1.
Evaluate: (0.07 x 0.02) ÷ 14.
0.01
0.001
0.0001
0.00001
Answer: C
Change the decimals to the form whole number x 10^{a} where a is the number of times you move the decimal point to get the whole number. If the decimal point is moved to the right, the sign is negative
0.07 = 7 x 10^{-2}
0.02 = 2 x 10^{-2}
⇒ (0.07 x 0.02) = 7 x 10^{-2} x 2 x 10^{-2}
(0.07 x 0.02) ÷ 14 = 7 x 10^{-2} x 2 x 10^{-2} ÷ 14
7 goes into 14, 2 times and 2 goes into itself 1
7 x 10^{-2} x 2 x 10^{-2} ÷ 14 = 1 x 10^{-2} x 1 x 10^{-2}/1
1 x 10^{-2} x 1 x 10^{-2}/1 = 1 x 10^{-2} x 10^{-2}
Applying a^{m} x a^{n} = a^{m + n}
⇒ 1 x 10^{-2} x 10^{-2} = 1 x 10^{-2+-2} = 1 x 10^{-4}
Now change the whole number x 10^{a} format back to decimal
Since the power in 10^{-4} is negative 4, you have to move the decimal point to the left four times
1 = 1.0 (Every number has .0 which is not written)
1.0 x 10^{-4} = 0.0001
2.
Vera is 11 years old and her brother is 9 years old. They shared 60 oranges in the ratio of their ages. How many more oranges does Vera get?
6
27
34
39
Answer: A
Each person's share from the ratio is calculated as:
(The ratio / Sum of ratios) x Total being shared
Vera:Brother = 11:9
Sum of ratios = 11+9 = 20
Vera's share = (11/20) x 60 = 11 x 3 = 33
NOTE: 20 goes into 60, 3 times and the 3 multiplies the 11 at the top and the result is 33
Brother's share = (9/20) x 60 = 9 x 3 = 27
To get how much more Vera got, subtract the brother's share from Vera's share to know how much more she got
Thus 33 - 27 = 6
3.
A trader sold 90 oranges at 3 for GH₵ 0.75. How much did she get from selling all the oranges?
GH₵ 22.50
GH₵ 67.50
GH₵ 75.00
GH₵ 225.50
Answer: A
Let x = amount for selling the 90 oranges
3 = 0.75
90 = x
Cross multiply. Left side multiplying right and right side multiplying left
3 x x = 90 x 0.75
Divide both sides by 3
3 x x/3 = 90 x 0.75/3
x = 90 x 0.75/3
Change the decimal to the format, whole number x 10^{a}
0.75 = 75 x 10^{-2} (When you move the decimal point to the right its negative to the number of times you moved)
⇒ x = 90 x 75 x 10^{-2}/3
3 goes into 90, 30 times
x = 30 x 75 x 10^{-2}
30 x 75 = 2250
⇒ x = 2250 x 10^{-2}
Now change the whole number x 10^{a} format back to decimal
Since the power in 10^{-2} is negative 2, you have to move the decimal point to the left two times
2250 = 2250.0 (Every number has .0 which is not written)
2250.0 x 10^{-2} = 22.50
⇒ x = GH₵ 22.50
4.
If P = {4, 8, 12, 16, 20}, Q = {16, 4, 12, k, 20} and P = Q, find the value of k.
20
16
8
4
Answer: C
Since the set P is equal to the set Q, thus P = Q, it means all the elements in P is in Q and all the elements in Q is also in P. So you simply have to find which element is in P but not in Q.
P = {4, 8, 12, 16, 20}
Q = {16, 4, 12, k, 20}
Since the elements of P are arranged from the smallest to the largest, let's arrange Q too as well
⇒ Q = {4, k, 12, 16, 20}
P = {4, 8, 12, 16, 20}
As you can see the missing element (k) in Q is 8
5.
A man travelled a distance of 8 km in an hour. How long will it take him to cover a distance of 12 km, travelling at the same speed?
1⅓ hrs
1½ hrs
1¾ hrs
2 hrs
Answer: B
Method I
Let x = the time taken to travel 12 km
8 km took 1hr ⇒ 8 km = 1hr
12 km will take xhr ⇒ 12 km = xhr
8 = 1
12 = x
Cross multiply both sides
x x 8 = 12 x 1
Divide both sides by 8
x8/8 = 12/8 = 3/2 = 1½
NOTE: 4 goes into 12, 3 times and into 8, 2 times
Method II
Speed = Distance/Time
Speed for 8 km in 1hr = 8km / 1hr = 8 km/h
Let x = time taken to cover the 12 km
Since the same speed was used to cover the 12 km, 12/x should give the same speed (8 km/h)
⇒ 12/x = 8
Multiply both sides by x
12 = 8 x x
Divide both sides by 8
12/8 = x
⇒ x = 3/2 = 1½
6.
Solve: (1-x)÷3 < 4.
x < -11
x > -11
x < 11
x >11
Answer: B
Solve: (1-x)÷3<4
Get rid of the denominators by multiplying both sides of the equation by L.C.M. Every number is divisible by 1.
The L.C.M of 3 and 1 is 3
Divide both sides by -1.
NOTE: When dividing x by negative, the sign must change x > -11
Alternatively you can take the x to the right hand side and the numbers to the left.
Which reads -11 is less than x
Hence if -11 is less than x means x is also greater than – 11.
Hence x > -11
Let’s test
Hence x > -11 is correct
Testing x < -11 is option too. Numbers less than -11 are -12, -13, -14, -15 etc.
when x = -12
7.
The longest chord of a circle is the
segment.
sector.
circumference.
diameter.
Answer: D
The chord of a circle can be defined as the line segment joining any two points on the circumference of the circle.
Diameter is the longest chord of a circle which passes through centre joining the two points on the circumference of a circle.
8.
Which of the following is not a quadrilateral?
Rhombus
Parallelogram
Rectangle
Triangle
Answer: D
A quadrilateral is a four-sided two-dimensional shape. Examples: square, rectangle, rhombus, trapezium, parallelogram and kite.
Triangle on the other hand has three sides and three angles also known as trigon. The root word "tri" means "three" and the root word "quad" means "four". Hence, a triangle is not a quadrilateral.
9.
