JAMB UTME - Mathematics - 1993

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}

Akọwa Nkọwa

The given equation is 2log3 y + log3 x2 = 4. We can simplify this equation using the laws of logarithms. First, we can use the power rule of logarithms to rewrite x2 as (x)2. Then, we can use the product rule of logarithms to combine the two logarithms on the left-hand side of the equation: 2log3 y + log3 (x)2 = log3 y2 + log3 (x)2 Now, we can use the sum rule of logarithms to combine the two logarithms on the right-hand side of the equation: log3 y2(x)2 = log3 (x2y2) We can now rewrite the original equation as: log3 (x2y2) = 4 Using the definition of logarithms, we know that log3 (x2y2) = 4 is equivalent to 3^4 = x2y2. Therefore, we have: x2y2 = 81 Taking the square root of both sides, we get: xy = ±9 Since y is a positive real number, we can discard the negative solution. Therefore, we have: xy = 9 Finally, we can solve for y: y = 9/x So, the answer is "±9/x".

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Sets

Logarithms

Lee Azịza

Ajụjụ 2 Ripọtì

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}

Akọwa Nkọwa

The shaded part exists on x

\cap

z but not in y

Gụọ akwụkwọ nkuzi na Logical Reasoning (WAEC)

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Logical Reasoning

Integration

Matrices And Determinants

Lee Azịza

Ajụjụ 3 Ripọtì

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

Akọwa Nkọwa

To find the area of a sector, we use the formula: Area of sector = (θ/360) x πr² Where θ is the angle of the sector in degrees, r is the radius of the circle, and π (pi) is a mathematical constant approximately equal to 3.14. In this problem, the radius is given as 3m and the angle of the sector is 60°. Substituting these values into the formula, we get: Area of sector = (60/360) x π(3)² = (1/6) x π(9) = (1/6) x 28.27 = 4.71m² (rounded to two decimal places) Therefore, the area of the sector of the circle with radius 3m and angle 60° is approximately 4.71m². The answer is option C.

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 4 Ripọtì

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

15o

25o

40o

50o

55o

Akọwa Nkọwa

Join PR
QRP = QPR
= 180 - 80 = 100/20 = 50o
SRP = SPR
= 180 - 30 = 150/2 = 75o
∴ SPQ = SPR - QPR
= 75 - 50 = 25o

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 5 Ripọtì

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

Akọwa Nkọwa

To calculate the angle of the sector with proportion 7, we first need to find the total of all the quantities in the proportion. Total quantity = 1 + 4 + 6 + 7 = 18 Next, we need to find the proportion of quantity 7 in relation to the total quantity. Proportion of quantity 7 = 7/18 To find the angle of the sector that represents this proportion, we need to use the formula: Angle of sector = Proportion of quantity * 360 degrees Angle of sector = (7/18) * 360 degrees Angle of sector = 140 degrees Therefore, the answer is option D, which is 140 degrees.

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

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 6 Ripọtì

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

68

^{\circ}

102

^{\circ}

107

^{\circ}

141

^{\circ}

Akọwa Nkọwa

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 $^{\circ}$

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

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Angles And Intercepts On Parallel Lines

Statistics

Lee Azịza

Ajụjụ 7 Ripọtì

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}

Akọwa Nkọwa

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 $^{\circ}$

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 8 Ripọtì

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

2

4

10

11

Akọwa Nkọwa

To calculate the standard deviation, we need to first find the mean (average) of the data. Mean = (7 + 8 + 9 + 10 + 11 + 12 + 13) / 7 = 10 Next, we need to find the difference between each data point and the mean, and square the differences. (7-10)^2 = 9 (8-10)^2 = 4 (9-10)^2 = 1 (10-10)^2 = 0 (11-10)^2 = 1 (12-10)^2 = 4 (13-10)^2 = 9 Then, we add up all of these squared differences and divide by the total number of data points. (9 + 4 + 1 + 0 + 1 + 4 + 9) / 7 = 28 / 7 = 4 Finally, we take the square root of this result to find the standard deviation. Standard deviation = √4 = 2 Therefore, the standard deviation of the given data is 2.

