JAMB UTME - Mathematics - 1985

Find the area of a regular hexagon inscribed in a circle of radius 8cm

16

\sqrt{3}

cm2

96

\sqrt{3}

cm2

192

\sqrt{3}

cm2

16

\sqrt{3}

cm2

33cm2

Awọn alaye Idahun

Area of a regular hexagon = 8 x 8 x sin 60o
= 32 x $\frac{\sqrt{3}}{2}$ =
Area = 16 $\sqrt{3}$ x 6 = 96 $\sqrt{3}$ cm2

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Areas

Wo Idahun

Ibeere 2 Ìròyìn

The bearing of a bird on a tree from a hunter on the ground is ₦72oE. What is the bearing of the hunter from the birds?

S 18o W

S 72o W

S 72o E

S 27o E

S 27o W

Bearings

Wo Idahun

Ibeere 3 Ìròyìn

Solve the following equation $\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$

(

\frac{5}{2}

, 1)

(5, -4)

(2, 1)

(1,

\frac{- 5}{2}

)

(

\frac{- 5}{2}

, 1)

Awọn alaye Idahun

$\frac{2}{2 r - 1}$ - $\frac{5}{3}$ = $\frac{1}{r + 2}$
$\frac{2}{2 r - 1}$ - $\frac{1}{r + 2}$ = $\frac{5}{3}$
$\frac{2 r + 4 - 2 r + 1}{2 r - 1 (r + 2)}$ = $\frac{5}{3}$
$\frac{5}{(2 r + 1) (r + 2)}$ = $\frac{5}{3}$
5(2r - 1)(r + 2) = 15
(10r - 5)(r + 2) = 15
10r2 + 20r - 5r - 10 = 15
10r2 + 15r = 25
10r2 + 15r - 25 = 0
2r2 + 3r - 5 = 0
(2r2 + 5r)(2r + 5) = r(2r + 5) - 1(2r + 5)
(r - 1)(2r + 5) = 0
r = 1 or $\frac{- 5}{2}$

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 4 Ìròyìn

If a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = 6x. Find a in its simplest form

x2 - 1

x2 + 1

x2 + 4

x2 - 4

1

Awọn alaye Idahun

a{ $\frac{x + 1}{x - 2} - \frac{x - 1}{x + 2}$ } = a{ $\frac{(x + 1) (x + 2) - (x - 1) (x - 2)}{(x - 2) (x + 2)}$ }
= 6
$\frac{6 x}{x^{2} - 4}$ = 6x
a = x2 - 4

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 5 Ìròyìn

Arrange the following numbers in ascending order of magnitude $\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

$\frac{6}{7}$ < 0.865 < $\frac{13}{15}$

$\frac{13}{15}$ < $\frac{6}{7}$ < 0.865

$\frac{6}{7}$ < $\frac{13}{15}$ < 0.865

0.865 < $\frac{6}{7}$ < $\frac{13}{15}$

Awọn alaye Idahun

$\frac{6}{7}$ , $\frac{13}{15}$ , 0.8650

In ascending order, we have 0.8571, 0.8650, 0.8666

i.e. $\frac{6}{7}$ < 0.8650 < $\frac{13}{15}$

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

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 6 Ìròyìn

If ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + .....Find $\frac{1}{e^{x}}$

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 -

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{2.4.3}

+ ..

1 +

\frac{x}{2}

+

\frac{x^{2}}{1.2}

+

\frac{x^{3}}{1.2.3}

+

\frac{x^{4}}{1.23.4}

+ .........

1 - x +

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{3}}{2^{4} .3}

+ .........

1 +

\frac{x}{2}

+

\frac{x^{2}}{{1.2}^{3}}

+

\frac{x^{4}}{1.2.6}

+ ........

