KUULCHAT
MATHEMATICS MOCK

#### OBJECTIVE TEST

1.

Evaluate: (0.07 x 0.02) ÷ 14.

A.

0.01

B.

0.001

C.

0.0001

D.

0.00001

Change the decimals to the form whole number x 10a 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 am x an = am + n

⇒ 1 x 10-2 x 10-2 = 1 x 10-2+-2 = 1 x 10-4

Now change the whole number x 10a 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?

A.

6

B.

27

C.

34

D.

39

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?

A.

GH₵ 22.50

B.

GH₵ 67.50

C.

GH₵ 75.00

D.

GH₵ 225.50

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 10a

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 10a 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.

A.

20

B.

16

C.

8

D.

4

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?

A.

1⅓ hrs

B.

1½ hrs

C.

1¾ hrs

D.

2 hrs

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.

A.

x < -11

B.

x > -11

C.

x < 11

D.

x >11

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

A.

segment.

B.

sector.

C.

circumference.

D.

diameter.

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?

A.

Rhombus

B.

Parallelogram

C.

Rectangle

D.

Triangle

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?

A.

27 km

B.

30 km

C.

240 km

D.

270 km

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.

A.

GH₵ 1,000.00

B.

GH₵ 1,200.00

C.

GH₵ 1,500.00

D.

GH₵ 1,800.00

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.

A.

6

B.

12

C.

24

D.

30

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 = 68o and angle SRT = 40o.

Use the information to answer the question below:

Find the value of angle PQR.

A.

40o

B.

68o

C.

72o

D.

108o

PARALLEL LINES ANGLE PROPERTIES

VERTICALLY OPPOSITE ANGLES

Vertically opposite angles are the same.

From the above diagram, angle TRS = angle PRQ = 40o

ANGLES IN A TRIANGLE

Angles in a triangle add up to 180

angle QPR + angle PRQ + angle PQR = 180

angle QPR = 68o

angle PRQ = 40o

68 + 40 + PQR = 180

108 + PQR = 180

PQR = 180 - 108

PQR = 72o

13.

Simplify 3(5a2 + 2c) - 2a(1 - 3a) - 6c.

A.

21a2 - 2a - 6c

B.

13a2 - 2a - 12c

C.

13a2 - 2a

D.

21a2 - 2a

Expand the brackets first

Concept

a(b + c) = a x b + a x c

a(b - c) = a x b - a x c

3(5a2 + 2c) = 3 x 5a2 + 3 x 2c

3(5a2 + 2c) = 15a2 + 6c

- 2a(1 - 3a) = - 2a x 1 - 2a x - 3a

NOTE: - x - = + and a x a = a2

- 2a(1 - 3a) = - 2a + 6a2

3(5a2 + 2c) - 2a(1 - 3a) - 6c = 15a2 + 6c - 2a + 6a2 - 6c

Group like terms

3(5a2 + 2c) - 2a(1 - 3a) - 6c = 15a2 + 6a2 - 2a + 6c - 6c

3(5a2 + 2c) - 2a(1 - 3a) - 6c = 21a2 - 2a + 0

3(5a2 + 2c) - 2a(1 - 3a) - 6c = 21a2 - 2a

14.

Find x when y = 37 for the above mapping

A.

6

B.

7

C.

8

D.

9

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

y2 - y1 = 9 - 5 = 4

y3 - y2 = 13 - 9 = 4

y4 - y3 = 17- 13 = 4

y5 - y4 = 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

Un = U1 + (n-1)d

Where Un is the value at the nth position, in the rule of mapping, the y value

U1 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.

A.

7352465.8

B.

7352.47

C.

7350

D.

735

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.

A.

68o

B.

75o

C.

112o

D.

124o

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 = 56o

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 180o

⇒ 56+y+56 = 180

y+56+56 = 180

y+112 = 180

y = 180 - 112

y = 68o

17.

Find the truth set of the inequality 2y + 5 < 4y - 5.

A.

{y:y > 5}

B.

{y:y < 5}

C.

{y:y > 1}

D.

{y:y > 0}

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?

A.

0

B.

1

C.

2

D.

4

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

A.

130.0

B.

134.7

C.

134.8

D.

135.0

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?

A.

1050

B.

1470

C.

1630

D.

1680

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

A.

GH₵ 300.00

B.

GH₵ 450.00

C.

GH₵ 480.00

D.

GH₵ 1,200.00

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?

A.

-2>y>2

B.

-2≤ y < 2

C.

-2 ≥ y > 2

D.

-2 < y ≤ 2

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.

A.

GH₵ 1,200.00

B.

GH₵ 1,500.00

C.

GH₵ 1,750.00

D.

GH₵ 1,800.00

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?

A.

B.

C.

D.

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?

A.

8

B.

15

C.

16

D.

30

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.

A.

GH₵ 64.00

B.

GH₵ 80.00

C.

GH₵ 84.00

D.

GH₵ 92.00

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]

A.

14 cm

B.

22 cm

C.

28 cm

D.

82 cm

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?

A.

2

B.

3

C.

4

D.

6

Elements of N are {1,2,4,6,7,9}. If you count the members you will get 6

Exercises

a) Find MN
b) Find MN
c) Elements of M
d) How many members are in the set M?

a) MN = {2,7}

Intersection (∩) is where the two circles join

b) MN = {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

A.

