This means that 170months = Tx

.Therefore,

T = 170x (i)

71 Solution, 2008

When the average age is increased byone month, we have the new averageage to be 171 months and

171 months =T + 15 years 2months

x+ 1.

We now convert 15 years 2 months tojust months.

15 years = 15× 12 months = 180 months

15 years 2months = 182 months

So we now have

171 months =T + 182months

x+ 1(ii)

Substitute (i) into equation (ii) to ob-tain

171 =170x+ 182

x+ 1

171(x+ 1) = 170x+ 182

171x+ 171 = 170x+ 182

171x− 170x = 182− 171

x = 11

Hence, x = 11, C

13. We note that for two sets A and B

n(A ∪B) = n(A) + n(B)− n(A ∩B)

n(U) = n(A ∪B) + n(A′ ∩B′)n(U) = n(A) + n(B)− n(A ∩B) + n(A′ ∩B′)

where U is the universal set contain-ing all the sets.

Now,

let U be the universal set, thenn(U) = 120

let Y be the pupils learning Yoruba thenn(Y ) = 75

let I be the pupils learning Igbo, thenn(I) = 60

Observe that Y ∩ I will be the pupilslearning both Yoruba and Igbo.

Let n(Y ∩ I) = x. Observe thatn(Y ′ ∩ I ′) which represents the num-ber of pupils learning neither Yorubanor Igbo is zero. Since all the 120pupils in the school learn one lan-guage or the other or both, thenn(A′ ∩B′) = 0.

From n(U) = n(A) + n(B)− n(A ∩B) + n(A′ ∩B′)120 = 75 + 60− x+ 0

120 = 135− xx = 135− 120

x = 15

Hence, 15 pupils learn both lan-guages. C.

14. Y ′ ∩ I represent the pupils learningIgbo only. Let n(Y ′ ∩ I) = y. We notethat:

n(I) = n(Y ′ ∩ I) + n(Y ∩ I)

60 = y + x

60 = y + 15

y = 60− 15 = 45

72 Solution, 2008

Hence, 45 pupils learn Igbo only,A.

15.

A =

(1 −12 3

)B =

(0 24 3

)A2 =

(1 −12 3

)(1 −12 3

)=

(−1 −48 7

)We note that(a bc d

)(e fg h

)=

(ae+ bg af + bhce+ dg cf + dh

)

AB =

(1 −12 3

)(0 24 3

)=

(−4 −112 13

)2A = 2

(1 −12 3

)=

(2 −24 6

)

A2 + AB − 2A

=

(−1 −48 7

)+

(−4 −112 13

)−(

2 −24 6

)=

(−1 −48 7

)+

(−6 18 7

)=

(−7 −316 14

)Hence

A2 + AB − 2A =

(−7 −316 14

),

No correct option.

16. Please see solution to Q19 of 2013 andQ17 of 2011 before proceeding.∫ 2

1

(x2 +

1

x

)dx

=

[x2+1

2 + 1+

lnx

1

]21

=

[x3

3+ lnx

]21

=

[23

3+ ln 2

]−[

13

3+ ln 1

]=

8

3+ ln 2− 1

3− ln 1

=8

3− 1

3+ ln 2− ln 1

=7

3+ ln 2− 0

=7

3+ ln 2

(Note: ln 1 = 0, ln 0 = −∞)

Hence,∫ 2

1

(x2 +

1

x

)dx =

7

3+ ln 2, B.

17. We note

Velocity, V =d(distance)

dtime

V =d( distance)

dt

Velocity is the time–derivative of dis-tance.

Also,

Acceleration, a =d(velocity)

dtime

a =d(V )

dt

Acceleration is the time–derivative ofvelocity.

73 Solution, 2008

Now, distance, S = t3−6t2+5t and ini-tial velocity is the velocity when time,t = 0, that is, the velocity with whichthe body move or start from rest (t =0).

V =dS

dt=

d

dt(t3 − 6t2 + 5t)

V = (3t2 − 12t+ 5)m/s

Initial Velocity = V (t = 0)

= 3(0)2 − 12(0) + 5

= 0− 0 + 5

= 5m/s

Hence, the initial velocity is 5ms−1,D.

18. We note that natural numbers are thenumbers we use in counting thingsand it starts from 1 to infinity, that is,

N = {1, 2, 3, 4, 5, 6, · · · }

Now, the set of natural numbers notgreater than 10 is

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} andD = {1, 3, 5, 7, 9}E = {2, 4, 6, 8, 10}P = {2, 3, 5, 7}

where D,E, P represent respectivelyodd numbers, even numbers andprime numbers.

D′ = {2, 4, 6, 8, 10}D′ ∩ P = {2, 4, 6, 8, 10} ∩ {2, 3, 5, 7}

D′ ∩ P = {2}

D′ is the complement set of D.

Hence,

D′ ∩ P = {2}, B.

19.

D ∩ E = {1, 3, 5, 7, 9} ∩ {2, 4, 6, 8, 10}= ∅ or {}

The answer is an empty set which iseither ∅ or {} and never {∅}, it is neverthe combination of the two symbols.,D.