Awọn alaye Idahun

ex = 1 + $\frac{x}{2}$ + $\frac{x^{2}}{1.2}$ + $\frac{x^{3}}{1.2.3}$ + $\frac{x^{4}}{1.23.4}$

Wo Idahun

Ibeere 7 Ìròyìn

Solve for (x, y) in the equation 2x + y = 4: x^2 + xy = -12

(6, -8): (-2, 8)

(3, -4): (-1, 4)

(8, -4): (-1, 4)

(-8, 6): (8, -2)

(-4, 3): (4, -1)

Solution Of Linear Equations

Wo Idahun

Ibeere 8 Ìròyìn

The ratio of the length of two similar rectangular blocks is 2 : 3. If the volume of the larger block is 351cm3, then the volume of the other block is

234.00 cm3

526.50 cm3

166.00 cm3

687cm3

Awọn alaye Idahun

Let x represent total vol. 2 : 3 = 2 + 3 = 5
$\frac{3}{5}$ x = 351
x = $\frac{351 \times 5}{3}$
= 585
Volume of smaller block = $\frac{2}{3}$ x 585
= 234.00

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Polynomials

Wo Idahun

Ibeere 9 Ìròyìn

In the figure, GHIJKLMN is a cube of side a. Find the length of HN.

\sqrt{3 a}

3a

3a2

a

\sqrt{2}

a

\sqrt{3}

Awọn alaye Idahun

HJ2 = a2 + a2 = 2a2

HJ = $\sqrt{2 a^{2}} = a \sqrt{2}$

HN2 = a2 + (a $\sqrt{2}$ )2 = a2 + 2a2 = 3a2

HN = $\sqrt{3 a^{2}}$

= a $\sqrt{3}$ cm

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Polynomials

Mensuration

Inequalities

Wo Idahun

Ibeere 10 Ìròyìn

What is the probability that a number chosen at random from the intergers between 1 and 10 inclusive is either a prime or a multiple of 3?

\frac{7}{10}

\frac{3}{5}

\frac{4}{5}

\frac{3}{10}

Awọn alaye Idahun

Simple Space: (1, 2, 3, 4, 5, 6, 7, 8, 9, 10 = 10)
Prime: (2, 3, 5, 7)
multiples of 3: (3, 6, 9)
Prime or multiples of 3: (2, 3, 5, 6, 7, 9 = 6)
Probability = $\frac{6}{10}$
= $\frac{3}{5}$

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Probability

Wo Idahun

Ibeere 11 Ìròyìn

Find x if log9x = 1.5

72.0

27.0

36.0

3.5

24.5

Logarithms

Wo Idahun

Ibeere 12 Ìròyìn

A solid sphere of radius 4cm has a mass of 64kg. What will be the mass of a shell of the same metal whose internal and external radii are 2cm and 3cm respectively?

5kg

16kg

19kg

6kg

Awọn alaye Idahun

$\frac{1 \sqrt{3}}{(\frac{1}{2})^{2}}$
= $\frac{4}{\sqrt{3}}$
= $\frac{\sqrt{3}}{\sqrt{3}}$
= $\frac{4 \sqrt{3}}{\sqrt{3}}$
m = 64kg, V = $\frac{4 π r^{3}}{3}$
= $\frac{4 π (4)^{3}}{3}$
= $\frac{256 π}{3}$ x 10-6m3
density(P) = $\frac{Mass}{Volume}$
= $\frac{64}{\frac{256 π}{3 \times 10^{- 6}}}$
= $\frac{64 \times 3 \times 10^{- 6}}{256}$
= $\frac{3}{4 \times 10^{- 6}}$
m = PV = $\frac{3}{4 π \times 10^{- 6}}$ x $\frac{4}{3}$ $π$ [32 - 22] x 10-6
$\frac{3}{4 \times 10^{- 6}}$ x $\frac{4}{3}$ x 5 x 10-6
= 5kg

Wo Idahun

Ibeere 13 Ìròyìn

22 $\frac{1}{2}$ % of the Nigerian Naira is equal to 17 $\frac{1}{10}$ % of a foreign currency M. What is the conversion rate of the M to the Naira?