On a map, 1⁄3 cm represents 5km. If two towns A and B are 18 cm apart on the map, what is the actual distance between them?
27 km
30 km
240 km
270 km
Answer: D
1⁄3 = 5 km
18 cm = ?
Let 18 cm = xkm and cross multiply to solve for the value of x
1⁄3 x x = 5 x 18
Applying the principles of multiplying fractions
Every number is being divided by 1 hence x = x/1
1⁄3 x x = 5 x 18
Get rid of the fraction by multiplying both sides of the equation by the L.C.M of the denominators.
The L.C.M of 3 and 1 is 3, hence multiply both sides of the equation by 3
x = 270
Alternatively
You can cross multiply and solve for the value of x
x x 1 = 90 x 3
x = 270
Alternatively, you can calculate how many times 1⁄3 can go into 18 and multiply the result by 5 km
Applying the principles of dividing fractions
Thus when dividing fractions, you reverse the fraction at the right and change the division to multiplication
Now you can use the principles of multiplying fractions to simplify
Every number is being divided by 1
10.
Kofi and Ama shared an amount of GH₵ 3,000.00 in the ratio 2:3. Find the amount received by Kofi.
GH₵ 1,000.00
GH₵ 1,200.00
GH₵ 1,500.00
GH₵ 1,800.00
Answer: B
Kofi’s ratio is 2 and Ama’s ratio is 3 from the Kofi : Ama = 2:3
Sum of ratios = 2 + 3 = 5
NOTE: 5 goes into 30, 6 times then you multiply 2 by 600 to get the 1,200
ASSIGNMENT
Calculate the amount received by Ama
SOLUTION
Ama received 1,800. Alternatively you can subtract 1200 (Amount received by Kofi) from 3000 (Total amount shared) to get Ama’s amount.
11.
Find the Least Common Multiple (L.C.M) of 2, 3 and 5.
6
12
24
30
Answer: D
Write down the multiples for each. The Least Common Multiple is the least number which occurs in all of the list of multiples
Multiples of 2
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34
Multiples of 3
3,6,9,12,15,18,21,24,27,30,33,36
Multiples of 5
5,10,15,20,25,30,35,40
The least number which appeared in all 3 multiples is 30, hence the L.C.M for 2,3 and 5 is 30
12.
In the diagram, QP is parallel to ST, angle QPR = 68^{o} and angle SRT = 40^{o}.
Use the information to answer the question below:
Find the value of angle PQR.
40^{o}
68^{o}
72^{o}
108^{o}
Answer: C
PARALLEL LINES ANGLE PROPERTIES
VERTICALLY OPPOSITE ANGLES
Vertically opposite angles are the same.
From the above diagram, angle TRS = angle PRQ = 40^{o}
ANGLES IN A TRIANGLE
Angles in a triangle add up to 180
angle QPR + angle PRQ + angle PQR = 180
angle QPR = 68^{o}
angle PRQ = 40^{o}
68 + 40 + PQR = 180
108 + PQR = 180
PQR = 180 - 108
PQR = 72^{o}
13.
Simplify 3(5a^{2} + 2c) - 2a(1 - 3a) - 6c.
21a^{2} - 2a - 6c
13a^{2} - 2a - 12c
13a^{2} - 2a
21a^{2} - 2a
Answer: D
Expand the brackets first
Concept
a(b + c) = a x b + a x c
a(b - c) = a x b - a x c
3(5a^{2} + 2c) = 3 x 5a^{2} + 3 x 2c
3(5a^{2} + 2c) = 15a^{2} + 6c
- 2a(1 - 3a) = - 2a x 1 - 2a x - 3a
NOTE: - x - = + and a x a = a^{2}
- 2a(1 - 3a) = - 2a + 6a^{2}
3(5a^{2} + 2c) - 2a(1 - 3a) - 6c = 15a^{2} + 6c - 2a + 6a^{2} - 6c
Group like terms
3(5a^{2} + 2c) - 2a(1 - 3a) - 6c = 15a^{2} + 6a^{2} - 2a + 6c - 6c
3(5a^{2} + 2c) - 2a(1 - 3a) - 6c = 21a^{2} - 2a + 0
3(5a^{2} + 2c) - 2a(1 - 3a) - 6c = 21a^{2} - 2a
14.
Find x when y = 37 for the above mapping
6
7
8
9
Answer: D
Find the rule of the mapping first
Check if the mapping is a linear/arithmetic sequence by finding the difference between each consecutive value of y. A linear sequence has the common difference between consecutive numbers the same
If it is linear, use the equation of linear formula to find the rule of the mapping
y_{2} - y_{1} = 9 - 5 = 4
y_{3} - y_{2} = 13 - 9 = 4
y_{4} - y_{3} = 17- 13 = 4
y_{5} - y_{4} = 21 - 17 = 4
As you can see, the rule of the mapping is a linear since the difference is the same, hence use the linear sequence formula to find the rule
U_{n} = U_{1} + (n-1)d
Where U_{n} is the value at the n^{th} position, in the rule of mapping, the y value
U_{1} is the first term, in the mapping above, 5
d is the common difference between each consecutive terms
n is the position, in the case of the above mapping, x
y = 5 + (x-1)4
y = 5 + (x-1) x 4
y = 5 + x x 4 -1x4
y = 5 + 4x-4
y = 4x+5-4
y = 4x+1
We can test with when x = 3
y = 4x+1
y = 4 x 3+1
y = 12+1 = 13
You can try the values of x to see if you will get the corresponding y value to prove the rule is correct
x | y = 4x+1 | y |
1 | y = 4x1+1 = 4+1 = 5 | 5 |
2 | y = 4x2+1 = 8+1 = 9 | 9 |
3 | y = 4x3+1 = 12+1 = 13 | 13 |
4 | y = 4x4+1 = 16+1 = 17 | 17 |
5 | y = 4x5+1 = 20+1 = 21 | 21 |
y = 4x+1
y = 37
37 = 4x+1
Make x the subject
37 - 1 = 4x
36 = 4x
Divide both sides by 4
36/4 = 4x/4
9 = x
15.
Express 7352.4658 correct to three significant figures.
7352465.8
7352.47
7350
735
Answer: C
The significant digits of a number are the digits that have meaning or contribute to the value of the number. Sometimes they are also called significant figures.
