1.
If P = {-2, 0, 2, 4, 6} and Q = {-3, -1, 0, 2, 3, 5}, find the set P ∩ Q.
{}
{-3, -1, 3, 5}
{-2, 4, 6}
{0, 2}
Answer: D
Intersection (∩) are elements that can be found in both sets
P = {-2, 0, 2, 4, 6}
Q = {-3, -1, 0, 2, 3, 5}
P ∩ Q = {0, 2}
2.
Simplify:
Answer: B
=
Notes:
1. 16 = 4 x 4 = 42
Expressed this way in order to cancel the denominator 2 of the power
2. 9 = 3 x 3 = 32
Expressed this way in order to cancel the denominator 2 of the power
3. 16 = 2 x 2 x 2 x 2 = 24
Expressed this way in order to cancel the denominator 4 of the power
=
From the law of indices,
=
=
Notes:
1. The power 2 cancel the denominator 2 of the power
2. The power 4 cancel the denominator of 4 of the
=
=
=
The affects each of the powers in the bracket.
=
The power 3 cancels the denominator 3 in
=
Notes:
1. The power -1 means inverse which means you reciprocate (the numerator becomes the denominator and the denominator becomes the numerator).
2. =
3. Every number is being divided by 1 but it is not written.
2 =
4. 2-1 =
= x
= =
3.
Find the first quartile of 7, 8, 7, 9, 11, 8, 7, 9, 6 and 8.
8.5
7.0
7.5
8.0
Answer: B
Arrange the numbers in ascending order.
6, 7, 7, 7, 8, 8, 8, 9, 9, 11
First Quartile (ungrouped data) = th value
Where n is the number of items.
n = 10
First Quartile (ungrouped data) = = 2.5th value
First Quartile = = 7.0
4.
A fair die is tossed twice. What is the probability of getting a sum of at least 10?
Answer: D
1. Calculate the total number of outcomes when a fair die is tossed twice: 6 × 6 = 36
At least 10 means 10 or more.
2. Find the favorable outcomes for a sum of at least 10: (4,6), (6,4) for 10; (5,5) for 10; (5,6), (6,5) for 11, (6,6) for 12
Number of favourable outcome is 6
Probability =
P(Sum is at least 10) = =
You can also draw a table for all the possible outcomes.
1 | 2 | 3 | 4 | 5 | 6 | |
1 | 1, 1 | 1, 2 | 1, 3 | 1, 4 | 1, 5 | 1, 6 |
2 | 2, 1 | 2, 2 | 2, 3 | 2, 4 | 2, 5 | 2, 6 |
3 | 3, 1 | 3, 2 | 3, 3 | 3, 4 | 3, 5 | 3, 6 |
4 | 4, 1 | 4, 2 | 4, 3 | 4, 4 | 4, 5 | 4, 6 |
5 | 5, 1 | 5, 2 | 5, 3 | 5, 4 | 5, 5 | 5, 6 |
6 | 6, 1 | 6, 2 | 6, 3 | 6, 4 | 6, 5 | 6, 6 |
Probability =
Number of samples = 6 x 6 = 36
Number of possible selection = 6
P(Sum is at least 10) = =
5.
Find the quadratic equation whose roots are and -.
6x2 - x - 1 = 0
3x2 + x - 1 = 0
6x2 + x - 1 = 0
3x2 + x + 1 = 0
Answer: A
If α and β are the roots, then
(x - α)(x - β) = 0
(x - )(x - -) = 0
Note: -- = +
(x - )(x + ) = 0
x(x + ) - (x + ) = 0
x2 + - - = 0
x2 + - - = 0
Get rid of the fractions by multiplying both sides by the L.C.M of the denominators 3, 2 and 6.
The L.C.M of 3, 2 and 6 is 6.
6 x x2 + 6 x - 6 x - 6 x = 0 x 6
6x2 + 2x - 3x - 1 = 0
6x2 - x - 1 = 0
6.
From a point T, a man moves 12 km due West and then moves 12 km due South to another point Q.
Calculate the bearing of T from Q.
225°
315°
045°
135°
Answer: C
Since two sides are equal, the right angle is isosceles and the base angles are the same.
Base angle = = 45°
Bearing starts from the north clockwise. If the angle is less than 100°, begin with a 0.
7.
In the diagram PQ is parallel to RS, ∠ QFG = 105° and ∠ FEG = 50°.
Use the diagram to answer the question below
Find the value of m.
55°
75°
105°
130°
Answer: A
∠ QFG + ∠ EFG = 180°
105° + ∠ EFG = 180°
∠ EFG = 180° - 105°
∠ EFG = 75°
∠ FEG + ∠ EGF + ∠ EFG = 180° (Sum of angles of a triangle)
50° + m + 75° = 180°
m + 125° = 180°
m = 180° - 125°
m = 55°
8.
The area of a sector of a circle and the length of its arc are 231 cm2 and 66 cm respectively. Calculate the radius of the circle
[Take π = ]
3.5 cm
7.0 cm
10.5 cm
14.0 cm
Answer: B
Area of a sector = x area of the circle
Area of a sector = x π x radius x radius
Where θ = angle subtended by the arc at the centre of the circle
Length of an arc = x circumference of the circle
Length of an arc = x 2 x π x radius
Where θ = angle subtended by the arc at the centre of the circle
The angle (θ) substended are the same.