1M = 1

\frac{15}{57}

N

1M = 38

\frac{1}{4}

N

1M = 1

\frac{18}{57}

N

1M = 384

\frac{3}{4}

N

Awọn alaye Idahun

N = 22 $\frac{1}{2}$ %, M = 17 $\frac{1}{10}$ %
M = $\frac{171}{10}$ %, N = $\frac{45}{2}$
$\frac{45}{2}$ x $\frac{10}{171}$
= $\frac{225}{171}$
= 1 $\frac{54}{171}$
= 1 $\frac{18}{57}$

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Ka ìwé ẹ̀kọ́ nípa Financial Arithmetic (WAEC)

Percentages

Financial Arithmetic

Wo Idahun

Ibeere 14 Ìròyìn

In the figure, PQ is the tangent from P to the circle QRS with SR as its diameter. If QRS = $θ$ $^{\circ}$ and RQP = $ϕ$ $^{\circ}$ , which of the following relationships between $θ$ $^{\circ}$ and $ϕ$ $^{\circ}$ is correct

θ

^{\circ}

+

ϕ

^{\circ}

= 902

ϕ

^{\circ}

= 902 - 2

θ

^{\circ}

θ

^{\circ}

=

ϕ

^{\circ}

ϕ

^{\circ}

= 2

θ

^{\circ}

θ

^{\circ}

+ 2

ϕ

^{\circ}

Awọn alaye Idahun

180 - $ϕ$ $^{\circ}$ = $θ$ $^{\circ}$ + $ϕ$ $^{\circ}$ (Sum of opposite interior angle equal to its exterior angle)

180 = 2 $ϕ$ + $θ$ $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Introductory Calculus

Trigonometry

Circles

Wo Idahun

Ibeere 15 Ìròyìn

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) is equal to

1

(

\sqrt{2} + 4 \sqrt{2}

)

(6

\sqrt{2}

8

Awọn alaye Idahun

( $\sqrt{2} + 4 \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ ) ( $\sqrt{3} + \sqrt{2}$ )
( $\sqrt{3} + \sqrt{2}$ + ( $\sqrt{3} + \sqrt{2}$ ) = 3 + ( $\sqrt{6} - \sqrt{6}$ ) - 2
= 1

Wo Idahun

Ibeere 16 Ìròyìn

In a class of 120 students, 18 of them scored an A grade in mathematics. If the section representing the A grade students on a pie chart has angle Zo at the centre of the circle, what is Z?

15

27

50

52

54

Awọn alaye Idahun

Total students = 120
grade = 18
Zo = $\frac{18}{120}$ x $\frac{360}{1}$
= 54o

Wo Idahun

Ibeere 17 Ìròyìn

If the quadratic function 3x2 - 7x + R is a perfect square, find R

\frac{49}{24}

\frac{49}{12}

\frac{49}{13}

\frac{49}{3}

\frac{49}{36}

Awọn alaye Idahun

3x2 - 7x + R. Computing the square, we have
x2 - $\frac{7}{3}$ = - $\frac{R}{3}$
( $\frac{x}{1} - \frac{7}{6}$ )2 = - $\frac{R}{3}$ + $\frac{49}{36}$
$\frac{- R}{3}$ + $\frac{49}{36}$ = 0
R = $\frac{49}{36}$ x $\frac{3}{1}$
= $\frac{49}{12}$

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 18 Ìròyìn

Find correct to two decimals places 100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$

100.02

1000.02

100.22

100.01

100.51

Awọn alaye Idahun

100 + $\frac{1}{100}$ + $\frac{3}{1000}$ + $\frac{27}{10000}$
$\frac{1000, 000 + 100 + 30 + 27}{10000}$ = $\frac{1, 000.157}{10000}$
= 100.02

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Number Bases

Areas

Wo Idahun

Ibeere 19 Ìròyìn

If the hypotenuse of right angled isosceles triangle is 2, what is the length of each of the other sides?

\sqrt{2}

\frac{1}{2}

22

1

2 - 1

Awọn alaye Idahun

45o = $\frac{x}{2}$ , Since 45o = $\frac{1}{\sqrt{2}}$
x = 2 x $\frac{1}{\sqrt{2}}$
= $\frac{2 \sqrt{2}}{2}$
= $\sqrt{2}$

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

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Angles And Intercepts On Parallel Lines

Statistics

Angles

Wo Idahun

Ibeere 20 Ìròyìn

The number of goals scored by a football team in 20 matches is shown below:

$\begin{array}{cc} No. of goals & 0 & 1 & 2 & 3 & 4 & 5 \\ No. of matches & 3 & 5 & 7 & 3 & 1 & 0 \end{array}$

What are the values of the mean and the mode respectively?

