Mathematics grade 8 Revision test Question 1 1.1 Determine the highest common factor (HCF) of 54 and 90. 1.2 Determine the lowest common multiple (LCM) of 14; 28 and 30. (3) [6] Question 2 2.1 Write the following ratios ratio in simplest form. 2.1.1 36:120 2.1.2 $$ 5 \mathrm{~m}: 30 \mathrm{~cm} $$ (3) 2.2 A diesel vehicle uses 12 litres diesel every 200 km. How many Kilometres can this vehicle travel with 18 litres of diesel? 2.3 Increase 45 in the ratio $$ 5: 9 $$. 2.4 Decrease 16 in the ratio $$ 4: 3 $$. (2) 2.5 The minimum wage for a worker in South Africa is R180 for 9 hours work. 2.5.1 What is the minimum wage per hour? (2) 2.5.2 How much will a worker earn if he/she works for 12 hours? 2.5.3 How long will a worker have to work for to earn R300? (2) 2.6 Tim and Brett went out for lunch. The bill was R470. They decided to share the bill in the ratio $$ 2: 3 $$ since Brett's meal was cheaper than Tim's. How much did each pay? (3) [18] Question 3 3.1 The exchange rate on a specific day is $$ \$ 1=R 18,74 $$. You would like to exchange R1 200 for dollars. How many dollars will you get? 3.2 You earn 8,2\% simple interest per annum on an investment of R5 000 for 10 years. Calculate the interest earned after 10 years and hence the value of the investment at the end of the 10 years. [6] TOTAL MARKS: 30

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### Question 1

#### 1.1 Determine the highest common factor (HCF) of 54 and 90.

To find the HCF, we can use the prime factorization method.

- **Prime factorization of 54:**
  - $$ 54 = 2 \times 3^3 $$

- **Prime factorization of 90:**
  - $$ 90 = 2 \times 3^2 \times 5 $$

Now, we take the lowest power of each common prime factor:

- For $$ 2 $$: $$ \min(1, 1) = 1 $$
- For $$ 3 $$: $$ \min(3, 2) = 2 $$

Thus, the HCF is:
$$
HCF = 2^1 \times 3^2 = 2 \times 9 = 18
$$

#### 1.2 Determine the lowest common multiple (LCM) of 14, 28, and 30.

To find the LCM, we can also use the prime factorization method.

- **Prime factorization of 14:**
  - $$ 14 = 2 \times 7 $$

- **Prime factorization of 28:**
  - $$ 28 = 2^2 \times 7 $$

- **Prime factorization of 30:**
  - $$ 30 = 2 \times 3 \times 5 $$

Now, we take the highest power of each prime factor:

- For $$ 2 $$: $$ \max(1, 2, 1) = 2 $$
- For $$ 3 $$: $$ \max(0, 0, 1) = 1 $$
- For $$ 5 $$: $$ \max(0, 0, 1) = 1 $$
- For $$ 7 $$: $$ \max(1, 1, 0) = 1 $$

Thus, the LCM is:
$$
LCM = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7
$$

Calculating this step by step:
$$
4 \times 3 = 12
$$
$$
12 \times 5 = 60
$$
$$
60 \times 7 = 420
$$

So, the LCM is $$ 420 $$.

### Question 2

#### 2.1 Write the following ratios in simplest form.

##### 2.1.1 $$ 36:120 $$

To simplify the ratio, we find the HCF of 36 and 120.

- HCF of 36 and 120 is $$ 12 $$.

Now, divide both numbers by their HCF:
$$
\frac{36}{12} : \frac{120}{12} = 3:10
$$

##### 2.1.2 $$ 5 \mathrm{~m}: 30 \mathrm{~cm} $$

First, convert $$ 30 \mathrm{~cm} $$ to meters:
$$
30 \mathrm{~cm} = 0.3 \mathrm{~m}
$$

Now, simplify the ratio:
$$
5:0.3
$$

To eliminate the decimal, multiply both sides by $$ 10 $$:
$$
50:3
$$

#### 2.2 A diesel vehicle uses 12 litres diesel every 200 km. How many Kilometres can this vehicle travel with 18 litres of diesel?

