JAMB UTME - Mathematics - 1987

What is the circumference of latitude 0oS if R is the radius of the earth?

R cos

θ

2

π

R cos

θ

R sin

θ

2

π

R sin

θ

Akọwa Nkọwa

Circumference of Latitude 0oS where R is the radius of the earth is = 2

π

R cos

θ

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Lee Azịza

Ajụjụ 2 Ripọtì

If a u2 - 3v2 and b = 2uv + v2 evaluate (2a - b)(a - b2), when u = 1 and v = -1

9

15

27

33

Akọwa Nkọwa

a = u2 - 3y2 = (1)2 - 3(-1) = -2
b = 2uu + v2 + v2
= 2(1)(-1) + (-1)2 = -1
∴ 2(2a - b)(a - b∴) = [2(-2) - 1](-2 - (-1)2)
= -3 - 3
= 9

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 3 Ripọtì

In the diagram, PQRS is a circle center O. PQR is a diameter and ∠PRQ = 40o. Calculate ∠QSR.

30o

40o

45o

50o

Circles

Lee Azịza

Ajụjụ 4 Ripọtì

The minimum value of $y$ in the equation $y=x^2-6x+8$

8

3

0

-1

5

Quadratic Equations

Lee Azịza

Ajụjụ 5 Ripọtì

A train moves from P to Q at an average speed of 90km/h and immediately returns from Q to P through the same route at an average speed of 45km/h. Find the average speed for the entire journey

55.00km/h

60.00km/h

67.50km\h

75.00km\h

Akọwa Nkọwa

Average speed from P to Q = 90km\h
Average speed from O to P = 45km/h
Average for the entire journey = 90 + 45

\frac{135}{2}

= 67.50 km/h

Lee Azịza

Ajụjụ 6 Ripọtì

Simplify without using tables $ \frac{\log_2 6}{\log_2 8} - \frac{\log_2 3}{\log_2 \frac{1}{2}} $

1

5

1

2

-1

2

log²3

log²7

Akọwa Nkọwa

log26 - log23 = log2 (

\frac{6}{3}

) = log22
log28 - 2

\frac{1}{2}

log28 - log2

\frac{1}{2}

log28 - log2

\frac{1}{4}

\frac{l o g_{2} 2}{l o g_{2} 32} = \frac{l o g_{2} 2}{5 l o g_{2} 2}

=

\frac{1}{5}

N.B. log22 = 1

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 7 Ripọtì

Solve the inequality x - 1 > 4(x + 2)

x > -3

x < -3

2 < x < 3

-3 < x < -2

Akọwa Nkọwa

x - 1 > 4(x + 2) = x - 1 > 4x + 8
4x + 8 < x - 1 = 4x - x < -1 -8

= 3x < -9

∴ x < -3

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Linear Inequalities

Areas

Lee Azịza

Ajụjụ 8 Ripọtì

An $ (n-2)^2 $ sided figure has n diagonals, find the number n of diagonals for a 25 sided figure

8

7

6

9

10

Variation

Lee Azịza

Ajụjụ 9 Ripọtì

Evaluate without using tables $ (0.008)^{\frac{1}{3}} \times (0.16)^{-\frac{3}{2}} $

625

8

625

1

8

Akọwa Nkọwa

(0.008) -

\frac{1}{3}

x (0.16) -

\frac{3}{2}

= (8 x 10-3)10-3 x (16 x 10-2)-

\frac{3}{2}

=

\frac{(2^{3})}{10^{3}}

- 3 x

\frac{(2^{4})}{10}

-

\frac{- 3}{2}

=

\frac{625}{8}

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Gụọ akwụkwọ nkuzi na Graphs Of Linear And Quadratic Functions (WAEC)

Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 10 Ripọtì

Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles.

