JAMB UTME - Mathematics - 2008

If 125x = 2010 find x

2

3

4

5

Akọwa Nkọwa

125x = 20
1xX2 + 2xX1 + 5xX0 = 20
X2 + 2X + 5 = 20
X2 + 2X - 15 = 0
(X + 5)( X - 3) = 0
X + 5 implies X = -5
X - 3 implies X = 3
But X cannot be negative
∴X = 3

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 2 Ripọtì

In the diagram < OPQ is

90

^{\circ}

53

^{\circ}

36

^{\circ}

26

^{\circ}

Akọwa Nkọwa

74 + x + x = 180

2x = 180 - 74

2x = 103

x = 53

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Angles

Lee Azịza

Ajụjụ 3 Ripọtì

\[ \begin{pmatrix} -2 & 1 \\ 2 & 3 \end{pmatrix} + \begin{pmatrix} p & q \\ r & s \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}. \] What is the value of r?

-

\frac{1}{8}

\frac{3}{8}

\frac{5}{8}

\frac{1}{4}

Akọwa Nkọwa

-2p + r = 1.......(i)
2p + 3r = 0.......(ii)
r - 1 + 2p ........(iii)
2p + 3(1 + 2p) = 0
2p + 3(1 + 2p) = 0
2p + 3 + 6p = 0
3 - 8p = 0

\to

8p = 3
p =

\frac{3}{8}

6 = 1 - 2

\frac{3}{8}

= 1 -

\frac{6}{8}

\frac{8 - 6}{8}

=

\frac{2}{8}

=

\frac{1}{4}

Lee Azịza

Ajụjụ 4 Ripọtì

Find the mean deviation of 2, 4, 5, and 9

1

2

5

7

Akọwa Nkọwa

X = 20/4
= 5
mean deviation = 8/4 = 2

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 5 Ripọtì

Find the range of values of x which satisfy the inequalities $4x - 7 \leq 3x$ and $3x - 4 \leq 4x$

-4

\leq

x

\leq

7

-7

\leq

x

\leq

4

x

\geq

-7

\leq

x

\leq

6

Akọwa Nkọwa

4X - 7

\leq

3X and 3X - 4

\leq

4X
4X - 3X

\leq

7 and 3X - 4X

\leq

4
X

\leq

7 and -X

\leq

4 = X

\geq

-4
Range -4

\leq

x

\leq

7

Gụọ akwụkwọ nkuzi na Variation (JAMB)

Gụọ akwụkwọ nkuzi na Inequalities (JAMB)

Variation

Inequalities

Lee Azịza

Ajụjụ 6 Ripọtì

Find the median of 4, 1, 4, 1, 0, 4, 4, 2 and 0

zero

1

2

3

Akọwa Nkọwa

4, 4, 1, 0, 4, 4, 2, 0
Arrange in ascending order:
0, 0, 1, 1, 2, 4, 4, 4, 4
Position of median = N+1 / 2 = 9+1 / 2 = 5th
Median = 2

Gụọ akwụkwọ nkuzi na Measures Of Location (JAMB)

Measures Of Location

Lee Azịza

Ajụjụ 7 Ripọtì

If p = varies inversely as the square of q and p=8 when q=4, find when p=32

+/-16

+/-8

+/-4

+/-2

Akọwa Nkọwa

P ∝ 1/q
P = k/q
K = q2P
= 428
∴P = 128/q
32 = 128/q
q2 = 128/32
q2 = 4
q = √4 = +/-2

Gụọ akwụkwọ nkuzi na Progression (JAMB)

Progression

Lee Azịza

Ajụjụ 8 Ripọtì

If $ \frac{1+\sqrt{2}}{1-\sqrt{2}} $ is expressed in the form of x+y√2 find the values of x and y

(-3, -2)

(-2, 3)

(3,2)

(2,-3)

