JAMB UTME - Mathematics - 1987

Ibeere 1 Ìròyìn

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

Wo Idahun

Ibeere 2 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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Polynomials

Wo Idahun

Ibeere 3 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Lengths And Perimeters (WAEC)

Lengths And Perimeters

Wo Idahun

Ibeere 4 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Euclidean Geometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Euclidean Geometry

Simple Operations On Algebraic Expressions

Wo Idahun

Ibeere 5 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 6 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 7 Ìròyìn

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

350k

200k

150k

50k

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Change Of Subject Of A Formula/Relation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Change Of Subject Of A Formula/Relation

Areas

Wo Idahun

Ibeere 8 Ìròyìn

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

Awọn alaye Idahun

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}

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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Sine, Cosine And Tangent Of An Angle

Circles

Wo Idahun

Ibeere 9 Ìròyìn

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

12cm

10cm

8cm

5cm

Awọn alaye Idahun

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

Wo Idahun

Ibeere 10 Ìròyìn

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

Awọn alaye Idahun

h1 =

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

=

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

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

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

= $\frac{4}{5}$

Ka ìwé ẹ̀kọ́ nípa Mensuration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Mensuration

Triangles And Polygons

Wo Idahun

Ibeere 11 Ìròyìn

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

30o

40o

45o

50o

Circles

Wo Idahun

Ibeere 12 Ìròyìn

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

Wo Idahun

Ibeere 13 Ìròyìn

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

Awọn alaye Idahun

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}

Ka ìwé ẹ̀kọ́ nípa Probability (WAEC)

Ka ìwé ẹ̀kọ́ nípa Measures Of Location (JAMB)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Probability

Measures Of Location

Quadratic Equations

Wo Idahun

Ibeere 14 Ìròyìn

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

Awọn alaye Idahun

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

π

R cos

θ

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Circles

Coordinate Geometry

Wo Idahun

Ibeere 15 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 16 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 17 Ìròyìn

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}}

Awọn alaye Idahun

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}}

Ka ìwé ẹ̀kọ́ nípa Change Of Subject Of A Formula/Relation (WAEC)

Ka ìwé ẹ̀kọ́ nípa Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

Wo Idahun

Ibeere 18 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 19 Ìròyìn

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

104

103

102

101

10-1

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Logarithms

Wo Idahun

Ibeere 20 Ìròyìn

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}}

Awọn alaye Idahun

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}

Wo Idahun

Ibeere 21 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 22 Ìròyìn

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}

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Variation (JAMB)

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Variation

Sequence And Series

Wo Idahun

Ibeere 23 Ìròyìn

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

625

8

625

1

8

Awọn alaye Idahun

(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}

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Ka ìwé ẹ̀kọ́ nípa Surds (radicals) (WAEC)

Ka ìwé ẹ̀kọ́ nípa Graphs Of Linear And Quadratic Functions (WAEC)

Fractions, Decimals And Approximations

Number Bases

Surds (radicals)

Graphs Of Linear And Quadratic Functions

Wo Idahun

Ibeere 24 Ìròyìn

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}

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

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

Awọn alaye Idahun

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}

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

Ka ìwé ẹ̀kọ́ nípa Trigonometry (JAMB)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Sine, Cosine And Tangent Of An Angle

Trigonometry

Coordinate Geometry

Wo Idahun

Ibeere 25 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 26 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Quadratic Equations

Wo Idahun

Ibeere 27 Ìròyìn

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

Awọn alaye Idahun

\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}

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 28 Ìròyìn

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]

Awọn alaye Idahun

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]

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 29 Ìròyìn

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

x > -3

x < -3

2 < x < 3

-3 < x < -2

Awọn alaye Idahun

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

= 3x < -9

∴ x < -3

Ka ìwé ẹ̀kọ́ nípa Linear Inequalities (WAEC)

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Linear Inequalities

Areas

Wo Idahun

Ibeere 30 Ìròyìn

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

Awọn alaye Idahun

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

Wo Idahun

Ibeere 31 Ìròyìn

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

Awọn alaye Idahun

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

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

Angles And Intercepts On Parallel Lines

Wo Idahun

Ibeere 32 Ìròyìn

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%

Awọn alaye Idahun

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%

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Ka ìwé ẹ̀kọ́ nípa Ratio, Proportions And Rates (WAEC)

Modular Arithmetic

Ratio, Proportions And Rates

Wo Idahun

Ibeere 33 Ìròyìn

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'

Awọn alaye Idahun

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'

Wo Idahun

Ibeere 34 Ìròyìn

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

Awọn alaye Idahun

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}

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

Ka ìwé ẹ̀kọ́ nípa Angles Of Elevation And Depression (WAEC)

Sine, Cosine And Tangent Of An Angle

Angles Of Elevation And Depression

Wo Idahun

Ibeere 35 Ìròyìn

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

Awọn alaye Idahun

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

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

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Positive And Negative Integers, Rational Numbers

Number Bases

Wo Idahun

Ibeere 36 Ìròyìn

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

-12, -2

-12, 12

12, -2

2, -2

Awọn alaye Idahun

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)

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 37 Ìròyìn

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

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Areas

Wo Idahun

Ibeere 38 Ìròyìn

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

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

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

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

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

Awọn alaye Idahun

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

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

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

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

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

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

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Solution Of Linear Equations

Algebraic Expressions

Wo Idahun

Ibeere 39 Ìròyìn

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

9

15

27

33

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 40 Ìròyìn

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

213

212

211

210

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Sequence And Series (WAEC)

Sequence And Series

Wo Idahun

Ibeere 41 Ìròyìn

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

Awọn alaye Idahun

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

\frac{3}{5}

Wo Idahun

Ibeere 42 Ìròyìn

In the figure, find the value of x

110o

100o

90o

80o

Awọn alaye Idahun

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

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Angles

Wo Idahun

Ibeere 43 Ìròyìn

Convert 241 in base 5 to base 8,

718

1078

1768

2418

Awọn alaye Idahun

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