210o

150o

105o

50o

Akọwa Nkọwa

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Monagon = 9
Where 3 are equal and 6 other angles = 1110o
(2 x 9 - 4)90o = (18 - 4)90o
14 x 90o = 1260o
9 angles = 12600, 6 angles = 110o
Remaining 3 angles = 1260o - 1110o = 150o
Size of one of the 3 angles

\frac{150}{3}

= 50o

Gụọ akwụkwọ nkuzi na Angles And Intercepts On Parallel Lines (WAEC)

Angles And Intercepts On Parallel Lines

Lee Azịza

Ajụjụ 11 Ripọtì

Simplify $ \frac{x^2-y^2}{2x^2+xy-y^2} $

$\frac{x + y}{2 x + y}$

$\frac{x + y}{2 x - y}$

$\frac{x - y}{2 x - y}$

$\frac{x - y}{2 x + y}$

Akọwa Nkọwa

$\frac{x^{2} - y^{2}}{2 x^{2} + x y - y^{2}}$

$2 x^{2} + x y - y^{2} = 2 x^{2} - x y + 2 x y - y^{2}$

= $x (2 x - y) + y (2 x - y)$

= $(x + y) (2 x - y)$

$\frac{x^{2} - y^{2}}{2 x^{2} + x y - y^{2}} = \frac{(x + y) (x - y)}{(x + y) (2 x - y)}$

= $\frac{x - y}{2 x - y}$

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Lee Azịza

Ajụjụ 12 Ripọtì

Reduce each number to two significant figures and then evaluate $ \frac{0.021741 \times 1.2047}{0.023789} $

0.8

0.9

1.1

1.2

Akọwa Nkọwa

\frac{0.021741 \times 1.2047}{0.023789}

=

\frac{0.022 \times 1.2}{0.024}

(to 216)
=

\frac{0.0264}{0.024}

= 1.1

Gụọ akwụkwọ nkuzi na Fractions, Decimals And Approximations (WAEC)

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 13 Ripọtì

If the length of a square is increased by 20% while while its width is decreased by 20% to form a rectangle, what is the ratio of the area of the rectangle to the area of the square?

6 : 5

25 : 24

5 : 6

24 : 25

Akọwa Nkọwa

Length and width of a square is 100%
Length increased by 20% and
Width decreased by 20% to form a rectangle
Length of rectangle = 120% to form a rectangle
Length of rectangle = 120% and
Width of rectangle = 80%
Area of rectangle = L x W
Area of square = W
Ratio of the area of the rectangle to the area of the square
A =

\frac{Area of rectangle}{Area of square}

\frac{120 \times 30}{100 \times 100}

=

\frac{96}{100}

= 24 : 25

Gụọ akwụkwọ nkuzi na Lengths And Perimeters (WAEC)

Lengths And Perimeters

Lee Azịza

Ajụjụ 14 Ripọtì

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

Akọwa Nkọwa

In PNO, ONP
= 180 - (35 + 86)
= 180 - 121
= 59
PRQ = QNP = 59(angles in the same segment of a circle are equal)

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 15 Ripọtì

Taking the period of day light on a certain day to be from 5.30a.m to 7.00p.m. Calculate the angle of a pie chart designed to show the periods of the day light and of darkness on that day

187o30' 172o30'

135o225'

202o30' 157o30'

195o165'

Akọwa Nkọwa

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval
24hrs make 1 day
∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins.
Angle of sector of daylight =

\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}

=

\frac{810}{1440}

x 360o
=

\frac{291600}{1440}

= 202.5o
202.5o = 202.30o
darkness period = 360o - 202o 30'
= 157o30'
= 202o30' 157o 30'

Lee Azịza

Ajụjụ 16 Ripọtì

Find the probability of selecting a figure which is a parallelogram from a square, a rectangle, a rhombus, a kite and a trapezium

3

5

2

5

4

5

1

5

Akọwa Nkọwa

A rectangle, a square and a rhombus are the only parallelogram
∴ Probability of a parallelogram =

\frac{3}{5}

Lee Azịza

Ajụjụ 17 Ripọtì

In the diagram above, QR // TS, QR : TS = 2.3, Find the ratio of the area of triangle PQR to the area of the trapezium QRST.