Which digits are significant?
There are some basic rules that tell you which digits in a number are significant:
• All non-zero digits are significant
• Any zeros between significant digits are also significant
• Trailing zeros to the right of a decimal point are significant
7352.4658 has 8 significant figures (7,3,5,2,4,6,5 and 8). Count from the left, three times to land on the three significant figure. If the next digit is 5 or more, you add 1 to the digit the count ends if not you keep that digit and replace the remaining digits by zeros till the decimal point.
FIGURES | ONE | TWO | THREE | REPLACE BY 0 | DECIMAL POINT | ||
DIGITS | 7 | 3 | 5 | 2 | . |
⇒ 7352.4658 = 7350 three significant figures
16.
Use the diagram below to answer the question below
Find the value of y.
68^{o}
75^{o}
112^{o}
124^{o}
Answer: A
Concepts
A triangle which has two of its sides equal is called an isosceles triangle
Properties of Isosceles Triangle
1. The two equal sides of an isosceles triangle are called the legs and the angle between them is called the vertex angle or apex angle.
2. The side opposite the vertex angle is called the base and base angles are equal.
3. The perpendicular from the vertex angle bisects the base and the vertex angle.
Applying property 2 to redraw the triangle in the question gives us the diagram below:
From the above properties, you will realize that angle CBA = BAC = 56^{o}
Considering the basic properties of angles in a triangle below:
Applying the concept which says that the sum of all the angles in a triangle should be 180^{o}
⇒ 56+y+56 = 180
y+56+56 = 180
y+112 = 180
⇒ y = 180 - 112
y = 68^{o}
17.
Find the truth set of the inequality 2y + 5 < 4y - 5.
{y:y > 5}
{y:y < 5}
{y:y > 1}
{y:y > 0}
Answer: A
2y + 5 < 4y - 5
5 + 5 < 4y - 2y
10 < 2y
Divide both sides by 2
10/2 < 2y/2
5 < y
You can change the y position and change the direction of the sign
⇒ y > 5
18.
How many lines of symmetry has a square?
0
1
2
4
Answer: D
A line of symmetry is a line that cuts a shape exactly in half.
This means that if you were to fold the shape along the line, both halves would match exactly. Equally, if you were to place a mirror along the line, the shape would remain unchanged.
If all the sides and angles of a polygon are equal then it is said to be a regular polygon. Like the equilateral triangle, square etc. All the regular polygons are symmetrical shapes. In the regular polygon, the number of lines of symmetry is the same as the number of its sides
Regular Polygon | Number of Sides | Number of Symmetry | Image |
Equilateral Triangle | 3 | 3 | |
Square | 4 | 4 | |
Regular Pentagon | 5 | 5 | |
Regular Hexagon | 6 | 6 |
19.
Express 134.78 correct to the nearest tenth
130.0
134.7
134.8
135.0
Answer: C
ROUNDING WHOLE NUMBERS
If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40
If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down.
Example: 33 rounded to the nearest ten is 30.
All the numbers to the right of the place you are rounding to become zeros.
ROUNDING FRACTIONS (DECIMAL POINTS)
Rounding fractions works exactly the same way as rounding whole numbers. The only difference is that instead of rounding to tens, hundreds, thousands, and so on, you round to tenths, hundredths, thousandths, and so on after the decimal point.
nth | Number of Decimal Places |
tenth | Round to 1 decimal place |
hundredths | Round to 2 decimal places |
Thousandths | Round to 3 decimal places |
ROUNDING TO DECIMAL PLACE(S)
Count the numbers after the decimal point to the number of places you are rounding to. If the next number after where you land is 5 or more, you add 1 to the number where the number of the decimal places land.
In the question above we are to express 134.78 correct to the nearest tenth, hence correct to 1 decimal place. When we count from the decimal point (.) one, it lands on 7 and the next number is 8. Since 8 is more than 5, we add 1 to where our decimal point ended (7) and we get 8 instead of 7.
Hence the 134.8 is the nearest tenth of 134.78
20.
The number of girls in a mixed school is 420. If the ratio of boys to girls in the school is 3:2, how many students are in the school?
1050
1470
1630
1680
Answer: A
A number for a particular ratio can be calculated as:
Let x = total number of students
Ratio of boys : girls = 3:2
Ratio for girls = 2
Sum of ratio = 3+2 = 5
Number of girls is equal to the ratio of girls over the sum of the ratios times the total number of students
Every number is being divided by 1
420 = 420/1
Rewrite the equation
Cross multiply to solve for the value of x
2x x 1 = 5 x 420
Divide both sides by 2
Cancel out
x = 5 x 210 = 1050
Total number of students = 1050
ALTERNATIVELY
The ratio for the total students is the sum of the ratios
Ratio for the total students = 3+2 = 5
Let x = total students
We know the ratio of girls and the number of girls
Equate the ratio to the known numbers
Ratio of girls = Number of girls
Total ratios = x (Total number of students)
Cross multiply to solve for the value of x
2 x x = 5 x 420
2x = 5 x 420
Divide both sides by 2
Cancel out
x = 5 x 210 = 1050
Total number of students = 1050
21.
A man earned an interest of GH₵ 240.00 in 4 years at 20% per annum simple interest. Calculate the principal
GH₵ 300.00
GH₵ 450.00
GH₵ 480.00
GH₵ 1,200.00
Answer: A
Interest = Principal x Time x Rate / 100
Interest = 240, Time = 4 and Rate = 20
⇒ 240 = Principal x 4 x 20 / 100
Multiply both sides by 100 and simplify 4 x 20 (80)
⇒ 240 x 100 = 80 x Principal
Divide both sides by 80 so that Principal can be alone at one side of the equation
240 x 100 / 80 = Principal
80 goes into 240, 3 times
⇒ 3 x 100 = Principal
⇒ Principal = 300
22.
Which of the following inequalities is represented on the number line?
-2>y>2
-2≤ y < 2
-2 ≥ y > 2
-2 < y ≤ 2
Answer: D
Plotting inequalities on number lines
Inequalities can be represented on a number line.