(1.75, 5)

(1.75, 2)

(1.75, 1)

(65, 74)

Awọn alaye Idahun

$\begin{array}{cc} x & f & f x \\ 0 & 3 & 0 \\ 1 & 5 & 5 \\ 2 & 7 & 14 \\ 3 & 4 & 12 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \end{array}$
$\sum f$ = 20
$\sum f x$ = 35
Mean = $\frac{\sum f x}{\sum f}$
= $\frac{35}{20}$
= $\frac{7}{4}$
= 1.75
Mode = 2
= 1.75, 2

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Probability

Statistics

Wo Idahun

Ibeere 21 Ìròyìn

Find the values of p for which the equation x2 - (p - 2)x + 2p + 1 = 0

(21, 0)

(0, 12)

(1, 2)

(3, 4)

(4, 5)

Quadratic Equations

Wo Idahun

Ibeere 22 Ìròyìn

Make c the subject of formula v = 1 - $\frac{a}{5}$ (b + $\frac{3 c}{7}$ )

[

\frac{7}{3}

+

\frac{5}{a}

(v - 1)] + b

[

\frac{7 b}{3}

+

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5}{a}

(v - 1)]

\frac{7}{3}

[b +

\frac{5 v - 1}{a}

]

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Wo Idahun

Ibeere 23 Ìròyìn

John gives one-third of his money to Janet who has ₦105.00. He then finds that his money is reduced to one-fourth of what Janet now has. Find how much money john has at first

₦45.00

₦48.00

₦52.00

₦60.00

₦53.00

Awọn alaye Idahun

Let x be John's money, Janet already had ₦105, $\frac{1}{3}$ of x was given to Janet
Janet now has $\frac{1}{3^{2}}$ x + 105 = $\frac{x + 315}{3}$
John's money left = $\frac{2}{3}$ x
= $\frac{\frac{1}{4} (x + 315)}{3}$
= $\frac{2}{3}$
24x = 3x + 945
∴ x = 45

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

Fractions, Decimals And Approximations

Wo Idahun

Ibeere 24 Ìròyìn

In $△$ XYZ, XKZ = 90O, XK = 15cm, XY = 35cm and YK = 8cm. Find the area of $△$ XYZ

190 sq.cm

20 sq.cm

210 sq.cm

160sq.cm

320sq.cm

Awọn alaye Idahun

By Pythagoras, KZ2 = 252 - 52

KZ2 = (25 + 15)(25 - 15) = 400

KZ = $\sqrt{400}$ = 20

area of XYZ = $\frac{1}{2} \times 28 \times 15$

= 210 sq. cm

Wo Idahun

Ibeere 25 Ìròyìn

a sum of money was invested at 8% per annum simple interest. If after 4 years the money amounts to N330.00. Find the amount originally invested

N180.00

n165.00

N150.00

N250.00

N200.00

Awọn alaye Idahun

S.I = $\frac{P T R}{100}$

T = 4yrs, R = 8%, a = N330.00

330 - P = $\frac{P T R}{100}$ , A = P + I

i.e. A = P + $\frac{P T R}{100}$

330 = P + $\frac{P (4) (8)}{100}$

33000 = 32P + 100p

132P = 33000

P = N250.00

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

Positive And Negative Integers, Rational Numbers

Wo Idahun

Ibeere 26 Ìròyìn

If 32y + 6(3y) = 27. Find y

3

-1

2

-3

1

Solution Of Linear Equations

Wo Idahun

Ibeere 27 Ìròyìn

In $△$ XYZ, XY = 13cm, YZ = 9cm, XZ = 11cm and XYZ = $θ$ . Find cos $θ$ o

\frac{4}{39}

\frac{43}{39}

\frac{209}{3}

\frac{43}{78}

Awọn alaye Idahun

cos $θ$ = $\frac{13^{2} + 9^{2} - 11^{2}}{2 (13) (9)}$
= $\frac{169 + 81 - 21}{26 \times 9}$
cos $θ$ = $\frac{129}{26 \times 9}$
= $\frac{43}{78}$

