JAMB UTME - Mathematics - 1983

Ibeere 1 Ìròyìn

A man drove for 4 hours at a certain speed, he then doubled his speed and drove for another 3 hours. Although he covered 600 kilometers. At what speed did he drive for the last 3 hours?

120km/hr

60km/hr

670km/hr

40km/hr

Awọn alaye Idahun

Speed =

\frac{d i s t a n c e}{t i m e}

let x represent the speed, d represent distance
x =

\frac{d}{4}

d = 4x
2x =

\frac{600 - d}{3}

6x = 600 - d
6x = 600 - 4x
10x = 600
x =

\frac{600}{10}

= 60km/hr

Wo Idahun

Ibeere 2 Ìròyìn

PQRS is a cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of $50^\circ$ with the tangent as shown in the figure. What is the size of QRS?

50

^{\circ}

40

^{\circ}

110v

80

^{\circ}

^{\circ}

Awọn alaye Idahun

< SPQ = 80

^{\circ}

< SPQ + < SRQ = 180(Supplementary)

80 + < QRS = 180 $^{\circ}$ - 80 $^{\circ}$

= 100 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Indices, Logarithms And Surds (JAMB)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Circles

Indices, Logarithms And Surds

Triangles And Polygons

Wo Idahun

Ibeere 3 Ìròyìn

In a class of 60 pupils, the statistical distribution of the numbers of pupils offering Biology, History, French, Geography and Additional mathematics is as shown in the pie chart. How many pupils offer Additional Mathematics?

15

10

18

12

20

Awọn alaye Idahun

2x - 24)

^{\circ}

+ (3x - 18)

^{\circ}

+ (2x + 12)

^{\circ}

+ (x + 12)

^{\circ}

+ x

^{\circ}

= 360

^{\circ}

9x = 360 $^{\circ}$ + 18 $^{\circ}$

x = $\frac{378}{9}$

= 42 $^{\circ}$ , if x = 42 $^{\circ}$ , then add maths = $\frac{2 x - 42}{360}$ x 60

= $\frac{2 \times 42 - 24}{360}$ x 60

= $\frac{84 - 24}{6}$

= 10

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 4 Ìròyìn

Solve the following equations 4x - 3 = 3x + y = x - y = 3, 3x + y = 2y + 5x - 12

x = 5, y = 2

x = 2, y = 5

x = 5, y = -2

x = -2, y = -5

x = -5, y = -2

Awọn alaye Idahun

4x - 3 = 3x + y = x - y = 3.......(i)
3x + y = 2y + 5x - 12.........(ii)
eqn(ii) + eqn(i)
3x = 15
x = 5
substitute for x in equation (i)
5 - y = 3
y = 2

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 5 Ìròyìn

Find the angle of the sectors representing each item in pie chart of the following data 6, 10, 14, 16, 26

15o, 25o, 35o, 40o, 65o

60o, 100o, 140o, 160o, 260o

6o, 10o, 14o, 16o, 26o

30o, 50o, 70o, 80o, 130o

none of the above

Awọn alaye Idahun

6 + 10 + 14 + 16 + 26 = 72

\frac{6}{72}

x 360
= 30o
Similarly others give 30o, 50o, 70o, 80o and 130o respectively

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Polynomials

Measures Of Dispersion

Wo Idahun

Ibeere 6 Ìròyìn

Factorize completely 81a4 - 16b4

(3a + 2b)(2a - 3b)(9a2 + 4b2)

(3a - 2b)(2a - 3b)(4a2 - 9b2)

(3a - 2b)(3a - 2b)(9a2 - 4b2)

(6a - 2b)(8a - 3b)(4a3 - 9b2)

Awọn alaye Idahun

81a4 - 16b4 = (9a2)2 - (4b2)2
= (9a2 + 4b2)(9a2 - 4b2)
N:B 9a2 + 4b2 = (3a - 2b)(3a - 2b)

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 7 Ìròyìn

If a rod of length 250cm is measured as 255cm long in error, what is the percentage error in the measurement?

55

10

5

4

2

Awọn alaye Idahun

% error =

\frac{Actual error}{real value}

x 100
=

\frac{5}{250}

x 100
= 2

Ka ìwé ẹ̀kọ́ nípa Percentages (WAEC)

Percentages

Wo Idahun

Ibeere 8 Ìròyìn

On a square paper of length 2.524375cm is inscribed square diagram of length 0.524375cm. Find the area of the paper not covered by the diagram. correct to 3 significant figures.