Area of a sector = x π x radius x radius
Area = 231 cm2
x π x radius x radius = 231
Make the angle (θ) the subject.
Get rid of the fraction by multiplying both sides by the denominator 360.
360 x x π x radius x radius = 360 x 231
θ x π x radius x radius = 360 x 231
Divide both sides by π x radius x radius
θ = --- eqn 1
Length of an arc = x 2 x π x radius
Length = 66 cm
x 2 x π x radius = 66
Make the angle (θ) the subject.
Get rid of the fraction by multiplying both sides by the denominator, 360.
360 x x 2 x π x radius = 360 x 66
θ x 2 x π x radius = 360 x 66
Divide both sides by 2 x π x radius
θ = --- eqn 2
--- eqn 1 = --- eqn 2
=
Cross multiply.
360 x 231 x 2 x π x radius = 360 x 66 x π x radius x radius
Divide both sides by 360 x π x radius to cancel out.
=
231 x 2 = 66 x radius
Divide both sides by 66
Radius = = 7 cm
9.
Find the truth set of the equation (x - 2)2 + 3 = (x + 1)2 - 6.
{-2}
{-1}
{1}
{2}
Answer: D
Concept
(a + b)2 = a2 + 2 x a x b + b2 = a2 + 2ab + b2
(a - b)2 = a2-2 x a x b + b2 = a2 -2ab + b2
Alternatively
(a + b)2 = (a + b)x(a + b) [Expand and simplify]
(a -b)2 = (a - b)x(a - b) [Expand and simplify]
(x - 2)2 = x2 - 2 x x x 2 + 22 = x2 -4x + 4
(x + 1)2 = x2 + 2 x x x 1 + 12 = x2 + 2x + 1
(x - 2)2 + 3 = (x + 1)2 - 6
x2 - 4x + 4 + 3 = x2 + 2x + 1 - 6
x2 - 4x + 7 = x2 + 2x - 5
x2 - x2 - 4x - 2x = -5 - 7
-6x = -12
Divide both sides by -6
x = = 2
10.
The sum of the interior angles of a regular polygon with k sides is (3k - 10) right angles.
Find the size of the exterior angle.
60°
40°
90°
120°
Answer: A
Sum of interior angles = (n - 2) x 180°
Where n = the number of sides of the polygon.
n = k.
Sum of interior of regular polygon with k sides = (k - 2) x 180
But the sum is given as (3k - 10) x 90°
Note: right angle = 90°
(k - 2) x 180 = (3k - 10) x 90
Divide both sides by 90
(k - 2) x 2 = (3k - 10) x 1
2k - 4 = 3k - 10
10 - 4 = 3k - 2k
6 = k
∴ the polygon has 6 sides.
Exterior angle of a regular polygon =
Where n is the number of sides of the polygon.
Number of sides of the polygon = 6
Exterior angle of the polygon = = 60°
11.
If the area of the trapezium MNOP is 300 cm2, find the value of x.
15 cm
12 cm
10 cm
20 cm
Answer: A
The area of the trapezium is the summation of the area of the triangle and the rectangle.
Area of triangle = x base x height
Area of a rectangle = Length x Breadth
Base of the triangle = 24 - 16 = 8
Height of the triangle = x
Breadth of rectangle = x
300 = x 8 x x + 16 x x
2 divides itself once and 8, 4 times.
300 = 4x + 16x
300 = 20x
Divide both sides by 20
x = = 15
12.
The diagram shows a circle centre O. Use it to answer the question below.
Find the value of x.
43°
47°
54°
86°
Answer: B
|OB| = |OC| = Radius of the circle. Hence triangle BOC is an isosceles triangle.
Isosceles triangles have the base angles equal.
∠ OBC = ∠ OCB
∠ ABD = ∠ OBC = 43°
∠ OBC = ∠ OCB = y = 43°
∠ OBC + ∠ OCB + ∠ BOC = 180° (Sum of angles in a triangle)
∠ OBC = ∠ OCB = 43°
43° + 43° + ∠ BOC = 180°
86° + ∠ BOC = 180°
∠ BOC = 180° - 86°
∠ BOC = 94°
Central angle is twice any inscribed angle subtended by the same arc.
x = = = 47°
13.
In the diagram, O is the centre of the circle. If ∠ NLM = 74°, ∠ LMN = 39° and ∠ LOM = x, find the value of x.
106°
113°
126°
134°
Answer: D
Central angle is twice any inscribed angle subtended by the same arc.
∠ LOM = 2 x ∠ LNM
∠ LNM = =
∠ LNM + ∠ NLM + ∠ LMN = 180° (Sum of angles in a triangle)
∠ LNM =
∠ NLM = 74°
∠ LMN = 39°
+ 74 + 39 = 180
Get rid of the fraction by multiplying both sides of the equation by the denominator 2.
x + 2 x 74 + 2 x 39 = 180 x 2
x + 148 + 78 = 360
x + 226 = 360
x = 360 - 226
x = 134
14.
When the point (4, 5) is rotated through an angle in the anticlockwise direction about the origin, its image is (-5, 4). What is the angle of rotation?
90o
270o
180o
300o
Answer: A
Rotation through 90o anticlockwise
When rotating a point 90o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-y,x).