4:9

4:5

1:3

2:3

Akọwa Nkọwa

h1 =

\frac{h_{1} + h_{2}}{1.5}

=

\frac{h_{1}}{h_{2}} = 2

$\frac{A_{△ P Q R}}{A_{T P (Q R S T)}} = \frac{\frac{1}{2} \times 2 \times h_{1}}{\frac{1}{2} \times h_{2} (2 + 3)}$

$\frac{2}{5} \times \frac{h_{1}}{h_{2}} = \frac{2}{5} \times 2$

= $\frac{4}{5}$

Gụọ akwụkwọ nkuzi na Mensuration (JAMB)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Mensuration

Triangles And Polygons

Lee Azịza

Ajụjụ 18 Ripọtì

In the figure, PQRS is a rectangle. If the shaded area is 72 sq. cm, find h.

12cm

10cm

8cm

5cm

Akọwa Nkọwa

From the diagram, PQRS is a rectangle

Area of shaded part = 72sq.cm

But 72 = 3h - 4 + 6h - 4 + 4h

= 72 - 16h - 8

= 16h - 72 + 8

=16h = 80

h =

\frac{80}{16}

= 5cm

Lee Azịza

Ajụjụ 19 Ripọtì

The figure is an example of the construction of a

perpendicular bisector of a given straight line QR

perpendicular from a given point to a given line QR

perpendicular to a line from a given point P on that line

given angle

Akọwa Nkọwa

QR is a given line and P is a given point. The construction is the perpendicular from a given point P to a given line QR

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Euclidean Geometry

Simple Operations On Algebraic Expressions

Lee Azịza

Ajụjụ 20 Ripọtì

The goals scored by 40 football teams from three league divisions are recorded below:

Number of goals	0	1	2	3	4	5	6
frequency	4	3	15	16	1	0	1

What is the total number of goals scored by all the terms?

21

40

91

96

Akọwa Nkọwa

Let x rep. number of goals

\begin{array}{cc} " f & " & f r e q u e n c y \\ x & f & f x \\ 0 & 4 & 0 \\ 1 & 3 & 3 \\ 2 & 15 & 30 \\ 3 & 16 & 48 \\ 4 & 1 & 4 \\ 5 & 0 & 0 \\ 6 & 1 & 6 \end{array}

\sum f x

= 91
Total number of goals scored is

\sum f x

= 91

Lee Azịza

Ajụjụ 21 Ripọtì

Simplify without using tables $ \frac{2\sqrt{14}\times3\sqrt{21}}{7\sqrt{24}\times2\sqrt{98}} $

3√14

4

3√2

4

3√14

28

3√2

28

82

Akọwa Nkọwa

\frac{2 \sqrt{14} \times 3 \sqrt{21}}{7 \sqrt{24} \times 2 \sqrt{98}}

=

\frac{6 \sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}

=

\frac{3 \sqrt{3}}{14 \sqrt{6}}

=

\frac{3 \sqrt{3}}{14 \sqrt{2} \times \sqrt{3}}

=

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

x

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

=

\frac{3 \sqrt{2}}{28}

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Positive And Negative Integers, Rational Numbers

Surds (radicals)

Lee Azịza

Ajụjụ 22 Ripọtì

The numbers 3, 2, 8, 5, 7, 12, 9 and 14 are the marks scored by a group of students in a class test. If P is scored by an group of students in a class test. If P is the mean and Q the median, the P + Q is