Use a hollow dot for:
< and >
Use a solid dot for:
≤ and ≥
Example
This shows x is greater than or equal to : -1
x ≥ -1
Example
This shows x is less than : 2
x < 2
Example
This shows that x is greater than or equal to -1 and x is less than 3 :
x ≥ -1 and x < 3
Which can be written as -1 ≤ x < 3, which reads -1 is less than or equal to x and x is less than 3
Notice that we can change the order of the first part and it still means the same thing.
x ≥ -1 means as -1 ≤ x
23.
Esi bought a television set for GH₵ 1,500.00. If she sold it at a profit of 20%, find the selling price.
GH₵ 1,200.00
GH₵ 1,500.00
GH₵ 1,750.00
GH₵ 1,800.00
Answer: D
Profit = Selling Price - Cost Price
Profit percent (%) is the amount of the profit expressed in terms of percentage. The profit is based on the cost price, hence the formula to profit % is (Profit/Cost Price) x 100
But profit = Selling Price - Cost Price
Hence profit % = ([Selling Price-Cost Price]/Cost Price) x 100
From the question, cost price (bought) = 1500
Profit % = 20%
Selling price = ?
Let the selling price be x
Get rid of the denominator by multiplying both sides of the equation by the L.C.M
Every number is divisible by 1, hence 20 = 20/1
The L.C.M of 1500 and 1 is 1500, hence multiply both sides by 1500 and cancel out
100 x (x-1500) = 20 x 1500
100 x x - 100 x 1500 = 30000
100x - 150000 = 30000
100x = 30000 + 150000
100x = 180000
Divide both sides by 100
x = 1800
24.
Mary had a chance to select a number from 1 to 20 randomly. What is the probability that the number is divisible by 3?
Answer: C
Since selection is from 1 to 20, number of sample space being selected from = 20
Probability is calculated by the formula below:
We have to list all the numbers divisible by 3 (multiples of 3) from 1 to 20 to know how many numbers divisible by 3 in the sample
Numbers divisible by 3 from 1 to 20 = 3,6,9,12,15 and 18
Hence 6 of the numbers are divisible by 3
Substite the number of numbers divisible by 3 and the total sample space into the probability formula
Cancel out
2 goes into 6, 3 times and 20, 10 times
25.
In a class of 23 students, the girls were 7 more than the boys. How many boys were in the class?
8
15
16
30
Answer: A
Let x = number of boys in the class
The girls were 7 more than the boys ⇒ Girls = Boys + 7 = x + 7
Class of 23 students = Number of boys + Number of girls
⇒ x + (x + 7) = 23
x + x + 7 = 23
2x = 23 - 7
2x = 16
Divide both sides by 2
2x/2 = 16/2
x = 8
⇒ there were 8 boys in the class
26.
Find the simple interest on GH₵ 600.00 saved for 2 years 8 months at 5% per anum.
GH₵ 64.00
GH₵ 80.00
GH₵ 84.00
GH₵ 92.00
Answer: B
Simple interest (I) is calculated by the formula below:
Principal (P) = GH₵ 600.00
Convert the months to years.
There are 12 months in a year
8 months = 8/12 = 2/3 years
Total years (T) = 2 + 2/3 = 8/3 years
Rate (R) = 5
Cancel out
Simple Interest = 2 x 8 x 5 = 16 x 5 = 80
27.
Find the diameter of a circle whose circumference is 88 cm. [Take π = 22⁄7]
14 cm
22 cm
28 cm
82 cm
Answer: C
Circumference of a circle = 2πr
Diameter = 2 x radius = 2r
⇒ Circumference = π x diameter
π = 22/7
Circumference = 88
⇒ 88 = 22/7 x diameter
Multiply both sides by 7
88 x 7 = 22 x diameter
Divide both sides by 22
88 x 7 / 22 = diameter
NOTE: 22 goes into 88, 4 times
4 x 7 = diameter
28 = diameter
⇒ the diameter is 28 cm
28.
In the Venn diagram M and N are the subsets of the universal set U.
Use this information to answer the question below.
How many members are in the set N?
2
3
4
6
Answer: D
Elements of N are {1,2,4,6,7,9}. If you count the members you will get 6
Exercises
a) Find M ∩ N
b) Find M ∪ N
c) Elements of M
d) How many members are in the set M?
Answers to exercises
a) M ∩ N = {2,7}
Intersection (∩) is where the two circles join
b) M ∪ N = {1,2,3,4,5,6,7,8,9}
Union (∪) you list all the elements (members) in both sets. If a member appear in both sets, its written only once
c) M = {2,3,5,7,8}
d) 5
29.
Simplify
Answer: D
Change the mixed fractions to improper fractions
Rule
Apply the division of fraction rule
Thus you reciprocate (down goes top and top down) the fraction at the right and change the division sign to multiplication
Now cancel out and apply the principle of multiplying fractions
3 goes into 6, 2 times and 21, 7 times
5 goes into itself, 1 time and 10, 2 times
30.
Simplify 2^{2} x 2^{7} ÷ 2^{4}
2^{-1}
2^{5}
2^{11}
2^{13}
Answer: B
Concept
a^{b} x a^{c} = a^{b+c}
a^{b} ÷ a^{c} = a^{b-c}
⇒ 2^{2} x 2^{7} ÷ 2^{4} = 2^{2+7} ÷ 2^{4}
⇒ 2^{2} x 2^{7} ÷ 2^{4} = 2^{9} ÷ 2^{4}
⇒ 2^{2} x 2^{7} ÷ 2^{4} = 2^{9 - 4}
⇒ 2^{2} x 2^{7} ÷ 2^{4} = 2^{5}
31.
Expand and simplify: (a-2)(2a+3)
a^{2} - a + 6
2a^{2} + 7a - 6
2a^{2} - a - 6
2a^{2} - 12a + 6
Answer: C
Concept
(a+b)(c+d) = a x (c+d) + b x (c+d)
a(b+c) = a x (b+c) = a x b + a x c
⇒ (a - 2)(2a+3) = a x (2a+3) - 2 x (2a+3)
a x (2a+3) = a x 2a + a x 3 = 2a^{2} + 3a
NOTE: a x a = a^{1} x a^{1} = a^{1+1} = a^{2}
-2(2a+3) = -2 x 2a + -2 x 3 = -4a - 6
⇒ (a - 2)(2a+3) = 2a^{2}+3a - 4a - 6
⇒ (a - 2)(2a+3) = 2a^{2}- a - 6
NOTE: 3a - 4a = -1a = - a
32.