6.00cm2

6.10cm2

6cmv

6.09cm2

4.00cm2

Awọn alaye Idahun

Area of the paper = area of square = L x B or S2
where s = S x k
Area of the paper = (2.524)2
area of the diagram = (0.524)2
area not covered = (2.524)2 - (0.524)2
= 6.370576 - 0.274576
= 6.096
= 6.10cm2 (2 s.f)

Ka ìwé ẹ̀kọ́ nípa Coordinate Geometry (WAEC)

Coordinate Geometry

Wo Idahun

Ibeere 9 Ìròyìn

Given that cos z = L , whrere z is an acute angle, find an expression for $ \frac{\cot z - \cosec z}{\sec z + \tan z} $

1-L

1+L

L²√3

1+L

1+L³

L²

L(L−1)

1−L+1√1−L²

Awọn alaye Idahun

Given Cos z = L, z is an acute angle
$\frac{cot z - cosec z}{sec z + tan z}$ = cos z
= $\frac{cos z}{sin z}$
cosec z = $\frac{1}{sin z}$
cot z - cosec z = $\frac{cos z}{sin z}$ - $\frac{1}{sin z}$
cot z - cosec z = $\frac{L - 1}{sin z}$
sec z = $\frac{1}{cos z}$
tan z = $\frac{sin z}{cos z}$
sec z = $\frac{1}{cos z}$ + $\frac{sin z}{cos z}$
= $\frac{1}{l}$ + $\frac{sin z}{L}$
the original eqn. becomes
$\frac{cot z - cosec z}{sec z + tan z}$ = $\frac{L - \frac{1}{sin z}}{1 + s i n \frac{z}{L}}$
= $\frac{L (L - 1)}{sin z (1 + sin z)}$
= $\frac{L (L - 1)}{sin z + 1 - c o s^{2} z}$
= sin z + 1
= 1 + $\sqrt{1 - L^{2}}$
= $\frac{L (L - 1)}{1 - L + 1 \sqrt{1 - L^{2}}}$

Ka ìwé ẹ̀kọ́ nípa Matrices And Determinants (JAMB)

Ka ìwé ẹ̀kọ́ nípa Modular Arithmetic (WAEC)

Matrices And Determinants

Modular Arithmetic

Wo Idahun

Ibeere 10 Ìròyìn

Simplify $ \log_{10} a^{\frac{1}{3}} + \frac{1}{4}\log_{10} a - \frac{1}{12}\log_{10} a^{7} $

1

\frac{7}{6}

log10 a

zero

10

Awọn alaye Idahun

log10 a

\frac{1}{3}

+

\frac{1}{4}

log10 a -

\frac{1}{12}

log10a7 = log10 a

\frac{1}{3}

+ log10

\frac{1}{4}

- log10 a

\frac{7}{12}

= log10 a

\frac{7}{12}

- log10 a

\frac{7}{12}

= log10 1 = 0

Ka ìwé ẹ̀kọ́ nípa Logarithms (WAEC)

Logarithms

Wo Idahun

Ibeere 11 Ìròyìn

Find x if (xbase4)2 = 100100base2

6

12

100

210

110

Awọn alaye Idahun

(x2)4 to base10 gives (1 x 25) + (1 x 25) + (1 x 2)
32 + 4 = 36
x2 = 36, x = 6
610 to base 4 =

\begin{array}{cc} 4 & 6 \\ 1 r 2 \end{array}

= 124

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 12 Ìròyìn

If M represents the median and D the mode of the measurements 5, 9, 3, 5, 7, 5, 8, then (M, D) is

(6, 5)

(5, 8)

(5, 7)

(5, 5)

(7, 5)

Probability

Linear Inequalities

Wo Idahun

Ibeere 13 Ìròyìn

The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart?