Akọwa Nkọwa

\frac{1 + \sqrt{2}}{1 - \sqrt{2}} \times \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{1 + (1 + \sqrt{2}) + \sqrt{2} (1 + \sqrt{2})}{1^{2} - (\sqrt{2})^{2}} = \frac{(1 + \sqrt{2} + \sqrt{2} + 2)}{1 - 2} = \frac{3 + 2 \sqrt{2}}{- 1} = - 3 - 2 \sqrt{2} ∴ X a n d Y = - 3 a n d - 2

Gụọ akwụkwọ nkuzi na Surds (radicals) (WAEC)

Surds (radicals)

Lee Azịza

Ajụjụ 9 Ripọtì

Calculate the distance between points L(-1, -6) and M(-3, -5)

√5

2√3

√20

√50

Akọwa Nkọwa

L

(\begin{matrix} x_{1} & y_{1} \\ - 1 & - 6 \end{matrix})

m L

(\begin{matrix} x_{2} & y_{2} \\ - 3 & - 5 \end{matrix})

D =

\sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

D =

\sqrt{(- 3 - (- 1)^{2} + (- 5 - (- 6)^{2}}

D =

\sqrt{(- 3 + 1)^{2} + (- 5 + 6)^{2}}

D =

\sqrt{(- 2)^{2} + 1^{2}}

D =

\sqrt{4 + 1}

D =

\sqrt{5}

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (JAMB)

Euclidean Geometry

Coordinate Geometry

Lee Azịza

Ajụjụ 10 Ripọtì

The probability of picking a letter T from the word OBSTRUCTION is

\frac{1}{11}

\frac{3}{11}

\frac{3}{11}

\frac{4}{11}

Akọwa Nkọwa

Obstruction
if a T is picked
p =

\frac{2}{11}

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Probability

Lee Azịza

Ajụjụ 11 Ripọtì

In how many ways can the letters of the word ACCEPTANCE be arranged?

\frac{10!}{2! 2! 3!}

\frac{10!}{2! 3!}

\frac{10!}{2! 2!}

\frac{10!}{3! 2! 2!}

Akọwa Nkọwa

Acceptance

\frac{10!}{3! 2! 2!}

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Gụọ akwụkwọ nkuzi na Probability (JAMB)

Representation Of Data

Permutation And Combination

Probability

Lee Azịza

Ajụjụ 12 Ripọtì

Make Q the subject of formula when $L = \frac{4}{3}M\sqrt{PQ}$

\frac{9 L^{2}}{16 M^{2}} P

\frac{3 L}{4 M \sqrt{P}}

\frac{\sqrt{3 L}}{4 M P}

\frac{3 L^{2}}{16 M^{2}} P

Akọwa Nkọwa

L = \frac{4}{3} M \sqrt{P Q} = \frac{3}{4 M} \times L = \sqrt{P Q} = {(\frac{3 L}{4 M})}^{2} = (\sqrt{P Q})^{2} = \frac{9 L^{2}}{16 M^{2}} = P Q = Q = \frac{9 L^{2}}{16 M^{2} P}

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 13 Ripọtì

The bar chart above shows the number of times the word a, and , in, it, the , to appear in a paragraph in a book. What is the ratio of the least frequent word?

1/4

1/3

2/3

3/4

Akọwa Nkọwa

Ratio of least to most = 3:12
= 3/12
= 1/4

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Representation Of Data

Integration

Lee Azịza

Ajụjụ 14 Ripọtì

A binary operation on the real set of numbers excluding -1 is such that for all m, n ∈ R, mΔn = m+n+mn. Find the identity element of the operation.

1

zero

-1/2

-1

Akọwa Nkọwa

mΔn = m+n+mn
Let e be the identity element
∴mΔe = eΔm = m
m+e+me = m
e+me = m-m
e+me = 0
e(1+m) = 0
e = 0 / (1+m)
e = 0

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 15 Ripọtì

If x - 3 is directly proportional to the square of y and x = 5 when y =2, find x when y = 6.