18

17

\frac{1}{2}

16

15

Akọwa Nkọwa

Re-arranging them 2, 3, 5, 7, 8, 9, 12, 14
Mean = P and median = Q
P =

\frac{2 + 2 + 5 + 7 + 8 + 9 + 12 + 14}{8}

\frac{60}{8}

= 7.5
median (Q) =

\frac{7 + 8}{2}

\frac{15}{2}

= 7.5
P + Q = 7.5 + 7.5 = 15

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 23 Ripọtì

Make $y$ the subject of the formula $Z = x^2 + \frac{1}{y^3}$

y =

\frac{1}{(z - x^{2})^{3}}

y =

\frac{1}{(z + x^{2}) \frac{1}{3}}

y =

\frac{1}{(z x^{2}) \frac{1}{3}}

y =

3 \sqrt{z - x^{2}}

Akọwa Nkọwa

Z = x2 +

\frac{1}{y^{3}}

Z - x2 =

\frac{1}{y^{3}}

y3 =

\frac{1}{Z x^{2}}

y =

3 \sqrt{\frac{1}{2} x^{2}}

∴ y =

3 \sqrt{Z - x^{2}}

Gụọ akwụkwọ nkuzi na Change Of Subject Of A Formula/Relation (WAEC)

Gụọ akwụkwọ nkuzi na Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

Lee Azịza

Ajụjụ 24 Ripọtì

Convert 241 in base 5 to base 8,

718

1078

1768

2418

Akọwa Nkọwa

2415 = 2 x 52 + 4 x 5 + 1 x 5 + 1 x 5o
50 + 20 + 1 = 7110
Convert 7110 to base 8
$\begin{array}{cc} 8 & 71 \\ 8 & 8 R 7 \\ 8 & 1 R 0 \\ 8 & 0 R 1 \end{array}$
= 1078

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 25 Ripọtì

The base of a pyramid is a square of side 8cm. If its vertex is directly above the centre, find the height, given that the edge is 4.3cm

6cm

5cm

4cm

3cm

Akọwa Nkọwa

Base of pyramid of a square of side 8cm vertex directly above the centre edge =

4 \sqrt{3}

cm
From the diagram, the diagonal of one base is AC2 = 82 + 82
Ac2 = 64 + 64 = 128
AC =

8 \sqrt{2}

but OC =

\frac{1}{2}

AC = 8

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

=

4 \sqrt{2}

cm
OE = h = height
h2 = (

4 \sqrt{3}

)2
16 x 2 - 16 x 2
48 - 32 = 16
h =

\sqrt{16}

= 4

Lee Azịza

Ajụjụ 26 Ripọtì

Find the eleventh term of the progression 4, 8, 16.....

213

212

211

210

Akọwa Nkọwa

a = 4, r =

\frac{4}{2}

\frac{8}{4}

= 2
n = 11
Tn = arn - 1
T11 = 4(2)11 - 1
4(2)10 = 212
since 4 = 22
= 212

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Sequence And Series

Lee Azịza

Ajụjụ 27 Ripọtì

Find the least length of a rod which can be cut into exactly equal strips, each of either 40cm or 48cm in length

120cm

240cm

360cm

480cm

Akọwa Nkọwa

The least length is

\frac{40}{48}

=

\frac{5}{8}

for the rod to be cut in exactly equal trips
Ratio

\frac{5}{6}

:

\frac{48}{40}

\frac{\frac{5}{6}}{\frac{40}{48}}

= 1

\frac{5}{6}

x

\frac{48}{40}

=

\frac{240}{240}

= 1
The least length = 240cm

Gụọ akwụkwọ nkuzi na Positive And Negative Integers, Rational Numbers (WAEC)

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Positive And Negative Integers, Rational Numbers

Number Bases

Lee Azịza

Ajụjụ 28 Ripọtì

By selling 20 oranges for ₦1.35 a trader makes a profit of 8%. What is his percentage gain or loss if he sells the same 20 oranges for ₦1.10?

