JAMB UTME - Mathematics - 1993

Question 1 Report

Make x the subject of the relation $ \frac{1+ax}{1-ax}=\frac{p}{q} $

\frac{p + q}{a (p - q)}

\frac{p - q}{a (p + q)}

\frac{p - q}{a p q}

\frac{p q}{a (p - q)}

Solution Of Linear Equations

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Question 2 Report

A man's initial salary is ₦540.00 a month and increases after each period of six months by ₦36.oo a month. Find his salary in the eighth month of the third year

₦828.00

₦756.00

₦720.00

₦684.00

Sequence And Series

Ratio, Proportions And Rates

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Question 3 Report

Calculate the standard deviation of the following data: 7, 8, 9, 10, 11, 12, 13.

2

4

10

11

Measures Of Dispersion

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Question 4 Report

Quantities in the proportions 1, 4, 6, 7 are to be represented in a pie chart. Calculate the angle of the sector with proportion 7

20o

80o

120o

140o

Positive And Negative Integers, Rational Numbers

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Question 5 Report

Simplify $ \frac{1}{p} - \frac{1}{q} + \frac{p}{q} - \frac{q}{p} $

\frac{1}{p - q}

\frac{- 1}{p + q}

\frac{1}{p q}

\frac{1}{p q (p - q)}

Answer Details

\frac{1}{p}

-

\frac{1}{q}

+

\frac{p}{q}

-

\frac{q}{p}

=

\frac{q - p}{p q}

+

\frac{p^{2} - q^{2}}{p q}

\frac{q - p}{p q}

x

\frac{p q}{p^{2} q^{2}}

=

\frac{q - p}{p^{2} q^{2}}

\frac{- (p - q)}{(p + q) (p - q)}

=

\frac{- 1}{p + q}

Read lesson note on Algebraic Expressions (WAEC)

Algebraic Expressions

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Question 6 Report

Solve without using tables $ \log_{5}(62.5) - \log_{5}\left(\frac{1}{2}\right) $

3

4

5

8

Answer Details

log5(62.5) - log5(

\frac{1}{2}

)
= log5

\frac{(62.5)}{\frac{1}{2}}

- log5(2 x 62.5)
= log5(125)
= log553 - 3log55
= 3

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Read lesson note on Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

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Question 7 Report

If $ \sqrt{x^2 + 9} = x + 1 $, solve for x

5

4

3

2

1

Answer Details

\sqrt{x^{2} + 9}

= x + 1
x2 + 9 = (x + 1)2 + 1
0 = x2 + 2x + 1 - x2 - 9
= 2x - 8 = 0
2(x - 4) = 0
x = 4

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Read lesson note on Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

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Question 8 Report

Divide the expression x3 + 7x2 - x - 7 by -1 + x2

-x3 + 7x2 - x - 7

-x3 = 7x + 7

x - 7

x + 7

Algebraic Expressions

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Question 9 Report

If $ \sin \theta = \cos \theta $, find $ \theta $ between 0^o and 360^o

45o, 225o

135o, 315o

45o, 315o

135o, 225o

Answer Details

sin

θ

= cos

θ

0

\leq

θ

\leq

360o
The acute angle where sin

θ

= cos

θ

= 45o
But at the fourth Quadrant Cos

θ

= +ve
at the 4th quadrant, value with respect to Q is
(360 - Q) where Q = acute angle
(360 - 45) = 315o
The two solution are 45o, 315o

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Trigonometry

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Question 10 Report

If the function f is defined by f(x + 2) = 2x2 + 7x = 5, find f(-1)

-10

-8

4

10

Algebraic Expressions

Differentiation

Quadratic Equations

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Question 11 Report

In the figure, the line segment ST is tangent to two circles at S and T. O and Q are the centres of the circles with OS = 5cm. QT = 2cm and OR = 14cm. Find ST

7

\sqrt{3}

cm

12.9cm

\sqrt{87}

cm

7cm

Answer Details

SQ2 + OS2 = OQ2 + 52 = 142

SQ2 = 142 - 52

196 - 25 = 171

ST2 + TQ2 = SQ2

ST2 + 22 = 171

ST2 = 171 - 4

= 167

ST = $\sqrt{167}$

= 12.92 = 12.9cm

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Modular Arithmetic

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Question 12 Report

Two chords PQ and RS of a circle when produced meet at K. If ∠KPS = 31o and ∠PKR = 42o, find ∠KQR

11o

73o

107o

138o

149o

Inequalities

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Question 13 Report

Class	Frequency
1−5	2
6−10	4
11−15	5
16−20	2
21−25	3
26−30	2
31−35	1
36−40	1

Find the median of the observation in the table given.