In other words, switch x and y and make y negative
Rotation through 270o anticlockwise
When rotating a point 270o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(y,-x).
This means, we switch x and y and make x negative
Rotation through 180o anticlockwise
When rotating a point 180o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-x,-y).
So all we do is make both x and y negative
15.
Calculate the gradient of the line which passes through the points (1, 4) and (-2, 6).
-
-
Answer: B
Gradient = =
Y2 = 6
Y1 = 4
X2 = -2
X1 = 1
Gradient = = = -
16.
If P = {x: 1 ≤ x ≤ 6} and Q = {x: 2 < x < 9} where x ∈ R, find P ∩ Q.
{x: 2 ≤ x < 6}
{x: 2 ≤ x ≤ 6}
{x: 2 < x ≤ 6}
{x: 2 < x < 6}
Answer: C
P = {x: 1 ≤ x ≤ 6}
Read as x is such that 1 is less than or equal to x and x is less than or equal to 6.
P = {1, 2, 3, 4, 5, 6}
Q = {x: 2 < x < 9}
Read as x is such that 2 is less than x and x is less than 9.
Q = {3, 4, 5, 6, 7, 8}
Intersection (∩) are elements that can be found in both sets
(P ∩ Q) = {3, 4, 5, 6}
(P ∩ Q) = {x: 2 < x ≤ 6}
Note: ≤ means less than or equal to, hence the number at the right of it is inclusive.
17.
Evaluate : 2 - 3 +
4 +
4 - 9
4 - 11
4 - 21
Answer: B
2 - 3 + = 2 - 3 +
Express each surds to be in the form
=
=
=
2 - 3 + = 2 - 3 +
From = x
2 - 3 + = 2 x x - 3 x x + x
2 - 3 + = 2 x 2 x - 3 x 5 x + 6 x
2 - 3 + = 4 - 15 + 6
2 - 3 + = 4 - 9
18.
Which of these statements about an acute-angled triangle is true?
It has three equal angles.
It has two equal sides.
It has all its angles less than 90°.
It has one angle less than 90°.
Answer: C
19.
Which of the following about parallelograms is true?
Opposite angles are supplementary
Opposite angles are complementary
Opposite angles are equal
Opposite angles are reflex angles
Answer: C
1. The opposite angles of a parallelogram are equal
∠ A = ∠ C; ∠ D = ∠ B.
2. All the angles of a parallelogram add up to 360°.
∠ A + ∠ B + ∠ C + ∠ D = 360°.
3. All the respective consecutive angles are supplementary.
∠ A + ∠ B = 180°; ∠ B + ∠ C = 180°; ∠ C + ∠ D = 180°; ∠ D + ∠ A = 180°
20.
The gradient of the line passing through the points (3, 6) and (x, 4) is - . Find the value of x.
3
8
6
5
Answer: B
Gradient = =
Y2 = 4
Y1 = 6
X2 = x
X1 = 3
Gradient =
Gradient =
But gradient = - =
=
Since the numerator at the left and right are the same, the denominators are also the same.
x - 3 = 5
x = 5 + 3
x = 8
21.
If 9 y + 1 = () y - 2, find y.
-
1
-4
Answer: A
9 = 3 x 3 = 32
27 = 3 x 3 x 3 = 33
a-b =
= = 3-3
32(y + 1) = 3-3(y - 2)
Since the bases are the same, the powers are also the same.
2(y + 1) = -3(y - 2)
2y + 2 = -3y + 6
Note: - x - = +
-3 x -2 = + 6
2y + 2 = -3y + 6
2y + 3y = 6 - 2
5y = 4
Divide both sides by 5
y =
22.
The interquartile range of a distribution is 7. If the 25th percentile is 16, find the upper quartile.
9
23
30
35
Answer: B
Upper quartile - Lower quartile (25th percentile) = Interquartile range
Lower quartile (25th percentile) = 16
Interquartile range = 7
Upper quartile - 16 = 7
Upper quartile = 7 + 16
Upper quartile = 23
23.
Fati buys milk at ₦ x per tin and sells each at a profit of ₦ y. If she sells 10 tins of milk, how much does she receive from the sales?
₦ 10(x + y)
₦ (10x + y)
₦ (x + 10y)
₦ (xy + 10)
Answer: A
Selling Price = Cost Price + Profit
Selling Price of each = ₦ (x + y)
For the 10 tins, selling price = 10 x ₦ (x + y)
Amount received for the 10 tins = ₦ 10(x + y)
24.
To arrive on schedule, a ferry is to cover a distance of 40 km at 50 km/h. If the ferry delays for 18 minutes before starting the journey, at what speed must it move so as to arrive on schedule?
70 km/h
80 km/h
90 km/h
100 km/h
Answer: B
Speed =
Distance = Speed x Time
Time =
The regular time to reach destination = = 0.8 hours
18 minutes = hours = 0.3 hours
Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.
The ferry must cover the 40 km in 0.5 hours.
Speed = = = 80 km/h
25.
The pie chart represents fruits on display in a grocery shop. If there are 60 oranges on display, how many apples are there?
40
80
90
70
Answer: C
Angle of sector of orange = 360° - (100° + 60° + 120°)
Angle of sector of orange = 360° - 280°
Angle of sector of orange = 80°
Angle of sector of apple = 120°
Number of oranges = 60
If 80° = 60
120° = = 90
There are 90 apples
26.