8%

10%

12%

15%

Akọwa Nkọwa

profit 8% of ₦1.35 =

\frac{8}{100}

x ₦1.35 = ₦0.08
Cost price = ₦1.35 - ₦0.10 = ₦1.25
If he sells the 20 oranges for ₦1.10 now
%loss =

\frac{actual loss}{Cost price}

x 100

\frac{125 - 1.10}{1.25}

x 100
=

\frac{0.15 \times 100}{1.25}

=

\frac{15}{1.25}

= 12%

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 29 Ripọtì

A man invests a sum of money at 4% per annum simple interest. After 3 years, the principal amounts to ₦7,000.00. Find the sum invested.

₦7,840.00

₦6,250.00

₦6,160.00

₦5,833.33

Modular Arithmetic

Areas

Lee Azịza

Ajụjụ 30 Ripọtì

If $ \tan \theta = \frac{m^2-n^2}{2mn} $ find $ \sec \theta $

m²+n²

(m²−n²)

m²+n²

2mn

mn

2(m²+n²⁾

m²n²

2m²−n²

Akọwa Nkọwa

Tan

θ

=

\frac{m^{2} - n^{2}}{2 m n}

\frac{Opp}{Adj}

by pathagoras theorem
= Hyp2 = Opp2 + Adj2
Hyp2 = (m2 - n2) + (2mn)2
Hyp2 = m4 - 2m2n4 - 4m2 - n2
Hyp2 = m4 + 2m2 + n2n
Hyp2 = (m2 - n2)2
Hyp2 =

\frac{m^{2} + n^{2}}{2 m n}

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Angles Of Elevation And Depression (WAEC)

Sine, Cosine And Tangent Of An Angle

Angles Of Elevation And Depression

Lee Azịza

Ajụjụ 31 Ripọtì

If $ P = \frac{2}{3}\left(\frac{1-r^2}{n^2}\right) $, find n when r = $ \frac{1}{3} $ and p = 1

3

2

1

3

2

3

Akọwa Nkọwa

If P =

\frac{2}{3}

(

\frac{1 - r^{2}}{n^{2}}

), find n when r =

\frac{1}{3}

and p = 1
p =

\frac{2 (1 - r^{2})}{3 n^{2}}

when r =

\frac{1}{3}

and p = 1
1 =

\frac{2}{3}

\frac{(1 - (\frac{1}{3})^{2})}{n^{2}}

n2 =

\frac{2 (3 - 1)}{3 \times 3}

n2 =

\frac{2 \times 2}{3 \times 3}

=

\frac{4}{9}

n =

\frac{4}{9}

=

\frac{2}{3}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Probability

Measures Of Location

Quadratic Equations

Lee Azịza

Ajụjụ 32 Ripọtì

If P varies inversely as V and V varies directly as R2, find the relationship between P and R given that R = 7 when P = 2.

P = 98R2

PR2 = 98

P =

\frac{1}{98 R^{2}}

P =

\frac{P R^{2}}{98}

Akọwa Nkọwa

P =

\frac{1}{v}

and vR2 = P =

\frac{k}{v}

......(i)
and v KR2 .......(ii)
(where k is constant)
Subst. for v in equation (i) = p =

\frac{1^{2}}{K R}

.....(ii)
when r = 7, p = 2
2 =

\frac{k}{7^{2}}

k = 2 x 49
= 98
Subt. foe k in ....(iii)
P =

\frac{98}{R^{2}}

PR2 = 98

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

Variation

Sequence And Series

Lee Azịza

Ajụjụ 33 Ripọtì

Given that 1/2 log10 P = 1, find the value Of P

104

103

102

101

10-1

Akọwa Nkọwa

1/2log10P
log10P1/2 = 1
P1/2 = 10
p = 102

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 34 Ripọtì

Below are the scores of group of students in a music test:

Scores	1	2	3	4	5	6	7	8	9
No of students	3	6	10	8	6	5	2	4	12

If CF(x) is the number of students with scores less than or equal to x, find CF(6)

40

38

33

5

Akọwa Nkọwa

CF (6) = 5 + 6 + 8 + 10 + 6 + 3
= 38

Lee Azịza

Ajụjụ 35 Ripọtì

The formula Q = 1.5 + 0.5n gives the cost Q(in Naira)of feeding n people one additional person