20. We note that

cos(A+B) = cosA cosB − sinA sinB

cos(A−B) = cosA cosB + sinA sinB

sin(A+B) = sinA cosB + cosA sinB

sin(A−B) = sinA cosB − cosA sinB

sin2A+ cos2A = 1

Now,

cos 2A = cos(A+ A)

= cosA cosA− sinA sinA

= cos2A− sin2A

sin 2A = sin(A+ A)

= sinA cosA+ sinA cosA

= 2 sinA cosA

Then,

cos 2A+sin 2A = cos2A−sin2A+2 sinA cosA

21. Please see solution to Q8 of 2009 be-fore proceeding

Total balls = 10 ballsRed = 3 red balls

White = 7 white balls

Probability of picking a red ball

=number of red balls

Total ball

74 Solution, 2008

P (red ball) =3

10

P (white ball) =7

10

Note that with replacement meansthat once a ball is picked from the bag,it is returned but when it is without re-placement, the remaining balls in thebag will be reduced by one, since it isnot returned.

Since we have only two options topick red or white ball, then

P (none of the balls is red) = P (white)× P (white)

P (not red) =7

10× 7

10

P (not red) =49

100= 0.49, D

22. But without replacement, the totalnumber of white balls will reduce byone and the total balls in the bag willalso reduce by one, after a ball ispicked so that

P (none of balls is red without replacement)

=7

10× 6

9

P (not red) =7

15= 0.46666

' 0.47, B.

23. A fair die has the digits, 1, 2, 3, 4, 5 and6 on its sides. We know that 2, 4 and 6are even numbers which makes a totalof 3 even number on the die.

P (even number) =3

6=

1

2

H TH HH HTT TH TT

24. A coin has two sides with head (H)and tail (T) on each side. The tablebelow shows the possible outcomeswhen two coins are thrown We haveonly four possible outcomes (HH ,HT , TH and TT ) outcomes with atleast one tail are (HT , TH and TT ).

P (at least one tail) =3

4, B.

25.x2 − 5x+ 4 = 0.

Recall:x2 − (α + β)x + αβ = 0 is the mostgeneral form of a quadratic equationin terms of its roots α and β. On solv-ing x2 − 5x+ 4 = 0, we have

x2 − 4x− x+ 4 = 0

x(x− 4)− 1(x− 4) = 0

(x− 1)(x− 4) = 0

(x− 1) = 0 or (x− 4) = 0

x = 1 or x = 4

Now, since α and β are the roots of theequation, then, we let α = 1 and β = 4(Case 1),

1

α− 1

β= 1− 1

4

=4− 1

4

=3

4

Also, let α = 4 and β = 1 (Case 2),

75 Solution, 2008

then1

α− 1

β=

1

4− 1

=1− 4

4

= −3

4

This shows that 1α− 1

β= 3

4or −3

4.

Hence,1

α− 1

β= ±3

4, C


Recall that for a regular polygon,


I◦ + E◦ = 180◦

and n =360◦

Exterior angle

where n is the number of sides.

Now,

Interior angle = 160◦

interior angle + Exterior angle = 180◦

160◦ + Exterior angle = 180◦

Exterior = 180◦ − 160◦ = 20◦

number of sides =360◦

Exterior angle

=360◦

20◦= 18 sides, A.


Recall that an octagon has 8 sides, thatis, n = 8.

Sum of interior angles in an octagon= (8− 2)× 180◦

= 6× 180◦

= 1080◦

Since an octagon has 8 sides it impliesit has 8 interior angles. If one of theinterior angles is 100◦, and the otherseven interior angles are equal say x◦,then

100◦ + x◦ + x◦ + x◦ + x◦ + x◦ + x◦ + x◦ = 1080◦

100◦ + 7x◦ = 1080◦

7x = 1080◦ − 100◦

7x◦ = 980◦

x◦ =980◦

7= 140◦

This implies that the other 7 equal in-terior angles are 140◦ each.

But since:


Exterior angle = 180◦ − Interior angle= 180◦ − 140◦ = 40◦

Hence, the exterior angle of each ofthe equal interior angles is 40◦ D.

8

Question, 2007

1. The interior angles of a pentagon are :180◦, 80◦, 118◦, 78◦ and x. The value ofx isA. 75◦ B. 108◦ C. 120◦ D. 134◦

2. All the vertices of an isosceles trian-gle lie on a circle and each of the baseangles of the triangle is 65◦. The anglesubtended at the centre of the circle bythe base of the triangle isA. 130◦ B. 115◦ C. 100◦ D. 65◦

3. A square tile measures 20cm by 20cm.How many of such tiles will cover afloor diameter of the sphere?A. 500 B. 400 C. 320 D. 250

4. The volume of a certian sphere is nu-merically equal to twice its surfacearea. The diameter of the sphere isA. 6 B. 9 C. 12 D.

√6

5. A bearing of 310◦, expressed as a com-pass bearing isA. N50◦W B. N40◦WC. S40◦W D. S50◦W

6. Which of the following specified setsof data is not necessarily sufficient forthe construction of a triangle?A. three anglesB. two sides and a right angleC. two sides and an included angleD. three sides

7. The average age of the three childrenin a family is 9 years. If the average

age of their parent is 39 years, the av-erage age of the whole family isA. 20 years B. 21 yearsC. 24 years D. 27 years

8. Simplify

1 +2

3− 3÷

[1 +

2

3of

6

7

]A. − 8

33B. 21

11C. 33

21D. −21

8

9. If1 +

1

1 + 11+ 1

x

= 5,

find xA. 3

7B. 7

3C. −3

7D. −7

3

10. Evaluate x in base 3 if 41x − 22x = 17xA. 11 B. 8 C. 12 D. 22

11. A woman buys 4 bags of rice for N56per bag and 3 bags of beans for N26per bag using the currency ’LONI’(N)in base 7. What is the total cost ofthe items in another currency ’MONI’(M) in base 8?A. M224 B. M114C. M340 D. M440

12. When the price of egg was raised byN2 an egg, the number of eggs whichcan be bought for N120 is reduced by5. The present price of an egg isA. N6 B. N7 C. N8 D. N10