11.5

12.5

14.0

14.5

Answer Details

Median = L1 + (

\frac{E f}{f m}

) - fo

\frac{\sum f}{2}

=

\frac{20}{2}

= 10, L1 = 10.5, fo = 6, fm = 5
Median = 10.5 +

\frac{(10 - 6)}{5}

5
= 10.5 + 4
= 14.5

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Representation Of Data

Measures Of Location

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Question 14 Report

Calculate the length in cm. of the area of a circle of diameter 8cm which subtends an angle of $ \frac{1}{2} $o at the centre of the circle

2

π

π

\frac{2}{3}

\frac{π}{2}

Answer Details

Diameter = 8cm
∴ Radius = 4cm
Length of arc =

\frac{θ}{360}

x 2

π

r but Q = 22

\frac{1}{2}

∴ Length

\frac{22 \frac{1}{2}}{360}

x 2 x

π

x 4
=

\frac{22 \frac{1}{2} \times 8 π}{360}

=

\frac{180}{360}

=

\frac{π}{2}

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Circles

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Question 15 Report

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

35o

37o

49o

59o

Circles

Triangles And Polygons

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Question 16 Report

simplify $ \frac{1}{\sqrt{3}-2} - \frac{1}{\sqrt{3}+2} $

3

\frac{2}{3}

7

-4

Answer Details

\frac{1}{\sqrt 3 - 2}

-

\frac{1}{\sqrt 3 + 2}

L.C.M = (3- 2) (3 + 2)
∴

\frac{1}{\sqrt{3 - 2}}

-

\frac{1}{\sqrt{3 - 2}}

=

\frac{\sqrt{3 + 2} - \sqrt{3 - 2}}{\sqrt{3 - 2} + \sqrt{3 - 2}}

$\frac{\sqrt 3 + 2 - \sqrt 3 + 2}{3 - 2 \sqrt 3 + 2 \sqrt 3 - 4}$ = $\frac{4}{3 - 2}$
= $\frac{4}{- 1}$
= -4

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Read lesson note on Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Question 17 Report

In the diagram, QPS = SPR, PR = 9cm. PQ = 4cm and QS = 3cm, find SR.

6

\frac{3}{4}

cm

3

\frac{3}{8}

cm

4

\frac{3}{8}

cm

2

\frac{3}{8}

cm

Answer Details

Using angle bisector theorem: line PS bisects angle QPR

QS/QP = SR/PR

3/4 = SR/g

4SR = 27

SR = $\frac{27}{4}$

= 6 $\frac{3}{4}$ cm

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Question 18 Report

Evaluate $ \frac{3524}{0.05} $ correct to 3 significant figures

705

70,000

70, 480

70, 500

Fractions, Decimals And Approximations

Number Bases

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Question 19 Report

In the diagram, $QP // ST$: $PQR = 34^\circ$ $QRS = 73^\circ$ and $RS = RT$. Find $SRT$

68

^{\circ}

102

^{\circ}

107

^{\circ}

141

^{\circ}

Answer Details

Construction joins R to P such that SRP = straight line

R = 180 $^{\circ}$ - 107 $^{\circ}$

< p = 180 $^{\circ}$ - (107 $^{\circ}$ - 34 $^{\circ}$ )

108 - 141 $^{\circ}$ = 39 $^{\circ}$

Angle < S = 39 $^{\circ}$ (corr. Ang.) But in $△$ SRT

< S = < T = 39 $^{\circ}$

SRT = 180 - (39 $^{\circ}$ + 39 $^{\circ}$ )

= 180 $^{\circ}$ - 78 $^{\circ}$

= 102 $^{\circ}$

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Angles And Intercepts On Parallel Lines

Statistics

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Question 20 Report

In the diagram, O is the centre of the circle and POQ a diameter. If POR = $96^\circ$, find the value of ORQ.