B.

C.

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 22 x 27 ÷ 24

A.

2-1

B.

25

C.

211

D.

213

Concept

ab x ac = ab+c

ab ÷ ac = ab-c

⇒ 22 x 27 ÷ 24 = 22+7 ÷ 24

⇒ 22 x 27 ÷ 24 = 29 ÷ 24

⇒ 22 x 27 ÷ 24 = 29 - 4

⇒ 22 x 27 ÷ 24 = 25

31.

Expand and simplify: (a-2)(2a+3)

A.

a2 - a + 6

B.

2a2 + 7a - 6

C.

2a2 - a - 6

D.

2a2 - 12a + 6

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 = 2a2 + 3a

NOTE: a x a = a1 x a1 = a1+1 = a2

-2(2a+3) = -2 x 2a + -2 x 3 = -4a - 6

⇒ (a - 2)(2a+3) = 2a2+3a - 4a - 6

⇒ (a - 2)(2a+3) = 2a2- a - 6

NOTE: 3a - 4a = -1a = - a

32.

Find the equation of the straight line passing through the points (-3,5) and (6,8)

A.

y = ⅓x

B.

y = ⅓x+6

C.

y = ⅓x-10

D.

y = ⅓x+14

Equation of a straight line, y = mx+c

where m is the gradient of the line

m = change in y / change in x =(y2-y1)/(x2-x1)

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 = ⅓x+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 m3, find the height.

A.

0.67 m

B.

1.5 m

C.

1.8 m

D.

5.0 m

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.

A.

90

B.

120

C.

150

D.

300

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 = 52

Step 3

L.C.M = 2 x 3 x 52 = 6 x 25 = 150

35.

In the diagram, line MN is parallel to line TU, line TS cuts line MN at O and ∠MOS = 115o. Find ∠OTU.

A.

65o

B.

55o

C.

45o

D.

25o

Concepts

Transversal Angles

Alternate Angles

MOS = ∠NOT = 115o(Transversal angles)

MOT = ∠OTU (Alternate angles)

MOT + ∠NOT = 180 (Angles on a straight line add up to 180)

But ∠NOT = ∠MOS = 115o

MOT + 115 = 180

MOT = 180 - 115

MOT = 65o

But ∠MOT = ∠OTU = 65o

⇒ ∠OTU = 65o

36.

Given that N = {x:x is a factor of 18} and M = {x:x is a multiple of 12}, find NM.

A.

{1,2,3,6}

B.

{1,2,3,6,12}

C.

{2,3,6,12,18}

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 NM 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

A.

x ≤ 1

B.

x ≥ 1

C.

x ≥ 1⅔

D.

x < 1½

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 = 1-3x2+4x, find y when x = -1

A.

-3

B.

C.

½

D.

3

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 [-])

a2 = a x a

2y = 1-3x2+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 = 1-3x2+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.

A.

,

,

B.

,

,

C.

,

,

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 = 32

7 = 7

To get the L.C.M, multiply each of the prime numbers with each having its highest power

L.C.M = 32 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?

A.

4

B.

6

C.

8

D.

16

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

#### THEORY QUESTIONS

1.

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 10b 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 ab x ac = ab + 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 10n, 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 103, 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 103 (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 103 x 10-7

Applying ab x ac = ab + c

⇒ 0.03858 x 0.02 = 7.716 x 103 + -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 = πr2h

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

2.

a)

If M = {Prime integers between 1 and 11} and N = {factors of 12}, find:
(i) MN
(ii) MN

b)

Simplify 45 ÷ 3 + 2 x 8 - 12 + 42

c)

In the diagram, x, y and z are angles on a straight line. If xo : zo = 2 : 3 and y = 80o, 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) MN

Union (∪) is the arrangement of the elements in both sets. If an element occurs in both sets, its written only once

MN = {1,2,3,4,5,6,7,12}

(ii) MN

The intersection of a set A with a B is the set of elements that are in both set A and B.

MN = {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 180o

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

xo : zo = 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

3.

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 2n

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), x1 = 1, y1 = 0

From the point (4,9), x2 = 4, y2 = 9

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

am x an = am + n

Every number is raised to the power 1

a = a1

2 = 21

32 = 2 x 2 x 2 x 2 x 2

32 = 21 x 21 x 21 x 21 x 21 = 21+1+1+1+1 = 25

8 = 21 x 21 x 21 = 21+1+1 = 23

4 = 21 x 21 = 21+1 = 22

32 x 8 x 4 x 2 = 25 x 23 x 22 x 21 = 25+3+2+1 = 211

4.

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)

5.

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

am x an = am + n

(8x2y3)(⅜ xy4) = 8x2y3 x ⅜ xy4

The 8 in the ⅜ fraction cancels the 8 multiplying

(8x2y3)(⅜ xy4) = x2y3 x 3 xy4

Apply am x an = am + n

(8x2y3)(⅜ xy4) = 3x2+1y3+4

NOTE: x = x1, thus every number or letter is raised to the power 1 which is not written

(8x2y3)(⅜ xy4) = 3x3y7

6.

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 13th position

Thus:

To find the 13th 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 = 13th position + 1 = 14

Median = 5 Marks

iii)

The mean can be calculated by the formula below:

fx = 121 and ∑f = 25