Gụọ akwụkwọ nkuzi na Measures Of Dispersion (JAMB)

Measures Of Dispersion

Lee Azịza

Ajụjụ 9 Ripọtì

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

Estimate the mode of the frequency distribution above.

13.2g

15.0g

16.8g

17.5g

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Linear Inequalities (WAEC)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Representation Of Data

Probability

Linear Inequalities

Statistics

Lee Azịza

Ajụjụ 10 Ripọtì

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

Lee Azịza

Ajụjụ 11 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Modular Arithmetic

Lee Azịza

Ajụjụ 12 Ripọtì

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

Akọwa Nkọwa

Surface area = 154cm2 (area of sphere)
4

π

r2 = 154
r

\sqrt{\frac{154}{4 π}}

= 3.50cm

Lee Azịza

Ajụjụ 13 Ripọtì

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

\frac{1}{2}

1

2

3

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 14 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Representation Of Data

Measures Of Location

Lee Azịza

Ajụjụ 15 Ripọtì

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

45o, 225o

135o, 315o

45o, 315o

135o, 225o

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Trigonometry

Lee Azịza

Ajụjụ 16 Ripọtì

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

Lee Azịza

Ajụjụ 17 Ripọtì

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}

Akọwa Nkọwa

(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6
= 9x2 = 12x + 4
6x2 - 10x + 4 = 0
6x2 - 6x - 4x + 4 = 0
6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0
x = 1 or

\frac{2}{3}

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 18 Ripọtì

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

5

1

3

6

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 19 Ripọtì

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

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 20 Ripọtì

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

5

4

3

2

1

Akọwa Nkọwa

\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

Gụọ akwụkwọ nkuzi na Fractions, Decimals, Approximations And Percentage (JAMB)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Fractions, Decimals, Approximations And Percentage

Quadratic Equations

Lee Azịza

Ajụjụ 21 Ripọtì

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

17

20

25

30

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 22 Ripọtì

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

Lee Azịza

Ajụjụ 23 Ripọtì

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}

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 24 Ripọtì

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

x + 3

x - 3

x - 5

x + 5

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 25 Ripọtì

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

3

4

5

8

Akọwa Nkọwa

log5(62.5) - log5(

\frac{1}{2}

)
= log5

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

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

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

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Positive And Negative Integers, Rational Numbers

Logarithms

Lee Azịza

Ajụjụ 26 Ripọtì

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

Akọwa Nkọwa

\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

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

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

Fractions, Decimals And Approximations

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 27 Ripọtì

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)}

Akọwa Nkọwa

\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}

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 28 Ripọtì

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

Akọwa Nkọwa

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

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 29 Ripọtì

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

Akọwa Nkọwa

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

Lee Azịza

Ajụjụ 30 Ripọtì

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

3

\frac{2}{3}

7

-4

Akọwa Nkọwa

\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

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 31 Ripọtì

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

12

10

9

8

Akọwa Nkọwa

A rectangular polygon has each interior angle to be 150o
let the polygon has n-sides
therefore, Total interior angle 150 x n = 150n
hence 150n = (2n - 4)90
150n = 180n - 360
360 = (180 - 150)n
30n = 360
n = 12

Gụọ akwụkwọ nkuzi na Polynomials (JAMB)

Polynomials

Lee Azịza

Ajụjụ 32 Ripọtì

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

Akọwa Nkọwa

Let the age of the teacher be T. According to the problem, the mean age of the 10 pupils is 16. Therefore, the sum of their ages is 10 x 16 = 160. When the teacher's age is added to it, the sum becomes 160 + T. Also, it is given that the new mean age after adding the teacher's age is 19. So, we have: (160 + T) / 11 = 19 Multiplying both sides by 11, we get: 160 + T = 209 Subtracting 160 from both sides, we get: T = 49 Therefore, the age of the teacher is 49.

