### Example 1:

**1.** a. Express 132_{four} in Binary

**Solution**

First, convert 132_{four} to Base 10

Place Value | ^{2} | ^{1} | ^{0} |

1 | 3 | 2 |

= 1 × 4^{2} + 3 × 4^{1} + 2 × 4^{0}

= 16 + 12 + 2

= 30_{10}

Now convert 30_{10} to base 2

**Answer** = 11110_{2}

**1.** b. Solve the equation \( \frac{3x \: – \: 2}{8} = \frac{8 \: + \: 5x}{2} \)

**Solution**

Multiply both sides by 8

\( \normalsize \frac{3x \: – \: 2}{8} \scriptsize \: \times \: 8 = \normalsize \frac{8 \: + \: 5x}{2} \scriptsize \: \times \: 8\) \( \normalsize \frac{3x \: – \: 2}{\not{8}} \scriptsize \: \times \: \not{\!8}^1 = \normalsize \frac{8 \: + \: 5x}{\not{2}} \scriptsize \: \times \: \not{\!8}^4\)(3x – 2) = 4(8 + 5x)

**Open the brackets**

3x – 2 = 32 + 20x

**Collect like terms**

3x – 20x = 32 + 2

-17x = 34

**Divide both sides by -17**

x = -2

### Example 2:

**2.** a. Calculate the size of each angle of a regular pentagon

**Solution**

Regular pentagon = 5 sides

Each interior angle = \( \frac {n \: – \: 2 \: \times\: 180^o}{n} \)

= \( \frac {5 \: – \: 2 \: \times\: 180^o}{5} \)

= \( \frac {3 \: \times\: 180^o}{5} \)

\( \frac{540^o}{5} \)

= 108º

**2.** b. The table shows the scores of students in a test.

Scores | 13 | 14 | 15 | 16 | 18 |

No of Students | 3 | 5 | 8 | 7 | 12 |

(i) What is the modal score?

(ii) How many students took part in the test?

(iii) Find the median score

**Solution**

**(i)** Modal score = highest frequency = 18

**(ii) **Number of students = 3 + 5 + 8 + 7 + 12 = 35 students

**(iii)** Median score (middle score)

First list out the scores

13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18

Total frequency = 35

median = \( \frac{n \: + \: 1}{2} \)

= \( \frac{35 \: + \: 1}{2} \)

= \( \frac{36}{2} \)

= 18

The median score is the 18th frequency or score which is 16

13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 16, **16**, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18

### Example 3:

**3.** Find the largest angle in the figure below

**Solution**

Sum of angles in a triangle = 180º

∴ 2xº + 3xº + 4xº = 180º

9xº = 180º

**Divide both sides by 9**

x = 20º

From the triangle the biggest angle has to be 4x

**Substitute x = 20º** into 4x

4x = 4 × 20º

= 80º

### Example 4:

**4.** The height of an equilateral triangle is 15 cm and its perimeter is 36 cm. Find the area of the triangle.

**Solution**

Permieter = 36 cm

∴ 36 = x + x + x

36 = 3x

3x = 36

**Divide both sides by 3**

\( \frac{3x}{3} = \frac{36}{3} \)

x = 12

Area of the triangle = \( \frac{1}{2} \scriptsize \: \times \: base \: \times \: height \)

= \( \frac{1}{2} \scriptsize \: \times \: 12 \: \times \: 15 \)= 6 × 15

= 90 cm^{2}

### Example 5:

**5.** Calculate /DC/ using the similar shapes given below

**Solution**

The two triangles are similar

∴ \( \frac{|PQ|}{|QR|} = \frac{|DC|}{|BC|} \)

= \( \frac{12}{8} = \frac{|DC|}{6} \)

**Cross multiply**

8 × DC = 12 × 6

8DC = 72

**Divide both sides by **8

\( \frac{8DC}{8} = \frac{72}{8} \)

DC = 9 cm

### Example 6:

**6. **What is the average rainfall in mm^{2} from the bar chart below?

**Solution**

Jan | 40 |

Feb | 80 |

Mar | 120 |

Apr | 60 |

May | 40 |

Average Rainfall

= \( \frac{40+80+120+60+40}{5} \)

= \( \frac{340}{5} \)

= 68 mm^{2}

### Example 7:

**7.** Find the missing angles l, m and n in that order from the diagram below

**Solution**

The shape is a trapezoid.

Opposite angles of a trapezoid are supplementary. Supplementary angles of a trapezoid is defined as the **sum of two opposite angles is 180**

∴ n + 125 = 180

n = 180 – 125 = 55º

Also,

(25 + m) + (95 + l) = 180

But m = l (**alternate angles**)

so let’s say l and m = x

∴ (25 + x) + (95 + x) = 180

25 + x + 95 + x = 180

25 + 95 + x + x = 180

120 + 2x = 180

2x = 180 – 120

2x = 60

x = \( \frac{60}{2} \)

x = 30

∴ l and m = 30º

**Answer:** l, m and n = 30º, 30º and 55º

### Example 8:

**8.** Simplify \( \frac{x^2 – 6x + 9}{x^2 – 9} \)

**Solution**

**First step: Factorise the numerator**

= \( \scriptsize x(x \: – \: 3) \: – \: 3(x \: – \: 3) \)

= (x – 3)(x – 3)

**Second step: Factorise the denominator**

x^{2} – 9

= x^{2} – 3^{2}

= (x – 3)(x + 3) (Difference of two squares)

∴ \( \frac{x^2 – 6x + 9}{x^2 – 9} = \frac{(x – 3)(x – 3)}{(x – 3)(x + 3)}\)= \( \frac{(x – 3)}{(x + 3)}\)

### Example 9:

**9. **In the diagram below, ABCD is a kite and ADEF is a rhombus. If BC = 4 cm and CD = 7 cm, find the perimeter of ABCDEF (Not drawn to scale)

**Solution**

We can redraw the diagram as shown below.

A kite has 2 pairs of adjacent equal sides

A rhombus has all sides equal

A perimeter of a shape is the length of its boundary or the total length around the shape

= 7 + 7 + 7 + 4 + 4 + 7

= 36 cm

**Note:** Perimeter is the total length around the shape so we ignore the side in the middle (7 cm)

### Example 10:

**10.** An exercise book and two pens cost ₦90.00. If the difference between the cost of an exercise book and a pen is ₦45.00, determine the cost of a pen.

**Solution**

Let the cost of an exercise book = x

the cost of a pen = y

From the question

An exercise book and two pens cost ₦90.00

x + 2y = 90 ……………..(1)

The difference between the cost of an exercise book and a pen is ₦45.00

x – y = 45 …………………(2)

Subtract equation 2 from 1

x + 2y = 90

x – y = 45

2y – (-y) = 90 – 45

2y + y = 45

3y = 45

y = \( \frac{45}{3} \)

y = ₦15

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