30

21

16

12

Akọwa Nkọwa

(x – 3) ∝ y2
X-3 = Ky2
K = X-3 / y2
= 5-2/22
= 2/4
= 1/2
∴X-3 = 1/2y2
X-3 = 1/2(6)2
X-3 = 1/2 x 30/1
X-3 = 18
X = 21

Lee Azịza

Ajụjụ 16 Ripọtì

Find the values of x and y respectively if

\[ \begin{pmatrix} 1 & 0\\ -1 & -1\\ 2 & 2 \end{pmatrix} + \begin{pmatrix} x & 1\\ -1 & 0\\ y & -2 \end{pmatrix} = \begin{pmatrix} -2 & 1\\ -2 & -1\\ -30 & 0 \end{pmatrix} \]

-3, -2

-5, -3

-2, -5

-3, -5

Akọwa Nkọwa

(\begin{matrix} 1 & 0 \\ - 1 & - 1 \\ 2 & 2 \end{matrix})

+

(\begin{matrix} x & 1 \\ - 1 & 0 \\ y & - 2 \end{matrix})

=

(\begin{matrix} - 2 & 1 \\ - 2 & - 1 \\ - 30 & 0 \end{matrix})

therefore, (x, y) = (-3, -5) respectively

Lee Azịza

Ajụjụ 17 Ripọtì

Find the area of the figure above

[π = 22/7]

12.5 cm2

75.0 cm2

78.5 cm2

84.8 cm2

Akọwa Nkọwa

Area of the figure = Area of rect + area of semi circle

= L \times h + \frac{1}{2} π r^{2} 5 \times 15 + \frac{1}{2} \times \frac{22}{7} \times {(\frac{5}{2})}^{2} = 75 + \frac{(22 \times 25)}{2 \times 7} = 75 + 9 \frac{23}{28} = 84.8 c m

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 18 Ripọtì

A binary operation on the set of real numbers excluding -1 is such that sor all $m, n \in R$, $m \Delta n = m + n + mn$. Find the identity element of the operation.

1

0

-

\frac{1}{2}

-1

Akọwa Nkọwa

m

Δ

n = m + n + mn
m

Δ

e = e

Δ

m = m
m

Δ

e = m
m + e + me = m
e + me = m - m
e + me = 0
e(1 + m) = 0
e =

\frac{0}{1 + m}

= 0

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Binary Operations

Lee Azịza

Ajụjụ 19 Ripọtì

A binary operation * is defined on the set of positive integers is such x*y = 2x-3y+2 for all positive integers x and y. The binary operation is

commutative and close on the set of positive integers

neither commutative nor closed on the set of ppositive integers

commutative but not closed on the set of positive integers

not cummutative but closed on the set of positive integers

Akọwa Nkọwa

X * Y = 2X - 3Y + 2
2*3 = 2(2) - 3(3) + 2
=4-9+2
= -3
But -3 does not belong to positive integer

Gụọ akwụkwọ nkuzi na Matrices And Determinants (JAMB)

Gụọ akwụkwọ nkuzi na Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

Lee Azịza

Ajụjụ 20 Ripọtì

Evaluate $ \int_{1}^{2} (6x^{2} - 2x)\,dx $

16

13

12

11

Akọwa Nkọwa

\int_{1}^{2} (6 x^{2} - 2 x) d x = [\frac{6 x^{3}}{3} - \frac{2 x^{2}}{2}]_{1}^{2} = [2 x^{3} - x^{2}]_{1}^{2}

= [2(2)3 - (2)2] – [2(1)3 - (1)2]
= [16-4] – [2-1]
= 12 – 1
= 11

Gụọ akwụkwọ nkuzi na Modular Arithmetic (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Modular Arithmetic

Quadratic Equations

Lee Azịza

Ajụjụ 21 Ripọtì

\[ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x\, dx \]

7

1

2

3

Akọwa Nkọwa

\int_{- \frac{π}{2}}^{\frac{π}{2}}

cosx

[s i n x]_{\frac{π}{2}}^{\frac{π}{2}}

= sin

\frac{π}{2}

- (-sin

\frac{π}{2}

)
= sin

\frac{π}{2}

+ sin

\frac{π}{2}

= 2 sin

\frac{π}{2}

= 2 sin 1.5714
= 2(0.2704)
= 0.5
= 1

Gụọ akwụkwọ nkuzi na Integration (JAMB)

Integration

Lee Azịza

Ajụjụ 22 Ripọtì

A student sitting on a tower 68 metres high observes his principal's car at the angle of depression of 20o. How far is the car from the bottom of the tower to the nearest metre?