11

^{\circ}

19

\frac{1}{3}

^{\circ}

33

\frac{1}{3}

^{\circ}

91

\frac{1}{3}

^{\circ}

91

^{\circ}

Awọn alaye Idahun

Adding the values of all the items together, it gives 70

Okro sector = $\frac{145}{270}$ x $\frac{360^{o}}{1}$

= 19.33 $^{\circ}$

= 19 $\frac{1}{3}$ $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Angles

Wo Idahun

Ibeere 14 Ìròyìn

If w varies inversely as V and U varies directly as w3, Find the relationship between u and v given that u = 1, when v = 2

u =

\frac{8}{v^{3}}

v =

\frac{8}{u^{2} v^{3}}

u = 8v3

v = 8u2

Awọn alaye Idahun

W

α

\frac{1}{v}

u

α

w3
w =

\frac{k_{1}}{v}

u = k2w3
u = k2(

\frac{k_{1}}{v}

)3
=

\frac{k_{2} k_{1}^{2}}{v^{3}}

k = k2k1k2
u =

\frac{k}{v^{3}}

k = uv3
= (1)(2)3
= 8
u =

\frac{8}{v^{3}}

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

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

Variation

Change Of Subject Of A Formula/Relation

Wo Idahun

Ibeere 15 Ìròyìn

If x + 2 and x - 1 are factors of the expression 1x3 + 2kx2 + 24, find the values of 1 and K

1 = -6, k = -9

1 = -2, k = 1

k = -1, 1 = -2

1 = 0, k = 1

k = 0,1 = 6

Awọn alaye Idahun

f(x) = Lx3 + 2kx2 + 24
f(-2) = -8L + 8k = -24
4L - 4k = 12
f(1):L + 2k = -24
L - 4k = 3
3k = -27
k = -9
1 = -6

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 16 Ìròyìn

If x is jointly proportional to the cube of y and the fourth power of z. In what ratio is x increased or decreased when y is halved and z is doubled?

4:1 increase

2:1 increase

1:1 no change

1:4 decrease

3:4 decrease

Variation

Progression

Wo Idahun

Ibeere 17 Ìròyìn

In a triangle PQT, QR = $ \sqrt{3}\text{ cm} $, PR = 3cm, PQ = $ 2\sqrt{3} $cm and PQR = 30^o. Find angles P and R

P = 60o and R = 90o

P = 30o and R = 120o

P = 90o and R = 60o

P = 60o and R 60o

P = 45o and R = 105o

Awọn alaye Idahun

By using cosine formula, p2 = Q2 + R2 - 2QR cos p
Cos P =

\frac{Q^{2} + R^{2} - p^{2}}{2 Q R}

=

\frac{(3)^{2} + 2 (\sqrt{3})^{2} - 3^{2}}{2 \sqrt{3}}

=

\frac{3 + 12 - 9}{12}

=

\frac{6}{12}

=

\frac{1}{2}

= 0.5
Cos P = 0.5
p = cos-1 0.5 = 60

^{\circ}

= < P = 60

^{\circ}

If < P = 60

^{\circ}

and < Q = 30

< R = 180 $^{\circ}$ - 90 $^{\circ}$
angle P = 60 $^{\circ}$ and angle R is 90 $^{\circ}$

Wo Idahun

Ibeere 18 Ìròyìn

The figure FGHK is a rhombus. What is the value of angle X?

30

^{\circ}

90

^{\circ}

150

^{\circ}

120

^{\circ}

60

^{\circ}

Awọn alaye Idahun

< HKF = 60

^{\circ}

, < KFG = 120

^{\circ}

< KFG = < KHG = x(opposite angles)

x = 120 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Angles

Triangles And Polygons

Wo Idahun

Ibeere 19 Ìròyìn

Without using tables find the numerical value of $ \log_{7} 49 + \log_{7}\left(\frac{1}{7}\right) $

1

2

3

7

9

Awọn alaye Idahun

log749 + log7

\frac{1}{7}

= log77
= 1

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 20 Ìròyìn

PQRS is a desk of dimensions $2\text{m} \times 0.8$ which is inclined at $30^\circ$ to the horizontal. Find the inclination of the diagonal PR to the horizontal

30

^{\circ}

35'

30

^{\circ}

32'

15

^{\circ}

36'

10

^{\circ}

42'

Awọn alaye Idahun

tan

θ

=

\frac{0.8}{2}

= (0.4)

$θ$ = tan-1(0.4)

From the diagram, the inclination of the diagonal PR to the horizontal is 10 $^{\circ}$ 42'

Ka ìwé ẹ̀kọ́ nípa Measures Of Dispersion (JAMB)