Find the equation of the straight line passing through the points (-3,5) and (6,8)
y = ⅓^{x}
y = ⅓^{x}+6
y = ⅓^{x}-10
y = ⅓^{x}+14
Answer: B
Equation of a straight line, y = mx+c
where m is the gradient of the line
m = change in y / change in x =(y_{2}-y_{1})/(x_{2}-x_{1})
NOTE: The first coordinate represents the x-axis and the second the y-axis
Thus in (-3,5), x = -3 and y = 5
In (6,8), x = 6 and y = 8
⇒ The gradient of the line = (8-5)/(6--3) = 3/(6+3) = 3/9 = 1/3
Use any of the points and the gradient to calculate the value of c
Using the point (-3,5)
⇒ 5 = ⅓ x -3 + c
⇒ 5 = -1 + c
⇒ 5 + 1= c
⇒ c = 6
Using the point (6,8)
⇒ 8 = ⅓ x 6 + c
⇒ 8 = 2 + c
⇒ 8 - 2= c
⇒ c = 6
Hence the equation will be y = ⅓^{x}+6
33.
A tank in the form of a cuboid has length 6 m and breadth 4 m. If the volume of the tank is 36 m^{3}, find the height.
0.67 m
1.5 m
1.8 m
5.0 m
Answer: B
Volume of a cuboid = Length x Breadth x Height
Dividing both sides by Length x Breadth to make Height the subject
⇒ Height = Volume of cuboid / (Length x Breadth)
⇒ The height of the cuboid = 36 / (6 x 4) = 3/2 = 1.5 m
NOTE: 6 goes into 36, 6 times and 2 goes into 6, 3 times and 4, 2 times
34.
Find the L.C.M of 10, 15 and 25.
90
120
150
300
Answer: C
List the multiples of each and select the least number which appears in all
Multiples of 10 = 10, 10+10 = 20, 20+10 = 30, 30+10 = 40, 50,60,70,80,90,100,110,120,130,140,150,160,...
Multiples of 15 = 15, 15+15 = 30, 30+15 = 45, 45+15 = 60, 75, 90, 105, 120,135,150,165,...
Multiples of 25 = 25, 25+25 = 50, 50+25 = 75, 75+25 = 100,125, 150, 175, 200...
Alternatively
HOW TO: FIND THE L.C.M USING THE PRIME FACTORS METHOD
Step 1. Find the prime factorization of each number.
Step 2. Write each number as a product of primes.
Step 3. The L.C.M is the product (multiplication) of each of the prime numbers in its highest power.
Step 1 & 2
10 = 2 x 5
15 = 3 x 5
25 = 5 x 5 = 5^{2}
Step 3
L.C.M = 2 x 3 x 5^{2} = 6 x 25 = 150
35.
In the diagram, line MN is parallel to line TU, line TS cuts line MN at O and ∠MOS = 115^{o}. Find ∠OTU.
65^{o}
55^{o}
45^{o}
25^{o}
Answer: A
Concepts
Transversal Angles
Alternate Angles
∠MOS = ∠NOT = 115^{o}(Transversal angles)
∠MOT = ∠OTU (Alternate angles)
∠MOT + ∠NOT = 180 (Angles on a straight line add up to 180)
But ∠NOT = ∠MOS = 115^{o}
∠MOT + 115 = 180
∠MOT = 180 - 115
∠MOT = 65^{o}
But ∠MOT = ∠OTU = 65^{o}
⇒ ∠OTU = 65^{o}
36.
Given that N = {x:x is a factor of 18} and M = {x:x is a multiple of 12}, find N ∩ M.
{1,2,3,6}
{1,2,3,6,12}
{2,3,6,12,18}
{}
Answer: D
x:x is read as x is such that x is ...
N contains the list of factors of 18. A factor is a number or algebraic expression that divides another number or expression evenly, thus no remainder.
Factors of 18 are 1,2,3,6,9 and 18 itself
Thus
1x18 = 18
2x9 = 18
3x6 = 18
Hence 1,2,3,6,9 and 18 are factors of 18
⇒ N = {1,2,3,6,9,18}
Multiples of 12 are numbers 12 can divide without a remainder (Simply the multiplication of 12). If you don't know the recitation of the multiples, simply add 12 to the preceding numbers to get the subsequent numbers starting from 12.
Multiples of 12
12,12+12=24,24+12 = 36,36+12 = 48,48+12 = 60,60+12 = 72,72+12 = 84,84+12 = 96,96+12 = 108,108+12 = 120,120+12 = 132,132+12 = 144,...
⇒ multiples of 12 are 12,24,36,48,60,72,84,96,108,120,132,144,...
⇒ M = {12,24,36,48,60,72,84,96,108,120,132,144,...}
∩ means intersection. The element(s) which you can find in both sets.
Hence N ∩ M should contain element(s) which can be found in both set N and M.
From the above, you can see that there is no element that can be found in N,{1,2,3,6,9,18} and at the same time be found in M,{12,24,36,48,60,72,84,96,108,120,132,144,...}. Hence the intersection of N and M is an empty set {}
37.
Solve: 3 - (3x+4) ≤ -4
x ≤ 1
x ≥ 1
x ≥ 1⅔
x < 1½
Answer: B
3 - (3x+4) ≤ -4
Concept
a(b+c) = axb+axc
-a(b+c) = -a x b + -axc
-(a+b) = -1(a+b) = -1xa + -1xb
⇒ - (3x+4) = -1(3x+4) = -1 x 3x + -1x4 = -3x-4
3 - (3x+4) ≤ -4
⇒ 3 - 3x-4 ≤ -4
⇒ 3 -4 - 3x ≤ -4
⇒ -1 - 3x ≤ -4
Moving the - 3x to the right side of the equation and the -4 to the left
⇒ -1+4 ≤ 3x
3 ≤ 3x
Divide both sides by 3
3/3 ≤ 3x/3
⇒ 1 ≤ x
You can reverse the order and reverse the inequality sign
⇒ x ≥ 1
38.