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Wo Idahun

Ibeere 28 Ìròyìn

Find the missing value in the table below

x	-4	-3	-2	-1	0	1	2	3
y=4?3x?x^3	80		18	8	4	0	-10	-32

-32

-14

40

22

37

Wo Idahun

Ibeere 29 Ìròyìn

In the figure, the area of the shaded segment is

3

π

9

\frac{\sqrt{3}}{4}

3

π - 3 \frac{\sqrt{3}}{4}

\frac{(\sqrt{3 - π)}}{4}

π + \frac{9 \sqrt{3}}{4}

Awọn alaye Idahun

Area of sector = $\frac{120}{360} \times π \times (3)^{2} = 3 π$

Area of triangle = $\frac{1}{2} \times 3 \times 3 \times \sin 120^{o}$

= $\frac{9}{2} \times \frac{\sqrt{3}}{2} = \frac{9 \sqrt{3}}{4}$

Area of shaded portion = 3 $π - \frac{9 \sqrt{3}}{4}$

= 3 $π - 3 \frac{\sqrt{3}}{4}$

Ka ìwé ẹ̀kọ́ nípa Volumes (WAEC)

Ka ìwé ẹ̀kọ́ nípa Loci (WAEC)

Volumes

Loci

Wo Idahun

Ibeere 30 Ìròyìn

Two points X and Y both on latitude 60oS have longitude 147oE and 153oW respectively. Find to the nearest kilometer the distance between X and Y measured along the parallel of latitude (Take 2 $π$ R = 4 x 104km, where R is the radius of the earth)

16667km

28850km

8333km

2233km

Awọn alaye Idahun

Length of an area = $\frac{θ}{360}$ × 2 $π$ r
Longitude difference = 147 + 153 = 300oN
distance between xy = $\frac{θ}{360}$ × 2 $π$ R cos60o
= $\frac{300}{360}$ × 4 × 104 × $\frac{1}{2}$
= 1.667 × 104km (1667 km)

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 31 Ìròyìn

In the figure, MNQP is a cyclic quadrilateral. MN and Pq are produced to meet at X and NQ and MP are produced to meet at Y. If MNQ = 86 $^{\circ}$ and NQP = 122 $^{\circ}$ find (x $^{\circ}$ , y $^{\circ}$ )

28

^{\circ}

, 36

^{\circ}

36

^{\circ}

, 28

^{\circ}

43

^{\circ}

, 61

^{\circ}

61

^{\circ}

, 43

^{\circ}

36

^{\circ}

, 43

^{\circ}

Awọn alaye Idahun

y $^{\circ}$ = 180 $^{\circ}$ - (86 $^{\circ}$ + 58 $^{\circ}$ )

180 - 144 = 36 $^{\circ}$

x $^{\circ}$ = 180 - (94 + 58)

180 -152 = 28

(x $^{\circ}$ , y $^{\circ}$ ) = (28 $^{\circ}$ , 36 $^{\circ}$ )

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Differentiation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Polynomials

Differentiation

Inequalities

Wo Idahun

Ibeere 32 Ìròyìn

The factors of 9 - (x2 - 3x - 1)2 are

-(x - 4)(x + 1) (x - 1)(x - 2)

(x - 4)(x - 2) (x - 1)(x + 1)

-(x - 2)(x + 1) (x - 2) (x - 1)

(x - 2)(x + 2) (x - 1)(x + 1)

Algebraic Expressions

Wo Idahun

Ibeere 33 Ìròyìn

If f(x - 2) = 4x2 + x + 7, find f(1)

12

27

7

46

17

Solution Of Linear Equations

Algebraic Expressions

Wo Idahun

Ibeere 34 Ìròyìn

In the equation below, Solve for x if all the numbers are in base 2: $\frac{11}{x}$ = $\frac{1000}{x + 101}$

101

11

110

111

10

Awọn alaye Idahun

$\frac{11}{x}$ = $\frac{1000}{x + 101}$ = 11(x + 101)
1000x = 11x + 1111
1000x - 11x = 1111
101x = 1111
x = $\frac{1111}{101}$
x = 11

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 35 Ìròyìn

Factorize abx2 + 8y - 4bx - 2axy

(ax - 4)(bx - 2y)

(ax + b)(x - 8y)

(ax - 2y)(bx - 4)

(bx - 4)(ax - 2y)

(abx - 4)(x - 2y)