Simplify: (x-1)-(x-1).
(x-1)
(2x+1)
(2x-1)
(1-2x)
Answer: C
(x-1)-(x-1)
Expand the brackets
(x-1)-(x-1) = - - +
Note: - x - = +
- x - 1 =
The Least Common Multiples (L.C.M) of the denominators (2,3 and 6) is 6
(x-1)-(x-1) =
(x-1)-(x-1) =
(x-1)-(x-1) =
(x-1)-(x-1) =
(x-1)-(x-1) = (2x - 1)
27.
Calculate the distance between points (3, -2) and (8, 10).
12 units
13 units
14 units
15 units
Answer: B
Distance =
Distance =
Note: - - = +
Distance =
Distance =
Distance =
Distance = = 13 units
28.
Find the value of ∠ RPT in the diagram.
20°
40°
50°
60°
Answer: B
∠ QRS = ∠ PQT = 40° (Corresponding angles)
∆ PQT is an isosceles triangle (Two sides equal).
Isosceles triangle has base angles equal.
∠ RPT = ∠ PQT = 40° (Base angles equal)
29.
If cos(3x + 28°) = sin (2x + 48°), 0° ≤ x ≤ 90°, find the value of x.
2.8°
15.2°
20.0°
22.0°
Answer: A
cos (3x + 28°) = sin (2x + 48°)
sin (90° - θ) = cos θ
cos (90° - θ) = sin θ
sin (90° + θ) = cos θ
cos (90° + θ) = -sin θ
Use the trigonometric values to change either cos to sin or sin to cos and equate the angles to solve for x.
Method I
Changing the cos to sin.
cos (3x + 28°) = sin (90° - [3x + 28°])
cos (3x + 28°) = sin (2x + 48°)
sin (90° - [3x + 28°]) = sin (2x + 48°)
Since both sides are sin, the angles are the same.
90° - (3x + 28°) = 2x + 48°
90° - 3x - 28° = 2x + 48°
90° - 28° - 48° = 2x + 3x
90° - 76° = 5x
14° = 5x
Divide both sides by 5.
x = = 2.8°
Method II
Changing the sin to cos.
sin (2x + 48°) = cos(90° - [2x + 48°])
cos (3x + 28°) = sin (2x + 48°)
cos (3x + 28°) = cos(90° - [2x + 48°])
Since both sides are cos, the angles are the same.
3x + 28° = 90° - (2x + 48°)
3x + 28° = 90° - 2x - 48°
3x + 2x = 90° - 48° - 28°
5x = 14°
Divide both sides by 5.
x = = 2.8°
30.
Make b the subject of the relation lb = (a + b)h.
Answer: A
lb = (a + b)h
Get rid of the fraction by multiplying both sides by the denominator 2
lb x 2 = h(a + b)
2lb = ah + bh
2lb - bh = ah
Factorize b out.
b(2l - h) = ah
Divide both sides by (2l - h)
b =
31.
In the diagram, EFGH are points on a circle centre P. If ∠ EFG = 142o and ∠ FGP = 63o, find ∠ FEP.
63o
76o
79o
89o
Answer: C
Central angle is twice any inscribed angle subtended by the same arc.
∠ EPG = 2 x ∠ EFG
∠ EPG = 2 x 142o
∠ EPG = 284o
∠ EPG + x = 360o (Sum of angles of a circle)
284o + x = 360o
x = 360o - 284o
x = 76o
Sum of interior angles of a quadrilateral is 360o
Note: sum of interior angles = (n - 2) x 180o, where n is the number of sides.
63o + 142o + ∠ FEP + x = 360o
63o + 142o + ∠ FEP + 76o = 360o
∠ FEP + 281o = 360o
∠ FEP = 360o - 281o
∠ FEP = 79o
32.
The sum of the interior angles of a polygon is 1260°. Find the number of sides.
9
6
7
8
Answer: A
Sum of interior angles = (n - 2) x 180°
Where n = number of sides of the regular polygon.
Sum of interior angles = 1260°
(n - 2) x 180° = 1260°
(n - 2) x 180 = 1260
Divide both sides by 180
n - 2 =
n - 2 = 7
n = 7 + 2
n = 9
33.
Find the value of x in the diagram.
16
44
36
28
Answer: D
Sum of exterior angles of a polygon = 360°
3x + 2x + x + 10 + 4x + 50 + 20 = 360
10x + 80 = 360
10x = 360 - 80
10x = 280
Divide both sides by 10
x = = 28
34.
A die is rolled once. Find the probability of obtaining a number less than 3.
Answer: D
Probability =
Numbers in a die rolled once = {1,2,3,4,5,6}
Numbers less than 3 = {1,2}
Number of possible selection = 2
Number of samples = 6
Probability of less than 3 = =
35.
What is the coefficient of x in the expansion of (4x2 + 3x - 1)(3x + 1)?
