JAMB UTME - Mathematics - 2008

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

Bayanin Amsa

(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 ,

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Quadratic Equations

Duba Amsa

Tambaya 2 Rahoto

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}

Bayanin Amsa

\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}

Duba Amsa

Tambaya 3 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Binary Operations (JAMB)

Binary Operations

Duba Amsa

Tambaya 4 Rahoto

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

√5

2√3

√20

√50

Bayanin Amsa

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}

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Karanta bayanin darasi akan Coordinate Geometry (JAMB)

Euclidean Geometry

Coordinate Geometry

Duba Amsa

Tambaya 5 Rahoto

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)

Bayanin Amsa

\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

Karanta bayanin darasi akan Surds (radicals) (WAEC)

Surds (radicals)

Duba Amsa

Tambaya 6 Rahoto

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

+/-16

+/-8

+/-4

+/-2

Bayanin Amsa

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

Karanta bayanin darasi akan Progression (JAMB)

Progression

Duba Amsa

Tambaya 7 Rahoto

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

N3,000

N600

N300

N150

Bayanin Amsa

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

Karanta bayanin darasi akan Financial Arithmetic (WAEC)

Financial Arithmetic

Duba Amsa

Tambaya 8 Rahoto

Add 11012,101112 and 1112

1110112

1101102

1010112

1010102

Bayanin Amsa

11012 + 101112 + 1112 = 101011

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 9 Rahoto

Express 1223456 to 3 significant figures

123000

124000

125000

126000

Fractions, Decimals And Approximations

Duba Amsa

Tambaya 10 Rahoto

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)

Bayanin Amsa

(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)

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 11 Rahoto

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

zero

1

2

3

Bayanin Amsa

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

Karanta bayanin darasi akan Measures Of Location (JAMB)

Measures Of Location

Duba Amsa

Tambaya 12 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Integration (JAMB)

Representation Of Data

Integration

Duba Amsa

Tambaya 13 Rahoto

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

Bayanin Amsa

(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

Duba Amsa

Tambaya 14 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Bearings (WAEC)

Karanta bayanin darasi akan Coordinate Geometry (WAEC)

Bearings

Coordinate Geometry

Duba Amsa

Tambaya 15 Rahoto

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

60o

80o

120o

160o

Bayanin Amsa

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

Karanta bayanin darasi akan Circles (WAEC)

Circles

Duba Amsa

Tambaya 16 Rahoto

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

4

16

32

64

Bayanin Amsa

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

Karanta bayanin darasi akan Logarithms (WAEC)

Logarithms

Duba Amsa

Tambaya 17 Rahoto

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

\frac{1}{11}

\frac{3}{11}

\frac{3}{11}

\frac{4}{11}

Bayanin Amsa

Obstruction
if a T is picked
p =

\frac{2}{11}

Karanta bayanin darasi akan Probability (WAEC)

Probability

Duba Amsa

Tambaya 18 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Solution Of Linear Equations (WAEC)

Karanta bayanin darasi akan Positive And Negative Integers, Rational Numbers (WAEC)

Solution Of Linear Equations

Positive And Negative Integers, Rational Numbers

Duba Amsa

Tambaya 19 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Matrices And Determinants (JAMB)

Karanta bayanin darasi akan Binary Operations (JAMB)

Matrices And Determinants

Binary Operations

Duba Amsa

Tambaya 20 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Variation (JAMB)

Karanta bayanin darasi akan Inequalities (JAMB)

Variation

Inequalities

Duba Amsa

Tambaya 21 Rahoto

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

-1/4

-1/2

1/4

1/2

Bayanin Amsa

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

Duba Amsa

Tambaya 22 Rahoto

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

128

216

215

210

Bayanin Amsa

^{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

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 23 Rahoto

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)

Bayanin Amsa

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

Karanta bayanin darasi akan Loci (JAMB)

Karanta bayanin darasi akan Loci (WAEC)

Karanta bayanin darasi akan Euclidean Geometry (JAMB)

Loci

Euclidean Geometry

Duba Amsa

Tambaya 24 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Change Of Subject Of A Formula/Relation (WAEC)

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Change Of Subject Of A Formula/Relation

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 25 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Binary Operations (JAMB)

