JAMB UTME - Mathematics - 1993

Ibeere 1 Ìròyìn

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

-x3 + 7x2 - x - 7

-x3 = 7x + 7

x - 7

x + 7

Algebraic Expressions

Wo Idahun

Ibeere 2 Ìròyìn

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

Awọn alaye Idahun

x

*

y = xy
x

*

2 = x2
x

*

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

Ka ìwé ẹ̀kọ́ nípa Binary Operations (JAMB)

Binary Operations

Wo Idahun

Ibeere 3 Ìròyìn

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

Wo Idahun

Ibeere 4 Ìròyìn

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

2

4

10

11

Measures Of Dispersion

Wo Idahun

Ibeere 5 Ìròyìn

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}

Awọn alaye Idahun

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

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

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Surds (radicals)

Wo Idahun

Ibeere 6 Ìròyìn

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

Wo Idahun

Ibeere 7 Ìròyìn

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)}

Awọn alaye Idahun

\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}

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 8 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Location (JAMB)

Representation Of Data

Measures Of Location

Wo Idahun

Ibeere 9 Ìròyìn

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

3

4

5

8

Awọn alaye Idahun

log5(62.5) - log5(

\frac{1}{2}

)
= log5

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

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

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

Wo Idahun

Ibeere 10 Ìròyìn

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

12

10

9

8

Polynomials

Wo Idahun

Ibeere 11 Ìròyìn

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

Awọn alaye Idahun

Surface area = 154cm2 (area of sphere)
4

π

r2 = 154
r

\sqrt{\frac{154}{4 π}}

= 3.50cm

Wo Idahun

Ibeere 12 Ìròyìn

change 7110 to base 8

1078

1068

718

178

Number Bases

Wo Idahun

Ibeere 13 Ìròyìn

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

Wo Idahun

Ibeere 14 Ìròyìn

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

Estimate the mode of the frequency distribution above.

13.2g

15.0g

16.8g

17.5g

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Representation Of Data

Probability

Linear Inequalities

Statistics

Wo Idahun

Ibeere 15 Ìròyìn

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

5

4

3

2

1

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals, Approximations And Percentage (JAMB)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

Wo Idahun

Ibeere 16 Ìròyìn

The angle between latitudes 30oS and 13oN is

17o

33o

43o

53o

Angles

Wo Idahun

Ibeere 17 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 18 Ìròyìn

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}

Awọn alaye Idahun

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 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 19 Ìròyìn

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

Wo Idahun

Ibeere 20 Ìròyìn

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

Wo Idahun

Ibeere 21 Ìròyìn

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

Wo Idahun

Ibeere 22 Ìròyìn

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Introductory Calculus

Trigonometry

Wo Idahun

Ibeere 23 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 24 Ìròyìn

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

Wo Idahun

Ibeere 25 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 26 Ìròyìn

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

17

20

25

30

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Wo Idahun

Ibeere 27 Ìròyìn

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}

Awọn alaye Idahun

< 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 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 28 Ìròyìn

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

3

4

12

17

Awọn alaye Idahun

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.

Ka ìwé ẹ̀kọ́ nípa Sine, Cosine And Tangent Of An Angle (WAEC)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Financial Arithmetic

Wo Idahun

Ibeere 29 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Integration

Wo Idahun

Ibeere 30 Ìròyìn

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

Wo Idahun

Ibeere 31 Ìròyìn

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

Wo Idahun

Ibeere 32 Ìròyìn

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

Wo Idahun

Ibeere 33 Ìròyìn

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

Wo Idahun

Ibeere 34 Ìròyìn

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

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Fractions, Decimals And Approximations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

Wo Idahun

Ibeere 35 Ìròyìn

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

\frac{1}{2}

1

2

3

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 36 Ìròyìn

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

Wo Idahun

Ibeere 37 Ìròyìn

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}

Awọn alaye Idahun

The shaded part exists on x

\cap

z but not in y

Ka ìwé ẹ̀kọ́ nípa Logical Reasoning (WAEC)

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Logical Reasoning

Integration

Matrices And Determinants

Wo Idahun

Ibeere 38 Ìròyìn

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

x + 3

x - 3

x - 5

x + 5

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 39 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Modular Arithmetic

Wo Idahun

Ibeere 40 Ìròyìn

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)

Awọn alaye Idahun

(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)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 41 Ìròyìn

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

Wo Idahun

Ibeere 42 Ìròyìn

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

705

70,000

70, 480

70, 500

Fractions, Decimals And Approximations

Number Bases

Wo Idahun

Ibeere 43 Ìròyìn

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

3

\frac{2}{3}

7

-4

Awọn alaye Idahun

\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

Ka ìwé ẹ̀kọ́ nípa Positive And Negative Integers, Rational Numbers (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Wo Idahun

Ibeere 44 Ìròyìn

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

45o, 225o

135o, 315o

45o, 315o

135o, 225o

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Trigonometry

Wo Idahun

Ibeere 45 Ìròyìn

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

15o

25o

40o

50o

55o

Algebraic Expressions

Wo Idahun

Ibeere 46 Ìròyìn

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

5

1

3

6

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 47 Ìròyìn

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

Wo Idahun

Ibeere 48 Ìròyìn

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}

Wo Idahun

Ibeere 49 Ìròyìn

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

68

^{\circ}

102

^{\circ}

107

^{\circ}

141

^{\circ}

Awọn alaye Idahun

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 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Angles And Intercepts On Parallel Lines (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Angles And Intercepts On Parallel Lines

Statistics

Wo Idahun

Ibeere 50 Ìròyìn

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}

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

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






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