JAMB UTME - Mathematics - 1993

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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Frage 2 Bericht

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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Frage 3 Bericht

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}

Antwort anzeigen

Frage 4 Bericht

The shaded portion in the Venn diagram is

x

\cap

z

x

^{\circ}

\cap

y

\cap

z

x

\cap

y

^{\circ}

\cap

z

x

\cap

y

\cap

z

^{\circ}

Antwortdetails

The shaded part exists on x

\cap

z but not in y

Lies die Unterrichtsnotiz zu Logical Reasoning (WAEC)

Lies die Unterrichtsnotiz zu Integration (JAMB)

Lies die Unterrichtsnotiz zu Matrices And Determinants (JAMB)

Logical Reasoning

Integration

Matrices And Determinants

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Frage 5 Bericht

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

Antwortdetails

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

Lies die Unterrichtsnotiz zu Integration (JAMB)

Integration

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Frage 6 Bericht

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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Frage 7 Bericht

The angle between latitudes 30oS and 13oN is

17o

33o

43o

53o

Angles

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Frage 8 Bericht

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

Antwort anzeigen

Frage 9 Bericht

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

Antwortdetails

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

Antwort anzeigen

Frage 10 Bericht

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}

Antwortdetails

< 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 $^{\circ}$

Lies die Unterrichtsnotiz zu Circles (WAEC)

Circles

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Frage 11 Bericht

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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Frage 12 Bericht

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}

Antwortdetails

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}

Lies die Unterrichtsnotiz zu Probability (WAEC)

Probability

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Frage 13 Bericht

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

17

20

25

30

Antwortdetails

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

Lies die Unterrichtsnotiz zu Integration (JAMB)

Integration

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Frage 14 Bericht

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)

Antwortdetails

(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)

Lies die Unterrichtsnotiz zu Quadratic Equations (WAEC)

Quadratic Equations

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Frage 15 Bericht

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

3

4

12

17

Antwortdetails

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.

Lies die Unterrichtsnotiz zu Sine, Cosine And Tangent Of An Angle (WAEC)

Lies die Unterrichtsnotiz zu Modular Arithmetic (WAEC)

Lies die Unterrichtsnotiz zu Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Financial Arithmetic

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Frage 16 Bericht

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}

Antwortdetails

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 $^{\circ}$

Lies die Unterrichtsnotiz zu Circles (WAEC)

Circles

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Frage 17 Bericht

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

45o, 225o

135o, 315o

45o, 315o

135o, 225o

Antwortdetails

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

Lies die Unterrichtsnotiz zu Trigonometry (JAMB)

Trigonometry

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Frage 18 Bericht

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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Frage 19 Bericht

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

Antwortdetails

x

*

y = xy
x

*

2 = x2
x

*

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

Lies die Unterrichtsnotiz zu Binary Operations (JAMB)

Binary Operations

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Frage 20 Bericht

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

68

^{\circ}

102

^{\circ}

107

^{\circ}

141

^{\circ}

Antwortdetails

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 $^{\circ}$

Lies die Unterrichtsnotiz zu Angles And Intercepts On Parallel Lines (WAEC)

Lies die Unterrichtsnotiz zu Statistics (WAEC)

Angles And Intercepts On Parallel Lines

Statistics

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Frage 21 Bericht

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

3

\frac{2}{3}

7

-4

Antwortdetails

\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

Lies die Unterrichtsnotiz zu Positive And Negative Integers, Rational Numbers (WAEC)

Lies die Unterrichtsnotiz zu Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

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Frage 22 Bericht

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

Antwort anzeigen

Frage 23 Bericht

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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Frage 24 Bericht

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

3

4

5

8

Antwortdetails

log5(62.5) - log5(

\frac{1}{2}

)
= log5

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

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

Lies die Unterrichtsnotiz zu Positive And Negative Integers, Rational Numbers (WAEC)

Lies die Unterrichtsnotiz zu Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