13. How long will it take a sum of moneyinvested at 8% simple interest to dou-ble the original sum?A. 8 years B. 10.5yearsC. 12years D. 12.5years

76

77 Question, 2007

14. The journey from Lagos to Ibadanusually takes a motorist 1 hour30minutes. By increasing his averagespeed by 20km/hr, the motorist saves15 minutes. His usual speed in km/hrisA. 100 B. 90 C. 85 D. 80

15. The smallest section of a rod whichcan be cut exactly into equal sections,each of either 30cm or 36cm in lengthisA. 90cm B. 180cmC. 360cm D. 540cm

16. If x = 0.0012+0.00074+0.003174, whatis the difference between x in 2 deci-mal places and x in 1 significant fig-ure?A. 0.01 B. 0.0051C. 0.1 D. 0.005

17. The angle of depression of two pointsA and B on a plane field from the topof a mast connected between A andB are 30◦ and 45◦ respectively. If Ais westward of B, find /AB/ if theheight of the mast is 1m from the field.A. 15

√3m B. 5(3 +

√3)m

C. 15(1 +√

3)m D. 15(√

3− 1)m

18. The radius of a circle is given as 10cmsubject to an error of 0.2cm. The errorin the area of the circle isA. 1

4% B. 1

5% C. 2% D. 4%

19. If θ is acute, evaluate

cos(90− θ) + sin(180− θ)cos(180− θ)− sin(90− θ)

A. tan θ B. − tan θC. cot θ D. − cot θ

20. In a survery of 100 students in an in-stitution. 80 students speak Yoruba,

22 speaks Igbo; while 6 speaks neitherYoruba nor Igbo. How many studentsspeak Yoruba and Igbo?A. 96 B. 8 C. 64 D. 12

21. A bag contains 5 yellow balls, 6 greenballs and 9 black balls. A ball is drawnfrom the bag. What is the probabilitythat it is black or yellow balls?A. 27

160B. 172

400C. 77

200D. 172

200

The table below shows the distribu-tion of weight measure for 100 stu-dents

Weight(kg) F60− 62 563− 65 1866− 68 4269− 71 2772− 74 8

22. Calculate the mean of the distributionto two decimal places.A. 64.45 B. 62.45C. 65.45 D. 65.45

23. Calculate the mode of the distributionto two decimal places.A. 67.33 B. 65.33C. 65.53 D. 67.35

24. T ′Q is a tangent to the circle ABCDT ,∠DTQ = 40◦, ∠ATT ′ = 30◦, then∠ATD isA. 70◦ B. 90◦ C. 250◦ D. 110◦

78 Solution, 2007

Solution, 2007

1. Please see solution of Q11 of 2012 be-fore proceeding.

Recall that a pentagon has 5 sides, thatis, n = 5

Sum of interior angles in a pentagon= (5− 2)× 180◦

= 3× 180◦

= 540◦

Since the interior angles in the pen-tagon are 180◦, 80◦, 118◦, 78◦ and x◦,then

180◦ + 80◦ + 118◦ + 78◦ + x◦ = 540◦

456◦ + x = 540◦

x = 540◦ − 456◦

x = 84◦

2. Note

∠BOC = 2∠BAC, {angle at the cen-ter= twice the angle at the circumfer-ence}.Now,

Considering the isosceles triangleABC,

65◦ + 65◦ + θ = 180◦

θ + 130◦ = 180◦

θ = 180◦ − 130◦

θ = 50◦

so the angle at the circumference, θ =50◦

Since ∠BOC = 2∠BAC, the

Angle at the center, ∠BOC = 2× 50◦

= 100◦

Hence, the angle subtended at thecenter of the circle by the base of thetriangle is 100◦, C.

3. Note that

Area of a square of side, l = l2,

Perimeter of a square of side, l = 4l,

Area of a rectangular floor of length, land breath B , is = l ×B.

Now, length of the rectangular flooris, l = 5m = 500cm and breath of thefloor is b = 4m = 400cm

Area of the rectangular floor = l × b= 500× 400

= 200000cm2

79 Solution, 2007

So,

Number of tiles

=Area of rectangular floor

Area of square tiles

=200000cm2

400cm2

= 500 tiles

Hence, 500 tiles will fit into the rectan-gular floor. A.

4. We note that

Volume of a sphere =4

3πr3,

Curve surface area of a sphere = 4πr2.

Now, the volume of the sphere equalstwice its surface area implies that

4

3πr3 = 2(4πr2)

4

3πr3 = 2× 4πr2

4

3�πr

3 = 8�πr2

r3

r2=

8× 3

4r = 6

But,

diameter = 2× radius= 2× 6

= 12 units, C

5. 310◦ = N50◦W , in compass bearing.We note that you measure the angle(310◦) from the North and move inclockwise direction. Each quadrant is90◦.

The angle must be located near thenorth or south as the case may be.A.

6. A triangle has 3 sides and 3 angles.In construction,at least one side of thetriangle must be given in order to con-struct the triangle. When you aregiven the three angles only, withoutany side, the data is not necessarilysufficient for the construction of thetriangle. A.

7. Please see solution to Q12 of 2008 be-fore proceeding. Recall:

Average age of n people

=Total age of npeople

n

Let the total age of the 3 children be T1then,

Average age of the children =T13

9 =T13

T1 = 3× 9 yearsT1 = 27 years.

Let the total age of the parents (2) beT2, then

Average age of parents =T22

39 years =T22

T2 = 39× 2 yearsT2 = 78 years

The whole family comprises the 3children and the 2 parents making 5

80 Solution, 2007

people in total and the total age ofthe whole family will be T1 + T2 =27 years + 78 years = 105 years.