Algebraic Expressions

Wo Idahun

Ibeere 36 Ìròyìn

Without using table, calculate the value of 1 + sec2 30o

2

\frac{1}{3}

\frac{2}{15}

\frac{5}{3}

3

\frac{1}{2}

Awọn alaye Idahun

1 + sec2 30o = sec 30o
= $\frac{2}{\sqrt{3}}$
$\frac{(2)^{2}}{3}$
= $\frac{4}{3}$
1 + sec2 30o = sec 30o
= 1 + $\frac{4}{\sqrt{3}}$
= 2 $\frac{1}{3}$

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

Ka ìwé ẹ̀kọ́ nípa Introductory Calculus (WAEC)

Ka ìwé ẹ̀kọ́ nípa Loci (JAMB)

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Sine, Cosine And Tangent Of An Angle

Introductory Calculus

Loci

Matrices And Determinants

Wo Idahun

Ibeere 37 Ìròyìn

If cos $θ$ = $\frac{\sqrt{3}}{2}$ and $θ$ is less than 90o. Calculate cos $\frac{90 - θ}{s i n^{2} θ}$

\frac{4}{\sqrt{3}}

\sqrt{\frac{3}{2}}

\sqrt{\frac{76}{2}}

4

\sqrt{\frac{76}{2}}

Awọn alaye Idahun

cos $\frac{90 - θ}{s i n^{2} θ}$
= $\frac{\tan θ}{s i n^{2} θ}$
Sin2 $θ$ = $\frac{1}{4}$
$\frac{c o s (90 - θ)}{s i n^{2} θ}$
= $\frac{1}{\sqrt{3}}$
= $\frac{4}{\sqrt{3}}$

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

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Wo Idahun

Ibeere 38 Ìròyìn

At what real value of x do the curves whose equations are y = x3 + x and y = x2 + 1 intersect?

-2

2

-1

9

1

Awọn alaye Idahun

y = x3 + x and y = x2 + 1
$\begin{array}{cc} x & - 2 & - 1 & 0 & 1 & 2 \\ Y = x^{3} + x & - 10 & - 2 & 0 & 2 & 10 \\ y = x^{2} + 1 & 5 & 2 & 1 & 2 & 5 \end{array}$
The curves intersect at x = 1

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Coordinate Geometry

Quadratic Equations

Wo Idahun

Ibeere 39 Ìròyìn

If three numbers P, Q, R are in ratio 6 : 4 : 5, find the value of $\frac{3 p - q}{4 q + r}$

3

2

3

2

3

18

Awọn alaye Idahun

P : Q : r = 6 : 4 : 5
5 = 6 + 4 + 5
= 15
P = $\frac{6}{15}$ , q = $\frac{4}{15}$ , r = $\frac{5}{15}$ = $\frac{1}{3}$
To find $\frac{3 p - q}{4 q + r}$
3p - q = 3 x $\frac{6}{15}$ - $\frac{4}{15}$
$\frac{18}{15}$ - $\frac{4}{15}$ = $\frac{14}{15}$
∴ 4q + r = 4 x $\frac{4}{15}$ + $\frac{5}{15}$
$\frac{16}{15}$ = $\frac{16}{15}$ + $\frac{5}{15}$
= $\frac{21}{15}$
$\frac{14}{15}$ x $\frac{15}{21}$ = $\frac{14}{21}$
= $\frac{2}{3}$

Wo Idahun

Ibeere 40 Ìròyìn

Which of these angles can be constructed using ruler and a pair of compasses only?

115o

125o

135o

145o

Volumes

Angles

Wo Idahun

Ibeere 41 Ìròyìn

PRSQ is a trapezium of area 14cm2 in which PQ||RS. If PQ = 4cm and SR = 3cm, Find the area of SQR in cm2

7.0

6.0

5.2

5.0

Awọn alaye Idahun

Area of trapezium = 14cm2
Area of trapezium = $\frac{1}{2}$ (a + b)h
14 = $\frac{1}{2}$ (4 + 3)h
14 = $\frac{7}{2}$ h
h = $\frac{14 \times 2}{7}$
= 4
Area of SQR = $\frac{1}{2}$ (3 x 4)
= $\frac{12}{2}$
= 6.0

Wo Idahun

Ibeere 42 Ìròyìn

In a restaurant, the cost of providing a particular type of food is partly constant and partially inversely proportional to the number of people. If cost per head for 100 people is 30k and the cost for 40 people is 60k, Find the cost for 50 people?