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Frage 25 Bericht

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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Frage 26 Bericht

change 7110 to base 8

1078

1068

718

178

Number Bases

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Frage 27 Bericht

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

Antwortdetails

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

Lies die Unterrichtsnotiz zu Modular Arithmetic (WAEC)

Modular Arithmetic

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Frage 28 Bericht

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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Frage 29 Bericht

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

Antwortdetails

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

Antwort anzeigen

Frage 30 Bericht

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

Antwort anzeigen

Frage 31 Bericht

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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Frage 32 Bericht

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

5

4

3

2

1

Antwortdetails

\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

Lies die Unterrichtsnotiz zu Fractions, Decimals, Approximations And Percentage (JAMB)

Lies die Unterrichtsnotiz zu Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

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Frage 33 Bericht

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

5

1

3

6

Antwortdetails

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

Lies die Unterrichtsnotiz zu Algebraic Expressions (WAEC)

Algebraic Expressions

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Frage 34 Bericht

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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Frage 35 Bericht

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

Antwort anzeigen

Frage 36 Bericht

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Introductory Calculus

Trigonometry

Antwort anzeigen

Frage 37 Bericht

$Weight(s)$	$0-10$	$10-20$	$20-30$	$40-50$
Number of coconuts	$10$	$27$	$19$	$6$	$2$

Estimate the mode of the frequency distribution above.

13.2g

15.0g

16.8g

17.5g

Antwortdetails

Mode = a +

\frac{(b - a) (F_{m} - F_{b})}{2 F_{m} - F_{a} - F_{b}}

=

L_{1} + \frac{Δ_{1} x^{c}}{Δ_{1} + Δ_{2}}

=

10 + \frac{(20 - 10) (27 - 10)}{2 (27) - 10 - 19}

= 10 +

\frac{170}{25}

= 10 + 6.8
= 16.8

Lies die Unterrichtsnotiz zu Representation Of Data (JAMB)

Lies die Unterrichtsnotiz zu Probability (WAEC)

Lies die Unterrichtsnotiz zu Linear Inequalities (WAEC)

Lies die Unterrichtsnotiz zu Statistics (WAEC)

Representation Of Data

Probability

Linear Inequalities

Statistics

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Frage 38 Bericht

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

Antwort anzeigen

Frage 39 Bericht

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

Antwort anzeigen

Frage 40 Bericht

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}

Antwortdetails

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

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

Lies die Unterrichtsnotiz zu Surds (radicals) (WAEC)

Surds (radicals)

Antwort anzeigen

Frage 41 Bericht

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)}

Antwortdetails

\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}

Lies die Unterrichtsnotiz zu Algebraic Expressions (WAEC)

Algebraic Expressions

Antwort anzeigen

Frage 42 Bericht

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

705

70,000

70, 480

70, 500

Fractions, Decimals And Approximations

Number Bases

Antwort anzeigen

Frage 43 Bericht

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

Antwortdetails

Surface area = 154cm2 (area of sphere)
4

π

r2 = 154
r

\sqrt{\frac{154}{4 π}}

= 3.50cm

Antwort anzeigen

Frage 44 Bericht

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

\frac{1}{2}

1

2

3

Antwortdetails

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

Lies die Unterrichtsnotiz zu Number Bases (WAEC)

Number Bases

Antwort anzeigen

Frage 45 Bericht

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

Antwortdetails

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

Lies die Unterrichtsnotiz zu Representation Of Data (JAMB)

Lies die Unterrichtsnotiz zu Measures Of Location (JAMB)

Representation Of Data

Measures Of Location

Antwort anzeigen

Frage 46 Bericht

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

Antwortdetails

\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

Lies die Unterrichtsnotiz zu Fractions, Decimals And Approximations (WAEC)

Lies die Unterrichtsnotiz zu Positive And Negative Integers, Rational Numbers (WAEC)

Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

Antwort anzeigen

Frage 47 Bericht

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

x + 3