If 2y = 1-3x^{2}+4x, find y when x = -1
-3
-½
½
3
Answer: A
Concept
- x - = + (When a negative number [-] multiplies another negative number[-], the sign becomes positive [+])
- x + = - (When a negative number [-] multiplies a positive number [+], the sign becomes negative [-])
a^{2} = a x a
2y = 1-3x^{2}+4x
When x = -1, substitute x by -1
2y = 1-3(-1)^{2}+4(-1)
(-1)^{2} = -1 x -1 = + 1
4(-1) = 4 x -1 = -4
2y = 1-3x^{2}+4x
⇒ 2y = 1-3(1)-4
2y = 1 -3 - 4
2y = 1 - 7
2y = - 6
Divide both sides by 2 to get rid of the 2 multiplying y
⇒ y = -6 / 2 = -3
39.
Arrange
,
and
in ascending order.
,
,
,
,
,
,
,
,
Answer: D
Make sure the denominators are the same for each fraction by multiplying both the numerator and denominators by the number of times the denominator goes into the L.C.M
The L.C.M of 3, 9 and 7 is 63
Thus:
3 = 3
9 = 3 x 3 = 3^{2}
7 = 7
To get the L.C.M, multiply each of the prime numbers with each having its highest power
L.C.M = 3^{2} x 7 = 9 x 7 = 63
Now express each fraction to have 63 as the denominator
Since the denominators are the same, you can compare them by the numerators
The question said in ascending order, hence lowest to highest
,
,
=
,
,
40.
What is the Highest Common Factor (HCF) of 24, 32 and 64?
4
6
8
16
Answer: C
A factor is a number that fits exactly into another number. For example, 2 is a factor of 4 because 2 goes into 4, 2 times
The highest common factor is the largest whole number which is shared by given numbers.
To find the highest common factor, list the factors of all the numbers and select the highest among them
Write down the multiples which gives the numbers you are looking for their factors. Pick each of the numbers from the list
24 | 32 | 64 |
1 x 24 | 1 x 32 | 1 x 64 |
2 x 12 | 2 x 16 | 2 x 32 |
3 x 8 | 4 x 8 | 4 x 16 |
4 x 6 | 8 x 8 | |
Factors | ||
1,2,3,4,6,8,12,24 | 1,2,4,8,16,32 | 1,2,4,8,16,32,64 |
You can verify if you've list all the factors, by picking the first and multiplying with the last then the second and the last but two and so on. If none is left without a pair, then you've listed all. Only numbers which multiplied by itself gives the number you are looking for its factor doesn't have a pair since multiplying itself gives the number
From the above list, 8 is the highest number which appears in all the list of factors of 24, 32 and 64
Hence the Highest Common Factor of 24, 32 and 64 is 8
a)
Solve:
b)
Multiply 0.03858 by 0.02, leaving the answer in standard form.
c)
A cylindrical container of height 28 cm and diameter 18 cm is filled with water. The water is then poured into another container with a rectangular base of length 27 cm and width 11 cm. Calculate the depth of the water in the container. (Take π = 22⁄7)
a)
Get rid of the denominator by multiplying both sides by the LCM of the denominators (3,1 and 1). LCM of the denominators is 3
⇒ 3 x (2x + 3)/3 + 3 x 2x = 3 x 10
⇒ 2x + 3 + 6x = 30
⇒ 2x + 6x = 30 - 3
⇒ 8x = 27
Divide both sides by 8
⇒ 8x/8 = 27/8
⇒ x = 27/8 or 3 ⅜
b)
0.03858 x 0.02
Change the decimals to a whole number in the form a x 10^{b} by moving the decimal points to get a whole number. Count how many moves. If to the right, the power to 10 is negative else if to the left positive.
0.03858 = 3858 x 10^{-5} (Moved to the right 5 times)
0.02 = 2 x 10^{-2} (Moved to the right 2 times)
⇒ 0.03858 x 0.02 = 3858 x 10^{-5} x 2 x 10^{-2}
⇒ 0.03858 x 0.02 = 3858 x 2 x 10^{-5} x 10^{-2}
Applying a^{b} x a^{c} = a^{b + c}
⇒ 0.03858 x 0.02 = 7716 x 10^{-5+-2}
⇒ 0.03858 x 0.02 = 7716 x 10^{-7}
Now write the result in standard form
Standard form, or standard index form, is a system of writing numbers which can be particularly useful for working with very large or very small numbers.
It is based on using powers of 10 to express how big or small a number is.
Standard form is written in the form of a x 10^{n}, where a is a number bigger than or equal to 1 and less than 10.
n can be any positive or negative whole number.
For example 1.2 x 10^{3}, 3.14 x 10^{-5}
The rules when writing a number in standard form is that first you write down a number between 1 and 10, then you write × 10(to the power of a number).
Thus the decimal point must be behind the first non-zero digit
If you move the decimal point to the left, the power (number of times of movement) of 10 will be positive and if to the right, negative
First express 7716 in standard form
7716 = 7716.0, thus every number has a .0 behind which is often not written.
The first non-zero digit in 7716 is 7 so we have to move the decimal point to left 3 times. Since its to the left, the power is going to be positive
⇒ 7716 = 7.716 x 10^{3} (The decimal is moved 3 times to the left)
0.03858 x 0.02 = 7716 x 10^{-7}
⇒ 0.03858 x 0.02 = 7.716 x 10^{3} x 10^{-7}
Applying a^{b} x a^{c} = a^{b + c}
⇒ 0.03858 x 0.02 = 7.716 x 10^{3 + -7} = 7.716 x 10^{-4}
c)
Concept
The same volume of water in the cylindrical container is poured into the rectangular container
⇒ The volume of the water in the cylindrical container = volume of water in the rectangular container
Volume of a cylinder = πr^{2}h
Where r = radius and h = height
Radius = Diameter/2 (Half of a diameter)
⇒ r = 18/2 = 9
Height = 28
⇒ volume of the water in the cylinder = 22⁄7 x 9 x 9 x 28 = 22 x 9 x 9 x 4
NOTE: 7 goes into 28, 4 times
Volume of a rectangular container = Length x Breadth(width) x height
Length = 27 and width = 11
Let height/depth of the water = d
Volume of the water in the rectangular container = 27 x 11 x d
Since the volume of the water in the rectangular container is the same in the cylindrical container,
⇒ 27 x 11 x d = 22 x 9 x 9 x 4
Divide both sides by 27 x 11
NOTE: Its better not to multiply yet so you know which ones can cancel each other doing the division to make the simplification easier
⇒ d = (22 x 9 x 9 x 4)/(27 x 11) = 2 x 1 x 3 x 4 = 24 cm
NOTE: 11 goes into 22, 2 times and 9 goes into 27, 3 times and 3 also goes into 9, 3 times
a)
If M = {Prime integers between 1 and 11} and N = {factors of 12}, find:
(i) M ∪ N
(ii) M ∩ N
b)
Simplify 45 ÷ 3 + 2 x 8 - 12 + 42
c)
In the diagram, x, y and z are angles on a straight line. If x^{o} : z^{o} = 2 : 3 and y = 80^{o}, find x.
a)
M = {Prime integers between 1 and 11}
A prime is a number that is greater than 1 and divisible only by itself and 1 (e.g. 2, 3, 5, 7, 11,...)