15k

20k

50k

40k

45k

Awọn alaye Idahun

C = a + k
$\frac{1}{N}$ = c
= $\frac{a N + k}{N}$
CN = aN + K
30(100) = a(100) + k
3000 = 100a + k.......(i)
60(40) = a(40) + k
2400 = 40a + k.......(ii)
eqn (i) - eqn (ii)
600 = 60a
a = 10
subt. for a in eqn (i) 3000 = 100(10) + K
3000 - 1000 = k
k = 2000
CN = 10N + 2000. when N = 50,
50C = 10(50) + 2000
50C = 500 + 2000
C = $\frac{2500}{50}$
= 50k

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Variation

Linear Inequalities

Wo Idahun

Ibeere 43 Ìròyìn

If two fair coins are tossed, what is the probability of getting at least one head?

\frac{1}{4}

\frac{1}{2}

1

\frac{43}{78}

\frac{3}{4}

Awọn alaye Idahun

Prob. of getting at least one head
Prob. of getting one head + prob. of getting 2 heads
= $\frac{1}{4}$ + $\frac{2}{4}$
= $\frac{3}{4}$

Wo Idahun

Ibeere 44 Ìròyìn

A bag contains 4 white balls and 6 red balls. Two balls are taken from the bag without replacement. What is the probability that they are both red?

\frac{2}{3}

\frac{2}{15}

\frac{1}{2}

\frac{1}{3}

Awọn alaye Idahun

P(R1) = $\frac{6}{10}$
= $\frac{2}{3}$
P(R1 n R11) = P(both red)
$\frac{3}{5}$ x $\frac{5}{9}$
= $\frac{1}{3}$

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Probability (JAMB)

Probability

Wo Idahun

Ibeere 45 Ìròyìn

Solve the simultaneous equations 2x - 3y + 10 = 10x - 6y = 5

x = 2

\frac{1}{2}

, y = 3

\frac{1}{2}

x =

\frac{1}{2}

, y =

\frac{3}{2}

x = 2

\frac{1}{4}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 3

\frac{1}{2}

x = 2

\frac{1}{3}

, y = 2

\frac{1}{2}

Awọn alaye Idahun

2x - 3y = -10; 10x - 6y = -5
2x - 3y = -10 x 2
10x - 6y = 5
4x - 6y = -20 .......(i)
10x - 6y = 5.......(ii)
eqn(ii) - eqn(1)
6x = 15
x = $\frac{15}{6}$
= $\frac{5}{2}$
x = 2 $\frac{1}{2}$
Sub. for x in equ.(ii) 10( $\frac{5}{2}$ ) - 6y = 5
y = 3 $\frac{1}{2}$

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 46 Ìròyìn

List all integers satisfying the inequality -2 $\leq$ 2 x -6 < 4

2, 3, 4, 5

2, 3, 4

2, 5

3, 4, 5

4, 5

Awọn alaye Idahun

-2 $\leq$ 2x - 6 < 4 = 2x - 6 < 4

= 2x < 10

= x < 5

2x $\geq$ -2 + 6 $\geq$

= x $\geq$ 2

∴ 2 $\leq$ x < 5 [2, 3, 4]

Ka ìwé ẹ̀kọ́ nípa Inequalities (JAMB)

Inequalities

Wo Idahun

Ibeere 47 Ìròyìn

Write h in terms of a, b, c, d if a = $\frac{b (1 - c h)}{a - d h}$

h = $\frac{a - b}{a d}$

h = $\frac{1 - b}{a d - b c}$

h = $\frac{(a - b)^{2}}{a d - b c}$

h =

\frac{a - b}{a d - b c}

h = $\frac{b - a}{a b - d c}$

Awọn alaye Idahun

a = $\frac{b (1 - c h)}{a - d h}$
a = $\frac{b - b c h}{1 - d h}$
= a - adh
= b - bch
a - b = bch + adn
a - b = adh
a - b = h(ad - bc)
h = $\frac{a - b}{a d - b c}$

Wo Idahun






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