-1
0
1
2
Answer: B
(4x2 + 3x - 1)(3x + 1) = (4x2 + 3x - 1)(3x + 1)
(4x2 + 3x - 1)(3x + 1) = 3x(4x2 + 3x - 1) + 1(4x2 + 3x - 1)
(4x2 + 3x - 1)(3x + 1) = 12x3 + 9x2 - 3x + 4x2 + 3x - 1
(4x2 + 3x - 1)(3x + 1) = 12x3 + 9x2 + 4x2 - 3x + 3x - 1
(4x2 + 3x - 1)(3x + 1) = 12x3 + 13x2 + 0x - 1
36.
Make U the subject of the relation: x = .
U =
U =
U =
U =
Answer: D
x =
Get rid of the fraction by multiplying each sides by the denominator (3U + 2).
x x (3U + 2) = 2U - 3
3xU + 2x = 2U - 3
3xU - 2U = -2x - 3
Factorize U out.
U(3x - 2) = -2x - 3
Divide both sides by (3x - 2).
U =
Factorize (-1) out.
U = x
-1 cancels -1 and you can rearrange -3x + 2 as 2 - 3x
U =
Alternatively,
3xU + 2x = 2U - 3
2x + 3 = 2U - 3xU
Factorize U out.
2x + 3 = U(2 - 3x)
Divide both sides by (2 - 3x)
= U
U =
37.
Simplify: (11two)2
1001two
1101two
101two
10001two
Answer: A
Method I
A number raised to the power 2 (square) means the number multiplied by itself.
1 | 1 | ||
1 | 1 | ||
x | 1 | 1 | |
1 | 1 | ||
+ | 1 | 1 | |
1 | 0 | 0 | 1 |
Note: 2 in base two is 10.
Method II
Change the binary (base two) to base 10 and simplify and change the result back to base two.
11two = 1 x 21 + 1 x 20
11two = 1 x 2 + 1 x 1
Note: any number raised to the power 0 is 1.
11two = 2 + 1 = 3
(11two)2 = 32 = 3 x 3 = 9
Change the 9 base 10 to base 2.
9 base 10 to base 2
You list the remainders from the bottom to the top.
9ten = 1001two
38.
If 2a = and = 3, evaluate a2 + b2.
48
90
160
250
Answer: B
=
=
64 = 26
= = 23
2a = 23
Since the bases are the same, the powers are also the same.
a = 3
= 3
b = 3 x a
b = 3 x 3
b = 9
a2 + b2 = 32 + 92
a2 + b2 = 9 + 81
a2 + b2 = 90
39.
In the diagram, O is the centre of the circle. SOQ is a diameter and ∠ SRP = 37°.
Find ∠ PSQ.
37°
53°
65°
127°
Answer: B
Notes:
Angles subtended by the diameter (or semi-circle) is 90°.
∠ QPS = 90°
Angles subtended by the same arc are the same.
From arc SP,
∠ PRS = ∠ PQS = 37° (Angles subtended by the same arc)
∠ PQS + ∠ QPS + ∠ PSQ = 180° (Sum of angles in a triangle)
37° + 90° + ∠ PSQ = 180°
127° + ∠ PSQ = 180°
∠ PSQ = 180° - 127°
∠ PSQ = 53°
40.
Abudu can do a piece of work in 6 days and Efah can do the same work in 3 days. What fraction of the work can both do together in a day?
Answer: D
Abudu takes 6 days to complete work.
Fraction of work done by Abudu in a day =
Efah takes 3 days to complete the same work.
Fraction of work completed by Efah in a day =
Fraction of work done by both in a day is the sum of fractions of work done by each.
Fraction of work done by both = +
Fraction of work done by both =
Fraction of work done by both =
Fraction of work done by both =
Alternatively
Let y the work to be done
Abudu takes 6 days to complete work.
Work completed by Abudu in a day = x y
Work completed by Abudu in a day =
Efah takes 3 days to complete the same work.
Work completed by Efah in a day = x y
Work completed by Efah in a day =
Work completed by both Abudu and Efah in a day is the sum of work done by each in a day.
Work done by both = +
Work done by both =
Work done by both =
Work done by both =
Fraction of work done in a day =
Total work to be done is y
Work done by both in a day =
Fraction of work done in a day = ÷ y
Every number is divisible by 1
Fraction of work done in a day = ÷
Reciprocate the fraction at the right of ÷ and change the ÷ to multiplication (x)
Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.
Fraction of work done in a day = x
The y cancels each other.
Fraction of work done in a day =
41.
Make x the subject of the relation:
E = + z.
x =
x =
x =
x =
Answer: C
E = + z
Multiply both sides the denominator 2y
E x 2y = x 2y + z x 2y
2Ey = kx2 + 2yz
kx2 = 2Ey - 2yz
Divide both sides by k.
x2 =
Factorize 2y out.
x2 =
Take square root of both sides.
x =
42.
Factorize: p2q2 - 6pqr + 9r2
(pq - 3r)2
(pq - 3r)(pq + 3r)
(pq + 3r)2
(pr + 3q)(pr - 3q)
Answer: A
Since 3 can go into 6 and 9, split -6pqr into -3pqr and -3pqr
p2q2 - 6pqr + 9r2 = p2q2 - 3pqr - 3pqr + 9r2
p2q2 - 6pqr + 9r2 = pq(pq - 3r) - 3r(pq - 3r)
p2q2 - 6pqr + 9r2 = (pq - 3r)(pq - 3r)
Since pq - 3r = pq - 3r, the multiplication makes it square
p2q2 - 6pqr + 9r2 = (pq - 3r)2
The equation above is similar to the general factorization rule of (a - b)2
(a - b)2 = a2 - 2ab + b2
Thus the first term square then the sign (either + or -) and 2 times the first and second terms + the second term square
Similarly (a + b)2 = a2 + 2ab + b2
The first term for the equation, p2q2 - 6pqr + 9r2 is pq and the second term is 3r and the sign is -
43.