Measures Of Dispersion

Wo Idahun

Ibeere 21 Ìròyìn

Find the mean of the following 24.57, 25.63, 24.32, 26.01, 25.77

25.12

25.30

25.26

25.50

25.75

Awọn alaye Idahun

\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}

mean =

\frac{126.3}{5}

= 25.26

Ka ìwé ẹ̀kọ́ nípa Integration (JAMB)

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Integration

Statistics

Wo Idahun

Ibeere 22 Ìròyìn

In the figure, find PRQ

66

\frac{1}{2}

^{\circ}

^{\circ}

125

^{\circ}

105

^{\circ}

65

^{\circ}

Awọn alaye Idahun

Angle subtended at any part of the circumference of the circle

\frac{125^{o}}{2}

at centre = 360

^{\circ}

- 235

^{\circ}

= 125

^{\circ}

$\bar{P Q R}$ = $\frac{125}{2}$

= 62 $\frac{1}{2}$ $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Representation Of Data (JAMB)

Ka ìwé ẹ̀kọ́ nípa Triangles And Polygons (WAEC)

Representation Of Data

Triangles And Polygons

Wo Idahun

Ibeere 23 Ìròyìn

Make T the subject of the equation $ \frac{av}{1-v} = \sqrt{\frac{2v+T}{a+2T}} $

T = $\frac{3 a v}{1 - v}$

T = $\frac{1 + v}{2 a^{2} v^{3}}$

T = $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} + (1 - v)^{2}}$

\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}

Awọn alaye Idahun

$\frac{a v}{1 - v}$ = $\sqrt{\frac{2 v + T}{a + 2 T}}$
$\frac{(a v)^{3}}{(1 - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
$\frac{a^{3} v^{3}}{(1^{3} - v)^{3}}$ = $\frac{2 v + T}{a + 2 T}$
= $\frac{2 v (1 - v)^{3} - a^{4} v^{3}}{2 a^{3} v^{3} - (1 - v)^{3}}$

Ka ìwé ẹ̀kọ́ nípa Variation (WAEC)

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

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Ka ìwé ẹ̀kọ́ nípa Quadratic Equations (WAEC)

Variation

Change Of Subject Of A Formula/Relation

Solution Of Linear Equations

Quadratic Equations

Wo Idahun

Ibeere 24 Ìròyìn

PQR is the diagram of a semicircle RSP with centre at Q and radius of length 3.5cm. If QPT = 60^o. Find the perimeter of the figure PTRS. $ \pi = \frac{22}{7} $

25cm

18cm

36cm

29cm

25.5cm

Awọn alaye Idahun

Circumference of PRS =

\frac{π}{2}

=

\frac{22}{7}

x

\frac{7}{1}

x

\frac{1}{2}

= 11cm
Side PT = 7cm, Side TR = 7cm
Perimeter(PTRS) = 11cm + 7cm + 7cm
= 25cm

Wo Idahun

Ibeere 25 Ìròyìn

In the figure, PQR is a straight line. Find the values of x and y.

^{\circ}

x = 15

^{\circ}

, y = 52.5

^{\circ}

x = 22.5

^{\circ}

, y = 45.0

^{\circ}

x = 56.25

^{\circ}

, y = 11.5

^{\circ}

Awọn alaye Idahun

\frac{2}{3}

+ 3y + 45

^{\circ}

= 180

^{\circ}

3x + 6y + 90 $^{\circ}$ = 360 $^{\circ}$

3x + 67 = 270.......(i) x 2, 5x + y + y = 180 $^{\circ}$

5x + 2y = 180 $^{\circ}$ .......(ii) x 6

6 x 12y = 540..........(iii)

30x + 12y = 1080.........(iv)

egn(iv) - eqn(iii)

24x = 540

x = 22.5 and y = 33.75 $^{\circ}$

Wo Idahun

Ibeere 26 Ìròyìn

Given a regular hexagon, calculate each interior angle of the hexagon

60o

30o

120o

45o

135o

Awọn alaye Idahun

Sum of interior angles of polygon = (2n - 4)rt < s

sum of interior angles of an hexagon
(2 x 6 - 4) x 90o = (12 - 4) x 90o
= 8 x 90o
= 720o
each interior angle will have $\frac{720^{o}}{6}$
= 120o