84

^{\circ}

48

^{\circ}

45

^{\circ}

42

^{\circ}

Answer Details

OQ = OR = radii

< ROQ = 180 - 86 = 84 $^{\circ}$

$△$ OQR = Isosceles

R = Q

R + Q + 84 = 180(angle in a $△$ )

2R = 96 since R = Q

R = 48 $^{\circ}$

ORQ = 48 $^{\circ}$

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Circles

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Question 21 Report

Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0

-2, 4

2, 4

-1,8

1, -8

Solution Of Linear Equations

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Question 22 Report

Find the area of the sector of a circle with radius 3m, if the angle of the sector is 60o

4.0m2

1m2

4.7m2

5.0m2

Circles

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Question 23 Report

PQRST is a regular pentagon and PQVU is a rectangle with U and V lying on TS and SR respectively as shown in the diagram. Calculate TUP

18

^{\circ}

54

^{\circ}

90

^{\circ}

108

^{\circ}

Triangles And Polygons

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Question 24 Report

If k + 1; 2k - 1, 3k + 1 are three consecutive terms of a geometric progression, find the possible values of the common ratio

0, 8

-1	5
	3

2, 3

1, -1

Number Bases

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Question 25 Report

The three sides of an isosceles triangle are length of lengths (x + 3), (2x + 3), (2x - 3) respectively. Calculate x.

5

1

3

6

Answer Details

2x + 3

\neq

2x - 3 for any value of x
∴ for the

△

to be isosceles, either
2x - 3 = x + 3 or 2x + 3 = x + 3
solve the two equations we arrive at
x = 6 or x = 0
When x = 6, the sides are 9, 15, 9
When x = 0, the sides are 3, 4, -3 since lengths of a

△

can never be negative then the value of x = 6

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Algebraic Expressions

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Question 26 Report

The mean of the ages of ten secondary school pupils is 16 but when the age of their teacher is added to it the men becomes 19. Find the age of the teacher

27

35

38

49

Integration

Statistics

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Question 27 Report

The bar chart shows the distribution of marks in a class test. How many students took the test?

17

20

25

30

Answer Details

Number of students that took the test =

\sum f

. Where

f

is the frequencies
= 2 + 5 + 0 + 3 + 4 + 0 + 0 + 0 + 1 + 0 + 2 = 17

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Integration

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Question 28 Report

In the diagram, PQRs is a circle with 0 as centre and PQ/RT. If RTS = $32^\circ$. Find PSQ

32

^{\circ}

45

^{\circ}

58

^{\circ}

90

^{\circ}

Answer Details

< PSO =

\frac{1}{2}

< SOQ =

\frac{1}{2}

(180) = 90

^{\circ}

< RTS = < PQS = 32 $^{\circ}$ (Alternative angle)

< PSQ = 90 - < PSQ = 90 $^{\circ}$ - 32 $^{\circ}$

= 58 $^{\circ}$

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Circles

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Question 29 Report

If ₦225.00 yields ₦27.00 in x years simple interest at the rate of 4% per annum, find x

3

4

12

17

Answer Details

To solve this problem, we use the formula for simple interest: I = P*r*t Where: I = Interest earned P = Principal amount r = Rate of interest per year (as a decimal) t = Time period in years In this case, we are given: P = ₦225.00 (the principal amount) I = ₦27.00 (the interest earned) r = 4% = 0.04 (the rate of interest per year) Substituting these values into the formula, we get: 27 = 225 * 0.04 * t Simplifying the equation, we have: 27 = 9t Dividing both sides by 9, we get: t = 3 Therefore, the answer is 3 years. Explanation: The interest earned on the principal amount of ₦225.00 is ₦27.00. This means that for every year, the interest earned is 27/225 = 0.12 or 12% of the principal amount. We know that the rate of interest per year is 4%, which is much lower than the interest earned per year of 12%. This means that it will take more than 1 year for the interest earned to reach the rate of interest. In fact, it takes 3 years for the interest earned to reach the rate of interest of 4%, which is why the answer is 3.