The height of a square base pyramid is thrice the length of a side of its base. If the base area is 324 cm2, find the volume of the pyramid.
17,496 cm3
5,832 cm3
324 cm3
972
Answer: B
Volume of a square base pyramid =
Area of a square = Length x Length = L2
Area of the base = 324 cm2
L2 = 324
Take square root of both sides.
L = = 18 cm
The height of a square base pyramid is thrice the length of a side of its base
Height = 3 x Length of its base
Length of base = 18 cm
Height = 3 x 18 cm = 54 cm
Volume of the square base pyramid = = 5,832 cm3
44.
The graph is for the relation: y = d - x.
Find the value of d.
2
1
0
-1
Answer: B
Pick a point on the line of the relation: y = d - x and substitute into the relation to find d.
y = d - x
Using the point (1, 0)
x = 1 and y = 0
0 = d - 1
1 = d
d = 1
OR
Using the point (0, 1)
x = 0, y = 1
1 = d - 0
1 = d
d = 1
OR
Using the point (-2, 3)
x = -2 and y = 3
3 = d - (-2)
Note: -- or -(-) = +
3 = d + 2
3 - 2 = d
1 = d
d = 1
45.
Find the value of x for which is not defined.
1
2
-1
Answer: D
For a relation to be not defined/undefined, the denominator must be equal to 0
x2 + 2x + 1 = 0
Express 2x as two numbers when you multiply you will get c (1) and when you add you will get b (2)
1 x 1 = 1 and 1 + 1 = 2 hence the number is 1
x2 + x + x + 1 = 0
x(x + 1) + 1(x + 1) = 0
(x + 1)(x + 1) = 0
When x + 1 = 0, x = -1
46.
A farmer cleared 40% of a piece of land the first day and 60% of the remainder the next day. What percentage of the land was remaining at the end of the second day?
30%
24%
20%
15%
Answer: B
Let y = area of the field
First day area cleared = 40% of y
First day area cleared = x y = 0.4y
Remaining uncleared area = y - 0.4y
Remaining uncleared area = 0.6y
Second day area cleared = 60% of remaining
Second day area cleared = x 0.6y
Second day area cleared = 0.36y
Remaining area after the second day = 0.6y - 0.36y
Remaining area after the second day = 0.24y
Percentage = x 100%
Percentage of the land remaining at the end of the second day = x 100%
Note: the ys cancel each other.
Percentage of the land remaining at the end of the second day = x 100% = 24%
47.
The mean of the numbers 2, 5, 2x and 7 is not greater than 5. Find the range of values of x.
x ≤ 3
x ≥ 3
x < 3
x > 3
Answer: A
Mean =
Mean =
The mean is not greater than 5 means the mean can be less than 5 or equal to 5.
≤ 5
≤ 5
Get rid of the fraction by multiplying both sides by the denominator, 4
4 x ≤ 5 x 4
14 + 2x ≤ 20
2x ≤ 20 - 14
2x ≤ 6
Divide both sides by 2.
x ≤
x ≤ 3
48.
Solve 4x2 - 16x + 15 = 0
x = -1 or -2
x = 1 or -1
x = 1 or 2
x = 1 or -2
Answer: C
4x2 - 16x + 15 = 0
Notes:
1. a x c = 4 x 15 = 60
2. Split b (-16) such that two numbers when you sum you will get b (-16) and when you multiply you will get ac (60). The two numbers are -10 and -6
3. -16x = -10x - 6x
4x2 -10x - 6x + 15 = 0
2x(2x - 5) - 3(2x - 5) = 0
(2x - 5)(2x - 3) = 0
When 2x - 5 = 0, x = = 2
When 2x - 3 = 0, x = = 1
49.
x | 6.20 | 6.85 | 7.5 |
y | 3.9 | 5.2 | 6.5 |
The points on a linear graph are as shown in the table. Find the gradient (slope) of the line.
1
2
2
Answer: C
Gradient = =
Choose any two points from the table.
P1(6.20, 3.90) and P2(6.85, 5.20)
Gradient = = = 2
50.
The foot of a ladder is 6 m from the base of an electric pole. The top of the ladder rests against the pole at a point 8 m above the ground. How long is the ladder?
7 m
10 m
12 m
14 m
Answer: B
Let x = the length of the ladder.
From Pythagorean theorem,
The hypothenuse (c) is the longest side. The side facing the right-angle (∟)
x2 = 62 + 82
x2 = 36 + 64
x2 = 100
Take square root of both sides.
x = = 10
The ladder is 10 m long.
(a)
Solve, correct to one decimal place, tan(θ + 25°) = 5.145, where ° ≤ θ ≤ 90°.
(b)
In the relation t = m:
(i)
make n the subject of the relation;
(ii)
find the positive value of n when t = 25, m = 5 and r = 4.
(a)
tan(θ + 25°) = 5.145
Take tan inverse (tan-1) of both sides.