184m

185m

186m

187m

Akọwa Nkọwa

Tan 20o =

\frac{68 m}{x}

x tan 20o = 68
x =

\frac{68}{t a n 20}

=

\frac{68}{0.364}

x = 186.8
= 187m

Gụọ akwụkwọ nkuzi na Bearings (WAEC)

Gụọ akwụkwọ nkuzi na Coordinate Geometry (WAEC)

Bearings

Coordinate Geometry

Lee Azịza

Ajụjụ 23 Ripọtì

If logx1/264 = 3, find the value of x

4

16

32

64

Akọwa Nkọwa

If logx1/264 = 3
(X 1/2)3 = 64
(X 1/2)3 = 4 3
X 1/2 = 4
X = 42
X = 16

Gụọ akwụkwọ nkuzi na Logarithms (WAEC)

Logarithms

Lee Azịza

Ajụjụ 24 Ripọtì

Find the number of ways of selecting 6 out of 10 subjects for an examination

128

216

215

210

Akọwa Nkọwa

^{10} C_{6} = \frac{10!}{(10 - 6)! 6!} = \frac{10!}{4! 6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4 \times 3 \times 2 \times 1 \times 6!} = 2

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 25 Ripọtì

Find the area of the figure given

12.5cm2

75.0cm2

78.5cm2

84.8cm2

Akọwa Nkọwa

Area of semicircle + Area of rectangle

A = $\frac{1}{2} π r^{2}$ + LB

A = $\frac{1}{2} \times \frac{22}{77} \times (\frac{5}{2})^{2} + (15 \times 5)$

= $\frac{1}{2} \times \frac{22}{7} \times \frac{25}{4} + 75$

A = $\frac{275}{28} + \frac{75}{1}$

$\frac{275 + 2100}{28} = \frac{2375}{28}$

A = 84.8cm2

Gụọ akwụkwọ nkuzi na Areas (WAEC)

Areas

Lee Azịza

Ajụjụ 26 Ripọtì

If x > 0, find the range of number x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x and 3x+1

3x+3

3x+1

2x+5

2x+1

Akọwa Nkọwa

x-3, 3x+2,x-1, 4x, 2x-1, x-2, 2x-2, 3x, 3x+1
Range = 3x+2 - (x-3)
= 3x+2 - x - 3
= 2x + 5

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Algebraic Expressions

Quadratic Equations

Lee Azịza

Ajụjụ 27 Ripọtì

What is the mean of the data t, 2t-1, t-2, 2t-1, 4t and 2t+2?

2t

2t-1

2/3t+1

2t-1/3

Akọwa Nkọwa

M e a n = \frac{t + 2 t - 1 + t - 2 + 2 t - 14 t + 2 t + 2}{6} M e a n = \frac{12 t - 2}{6} = \frac{2 (6 t - 1)}{6} = \frac{6 t - 1}{3} = \frac{6 t}{3} - \frac{1}{3} = 2 t - \frac{1}{3}

Gụọ akwụkwọ nkuzi na Representation Of Data (JAMB)

Representation Of Data

Lee Azịza

Ajụjụ 28 Ripọtì

If 2x2 - kx - 12 is divisible by x-4, Find the value of k.

4

5

6

7

Akọwa Nkọwa

2x2 - kx - 12 is divisible by x-4
implies x is a factor ∴ x = 4
f(4) implies 2(4)2 - k(4) - 12 = 0
32 - 4k - 12 = 0
-4k + 20 = 0
-4k = -20
k = 5

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 29 Ripọtì

calculate the simple interest on N6,500 for 8 years at 5% per annum.

N3,000

N600

N300

N150

Akọwa Nkọwa

I = \frac{P \times T \times R}{100} = \frac{7500 \times 8 \times 5}{100} = N 3000

Gụọ akwụkwọ nkuzi na Financial Arithmetic (WAEC)

Financial Arithmetic

Lee Azịza

Ajụjụ 30 Ripọtì

If $ \log X^{\frac{1}{2}} $ 64 = 3, find the value of x.