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Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Financial Arithmetic

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Question 30 Report

From the figure, calculate TH in centimeters

\frac{5}{\sqrt{3} + 1}

\frac{5}{\sqrt{3} - 1}

\frac{5}{\sqrt{3}}

\frac{\sqrt{3}}{5}

Answer Details

TH = tan 45

^{\circ}

, TH = QH

$\frac{T H}{5 + Q H}$ = tan 30 $^{\circ}$

TH = (b + QH) tan 30 $^{\circ}$

QH = 56 (5 + QH) $\frac{1}{\sqrt{3}}$

QH(1 - $\frac{1}{\sqrt{3}}$ ) = $\frac{5}{\sqrt{3}}$

QH = $\frac{5 \sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}$

= $\frac{5}{\sqrt{3} - 1}$

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Surds (radicals)

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Question 31 Report

Integrate with respect to x

$\frac{x - x^{2}}{x^{4}}$ + k

\frac{4}{x^{4}} - \frac{3 + k}{x^{3}}

$\frac{1}{x} - \frac{1}{2 x^{2}}$ + k

$\frac{1}{3 x^{2}} - \frac{1}{2 x}$ + k

Answer Details

$\int \frac{1 - x}{x^{3}}$
= $\int_{x^{3}}^{1} - \int_{x^{3}}^{x}$
= x-3 dx - x-2dx
= $\frac{1}{2 x^{2}} + \frac{1}{x}$

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Integration

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Question 32 Report

The chances of three independent events X, Y, Z occurring are $ \frac{1}{2} $, $ \frac{2}{3} $, $ \frac{1}{4} $ respectively. What are the chances of Y and Z only occurring?

\frac{1}{8}

\frac{1}{24}

\frac{1}{12}

\frac{1}{4}

Answer Details

Chance of x =

\frac{1}{2}

Change of Y =

\frac{2}{3}

Chance of Z =

\frac{1}{4}

Chance of Y and Z only occurring
= Pr (Y ? Z ? Xc)
where Xc = 1 - Pr(X)
1 =

\frac{1}{2}

= 1

\frac{1}{2}

= Pr(Y) x Pr(Z) x Pr(Xc)
=

\frac{2}{3}

x

\frac{1}{4}

x

\frac{1}{2}

=

\frac{1}{12}

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Probability

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Question 33 Report

Solve for y in the equation 101 x 5(2x - 2) x 4(y - 1) = 1

\frac{3}{4}

\frac{5}{4}

\frac{2}{3}

5

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Question 34 Report

Solve the following equation (3x - 2)(5x - 4) = (3x - 2)2

\frac{- 2}{3}

, 1

1

\frac{2}{3}

, 1

\frac{2}{3}

,

\frac{4}{5}

Algebraic Expressions

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Question 35 Report

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Introductory Calculus

Trigonometry

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Question 36 Report

If two angles of a triangle are 30o each and the longest side is 10cm. Calculate the length of each of the other sides

5cm

4cm

3√3 cm

10

\frac{3}{3}

cm

Answer Details

102 = 2x2 - 2x2 cos 120o (Cosine rule)
100 = 242 - 2x2 - 2x2 x -

\frac{1}{2}

100 = 3x2 + x2
= 3x2
x =

\sqrt{\frac{100}{3}}

=

\frac{10}{\sqrt{3}}

x

\frac{3}{\sqrt{3}}

x = 10

\frac{3}{3}

cm

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Question 37 Report

If $9\left(x - \frac{1}{2}\right)^3x^2$

\frac{1}{2}

1

2

3

Answer Details

9(x -

\frac{1}{2}

) 3x2 = 32(x -

\frac{1}{2}

) = 3x2
∴ 2(x -

\frac{1}{2}

) = x2
2x - 1 = x2
hence x2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1