We can set up a proportion:
$$
\frac{12 \text{ litres}}{200 \text{ km}} = \frac{18 \text{ litres}}{x \text{ km}}
$$

Cross-multiplying gives:
$$
12x = 200 \times 18
$$
$$
12x = 3600
$$
$$
x = \frac{3600}{12} = 300 \text{ km}
$$

#### 2.3 Increase 45 in the ratio $$ 5: 9 $$.

To increase $$ 45 $$ in the ratio $$ 5:9 $$, we first find the total parts:
$$
5 + 9 = 14 \text{ parts}
$$

Now, find the value of one part:
$$
\text{Value of one part} = \frac{45}{5} = 9
$$

Now, calculate the increased amount:
$$
\text{Increased amount} = 9 \times 9 = 81
$$

#### 2.4 Decrease 16 in the ratio $$ 4: 3 $$.

To decrease $$ 16 $$ in the ratio $$ 4:3 $$, we find the total parts:
$$
4 + 3 = 7 \text{ parts}
$$

Now, find the value of one part:
$$
\text{Value of one part} = \frac{16}{4} = 4
$$

Now, calculate the decreased amount:
$$
\text{Decreased amount} = 4 \times 3 = 12
$$

#### 2.5 The minimum wage for a worker in South Africa is R180 for 9 hours work.

##### 2.5.1 What is the minimum wage per hour?

$$
\text{Minimum wage per hour} = \frac{180}{9} = R20
$$

##### 2.5.2 How much will a worker earn if he/she works for 12 hours?

$$
\text{Earnings for 12 hours} = 20 \times 12 = R240
$$

##### 2.5.3 How long will a worker have to work for to earn R300?

Let $$ x $$ be the number of hours worked:
$$
20x = 300
$$
$$
x = \frac{300}{20} = 15 \text{ hours}
$$

#### 2.6 Tim and Brett went out for lunch. The bill was R470. They decided to share the bill in the ratio $$ 2: 3 $$.

Total parts = $$ 2 + 3 = 5 $$.

Now, find the value of one part:
$$
\text{Value of one part} = \frac{470}{5} = 94
$$

Now, calculate how much each pays:
- Tim's share: $$ 2 \times 94 = R188 $$
- Brett's share: $$ 3 \times 94 = R282 $$

### Question 3

#### 3.1 The exchange rate on a specific day is $$ \$ 1=R 18,74 $$. You would like to exchange R1 200 for dollars. How many dollars will you get?

$$
\text{Dollars} = \frac{1200}{18.74} \approx 64.1
$$

#### 3.2 You earn 8.2% simple interest per annum on an
### Question 1 1. **Highest Common Factor (HCF) of 54 and 90:** - HCF is 18. 2. **Lowest Common Multiple (LCM) of 14, 28, and 30:** - LCM is 420. ### Question 2 1. **Simplify the ratios:** - $$ 36:120 = 3:10 $$ - $$ 5 \mathrm{~m}: 30 \mathrm{~cm} = 50:3 $$ 2. **Fuel Consumption:** - 18 litres of diesel allows the vehicle to travel 300 km. 3. **Increase and Decrease:** - Increase 45 in the ratio $$ 5:9 $$ results in 81. - Decrease 16 in the ratio $$ 4:3 $$ results in 12. 4. **Minimum Wage:** - Minimum wage per hour is R20. - Earnings for 12 hours are R240. - To earn R300, the worker needs to work 15 hours. 5. **Sharing the Bill:** - Tim pays R188. - Brett pays R282. ### Question 3 1. **Exchange R1 200 for dollars:** - Approximately \$64.10. 2. **Simple Interest Calculation:** - Interest earned after 10 years: R5 000 × 8.2% × 10 = R4,100. - Total investment value: R5 000 + R4,100 = R9,100.

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