350k

200k

150k

50k

Akọwa Nkọwa

Q = 1.5 + 0.5n gives the cost 1(in Naira) of feeding n people for a week. Extra cost of feeding one additional person = n = 1
Subt. for n in the formula Q = 1.5 + 0.5(1)
= 15 + 0.5(2)
Q = ₦2
= 200k

Gụọ akwụkwọ nkuzi na Change Of Subject Of A Formula/Relation (WAEC)

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Change Of Subject Of A Formula/Relation

Areas

Lee Azịza

Ajụjụ 36 Ripọtì

Find the values of m which make the following quadratic function a perfect square. x2 + 2(m + 1)x + m + 3

-1, 1

-1, 2

1, -2

2, -2

Akọwa Nkọwa

x2 + 2(m + 1)x + m + 3 from (A + B)2 = a2 + 2AB + b2
2B =

\frac{2 (m + 1)}{2}

= m + 1 .......(i)
Also B2 = m + 3..........(ii)
Subst. for B in equation ..........(iii)
= (m + 1)2 = m + 3 = m2 + 2m + 1 = m + 3
= m2 + 2m - m + 1 - 3 = 0
m2 + m - 2 = 0
(m + 2)(m - 1) = 0 or m - 1 = 0
m = -2 or 1

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 37 Ripọtì

Two brothers invested a total of ₦5,000.00 on a farm project, the farm yield was sold for ₦15,000.00 at the end of the season. If the profit was shared in the ratio 2 : 3, what is the difference in the amount to profit received by the brothers?

₦2,000.00

₦4,000.00

₦6,000.00

₦10,000.00

Akọwa Nkọwa

Total amount invested by A and B = ₦5,000
farm yield was sold for ₦15,000.00
profit = 15,000.00 - 5,000.00
= ₦10,000.00
Profit was shared in ratio 2 : 3
2 + 3 = 5
A received

\frac{2}{5}

of profit =

\frac{2}{5}

x 10,000 = ₦4,000.00
A receive

\frac{3}{5}

of profit =

\frac{3}{5}

x 10,000 = ₦6,000.00
Difference in profit received = ₦6,0000 - ₦4,000.00
= ₦2,000.00

Lee Azịza

Ajụjụ 38 Ripọtì

What is the locus or the mid-point of all the chords of length 6cm with circle of radius 5cm and with center 0?

A circle of radius 4 cm and the center 0

The perpendicular bisector of the chords

A straight line passing through 0

A circle of radius 6 cm and with center 0

Akọwa Nkọwa

x2 + 33 = 52
x2 + 9 = 25
x2 = 25 – 9
x2 = 16

x = √ 16
= 4 cm
∴ The locus is a circle of radius 4 cm with the center 0

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 39 Ripọtì

Four boys and ten girls can cut a field in 5 hours if the boys work at $ \frac{5}{4} $ the rate at which the girls work. How many boys will be needed to cut the field in 3 hours?

180

60

25

20

Akọwa Nkọwa

Let x represents number of boys that can work at

\frac{5}{4}

the rate at which the 10 girls work
For 1hr. x boys will work for

\frac{\frac{1}{5}}{4}

x 10
x =

\frac{4}{5}

x 10 = 8 boys
8 boys will do the work of ten girls at the same rate 4 + 8 = 12 boys cut the field in 5 hrs
For 3 hrs,

\frac{12 \times 5}{3}

boys will be needed = 20 boys

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 40 Ripọtì

From two points x and y, 8m apart, and in line with a pole, the angle of elevation of the top of the pole are 30o and 60o respectively. Fins the height of the pole assuming that x, y and the foot of the pole are the same horizontal plane and x and y are on the same side of the pole.