Average age of the whole family

=Total ages

5

=105 years

5= 21 years

Hence, the average age of the wholefamily is 21 years, B.

8.

1 +2

3− 3÷

(1 +

2

3of

6

7

).

Using BODMAS, which shows the se-quence any question like the formabove will be solved.

B = We solve the fractions in bracket beforeany other

O = We deal with the ′of ′ part whichmeans multiplication, next

D = Division is nextM = Multiplication followsA = Addition followsS = Subtraction is treated last

We treat the fractions in bracket first.For the fractions in bracket, we stillhave two signs + and of but of comesfirst in BODMAS. So

2

3of

6

7=

2

3× 6

7=

4

7

= 1 +2

3− 3÷

(1 +

4

7

)= 1 +

2

3− 3÷ 11

7

Division comes next in BODMAS, sowe solve it next

= 1 +2

3− 3× 7

11

= 1 +2

3− 21

11

Next, Addition comes before subtrac-tion in BODMAS, so

= 1 +2

3− 21

11

=3 + 2

3− 21

11

=5

3− 21

11

= − 8

33, A.

9.

1 +1

1 + 11+ 1

x

= 5

1

1 + 11+ 1

x

= 5− 1 = 4

1

1 + 11+ 1

x

=4

1

1 = 4

(1 +

1

1 + 1x

)1 = 4 +

4

1 + 1x

1− 4 =4

1 + 1x

−3

1=

4

1 + 1x

81 Solution, 2007

We have,

−3

(1 +

1

x

)= 4

−3− 3

x= 4

−3

x= 4 + 3 = 7

−3

x= 7

On cross–multiplication, we have

(−3)(1) = (7)(x)

−3 = 7x

x = −3

7

Hence,

x = −3

7, C

10.41x − 22x = 17x

To find x, we first convert the num-bers from base x to numbers in base10, then we convert the x, we obtainedin base 10 to number in base 3. To con-vert from any base (x in this case) tobase 10 follow these steps

41x = 4× x1 + 1× x0

= 4x+ 1× 1

= (4x+ 1)10

22x = 2× x1 + 2× x0

= 2x+ 2× 1

= (2x+ 2)10

17x = 1 + x1 + 7× x0

= x+ 7

So that,

41x − 22x = 17x becomes

4x+ 1− (2x+ 2) = x+ 7

4x+ 1− 2x− 2 = x+ 7

2x− 1 = x+ 7

2x− x = 7 + 1

x = 8

This mean that the question was orig-inally in base 8. Now to convert x = 8in base 10 to x in base 3, we followthese steps: We divide the number

3 83 2 remainder 2

0 remainder 2

we want to convert from base 10 bythe base we desire (here 3) and takenote of the remainders of this divi-sion counted upwards, from the low-est placed reminder, that is, x = 810 =223

Hence,

x = 22three, D

11. Note: Please see solution to Q1 of 2012before proceeding.

”LONI” (N) is a currency in base7, which mean that all estimatingwith ”LONI” will be in base 7.”MONI”(M) is a currency in base 8,which means that all calculations with”MONI” will be in base 8

If one bag of rice costs N567

4 bags of rice cost = N567 × 4 = N3237

(that is, 6 × 4 = 24 = 7(3) + 3, write3(the reminder), carry 3, 5×4 = 20; +3

82 Solution, 2007

(carried) = 23 = 7(3) + 2, write 2 (thereminder), carry 3 and write the 3 car-ried.)

If one bag of beans costs N267

3 bags of beans cost = N267 × 3 = N1147

The total cost of the items in ”LONI”(N) is= N3237 +N1147 = N4407

We convert 4407 to base 8 in orderto obtain the cost of the items in”MONI”(M) currency.

We first convert 4407 to base 10 as fol-lows:

4407 = 4× 72 + 4× 71 + 0× 70

= 4× 49 + 4× 7 + 0× 1

= 196 + 28 + 0

= 22410

We now convert 22410 to a number inbase 8. Hence,

8 2248 28 rm 08 3 rm 4

0 rm 3

N4407 = 22410 = M3408, C

12. Let the initial price of an egg be Nx.Let the number of eggs that can bebought with N120 for the price of Nxper egg be y, then

N120

Nx= y (i)

Now, when the price is raised by N2,the price becomes = Nx+N2 = N(x+

2) and the number of eggs which canbe bought withN120 reduced by 5 im-plies = (y − 5) number of eggs, then

N120

N(x+ 2)= y − 5 (ii)

Substitute equation (i) into equation(ii),we have

120

x+ 2=

120

x− 5

120

x+ 2=

120− 5x

x

On cross multiplying, we have

120(x) = (120− 5x)(x+ 2)

120x = 120x+ 2(120)− 5x(x)− 5x(2)

120x = 120x+ 240− 5x2 − 10x

120x− 120x+ 5x2 + 10x− 240 = 0

5x2 + 10x− 240 = 0

Dividing through by 5 gives

x2 + 2x− 48 = 0

Solving this quadratic equation gives;

x2 + 8x− 6x− 48 = 0

x(x+ 8)− 6(x+ 8) = 0

(x− 6)(x+ 8) = 0

(x− 6) = 0 or x+ 8 = 0

x = 6 or x = −8

Since price Nx cannot be negative,then x = N6. But x = N6 was theinitial price and the present price is

N(x+ 2) = N(6 + 2) = N8, C.