Note: 1 is not a prime number
M = {2,3,5,7}
NOTE: Between excludes the beginning and ending numbers so 11 is not part of the elements of set M
N = {factors of 12}
Factor, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder.
N = {1,2,3,4,6,12}
(i) M ∪ N
Union (∪) is the arrangement of the elements in both sets. If an element occurs in both sets, its written only once
⇒ M ∪ N = {1,2,3,4,5,6,7,12}
(ii) M ∩ N
The intersection of a set A with a B is the set of elements that are in both set A and B.
⇒ M ∩ N = {2,3}
b)
Simplify 45 ÷ 3 + 2 x 8 - 12 + 42
Applying BODMAS (Bracket Of Division Multiplication Addition Substraction), the division (÷) needs to be solved first
45 ÷ 3 + 2 x 8 - 12 + 42 = 45/3 + 2 x 8 - 12 + 42
45 ÷ 3 + 2 x 8 - 12 + 42 = 15 + 2 x 8 - 12 + 42
Applying BODMAS, multiplication needs to be solved first
45 ÷ 3 + 2 x 8 - 12 + 42 = 15 + 16 - 12 + 42
You can now simplify
45 ÷ 3 + 2 x 8 - 12 + 42 = 15 + 16 + 42 - 12
45 ÷ 3 + 2 x 8 - 12 + 42 = 73 - 12
45 ÷ 3 + 2 x 8 - 12 + 42 = 61
c)
The sum of all angles on a straight line add up to 180^{o}
⇒ x + y + z = 180
y = 80
⇒ x + 80 + z = 180
⇒ x + z = 180 - 80
⇒ x + z = 100
Solving for x using substitution
x + z = 100
z = 100 - x
x^{o} : z^{o} = 2 : 3
⇒ x/z = 2 / 3
But z = 100 - x
⇒ x/(100 - x) = 2 / 3
Cross multiply, thus the numerator at the left multiplies the denominator at the right and the denominator at the left multiplies the numerator at the right
3 x x = 2 x (100 - x)
3x = 200 - 2x
Move - 2x to the left
3x + 2x = 200
5x = 200
Divide both sides by 5
5x/5 = 200/5 ⇒ x = 40
NOTE 5 goes into 200, 40 times
Solving for x using ratio concept
Number of x = Ratio of x / sum of ratios x Total Number
⇒ x = 2/(2+3) x 100
NOTE: The total is 100 because x + z = 100
⇒ x = 2/5 x 100 = 2 x 20 = 40
NOTE: 5 goes into 100, 20 times
a
x |
1 |
2 |
3 |
4 |
5 |
↓ |
↓ |
↓ |
↓ |
↓ |
↓ |
y |
0 |
3 |
6 |
9 |
12 |
The mapping shows the relationship between x and y.
i)
using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 2 units on the y-axis, draw two perpendicular axes 0x and 0y on a graph sheet for 1 ≤ x ≤ 5 and 0 ≤ x ≤ 14;
ii)
plot the point for each ordered pair, (x, y).
iii)
join the points with a straight line;
iv)
using the graph, find the gradient of the line in (a)(iii);
v)
use the graph to find the equation of the line in (a)(iii).
b
Simplify: 32 x 8 x 4 x 2, leaving the answer in the form 2^{n}
a
The mapping shows the relationship between x and y.
i-iii)
iv)
The gradient of a line is calculated by the formula below:
You can pick any two points on the line to use to calculate for the gradient
For any point, the first cordinate is x and the second y, thus p(x,y)
Using the points (1,0) and (4,9)
From the point (1,0), x_{1} = 1, y_{1} = 0
From the point (4,9), x_{2} = 4, y_{2} = 9
Insert into the gradient formula
v)
To find the equation of a line when the gradient is known, use the cordinates (x,y) and any point on the line to find the gradient using the gradient formula and equate it to the known gradient and make y the subject of the equation.
Using the points (4,9) and (x,y) and the known gradient 3
Every number is being divided by 1, hence 3 = 3/1
Rewrite the equation
Cross multiply
(y - 9) x 1 = (x - 4) x 3
Expand the bracket
y x 1 - 9 x 1 = x x 3 - 4 x 3
y - 9 = 3x - 12
Make y the subject
y = 3x - 12 + 9
y = 3x - 3
b
32 x 8 x 4 x 2
Concept
a^{m} x a^{n} = a^{m + n}
Every number is raised to the power 1
a = a^{1}
2 = 2^{1}
32 = 2 x 2 x 2 x 2 x 2
32 = 2^{1} x 2^{1} x 2^{1} x 2^{1} x 2^{1} = 2^{1+1+1+1+1} = 2^{5}
8 = 2^{1} x 2^{1} x 2^{1} = 2^{1+1+1} = 2^{3}
4 = 2^{1} x 2^{1} = 2^{1+1} = 2^{2}
32 x 8 x 4 x 2 = 2^{5} x 2^{3} x 2^{2} x 2^{1} = 2^{5+3+2+1} = 2^{11}
The table shows the number of marbles students sent to class for Mathematics lesson.
Number of Marbles(x) | Number of Students(f) | fx |
1 | 4 | - |
2 | 5 | - |
3 | - | 42 |
4 | 9 | - |
5 | - | 30 |
6 | 2 | 12 |
(a)
Copy and complete the table.
(b)
How many:
(i)
students were in the class?