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Gụọ akwụkwọ nkuzi na Statistics (WAEC)

Integration

Statistics

Lee Azịza

Ajụjụ 33 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 Circles (WAEC)

Gụọ akwụkwọ nkuzi na Triangles And Polygons (WAEC)

Circles

Triangles And Polygons

Lee Azịza

Ajụjụ 34 Ripọtì

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)}

Akọwa Nkọwa

To make x the subject of the given relation, we need to isolate x on one side of the equation. First, we can simplify the expression by finding the common denominator of the two fractions in the numerator: (1 + ax)/(1 - ax) = (1 - ax + ax + a^2x)/(1 - ax) = (1 + a^2x)/(1 - ax) Now we can cross-multiply to get rid of the denominator: (1 + a^2x) = pq - axpq Next, we can move the term with x to the left-hand side of the equation: a^2x + axpq = pq - 1 Finally, we can factor out x from the two terms on the left-hand side: x(a^2 + apq) = pq - 1 Therefore, we can isolate x by dividing both sides by (a^2 + apq): x = (pq - 1)/(a^2 + apq) So the correct option is: p-q / a(p+q) which is not among the given options.

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 35 Ripọtì

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

Akọwa Nkọwa

Initial salary = ₦540
increment = ₦36 (every 6 months)
Period of increment = 2 yrs and 6 months
amount(increment) = ₦36 x 5 = ₦180
The man's new salary = ₦540 = ₦180
= ₦720.00

Gụọ akwụkwọ nkuzi na Sequence And Series (WAEC)

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

Sequence And Series

Ratio, Proportions And Rates

Lee Azịza

Ajụjụ 36 Ripọtì

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)

Akọwa Nkọwa

(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)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 37 Ripọtì

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

Akọwa Nkọwa

10y x 5(2y - 2) x 4(y - 1) = 1
but 10y - (5 x 2)y = 5y x 2y
= (Law of indices)
5y x 2y x 5(2y - 2) x 4(y - 1) = 1
but 4(y - 1) = 22(y - 1)
= 2y - 2 (Law of indices)
5y x 5(2y -2) x 2(- 2) = 1
5(3y -2) x 2y x 2(2y -2) = 1
= 5(3y -2) x 2(3y -2) = 1
But ao = 1
10(3y -2) = 10o
3y - 2 = 0
? y =

\frac{2}{3}

Lee Azịza

Ajụjụ 38 Ripọtì

Solve the following simultaneous equation for x. x2 + y - 5 = 0, y - 7x + 3 = 0

-2, 4

2, 4

-1,8

1, -8

Akọwa Nkọwa

x2 + y - 5 = 0.....(i)
y - 7x + 3 = 0.........(ii)
y = 7x - 3, substituting the value of y in equation (i)
x2 + (7x - 3) - 5 = 0
x2 + 7x + 3 = 0
(x + 8)(x - 1) = 0
x = -8 or 1

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

Solution Of Linear Equations

Lee Azịza

Ajụjụ 39 Ripọtì

If sin x = cos 50o, then x equals

40o

45o

50o

90o

130o

Akọwa Nkọwa

We know that sin x = cos 50°. To solve for x, we need to find the angle whose sine is equal to cos 50°. We can use the trigonometric identity that sin (90° - θ) = cos θ. So, sin x = cos 50° can be rewritten as sin x = sin (90° - 50°). Using the identity, we have: sin x = sin 40° Now, we need to find the angle x whose sine is equal to sin 40°. Since sine is a periodic function, there are multiple angles whose sine is equal to a given value. One such angle is x = 40°. However, sine is also negative in the third and fourth quadrants. In the third quadrant, x = 180° - 40° = 140° and in the fourth quadrant, x = 360° - 40° = 320° also satisfy the equation sin x = sin 40°. However, since x has to be between 0° and 360°, we can eliminate the30°. Therefore, the possible values of x are 40°, 140°, and 320°. However, since we know that sin x = cos 50° and cos is positive in the first quadrant, x cannot be in the third or fourth quadrants. Therefore, the only possible value of x is x = 40°. Hence, the answer is x = 40°.