Simple interest,

I =Principal× Rate× Time

100=PRT

100

83 Solution, 2007

R = 8%, let the time be x, T = xyears.When the money invested (Principal)doubles, then it means that simple in-terest= Principal, that is, I = P

I = P =PRT

100, which implies that 1 =

RT

100RT = 100

8× T = 100, which implies that

T =100

8= 12.5 years, D.

14. Please see the solution to Q7 of 2008before proceeding.Recall:

Speed (km/hr) =distance(km)

time(hrs)

Let the usual speed be x km/hr. Thedistance in km from Lagos to Ibadanis fixed and constant, no matter whatspeed was used, so let the distance bez1.

Now,

distance (km) = speed (km/hr)× time (hrs)

at the usual speed x km/hr,

distance, z1 = x× 1hr 30mins

We convert 1 hr 30 mins to hours;

60mins = 1hr

30mins =1

2hrs

1hr 30mins = 1 +1

2hrs

=3

2hrs

so that,

Z1 = x× 3

2=

3

2x km (i)

By increasing his average speed by20 km, the new speed becomes, (x +20)km/hrs, and since the motoristssaves 15 minutes (15mins = 1

4hrs) do-

ing so, the new time, t becomes,

t =

(3

2− 1

4

)hours =

5

4hours.

The distance covered for thesechanges in speed x is still the distancefrom Lagos to Ibadan, which is z1 .

So,

z1 = new speed× new time

= (x+ 20) km/hr × 5

4hours

z1 =5

4(x+ 20) km (ii)

Combining equations (i) and (ii) gives

z1 =3

2x =

5

4(x+ 20) km

(3x)4 = 2[5(x+ 20)

]12x = 10(x+ 20)

12x = 10x+ 200

12x− 10x = 200

2x = 200

x = 100 km/hrs

Hence, the usual speed is 100 km/hrA.

15. This means we should find the lengthof the smallest rod which can be cutinto smaller, equal, sections of length30cm or 36cm.

To do this, we simply find thelowest common multiple of the twolengths which is obtained as follows

84 Solution, 2007

2 30 362 15 183 15 93 5 35 5 1

1 1

LCM(30, 36) = 2× 2× 3× 3× 5

= 4× 9× 5

= 36× 5

= 180cm

So, the length of the smallest rodwhich can be cut into smaller, equalsections of length 30cm or 36cm is180cm, B

16.0.0012

+ 0.00074+ 0.003174

0.005114

The answer x = 0.005114 to 2 decimalplaces is 0.01 (that is, two digits afterthe decimal point and round up thethird digit). The answer x = 0.005114to 1 significant figure is 0.005 (that is,the first natural number in the num-ber and round up the next). The dif-ference is = 0.01− 0.005 = 0.005, D.

17. Note the diagrams

From the alternating angles beingequal, we have

Note: |AB| = |AC|+ |CB|.So considering4ACD,

tan 30◦ =15m

|AC|

|AC| = 15m

tan 30◦= 15m÷

√3

3

|AC| = 15× 3√3

=45√

3

On rationalizing,

|AC| = 45√3×√

3√3

=45√

3

3= 15

√3m

Considering,4BCD,

tan 45◦ =15m

|CB|

|CB| = 15m

tan 45◦= 15÷ 1 = 15m

Now,

|AB| = |AC|+ |CB|= 15

√3m+ 15m

= 15(√

3 + 1)m

Hence, |AB| = 15(1 +√

3)m, C

18.Area of a circle = πr2,

where r is the radius of the circle. LetA represents the area and 4A the er-ror in the area of the circle. Let r repre-sent the radius and4r the error in theradius of the circle. r = 10cm, 4r =0.2cm

A = πr2 = π(10cm)2 = π(100cm2) = 100πcm2

4A4r≈ 2πr

85 Solution, 2007

(i.e the derivative of A = πr2 with re-spect to r.)

4A = 2πr4 r = 2× π × 10× 0.2

= 4πcm2

% error in the area =Error in areaOriginal area

× 100

=4AA× 100

1

=4�πcm2

��100�πcm2×��100

1= 4%

Hence, the % error in the area of thecircle is 4%, D.


Note these facts{cos(90− θ) = sin θsin(90− θ) = cos θ

Now, (180◦ − θ) is an angle in the sec-ond quadrant and only sine is positivein the 2nd quadrant, others are nega-tive, so

sin(180◦ − θ) = sin θ

cos(180◦ − θ) = − cos θ

tan(180◦ − θ) = − tan θ

Using all the above information wehave

cos(90◦ − θ) + sin(180◦ − θ)cos(180◦ − θ)− sin(90◦ − θ)

=sin θ + sin θ

− cos θ − cos θ

=2 sin θ

−2 cos θ

= − sin θ

cos θ= − tan θ, B.

(Note that tan θ = sin θ

cos θ, cot θ =

1tan θ

= cos θsin θ

).


Let the universal set be U , then n(U) =100. Let Y represents the studentswho speak Yoruba, then n(Y ) = 80.Let I represents the students whospeak Igbo, then n(I) = 22, then Y ′∩I ′represents the students speaking nei-ther Yoruba nor Igbo, and n(Y ′∩ I ′) =6. Also Y ∩ I represents the studentsspeaking both Yoruba and Igbo.

Let n(Y ∩ I) = x and recall that

n(U) = n(Y )+n(I)−n(Y ∩I)+n(Y ′∩I ′).

So

100 = 80 + 22− x+ 6

100 = 102− x+ 6

100 = 108− xx = 108− 100 = 8

Hence, x = 8 students, speaks bothYoruba and Igbo. B.