θ + 25° = tan-1(5.145)
θ + 25° = 79°
θ = 79° - 25°
θ = 54°
(b)
(i)
t = m
Get rid of the square root by squaring both sides of the equation.
t2 = m2 x (n2 + 4r)
Divide both sides by m2
= n2 + 4r
- 4r = n2
Every number is being divided by 1
4r =
- = n2
= n2
= n2
Get rid of the square by taking square root of both sides.
n =
(ii)
n =
When t = 25, m = 5 and r = 4
n =
n =
n = = = 3
(a)
Copy and complete the following table for the relation: y = 2(x + 2)2 - 3 for -5 ≤ x ≤ 2.
x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
y | -1 | -3 | 5 |
(b)
Using scales of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the y-axis, draw the graph of the relation y = 2(x + 2)2 - 3 for -5 ≤ x ≤ 2.
(c)
Use the graph to find the solution of:
(i)
2(x + 2)2 = 3;
(ii)
2(x + 2)2 = 5.
(d)
For what values of x, from the graph is y increasing in the interval?
(a)
y = 2(x + 2)2 - 3
When x = -5
y = 2(-5 + 2)2 - 3
y = 2(-3)2 - 3
y = 2 x 9 - 3
y = 18 - 3
y = 15
When x = -4
y = 2(-4 + 2)2 - 3
y = 2(-2)2 - 3
y = 2 x 4 - 3
y = 8 - 3
y = 5
When x = -3
y = 2(-3 + 2)2 - 3
y = 2(-1)2 - 3
y = 2 x 1 - 3
y = 2 - 3
y = -1
When x = -2
y = 2(-2 + 2)2 - 3
y = 2(0)2 - 3
y = 2 x 0 - 3
y = 0 - 3
y = -3
When x = -1
y = 2(-1 + 2)2 - 3
y = 2(1)2 - 3
y = 2 x 1 - 3
y = 2 - 3
y = -1
When x = 0
y = 2(0 + 2)2 - 3
y = 2(2)2 - 3
y = 2 x 4 - 3
y = 8 - 3
y = 5
When x = 1
y = 2(1 + 2)2 - 3
y = 2(3)2 - 3
y = 2 x 9 - 3
y = 18 - 3
y = 15
When x = 2
y = 2(2 + 2)2 - 3
y = 2(4)2 - 3
y = 2 x 16 - 3
y = 32 - 3
y = 29
x | -5 | -4 | -3 | -2 | -1 | 0 | 1 | 2 |
y | 15 | 5 | -1 | -3 | -1 | 5 | 15 | 29 |
(b)
Notes:
1. Each grid value =
2. Each grid value on the y-axis = = 0.5
3. Number of grids =
4. Number of grids on the y-axis for 1 = = 2
5. Number of grids on the y-axis for 3 = = 6
6. Number of grids on the y-axis for 29 = = 58
(c)
Notes:
1. Each grid value =
2. Each grid value on the x-axis = = 0.1
(i)
Express the relation in such a way that one side equals the given relation.
2(x + 2)2 = 3
2(x + 2)2 - 3 = 0
The solution are the values of x where the line y = 0 cuts the curve.
x = -3.2 ± 0.1 or -0.8 ± 0.1
(ii)
Express the relation in such a way that one side equals the given relation.
2(x + 2)2 = 5
Subtract 3 from each sides.
2(x + 2)2 - 3 = 5 -3
2(x + 2)2 - 3 = 2
The solution are the values of x where the line y = 2 cuts the curve.
x = -3.6 ± 0.1 or -0.4 ± 0.1
(d)
The values of x for which y is increasing from the graph is −2 < x ≤ 2
In a military camp, 50 officers had a choice of beans, plantain and rice. Of these officers, 21 chose beans, 24 plantains and 18 rice. Also 3 chose beans only, 9 plantain only, 2 rice only and 5 chose all three kinds of food.
(a)
Illustrate the information on a Venn diagram.
(b)
Use the Venn diagram to find the number of officers who chose:
(i)
plantain and beans only;
(ii)
exactly two kinds of food;
(iii)
none of the three kinds of food.
(a)
Let U = {Officers}
n(U) = 50
Let B = {Officers who chose beans}
n(B) = 21
Let R = {Officers who choose rice}
n(R) = 18
x = number of officers who chose beans and plantain only
y = number of officers who chose beans and rice only
z = number of officers who chose rice and plantain only
a = number of officers who chose none of the three kinds of food
Covering the rice (R) region
18 + 3 + x + 9 + a = 50
x + a + 30 = 50
x + a = 50 - 30
x + a = 20 ------ eqn 1
Covering the beans (B) region
21 + a + 2 + z + 9 = 50
a + z + 32 = 50
a + z = 50 - 32
a + z = 18 ------ eqn 2
Covering the plantain (P) region
24 + 3 + y + 2 + a = 50
y + a + 29 = 50
y + a = 50 - 29
y + a = 21 -------- eqn 3
Sum of the beans (B) region should give 21.
3 + y + 5 + x = 21
y + x + 8 = 21
y + x = 21 - 8
y + x = 13 ------- eqn 4
Get rid of a by subtracting --- eqn 1 from ---- eqn 3
y - x = 1 -------- eqn 5
Solve ---- eqn 4 and ---- eqn 5 simultaneously.