4

16

32

64

Akọwa Nkọwa

l o g X^{\frac{1}{2}}

64 = 3, find the value of X
recall that X = 10p
log10X = P
64 =

X^{\frac{1}{2}}

43 =

X^{\frac{1}{2}}

4^{3 \times \frac{1}{2}}

=

X^{\frac{1}{2} \times \frac{1}{3}}

4 =

X^{\frac{1}{2}}

(4)2 =

X^{\frac{1}{2} \times 2}

x = 16

Lee Azịza

Ajụjụ 31 Ripọtì

simplify $16^{-\frac{1}{2}} \times 4^{-\frac{1}{2}} \times 27^{\frac{1}{3}}$

3/8

2/3

3/4

3/2

Akọwa Nkọwa

16^{\frac{? 1}{2}} \times 4^{\frac{? 1}{2}} \times 27^{\frac{1}{3}} = \frac{1}{16^{\frac{1}{2}}} \times \frac{1}{4^{\frac{1}{2}}} \times 27^{\frac{1}{3}} = \frac{1}{\sqrt{16}} \times \frac{1}{\sqrt{4}} \times s q r t [3] 27 = \frac{1}{4} \times \frac{1}{2} \times \frac{3}{1} = \frac{3}{8}

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

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

Number Bases

Simple Operations On Algebraic Expressions

Lee Azịza

Ajụjụ 32 Ripọtì

In the diagram, find the size of the angle marked ao

60o

80o

120o

160o

Akọwa Nkọwa

2 x s = 280o(Angle at centre = 2 x < at circum)

S = $\frac{280^{o}}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Circles

Lee Azịza

Ajụjụ 33 Ripọtì

Factorize complete;y (4x+3y)2 - (3x-2y)2

(x+5y)(7x+y)

(x+5y)(7x-y)

(x-5y)(7x+y)

(x-5y)(7x-y)

Akọwa Nkọwa

(4x+3y)2 - (3x-2y)2
(4x+3y+3x-2y)(4x+3y-(3x-2y))
(4x+3y+3x-2y)(4x+3y-3x+2y)
(x+5y)(7x+y)

Gụọ akwụkwọ nkuzi na Algebraic Expressions (WAEC)

Algebraic Expressions

Lee Azịza

Ajụjụ 34 Ripọtì

Find the gradient of a line which is perpendicular to the line with the equation 3x + 2y + 1 = 0

3/2

2/3

-2/3

-3/2

Akọwa Nkọwa

3X + 2Y + 1 = 0
2Y = -3X - 1

\frac{- 3}{2} X - \frac{1}{2}

Gradient of 3X + 2Y +1 = 0 is -3/2
Gradient of a line perpendicular to 3X + 2Y + 1 = 0

= - 1 \div \frac{3}{2} = - 1 \times \frac{- 2}{3} = \frac{2}{3}

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

Graphs Of Linear And Quadratic Functions

Lee Azịza

Ajụjụ 35 Ripọtì

Find the minimum value of the function y = x(1+x)

-1/4

-1/2

1/4

1/2

Akọwa Nkọwa

y = x(1+x)
= x + x2
dy/dx = 1 + 2x
As dy/dy → 0
1 + 2x = 0
2x = -1
X = -1/2
Y = x(1+x)
= -1/2(1 - 1/2) at x = -1/2
= -1/2(1/2)
= -1/4

Lee Azịza

Ajụjụ 36 Ripọtì

In the diagram above ∠OPQ is

90

^{\circ}

53

^{\circ}

36

^{\circ}

26

^{\circ}

Akọwa Nkọwa

a = a(base ∠s of Iss Δ)
∴ a+a+74 = 180
2a + 74 = 180
2a = 180-74
2a = 106
a = 53
∴∠OPQ = 53

^{\circ}

Gụọ akwụkwọ nkuzi na Circles (WAEC)

Gụọ akwụkwọ nkuzi na Angles (WAEC)