21. Please see solution to Q14 of 2012, andQ8 of 2009 before proceeding.

Yellow balls = 5

Green balls = 6

Black balls = 9

Total balls = 5 + 6 + 9 = 20 balls

Probability of picking a Black ball

=Number of Black ballsTotal number of balls

P (black ball) =9

20

P (yellow ball) =5

20=

1

4

86 Solution, 2007

We note that in probability or mathe-matics, ′or′, means addition and ′and′

means multiplication.

Now, the probability of picking ablack ball or yellow ball is,

P (black or yellow ball)= P (black ball) + P (yellow ball)

=9

20+

1

4

=14

20

=7

10Hence, the probability that it is a blackor yellow ball is 7

10.

22. We note that

Mean, x̄ =

n∑i=1

fixi

n∑i=1

fi

where x̄ represents the mean, xi is theindividual class mark and fi is the fre-quency of each class mark.

Class mark

=Lower class limit + Upper class limit

2Example; for the class interval 60− 62of the weight in kg

Class mark =60 + 62

2=

122

2= 61 kg

Now,5∑i=1

fixi = 305 + 1152 + 2814 + 1890 + 584

= 67455∑i=1

fi = 5 + 18 + 42 + 27 + 8

= 100

Class Class frequencyInterval (kg) Mark(xi) (fi) fixi60− 62 61 5 30563− 65 64 18 115266− 68 67 42 281469− 71 70 27 189072− 74 73 8 584

Mean, x̄ =

5∑i=1

fixi

5∑i=1

fi

=6745

100= 67.45 kg

Hence, the mean of the distribution totwo decimal places is 67.45 kg, C.

23. Note: Mode is the weight with thehighest frequency.

Mode = L1 +

(dfb

dfb + dfa

)× c

whereL1 is the lower class boundary of themodal class.dfb is the difference between the fre-quency of the modal class and the fre-quency before it.dfa is the difference between the fre-quency of the modal class and the fre-quency after it.c is the class size. Modal class isthe class interval with the highest fre-quency.

Now, from the table, the highest fre-quency is 42, f = 42. The correspond-ing class interval to frequency 42 is66 − 68. To obtain the class boundaryfrom the class interval, we subtract 0.5from the lower class limit 66 and add0.5 to the upper class limit 68 to ob-tain:

Modal class Boundary = 65.5− 68.5

87 Solution, 2007

Now,

L1 = 65.5

dfb = 42− 18 = 24

dfa = 42− 27 = 15

We note that 18 is the frequency beforethe modal frequency 42 and 27 is thefrequency after the modal frequency42. The class size c is the difference be-tween the upper class boundary andthe lower class boundary, i.e

c = 68.5− 65.5 = 3

Now,

Mode = 65.5 +

(24

24 + 15

)× 3

= 65.5 +

(24

39

)× 3

= 65.5 +

(24

13

)= 65.5 + 1.846

= 67.346

Mode= 67.35 to 2 decimal places.

24. Observe that ∠ATT ′ = 30◦, ∠DTQ =40◦ and ∠ATD are all in a straightline T ′TQ. But the sum of angles ina straight line is 180◦,

∴

∠ATT ′ + ∠DTQ+ ∠ATD = 180◦

30◦ + 40◦ + ∠ATD = 180◦

70◦ + ∠ATD = 180◦

∠ATD = 180◦ − 70◦

∠ATD = 110◦

Hence, ∠ATD is 110◦ D

9

Question, 2006

1. Solve for p in the following equationgiven in base two 11(P+110) = 1001PA. 10 B. 11 C. 110 D. 111

2. Factorize 16(3x+ 2y)2 − 25(a+ 2b)2

A. (12x+8y−5a−10b)(12x+8y−5a−10b)B. 20(3x+2y−a−2b)(3x+2y+a+2b)C. (12x+8y+5a+10b)(12x+8y−5a−10b)D. 20(3x+2y+a+2b)(3x+2y+a+2b)

3. A cone has a base radius 4cm andheight 3cm. The area of its curved sur-face isA. 12πcm2 B. 24πcm2

C. 20πcm2 D. 15πcm2

4. Let log y + log 3 = 3, then y isA.(10x

)2 B.(x10

)2C.(x10

)−2 D.(10x

)−12

5. Ifx2

a2− y2

b2= 1

then y isA. ± b

a

√a2 − x2 B. ± b

a

√x2 − a2

C. ±ab

√a2 − x2 D. ±a

b

√x2 − a2

6. A cyclist rode for 30 minutes atxkm/hr and due to a breakdown hehad to push the bike for 2 hrs at (x −5) km/hr . If the total distance coveredis less than 60 km, what is the range ofvalues for x?A. x < 14 B. x < 20C. x < 29 D. x < 28

7. A business invested a total ofN200, 000 in two companies whichpaid dividends of 5% and 7% re-spectively. If he received a total ofN11, 600 dividend, how much did heinvest at 5%?A. N160, 000 B. N140, 000C. N120, 000 D. N80, 000

8. In a class, 37 student take at least oneof chemistry, Economics and Govern-ment, 8 students take Chemistry, 19take Economics and 25 take Govern-ment. 12 students take Economics andGovernment but nobody takes Chem-istry and Economics. How many stu-dents take both chemistry and Gov-ernment?A. 3 B. 4 C. 5 D. 6

9. Z is partly constant and partly variesinversely as the square of d. When d =1, z = 11 and when d = 2, z = 5. Findthe value of z when d = 4A. 2 B. 3.5 C. 5 D. 5.5

10. Expand the expression

(x2 − 2x− 5)(x2 + x+ 1)

A. x4 − 4x3 − 5x− 3B. −x3 − 4x2 + 5x− 3C. x4 − x3 − 4x2 − 5x− 3D. x4 − 4x2 − 5x− 3

Suppose we have matrices

A =

(1 −12 3

)and B =

(0 24 3

)

88

89 Question, 2006

11. Find A2 + AB − 2A

A.(−5 −912 14

)B.(−1 −48 7

)C.(−4 −412 13

)D.(

0 −4−8 −6

)12. The inverse of matrix B is

A.1

8

(−3 24 0

)B.(−3 24 0

)C.