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Angles

Wo Idahun

Ibeere 27 Ìròyìn

If $ \left(\frac{2}{3}\right)^m \left(\frac{3}{4}\right)^n = \frac{252}{729} $, find the values of m and n

m = 4, n = 2

m = -4, n = -2

m = -4, n =2

m = 4, n = -2

m = -2, n = 4

Awọn alaye Idahun

(

\frac{2}{3}

)m (

\frac{3}{4}

)n =

\frac{252}{729}

\frac{2^{m}}{4^{n}}

x

\frac{3^{n}}{3^{m}}

= 283-6
2m - 2n x 3n - m = 28 x 3-6
m - 2n = 8........(i)
-m + n = -6........(ii)
Solving the equations simultaneously
m = 4, n = -2

Ka ìwé ẹ̀kọ́ nípa Number Bases (WAEC)

Number Bases

Wo Idahun

Ibeere 28 Ìròyìn

If O is the centre of the circle in the figure, find the value of x

50

260

100

65

130

Awọn alaye Idahun

From the diagram; The value of x = 360

^{\circ}

- 2(130

^{\circ}

)

= 360 - 260

= 100 $^{\circ}$

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

Circles

Wo Idahun

Ibeere 29 Ìròyìn

Solve the simultaneous equations for x in x2 + y - 8 = 0, y + 5x - 2 = 0

-28, 7

6, -28

6, -1

-1, 7

3, 2

Awọn alaye Idahun

x2 + y - 8 = 0, y + 5x - 2 = 0
Rearranging, x2 + y = 8.....(i)
5x + y = 2.......(ii)
Subtract eqn(ii) from eqn(i)
x2 - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6, -1

Ka ìwé ẹ̀kọ́ nípa Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Wo Idahun

Ibeere 30 Ìròyìn

Which of the following equations represents the graph above?

y = 2 + 7x + 4x2

y = 2 - 7x + 4x2

y = 2 + 7x - 4x2

y = 2 - 7x - 4x2

Awọn alaye Idahun

x = -2 or x =

\frac{1}{4}

(x + 2)(4x - 1) = 0

y = 2 - 7x - 4x2

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 31 Ìròyìn

If $0.0000152 \times 0.042 = A \times 10^8$, where $1 \le A < 10$, find A and B

A = 9, B = 6.38

A = 6.38, B = -9

A = -6.38, B = -9

A = -9, B = -6.38

Awọn alaye Idahun

0.0000152 x 0.042 = A x 108

1 $\leq$ A < 10, it means values of A includes 1 - 9

0.0000152 = 1.52 x 10-5

0.00042 = 4.2 x 10-4

1.52 x 4.2 = 6.384

10-5 x 10-4

= 10-5-4

= 10-9

= 6.38 x 10-9

A = 6.38, B = -9

Ka ìwé ẹ̀kọ́ nípa Statistics (WAEC)

Statistics

Wo Idahun

Ibeere 32 Ìròyìn

One interior angle of a convex hexagon is 170o and each of the remaining interior angles is equal to xo. Find x

120o

110o

105o

102o

100o

Awọn alaye Idahun

An hexagon polygon is a six sided polygon
n = 6
sum of all interior angles of hexagon polygon will be (2n - 4) x 90o
= [2 x 6 - 4] x 90
= 720o
If one angle is 170o and each of the remaining five angles is 5xo
5xo + 170o = 720o
= 5x + 550
x =

\frac{550}{5}

= 110o

Ka ìwé ẹ̀kọ́ nípa Polynomials (JAMB)

Polynomials

Wo Idahun

Ibeere 33 Ìròyìn

The scores of set of final year students in the first semester examination in a paper are 41, 29, 55, 21, 47, 70, 70, 40, 43, 56, 73, 23, 50, 50. Find the median of the scores.

47

50

48

\frac{1}{2}

48

49

Awọn alaye Idahun

By re-arranging 21, 23, 29, 40, 41, 43| 47, 50| 50, 55, 56, 70, 70, 73
The median =

\frac{47 + 50}{2}

\frac{97}{2}

= 48

\frac{1}{2}

Wo Idahun

Ibeere 34 Ìròyìn

The lengths of the sides of a right angled triangle are (3x + 1)cm, (3x - 1)cm and xcm. Find x

2

6

18

12

5

Awọn alaye Idahun

(3x + 1)2 = (3x - 1)2 + 2 (Pythagoras's theorem)
9x2 + 6x + 1 = 9x - 6x2 + 1 + x2
x2 - 12x = 0
x(x - 12) = 0
x = 0 or 12

Ka ìwé ẹ̀kọ́ nípa Algebraic Expressions (WAEC)

Algebraic Expressions

Wo Idahun

Ibeere 35 Ìròyìn

In a sample survey of a University Community, the following table shows the percentage distribution of the number of members per household:

No. of members per household	1	2	3	4	5	6	7	8	Total
No. of households	3	12	15	28	21	10	7	4	100

What is the median?