---- eqn 4 + ----- eqn 5
2y = 14
Divide both sides by 2
y = = 7
Substitute y = 7 into ---- eqn 5
7 - x = 1
7 - 1 = x
6 = x
x = 6
Substitute x = 6 into ------ eqn 1
6 + a = 20
a = 20 - 6
a = 14
Substituting a = 14 into --- eqn 2
14 + z = 18
z = 18 - 14
z = 4
Redraw the Venn diagram and substitute the values of the variables
(b)
(i)
plantain and beans only = 6
(ii)
exactly two kinds of food = 6 + 7 + 4 = 17
(iii)
none of the three kinds of food = 14
(a)
Given that P = , Q = , R = and PQ = R, find the values of x, y and z
(b)
(i)
Draw on a graph paper, using a scale of 2 cm to 1 unit on both axes, the lines x = 1; y = 2; and x + y = 5.
(ii)
Shade the region which satisfies simultaneously the inequalities x + y ≤ 5; y ≥ 2 and x ≥ 1
(a)
Multiply the elements of each row of the first matrix by the elements of each column in the second matrix (element by element) and add the products
x =
PQ = =
PQ =
PQ = R
=
-3y - 32 = -59
-3y = -59 + 32
-3y = -27
Divide both sides by -3
y = = 9
-15 + 2x = -27
2x = -27 + 15
2x = -12
Divide both sides by 2
x = = -6
z = -17
(b)
x + y = 5
When x = 0
y = 5
Coordinates = (0, 5)
When y = 0
x = 5
Coordinates = (5, 0)
(a)
Abiola starts a business with $1,250.00. Frances joins the business later with a capital of $1,875.00. At the end of the first year, profits are shared equally between Abiola and Frances. When did Frances join the business?
(b)
In the diagram, CE||AD and ED = EB = AB, find the value of:
(i)
n;
(ii)
m.
(a)
Let's take it that Frances joined the business n months after Abiola started it.
Abiola invested $1,250 for 12 months for the same profit as Frances who invested $1875 for n months.
Since the profit are shared equally, then Abiola's contribution is the same as Frances' contribution.
Contribution = Amount invested x Number of months in the business
Abiola's contribution = $1,250 x 12
Number of months that Frances was in the business = 12 - n
Frances' contribution = $1875 x (12 - n)
1250 x 12 = 1875 x (12 - n)
Divide both sides by 1875
12 - n =
12 - n = 8
12 - 8 = n
n = 4 months
∴ Frances joined the business 4 months after Abiola started.
(b)
∆ ABE and ∆ BED are isosceles (Two sides equal)
Isosceles triangles have the base angles equal.
(i)
∠ BAE = ∠ CEA = 40° (Alternate angles)
∠ AEB = ∠ BAE = 40° (Base angles equal)
∠ AEB + ∠ BAE + ∠ ABE = 180°(Sum of angles in a triangle)
40° + 40° + n° = 180°
80 + n = 180
n = 180 - 80
n = 100
(ii)
∠ ABE + ∠ DBE = 180° (Sum of angles on a straight line)
n° + ∠ DBE = 180°
But n = 100
100° + ∠ DBE = 180°
∠ DBE = 180° - 100°
∠ DBE = 80°
∠ DBE = ∠ BDE = 80°(Base angles equal)
∠ DBE + ∠ BDE + ∠ BED = 180°(Sum of angles in a triangle)
80° + 80° + m° = 180°
160° + m° = 180°
m° = 180° - 160°
m° = 20°
m = 20
(a)
A cone and a pyramid have equal heights and volumes. If the base area of the pyramid is 154 cm2, find the radius of the cone.
[Take π = ]
(b)
A spherical bowl of radius r cm is a quarter full when 6 litres of water is poured into it.
Calculate, correct to three significant figures, the diameter of the bowl.
[Take π = ]
(a)
Volume = π x r2 x
Where r = the radius and h = the height of the cone.
Volume = x Base area x Height
Let h = height of the pyramid and the cone.
Since the volumes are the same, the two formula can be equated.
π x r2 x = x Base area x h
Base area of the pyramid = 154
π x r2 x = x 154 x h
Divide both sides by h to get rid of the h
π x r2 x = x 154
Get rid of the fractions by multiplying both sides by the denominator 3
π x r2 = 154
π =
x r2 = 154
Get rid of the fraction by multiplying both sides by the denominator 7
22 x r2 = 154 x 7
Divide both sides by 22
r2 =
r2 = 49
Take square roots of both sides.
r = = 7
The radius of the cone is 7 cm.
(b)
Volume = x π x r3
Where r = the radius of the sphere.
Volume of water poured = 6 litres = 6 x 1000 cm3 = 6000 cm3
The water is a quarter full. Hence the volume of water is of the full volume of the sphere
π =
x x x r3 = 6000
x r3 = 6000
x r3 = 6000
Get rid of the fraction by multiplying both sides by the denominator 84.
88 x r3 = 6000 x 84
Divide both sides by 88
r3 =
r3 = 5727.27
Take cube root of both sides.
r =
r = 17.8916044472 cm
Diameter is 2 times the radius
Diameter = 2 x 17.8916044472 cm
Diameter = 35.7832088944 cm ≈ 35.8 cm(three significant figures)