\sqrt{6}

4

\sqrt{3}

√3

\frac{12}{\sqrt{3}}

Akọwa Nkọwa

From the diagram, WYZ = 60o, XYW = 180o - 60o
= 120o
LX = 30o
XWY = 180o - 120o + 30o
XWY = 30o
WXY = XYW = 30o
Side XY = YW
YW = 8m, sin 60o =

\frac{3}{2}

? sin 60o =

\frac{h}{Y W}

, sin 60o =

\frac{h}{8}

h = 8 x

\frac{3}{2}

= 4

\sqrt{3}

Lee Azịza

Ajụjụ 41 Ripọtì

The sine, cosine and tangent of 210o are respectively

\frac{1}{2}

,

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

,

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

\frac{1}{2}

,

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

,

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

\frac{1}{2}

,

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

,

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

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

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

Akọwa Nkọwa

210o = 180o - 210o = 30o
From ratio of sides, sin -30o = -

\frac{1}{2}

Cos 210o = 180o - 210o = -30o
= cos -30o =

\frac{- 3}{2}

But tan 30o =

\frac{1}{3}

, rationalizing this
=

\frac{1}{3}

x

\frac{3}{3}

=

\frac{3}{3}

∴ =

\frac{- 1}{2}

,

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

,

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

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Coordinate Geometry

Lee Azịza

Ajụjụ 42 Ripọtì

A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room

5

17

8

17

8

15

12

17

Akọwa Nkọwa

ABCD is the floor, by pythagoras
AC2 = 144 + 81 =

\sqrt{225}

AC = 15cm
Height of room is 8m, diagonal of floor is 15m
∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152
cosine =

\frac{adj}{hyp}

=

\frac{15}{17}

Gụọ akwụkwọ nkuzi na Sine, Cosine And Tangent Of An Angle (WAEC)

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Sine, Cosine And Tangent Of An Angle

Circles

Lee Azịza

Ajụjụ 43 Ripọtì

In the figure, find the value of x

110o

100o

90o

80o

Akọwa Nkọwa

Z = X = Y = 180o .........(i)

2Z + Y + Y = 190 = 2Z + 2Y = 180

Z + Y = 90o........(ii)

hence x + (z + y) = 180

x + 90o = 180o

x = 180o - 90o

= 90o

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza

Ajụjụ 44 Ripọtì

Factorize 62x + 1 + 7(6x) - 5

[3(6x) - 5][2(6x)m + 1]

[3(6x) + 5][2(6x) - 1]

[(6x) - 5][3(6x) + 1]

[2(6x) + 5][3(6x) - 1]

Akọwa Nkọwa

62x + 1 + 7(6x) - 5
Let 6x = y....(i)
6yx + 7y - 5 = 0
(3y + 5)(2y - 1) = 0.....(i)
Subt. for y from eqn. (i) in eqn. (ii)
= [3(6x) + 5][2(6x) - 1]

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 45 Ripọtì

A rectangular lawn has an area of 1815 square yards. if its length is 50 meters, find its width in meters given that 1 metre equals 1.1 yards

39.93

35.00

33.00

30.00

Akọwa Nkọwa

1m = 1.1 yard, length(L)= 50m
= (50 x 1.1)yards
= 55 yards
Area(A) = length(L) y width (W)
1815 = 55y width (W)
width (w) =

\frac{1815}{55}

= 33 years

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 46 Ripọtì

Find the values of y which satisfy the simultaneous equations x + y = 5, x2 - 2y2 = 1

-12, -2

-12, 12

12, -2

2, -2

Akọwa Nkọwa

x + y = 5.......(i)
x2 - 2y2 = 1.......(ii)
x = 5 - y.........(iii)
Subst. for x in eqn.(ii) = (5 - y)2 - 2y2 = 1
25- 10y + y2 - 2y2 = 1
25 - 1 = y2 + 10y
y2 + 10y2 - 24 = 0
(y + 12)(y - 2) = 0
Then Either y + 12 = 0 or y - 2 = 0
= (12, -2)

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza






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