Circles

Angles

Lee Azịza

Ajụjụ 37 Ripọtì

if X = {n2 + 1 : n = 0, 2, 3} and y = {n + 1 : n = 2, 3, 4, 5}, find x ∩ y

{1, 3}

{5,10}

Φ

(4, 6)

Akọwa Nkọwa

X = {n2 + 1 : n = 0, 2, 3}
y = {n + 1 : n = 2, 3, 4, 5}
find x ∩ y
x = (n2 + 1)(22 + 1)(32 + 1)
y = (2 + 1) (3 + 1)(5 + 1)
x = {1, 5, 10}
y = {3, 4, 6}
x ∩ y =

Φ

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Sets

Lee Azịza

Ajụjụ 38 Ripọtì

The result of rolling a fair die 150 times is as summarized in the table given.

Number	1	2	3	4	5	6
Frequency	12	18	$x$	30	$2x$	45

What is the probability of obtaining a 5?

\frac{3}{10}

\frac{1}{5}

\frac{1}{6}

\frac{1}{10}

Akọwa Nkọwa

\begin{array}{cc} N u m b e r & 1 & 2 & 3 & 4 & 5 & 6 \\ F r e q u e n c y & 12 & 18 & x & 30 & 2 x & 45 \end{array}

12 + 18 + x + 30 + 2x + 45 = 150
3x + 105 = 150
3x = 150 - 105
3x = 45
x =

\frac{45}{3}

x = 15
probability of 5 =

\frac{30}{150}

=

\frac{1}{5}

Lee Azịza

Ajụjụ 39 Ripọtì

The solution of the quadratic inequality $x^2 + x - 12 \ge 0$ is

x

\geq

3 or x

\geq

-4

x

\leq

3 or x

\leq

-4

x

\geq

3 or x

\leq

-4

x

\geq

-3 or x

\leq

4

Akọwa Nkọwa

(x2 + x - 12)

\geq

0 , (x - 3)(x + 4)

\geq

0
For the condition to hold, each of (x - 3) and (x + 4) must be of the same sign
.i.e. x - 3

\geq

0 and x + 4

\geq

0
or x - 3

\leq

0 and x + 4

\leq

0
when x

\geq

3, the condition is satisfied
when x

\geq

-4, the condition is not satisfied.
when x

\leq

3, the condition is not satisfied
when x

\leq

-4 , the condition is not satisfied. Thus, the solution of the inequality is x

\geq

3 or x

\leq

-4 ,

Gụọ akwụkwọ nkuzi na Quadratic Equations (WAEC)

Quadratic Equations

Lee Azịza

Ajụjụ 40 Ripọtì

Express 1223456 to 3 significant figures

123000

124000

125000

126000

Fractions, Decimals And Approximations

Lee Azịza

Ajụjụ 41 Ripọtì

If X = {n2 + 1:n = 0,2,3} and Y = {n+1:n=2,3,5}, find X∩Y.

{1,3}

{5,10}

∅

{4,6}

Akọwa Nkọwa

X = {1,5,10}
Y = {3,4,6}
X∩Y = ∅

Gụọ akwụkwọ nkuzi na Sets (JAMB)

Sets

Lee Azịza

Ajụjụ 42 Ripọtì

The locus of a point equidistant from two points p(6,2) and R(4,2) is a perpendicular bisector of PR passing through

(2,5)

(5,2)

(1,0)

(0,1)

Akọwa Nkọwa

X = \frac{6 + 2}{2} = \frac{10}{2} = 5 Y = \frac{2 + 2}{2} = \frac{4}{2} = 2 ∴ X, Y = (5, 2)

Gụọ akwụkwọ nkuzi na Loci (JAMB)

Gụọ akwụkwọ nkuzi na Loci (WAEC)

Gụọ akwụkwọ nkuzi na Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Lee Azịza