Gụọ akwụkwọ nkuzi na Introductory Calculus (WAEC)

Gụọ akwụkwọ nkuzi na Trigonometry (JAMB)

Introductory Calculus

Trigonometry

Lee Azịza

Ajụjụ 40 Ripọtì

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}

Akọwa Nkọwa

To solve this problem, we need to first determine the set of possible numbers that could be selected at random between 20 and 30. Since the problem states that both 20 and 30 are inclusive, the set of possible numbers is: {20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30} To find the probability that the selected number is prime, we need to determine the number of prime numbers in the set of possible numbers, and then divide that by the total number of possible numbers. Prime numbers are numbers that are only divisible by 1 and themselves. The prime numbers in the set of possible numbers are: { 23, 29 } Therefore, the probability that the selected number is prime is 2/11 or approximately 0.18.

Lee Azịza

Ajụjụ 41 Ripọtì

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

Akọwa Nkọwa

x

*

y = xy
x

*

2 = x2
x

*

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

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 42 Ripọtì

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

3

4

12

17

Akọwa Nkọwa

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.

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

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Sine, Cosine And Tangent Of An Angle

Modular Arithmetic

Financial Arithmetic

Lee Azịza

Ajụjụ 43 Ripọtì

The angle between latitudes 30oS and 13oN is

17o

33o

43o

53o

Akọwa Nkọwa

The angle between 2 latitudes one in northern hemisphere and the other in southern hemisphere and the other in southern hemisphere is the sum of their latitudes.
∴ Total angle difference = (30 + 13) = 43o

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza

Ajụjụ 44 Ripọtì

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

-x3 + 7x2 - x - 7

-x3 = 7x + 7

x - 7

x + 7

Algebraic Expressions

Lee Azịza

Ajụjụ 45 Ripọtì

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

705

70,000

70, 480

70, 500

Akọwa Nkọwa

To evaluate the given expression to 3 significant figures, we need to round the result to the nearest third digit after the decimal point. To do this, we can perform the division and then round the result. The expression is: 3524 / 0.05 = 70,480 To round this to 3 significant figures, we need to consider the digits up to the third digit after the decimal point, which is 4. Since 4 is less than 5, we round down the previous digit, which is 8. Therefore, the final answer is: 70,500 So, is the correct answer.

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

Gụọ akwụkwọ nkuzi na Number Bases (JAMB)

Fractions, Decimals And Approximations

Number Bases

Lee Azịza

Ajụjụ 46 Ripọtì

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}

Akọwa Nkọwa

< 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 $^{\circ}$

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 47 Ripọtì

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}

Akọwa Nkọwa

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

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

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 48 Ripọtì

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

Akọwa Nkọwa

QPS - QRK = 31o
QRK + RKQ + KQR = 180
31 + 42 + KQR = 180o
KQR = 180 - 73 = 107o

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Inequalities

Lee Azịza

Ajụjụ 49 Ripọtì

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}

Akọwa Nkọwa

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}

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 50 Ripọtì

change 7110 to base 8

1078

1068

718

178

Akọwa Nkọwa

To convert the number 71 to base 8, we need to divide it repeatedly by 8 and note down the remainders until the quotient becomes 0. The remainders in reverse order will give us the equivalent number in base 8. Here's how we can do it: - 71 ÷ 8 = 8 remainder 7 - 8 ÷ 8 = 1 remainder 0 - 1 ÷ 8 = 0 remainder 1 So, the equivalent number in base 8 is 107. Therefore, the correct answer is option A: 1078.

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

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






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