Binary Operations

Duba Amsa

Tambaya 26 Rahoto

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

16

13

12

11

Bayanin Amsa

\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

Karanta bayanin darasi akan Modular Arithmetic (WAEC)

Karanta bayanin darasi akan Quadratic Equations (WAEC)

Modular Arithmetic

Quadratic Equations

Duba Amsa

Tambaya 27 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Graphs Of Linear And Quadratic Functions (WAEC)

Graphs Of Linear And Quadratic Functions

Duba Amsa

Tambaya 28 Rahoto

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

x ≤ 2, x ≥ 3

x ≤ 3, x ≥ 2

x ≤ -2, x ≥ -3

x ≤-3, x ≥ 2

Bayanin Amsa

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

Karanta bayanin darasi akan Progression (JAMB)

Progression

Duba Amsa

Tambaya 29 Rahoto

If 125x = 2010 find x

2

3

4

5

Bayanin Amsa

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

Karanta bayanin darasi akan Number Bases (WAEC)

Number Bases

Duba Amsa

Tambaya 30 Rahoto

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

7

1

2

3

Bayanin Amsa

\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

Karanta bayanin darasi akan Integration (JAMB)

Integration

Duba Amsa

Tambaya 31 Rahoto

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

Bayanin Amsa

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}

Karanta bayanin darasi akan Representation Of Data (JAMB)

Representation Of Data

Duba Amsa

Tambaya 32 Rahoto

Find the capacity in liters of a cylindrical well of radius 1 meter and depth 14 meters

[π = 22/7]

44,000 liters

4,400 liters

440 liters

44 liters

Bayanin Amsa

V = πr2h
1m = 100cm
14cm = 1400cm

∴ V = \frac{22}{7} \times \frac{100 x 100 x 1400}{1000} = 44, 000 l i t e r s

Karanta bayanin darasi akan Volumes (WAEC)

Karanta bayanin darasi akan Inequalities (JAMB)

Volumes

Inequalities

Duba Amsa

Tambaya 33 Rahoto

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

4

5

6

7

Bayanin Amsa

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

Karanta bayanin darasi akan Algebraic Expressions (WAEC)

Algebraic Expressions

Duba Amsa

Tambaya 34 Rahoto

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)

Bayanin Amsa

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

Karanta bayanin darasi akan Sets (WAEC)

Sets

Duba Amsa

Tambaya 35 Rahoto

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

-

\frac{1}{8}

\frac{3}{8}

\frac{5}{8}

\frac{1}{4}

Bayanin Amsa

-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}

Duba Amsa

Tambaya 36 Rahoto

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

10! / (2!2!3!)

10! / ( 2!3!)

10! / (2!2!)

10!

Bayanin Amsa

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

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Permutation And Combination

Duba Amsa

Tambaya 37 Rahoto

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

Bayanin Amsa

(\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

Duba Amsa

Tambaya 38 Rahoto

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}

Bayanin Amsa

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}

Duba Amsa

Tambaya 39 Rahoto

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

3/8

2/3

3/4

3/2

Bayanin Amsa

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}

Karanta bayanin darasi akan Number Bases (WAEC)

Karanta bayanin darasi akan Simple Operations On Algebraic Expressions (WAEC)

Number Bases

Simple Operations On Algebraic Expressions

Duba Amsa

Tambaya 40 Rahoto

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!}

Bayanin Amsa

Acceptance

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

Karanta bayanin darasi akan Representation Of Data (JAMB)

Karanta bayanin darasi akan Permutation And Combination (JAMB)

Karanta bayanin darasi akan Probability (JAMB)

Representation Of Data

Permutation And Combination

Probability

Duba Amsa

Tambaya 41 Rahoto

Find the area of the figure given

12.5cm2

75.0cm2

78.5cm2

84.8cm2

Bayanin Amsa

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

Karanta bayanin darasi akan Areas (WAEC)

Areas

Duba Amsa

Tambaya 42 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Probability (WAEC)

Karanta bayanin darasi akan Sets (WAEC)

Probability

Sets

Duba Amsa

Tambaya 43 Rahoto

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

4

16

32

64

Bayanin Amsa

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

Duba Amsa

Tambaya 44 Rahoto

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