Ajụjụ 43 Ripọtì

Find the minimum value of X2 - 3x + 2 for all real values of x

-

\frac{1}{4}

-

\frac{1}{2}

\frac{1}{4}

\frac{1}{2}

Akọwa Nkọwa

y = X2 - 3x + 2,

\frac{d y}{d x}

= 2x - 3
at turning pt,

\frac{d y}{d x}

= 0
∴ 2x - 3 = 0
∴ x =

\frac{3}{2}

\frac{d^{2} y}{d x^{2}}

=

\frac{d}{d x}

(

\frac{d}{d x}

)
= 270
∴ ymin = 2

\frac{3}{2}

- 3

\frac{3}{2}

+ 2
=

\frac{9}{4}

-

\frac{9}{2}

+ 2
= -

\frac{1}{4}

Lee Azịza

Ajụjụ 44 Ripọtì

Solve the quadratic inequalities x2 - 5x + 6 ≥0

x ≤ 2, x ≥ 3

x ≤ 3, x ≥ 2

x ≤ -2, x ≥ -3

x ≤-3, x ≥ 2

Akọwa Nkọwa

x2 - 5x + 6 = 0
(X-2)(X-3) = 0
X-2 = 0 implies X = 2
X-3 = 0 implies X = 3
∴ x ≤ 2, x ≥ 3

Gụọ akwụkwọ nkuzi na Progression (JAMB)

Progression

Lee Azịza

Ajụjụ 45 Ripọtì

On a pie chart, there are six sectors of which four angles are 30o, 45o, 60o, 90o and the remaining two angles are in the ratio 2 : 1. Find the smaller angles of the remaining two angles

15o

30o

45o

60o

Lee Azịza

Ajụjụ 46 Ripọtì

In how many ways can the letters of the word ACCEPTANCE be arranged?

10! / (2!2!3!)

10! / ( 2!3!)

10! / (2!2!)

10!

Akọwa Nkọwa

ACCEPTANCE = 10 Letters
A = 2 letters
C = 3 letters
E = 2 letters
Can be arranged in 10! / (2!3!2!) ways

Gụọ akwụkwọ nkuzi na Permutation And Combination (JAMB)

Permutation And Combination

Lee Azịza

Ajụjụ 47 Ripọtì

Add 11012,101112 and 1112

1110112

1101102

1010112

1010102

Akọwa Nkọwa

11012 + 101112 + 1112 = 101011

Gụọ akwụkwọ nkuzi na Number Bases (WAEC)

Number Bases

Lee Azịza

Ajụjụ 48 Ripọtì

Evaluate $ \frac{\left(\frac{3}{8} \div \frac{1}{2} + \frac{1}{2}\right)}{\left(\frac{1}{8} \times \frac{2}{3} + \frac{1}{3}\right)} $

1/4

1/3

1/2

1

Akọwa Nkọwa

\frac{(\frac{3}{8} \div \frac{1}{2} + \frac{1}{2})}{(\frac{1}{8} \times \frac{2}{3} + \frac{1}{3})} = \frac{(\frac{3}{8} \times \frac{2}{1} + \frac{1}{2})}{(\frac{1}{8} \times \frac{2}{3} + \frac{1}{3})} = \frac{(\frac{3}{4} - \frac{1}{3})}{(\frac{1}{12} + \frac{1}{3})} = (\frac{9 - 4}{12}) \div (\frac{1 + 4}{12}) = (\frac{5}{12}) \div (\frac{5}{12}) = \frac{5}{12} \times \frac{12}{5} = 1

Gụọ akwụkwọ nkuzi na Solution Of Linear Equations (WAEC)

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

Solution Of Linear Equations

Positive And Negative Integers, Rational Numbers

Lee Azịza

Ajụjụ 49 Ripọtì

A book seller sells Mathematics and English books. If 30 customers buy Mathematics books, 20 customers buy English books and 10 customers buy the two books, How many customers has he altogether.

30

40

50

60

Akọwa Nkọwa

n(M) only = 30-10 = 20
n(E) only = 20-10 = 10
n(M∩E) = 10
∴M∪E = 20+10+10
= 40

Gụọ akwụkwọ nkuzi na Probability (WAEC)

Gụọ akwụkwọ nkuzi na Sets (WAEC)

Probability

Sets

Lee Azịza

Ajụjụ 50 Ripọtì