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Number Bases

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Question 38 Report

Which of the following is a factor of 15 + 7x - 2x2

x + 3

x - 3

x - 5

x + 5

Answer Details

Factorize 15 + 7x - 2x2
(5 - x)(3 + 2x); suppose 15 + 7x = 2x2 = 0
∴ (5 - x)(3 + 2x) = 0
x = 5 or x = -

\frac{3}{2}

Since 5 is a root, then (x - 5) is a factor

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Algebraic Expressions

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Question 39 Report

A binary operation $ \ast $ is defined on a set of real numbers by $x \ast y = x^y$ for all real values of $x$ and $y$. If $x \ast 2 = x$. Find the possible values of $x$

0, 1

1, 2

2, 2

0, 2

Answer Details

x

*

y = xy
x

*

2 = x2
x

*

2 = x
∴ x2 - x = 0
x(x - 1) = 0
x = 0 or 1

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Binary Operations

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Question 40 Report

Evaluate $ \left(x + \frac{1}{x} + 1\right)^2 - \left(x + \frac{1}{x} + 1\right)^2 $

4x2

(

\frac{2}{x}

+ 2)2

4

4(1 + x)

Answer Details

(x +

\frac{1}{x}

+ 1)2 - (x +

\frac{1}{x}

+ 1)2
= (x +

\frac{1}{x}

+ 1 + x +

\frac{- 1}{x}

- 1)(x -

\frac{1}{x}

+ 1 - x +

\frac{1}{x}

+ 1)
= (2x) (2 +

\frac{2}{x}

) = 2x x 2(1 +

\frac{1}{x}

)
4x (1 +

\frac{1}{x}

) = 4x + 4
= 4(1 + x)

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Quadratic Equations

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Question 41 Report

The angle between latitudes 30oS and 13oN is

17o

33o

43o

53o

Angles

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Question 42 Report

find the radius of a sphere whose surface area is 154cm² $ \left(\pi=\frac{22}{7}\right) $

7.00cm

3.50cm

3.00cm

1.75cm

Answer Details

Surface area = 154cm2 (area of sphere)
4

π

r2 = 154
r

\sqrt{\frac{154}{4 π}}

= 3.50cm

View Answer

Question 43 Report

In the diagram above, |PQ| = |QR|, |PS| = |RS|, ∠PSR = 30o and ∠PQR = 80o. Find ∠SPQ.

15o

25o

40o

50o

55o

Algebraic Expressions

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Question 44 Report

If 2log3 y + log3 x2 = 4, then y is

4 - log3

\frac{4}{l o g_{3} x}

\frac{4}{x}

\pm

\frac{9}{x}

Sets

Logarithms

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Question 45 Report

A rectangular polygon has 150o as the size of each interior angle. How many sides has the polygon?

12

10

9

8

Polynomials

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Question 46 Report

A number is selected at random between 20 and 30, both numbers inclusive. Find the probability that the number is a prime

\frac{2}{11}

\frac{5}{11}

\frac{6}{11}

\frac{8}{11}

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Question 47 Report

If x is negative, what is the range of values of x within which $ \frac{x+1}{3} > \frac{1}{X+3} $

3 < x < 4

-4 < x < -3

-2 < x < -1

-3 < x < 0

Answer Details

\frac{x + 1}{3}

>

\frac{1}{X + 3}

=

\frac{x + 1}{3}

>

\frac{x + 3}{X + 3}

= (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3)
x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0
= x(x + 3)9x + 4) > 0
Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0
= x > -4 (solution only)
Case 2 (+, -, -) = x > 0, x + 4 < 0

= x > 0, x < -3, x < -4 = x < -3(solution only)

Case 3 (-, +, -) = x < 0, x > -3, x < -4 = x < -0, -4 < x < 3(solutions)

Case 4 (-, -, +) = x < 0, x + 3 < 0, x + 4 > 0
= x < 0, x < -5, x > -4 = x < -0, -4 < x < -3(solution)

combining the solutions -4 < x < -3

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Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

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Question 48 Report