4

3

5

4.5

none

Awọn alaye Idahun

\begin{array}{cc} x & f \\ 1 & 3 \\ 2 & 13 \\ 3 & 15 \\ 4 & 28 \\ 5 & 21 \\ 6 & 10 \\ 7 & 7 \\ 8 & 4 \end{array}

Median is half of total frequency = 50th term
4 falls in the range = 4

Wo Idahun

Ibeere 36 Ìròyìn

The scores of 16 students in a mathematics test are 65, 65, 55, 60, 60, 65, 60, 70, 75, 70, 65, 70, 60, 65, 65,
70. What is the sum of the median and modal scores?

125

130

140

150

Measures Of Dispersion

Statistics

Wo Idahun

Ibeere 37 Ìròyìn

The value of $ (0.303)^3 - (0.02)^3 $ is

0.019

0.0019

0.00019

0.000019

0.0278

Wo Idahun

Ibeere 38 Ìròyìn

PQRS is a cyclic quadrilateral. If ∠QPS = 75o, what is the size of ∠QRS?

180o

150o

105o

75o

Circles

Wo Idahun

Ibeere 39 Ìròyìn

In a figure, PQR = 60o, PRS = 90o, RPS = 45o, QR = 8cm. Determine PS

2

\sqrt{3}

cm

4

\sqrt{6}

cm

2

\sqrt{6}

cm

8

\sqrt{6}

cm

8cm

Awọn alaye Idahun

From the diagram, sin 60o =

\frac{P R}{8}

PR = 8 sin 60 =

\frac{8 \sqrt{3}}{2}

= 4

\sqrt{3}

Cos 45o =

\frac{P R}{P S}

=

\frac{4 \sqrt{3}}{P S}

PS Cos45o = 4

\sqrt{3}

PS = 4

\sqrt{3}

x 2
= 4

\sqrt{6}

Ka ìwé ẹ̀kọ́ nípa Circles (WAEC)

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

Ka ìwé ẹ̀kọ́ nípa Areas (WAEC)

Ka ìwé ẹ̀kọ́ nípa Angles (WAEC)

Circles

Angles Of Elevation And Depression

Areas

Angles

Wo Idahun

Ibeere 40 Ìròyìn

A construction company is owned by two partners X and Y and it is agreed that their profit will be divided in the ratio 4:5. At the end of the year, Y received ₦5,000 more than X. What is the total profit of the company for the year?

₦20, 000

₦25,000

₦30,000

₦15,000

₦45, 000

Awọn alaye Idahun

Total sharing ratio is 9
X has 4, Y has 4 + 1
If 1 is ₦5000
Total profit = 5000 x 9
= ₦45,000

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

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

Sine, Cosine And Tangent Of An Angle

Ratio, Proportions And Rates

Wo Idahun

Ibeere 41 Ìròyìn

y varies partly as the square of x and partly as the inverse of the square root of x. Write down the expression for y if y = 2 when x = 1 and y = 6 when x = 4.

y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$

y = x2 + $\frac{1}{\sqrt{x}}$

y = x2 + $\frac{1}{x}$

y = $\frac{x^{2}}{31} + \frac{1}{31 \sqrt{x}}$

Awọn alaye Idahun

y = kx2 + $\frac{c}{\sqrt{x}}$
y = 2 when x = 1
2 = k + $\frac{c}{1}$
k + c = 2
y = 6 when x = 4
6 = 16k + $\frac{c}{2}$
12 = 32k + c
k + c = 2
32k + c = 12
= 31k + 10
k = $\frac{10}{31}$
c = 2 - $\frac{10}{31}$
= $\frac{62 - 10}{31}$
= $\frac{52}{31}$
y = $\frac{10 x^{2}}{31} + \frac{52}{31 \sqrt{x}}$