(ii)
marbles were brought altogether
(iii)
marbles did most of the students bring
(c)
Calculate, correct to the nearest whole number, the mean number of the marbles brought for the lesson.
(a)
fx = f x x
f = fx / x
x = fx / f
For fx = f x x
When x = 1, and f = 4, fx = 1 x 4 = 4
When x = 2, and f = 5, fx = 2 x 5 = 10
When x = 4, and f = 9, fx = 4 x 9 = 36
For f = fx / x
When x = 3 and fx = 42, f = 42 / 3 = 14
When x = 5 and fx = 30, f = 30 / 5 = 6
Number of Marbles(x) | Number of Students(f) | fx |
1 | 4 | 4 |
2 | 5 | 10 |
3 | 14 | 42 |
4 | 9 | 36 |
5 | 6 | 30 |
6 | 2 | 12 |
(b)
How many:
(i)
Total number of students = summation of all the frequencies (f), thus ∑f
Total students = 4+5+14+9+6+2 = 40
(ii)
Total marbles brought altogether = summation of fx (∑fx)
Total marbles = 4+10+42+36+30+12 = 134
(iii)
The most marbles brought is the one (x) with the highest frequency (f) which is 3, having frequency of 14
(c)
Number of Marbles(x) | Number of Students(f) | fx |
1 | 4 | 4 |
2 | 5 | 10 |
3 | 14 | 42 |
4 | 9 | 36 |
5 | 6 | 30 |
6 | 2 | 12 |
∑f = 40 | ∑fx = 134 |
∑fx = 134
∑f = 40
The nearest whole number of 3.35 is 3
Hence the mean = 3 (nearest whole number)
a
Solve:
.
b
The ratio of boys to girls in a school is 12:25. If there are 120 boys.
i) how many girls are in the school?
ii) what is the total number of boys and girls in the school?
c
Simplify:
a
Method I
Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (5,4,1). The L.C.M is 20
20 x (4x+5)/5 + 20 x (x-3)/4 = -1 x 20
4 x (4x+5) + 5 x (x-3) = -20
Applying a(b+c) = a x b + a x c
a(b-c) = a x b - a x c
4 x 4x + 4 x 5 + 5 x x-5 x 3 = -20
16x + 20 + 5x-15 = -20
16x + 5x + 20 + -15 = -20
21x + 5 = -20
21x = -20-5
21x = -25
Divide both sides by 21
21x/21 = -25/21
x = -25/21 or -1 4⁄21
Method II
Simplify the fractions at the left
21x + 5 = -20
21x = -20-5
21x = -25
Divide both sides by 21
21x/21 = -25/21
x = -25/21 or -1 4⁄21
b
Method I
Value of top ratio : Value of down ratio = Top ratio : Down ratio
No. of boys : No. of girls = Ratio of boys : Ratio of girls
No. of boys = 120
Ratio of boys = 12
Ratio of girls = 25
120 : No. of girls = 12 : 25
Let x = No. of girls
120 : x = 12 : 25
Change the ratios to fraction
120/x = 12/25
Cross multiply, the numerator at the left side multiplies the denominator at the right side and the denominator at the left side multiplies the numerator at the right side
120 x 25 = x x 12
Divide both sides by 12
120 x 25 / 12 = x x 12 / 12
12 goes into 120, 10 times
10 x 25 = x
250 = x
x = 250
∴ there are 250 girls in the school
ii) Total number of students = No. of boys + No. of girls = 120 + 250 = 370 boys and girls
Method II
Value of a ratio = (The ratio / sum of ratios) x The Total
No. of boys = (Ratio of boys/Sum of ratios) x Total No. of students
Let x = Total no. of students
No. of boys = 120
Ratio of boys = 12
Sum of ratios = 12 + 25 = 37
120 = 12/37 x x
Multiply both sides by 37
120 x 37 = 12 x x
Divide both sides by 12
120 x 37 / 12 = x
12 goes into 120, 10 times
10 x 37 = x
370 = x
Total no. of students = 370 boys and girls
No. of girls = Total no. of students - No. of boys
No. of girls = 370 - 120 = 250
c
Concepts
(a)(b) = a x b
a^{m} x a^{n} = a^{m + n}
(8x^{2}y^{3})(⅜ xy^{4}) = 8x^{2}y^{3} x ⅜ xy^{4}
The 8 in the ⅜ fraction cancels the 8 multiplying
(8x^{2}y^{3})(⅜ xy^{4}) = x^{2}y^{3} x 3 xy^{4}
Apply a^{m} x a^{n} = a^{m + n}
(8x^{2}y^{3})(⅜ xy^{4}) = 3x^{2+1}y^{3+4}
NOTE: x = x^{1}, thus every number or letter is raised to the power 1 which is not written
(8x^{2}y^{3})(⅜ xy^{4}) = 3x^{3}y^{7}
The marks obtained by students in a class test were
4 |
8 |
7 |
6 |
7 |
2 |
1 |
7 |
4 |
7 |
3 |
7 |
6 |
4 |
3 |
7 |
5 |
2 |
7 |
2 |
5 |
4 |
8 |
3 |
2 |
a
Construct a frequency distribution table for the data.
b
Find the:
i)
mode of the distribution
ii)
median mark of the test;
iii)
mean mark.
a
b
i)
The mode is the one the appears the most. Thus the mark with the highest frequency
Mode = 7 marks
ii)
The median is the middle value of an ordered set of data.
In a frequency table, if there is an odd number of observations, the median is the middle number.
If there is an even number of observations, the median will be the mean of the two central numbers.
For an odd observation, the median position is found by the formula below:
For an even number of observations, the median is the average of
and
Thus the formula below:
Since the number of observations (∑f) = 25, which is odd number, the median is the middle, which is the 13^{th} position
Thus:
To find the 13^{th} position, you have to add up the frequencies until you get to the mark which when added the sum will be 13 or more
Marks | Frequency | Adding Downward |
1 | 1 | 1 |
2 | 4 | 1+4 = 5 |
3 | 3 | 5+3 = 8 |
4 | 4 | 8+4 = 12 |
5 | 2 | 12 + 2 = 13^{th} position + 1 = 14 |
Median = 5 Marks
iii)
The mean can be calculated by the formula below:
∑fx = 121 and ∑f = 25