1

8

(3 −4−2 0

)D.(

1 00 1

)13. The indefinite integral of the function

f(x) = x cosxfor any constant k isA. − cosx+sin x+k B. x sinx−cosxC. x sinx+ cosx+ k D. x+ sinx+cosx+ k

14. Evaluate the integral∫ 2

1

(x2 +1

x)dx

A. 83

+ ln 2 B. 73

+ ln 2C. 7

3− ln 3 D. 8

3

15. The trigonometric expressioncos 2A+ sin 2A can be written asA. cosA(cosA− sinA)B. cos2A+ sin2A− 2 sinA cosAC. 2 sinA cosA + cos2A + sin2AD. cos2A+ sin2A− 2 sinA cosA

Suppose D,E and P are subsets ofa universal set U . Let U be the setof natural numbers not greater than10, while D,E and P are respectivelythe set of odd numbers, even numbersand prime numbers. For any set X , itscomplementary X ′ and ∅ denote theempty set.

16. Display the set D′ ∩ PA. {3, 5, 7} B. {2}C. {4, 6, 8, 10} D. {2, 3, 5, 7}

17. Find D ∩ EA. {2} B. {2, 3}C. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} D. {∅}

A big contains 10 balls of which 3 arered and 7 are white. Two balls aredrawn at random. Find the probabil-ity that none of the balls is red if thedraw is

18. With replacementA. 0.9 B. 1 C. 0.4 D. 0.49

19. Without replacementA. 0.1 B. 0.47 C. 0.42 D. 0.21

20. A regular polygon has each of its an-gle as 160◦. What is the number of sideof the polygon?A. 36 B. 9 C. 18 D. 20

21. A girl walks 30m from a point P on abearing of 040◦ to a point Q. She thenwalks 30m on a bearing of 140◦ to apoint R. The bearing of R from P isA. 90◦ B. 50◦ C. 45◦ D. 40◦

22. How many different three digits num-bers can be formed using the integers1 to 6 if no integer occurs twice in anumber?A. 24 B. 120 C. 60 D. 48

23. In how many different ways can theletters of the word GEOLOGY be ar-ranged in order?A. 720 B. 1260 C. 2520 D. 5040

90 Solution, 2006

Solution, 2006

1. Please see solution to Q1 of 2012, Q10of 2011 and Q10 and Q11 of 2007 be-fore proceeding.

112(P2 + 1102) = (10012)P2

This equation is in base 2. On expand-ing,

112P2 + 112(110)2 = (10012)P2

But,1102

×112

110110

100102

So we have

112P2 + 100102 = 10012P2

On collecting like terms, we obtain

100102 = 10012P2 − 112P2

10012P2 − 112P2 = 100102

(10012 − 112)P2 = 100102

But10012

−112

1102

(That is, starting from the last digits,1 − 1 = 0, 0 − 1 is not possible, per-haps, so we borrow a 1 from the nextnumber which is not 0, but from thefirst digit 1, so that the zero besidethe 1 becomes 2, borrowed 1 from this2 and we have one left and the bor-rowed 1 added to the digit 0 to make2 and now 2−1 = 1. Then the remain-ing one from the two is written down,

since there is nothing to subtract fromit. So we have 1102.

Now,

(1102)P2 = 100102

P2 =100102

1102

=1810

610

P2 = 310 = 112,

Hence, P2 = 112, B

2. The question is the same as Q2 of2008. Please see solution

3. The question is the same as Q3 of2008.



6. The question is the same as Q7 of 2008

7. Question is similar to Q8 of 2010, butwith a little modification. Please seesolution before proceeding.

Recall that the amount invested at 7%was N80, 000

Now, we find the amount investedat 5%. Since the total amount in-vested was N200, 000 in two compa-nies which paid dividends of 5% and7% respectively, then the amount in-vested at 5% = N200, 000− Amountinvested at 7%

Amount at 5% = N200, 000−Amount at 7%

= N200, 000−N80, 000

= N120, 000

Hence, the amount invested at 5% isN120, 000, C.

91 Solution, 2006

8. Please see solution to Q13 and Q14 of2008 before proceeding.

Let U represent the students in theclass, then n(U) = 37. Let C representthe students taking chemistry, thenn(C) = 8. Let E represent the stu-dents taking Economics, then n(E) =19. Let G represent the students tak-ing Governments, then n(G) = 25.

Observe that E ∩ G will representstudents taking both Economics andGovernment, then n(E ∩G) = 12.

Also C ∩E represents students takingboth Chemistry and Economics, thenn(C ∩ E) = 0.

AlsoC∩Gwill represent students tak-ing both Chemistry and Government.Let n(C ∩G) = x.

Observe that all of the students takeat least one of the 3 subjects, whichmeans n(C ′ ∩ E ′ ∩ G′) = 0. Sincen(C∩E) = 0, then number of studentstaking all three subjects are also zero,that is, n(C ∩ E ∩G) = 0.

Note: For three sets A,B,C in the uni-versal set U

n(U) = n(A) + n(B) + n(C)

− n(A ∩B)− n(A ∩ C)− n(B ∩ C)

+ n(A ∩B ∩ C) + n(A′ ∩B′ ∩ C ′)

In our case,C,E,G in the universal setU ,

n(U) = n(C) + n(E) + n(G)

− n(C ∩ E)− n(E ∩G)− n(C ∩G)

+ n(C ∩ E ∩G) + n(C ′ ∩ E ′ ∩G′)

which implies that

37 = 8 + 17 + 25− 0− 12− x+ 0 + 0

37 = 52− 12− x37 = 40− xx = 40− 37

x = 3

Hence, the number of students takingboth Chemistry and Government is 3Students, A.

9. Please see the solution to Q11 of 2009before proceeding.

Note:

(a) When they say A is partly con-stant and partly varies, directlyas B, this means

i. A = k, where k is a constantand

ii. A ∝ B (direct variation)A = QB where Q is the con-stant of proportionality. Thetwo clauses together gives,

A = k +QB

where k and Q are constants.

(b) When they say A is partly con-stant and partly varies inverselyas B, this means

i. A = K, where k is a constantii. A ∝ 1

B(Inverse varia-

tion)A = Q

Bwhere Q is the con-

stant of proportionality.

A = k +Q

B

where k and Q are constants.

92 Solution, 2006

In our case:

Z = K +Q

d2

where K and Q are constants.

When d = 1, z = 11, we have

11 = K +Q

12= k +Q = 11 (i)

When d = 2, z = 5, we have

5 = k +Q

22= K +

Q

4= 5 (ii)

We solve the simultaneous equationsusing elimination method, as follows;

Subtract (ii) from (i), gives

Q− Q

4= 6

3

4Q = 6 =⇒ 3Q = 6× 4

Q =24

3= 8

From equation (i),K +Q = 11, so that

K + 8 = 11

k = 11− 8 = 3

Which implies K = 3, Q = 8.

Now, the equation connecting Z andd2 is

Z = 3 +8

d2

To find Z when d = 4, we have,

Z = 3 +8

42= 3 +

8

16= 3 +

1

2

Z =7

2= 3.5

Hence, Z = 3.5, B.

10. Note:

(a+ b)(c+ d) = a(c+ d) + b(c+ d)

= ac+ ad+ bc+ bd

Now,

(x2 − 2x− 3)(x2 + x+ 1)

= x2(x2 + x+ 1)− 2x(x2 + x+ 1)− 3(x2 + x+ 1)

= x4 + x3 + x2 − 2x3 − 2x2 − 2x− 3x2 − 3x− 3

Collecting like terms give,= x4 + x3 − 2x3 + x2 − 2x2 − 3x2 − 2x− 3x− 3

= x4 − x3 − 4x2 − 5x− 3

Hence,

(x2−2x−3)(x2+x+1) = x4−x3−4x2−5x−3, C

11. Question is the same as Q15 of 2008.Kindly see the solution in the solutionto Q15 of 2008.

12. Note: If

A =

(a bc d

)is a matrix, then the determinant of A,|A| = ad− bc and the adjoint of A,

Adj(A) =

(d −b−c a

)and the Inverse of A, is

A−1 =AdjA|A|

=1

ad− bc

(d −b−c a

)Now;

B =

(0 24 3

)|B| = 0(3)− 2(4)

= 0− 8

= −8

93 Solution, 2006

Adj (B) =

(3 −2−4 0

)

B−1 =1

−8

(3 −2−4 0

)B−1 =

−1

8

(3 −2−4 0

)=

1

8

(−3 24 0

)Hence, the inverse of B,

B−1 =1

8

(−3 24 0

), A

13. Please see the solution to Q17 of 2010before proceeding.

Using similar integration by part.Since, we have to integrate a productof two functions x cosx = f(x)g(x).

Recall: ∫udv = uv −

∫vdu

Now, ∫x cosxdx

Let

u = x, dv = cosxdx

du

dx= 1 =⇒ du = 1dx, v =

∫cosxdx

then du = 1dx, v = sinx

which implies∫x cosxdx = x sinx−

∫sinx · 1dx

= x sinx− (− cosx) +K

Hence,∫x cosxdx = x sinx+cosx+K, C.

14. Question is the same as Q16 of 2008.








The bearing of R from P is the sum ofangles from the north of P to the lineconnecting P to R, which is 90◦

Also, the point R is directly east ofP since they both have the same dis-tance from Q, so its bearing is 90◦,A.

22. We note that the number of permuta-tion

nPr =n!

(n− r)!where n! = n × (n − 1)(n − 2) · · · 1.Example

4! = 4× (4− 1)× (4− 2)× (4− 3)

= 4× 3× 2× 1

= 24

nPr is used to know the different wayswe can have the arrangement of r dig-its out of n digits.

94 Solution, 2006

Now, in the question; r = 3, n = 6

6P3 =6!

(6− 3)!

=6× 5× 4× 3× 2× 1

3!

=6× 5× 4× 3× 2× 1

3× 2× 1

= 6× 5× 4

= 120 ways

Hence, there are 120 different 3 dig-its numbers that can be formed usingnumbers from 1 to 6. B

23. We note that to arrange a total of 10letters with 3 different letters appearstwice (say), then the number of waysof arrangement is

=10!

2!2!2!ways

In our case; GEOLOGY has 7 lettersin total with G and O both appearingtwice, we have then

Arrangements =7!

2!2!

=7× 6× 5× 4× 3× 2× 1

(2× 1)(2× 1)

= 7× 6× 5× 2× 3

= 210× 6

= 1260 ways B

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Mathematics Past Question and Answer for Pre-University Students