QUESTION 1

### QUESTION 1

#### 1.1
To calculate the total amount of money available after 3 years using simple interest, we use the formula:

\[
\text{Total Amount} = P + I
\]
where:
- \( P \) is the principal amount (initial investment),
- \( I \) is the interest earned.

The interest earned (\( I \)) can be calculated with the formula:

\[
I = P \times r \times t
\]
where:
- \( r \) is the rate of interest per year (as a decimal),
- \( t \) is the time in years.

Given:
- \( P = R85,000 \)
- \( r = 11\% = 0.11 \)
- \( t = 3 \) years

Calculating the interest:

\[
I = 85,000 \times 0.11 \times 3 = 85,000 \times 0.33 = R28,050
\]

Now, calculating the total amount:

\[
\text{Total Amount} = 85,000 + 28,050 = R113,050
\]

#### 1.2
To find the number of years it will take for Ali's investment to grow from \( R45,000 \) to \( R90,000 \) with a simple interest rate of \( 12\% \) per annum, we can rearrange the simple interest formula:

\[
A = P + I
\]

Where:
- \( A \) is the final amount,
- \( P \) is the principal (initial investment),
- \( I = P \times r \times t \).

We have:

\[
90,000 = 45,000 + (45,000 \times 0.12 \times t)
\]

Subtract the principal from both sides:

\[
90,000 - 45,000 = 45,000 \times 0.12 \times t
\]
\[
45,000 = 5,400 \times t
\]

Now, solving for \( t \):

\[
t = \frac{45,000}{5,400} \approx 8.33
\]

So it will take approximately **8.33 years** for the investment to grow to \( R90,000 \).

#### 1.3
If Noluthando bought a car 5 years ago for triple its current value, we can denote the current value of the car as \( x \). Therefore, the initial value is \( 3x \).

After 5 years, if the car depreciated at a simple interest rate of \( r\% \), we have:

\[
\text{Current Value} = \text{Initial Value} - \text{Depreciation}
\]

The depreciation can be calculated using:

\[
\text{Depreciation} = 3x \times \frac{r}{100} \times 5
\]

Substituting back:

\[
x = 3x - 3x \times \frac{r}{100} \times 5
\]

Rearranging gives us:

\[
x = 3x \left(1 - \frac{5r}{100}\right)
\]

Dividing through by \( x \) (where \( x ≠ 0 \)) gives:

\[
1 = 3 \left(1 - \frac{5r}{100}\right)
\]
\[
1 = 3 - \frac{15r}{100}
\]
\[
\frac{15r}{100} = 2
\]
\[
15r = 200
\]
\[
r = \frac{200}{15} \approx 13.33
\]

Thus, the yearly simple interest rate at which the car depreciated is approximately **13.3%**.

### **QUESTION 2**

For the calculations involving the triangles, we will use trigonometric identities.

#### 2.1.1 KL
In triangle \( KLM \) with angle \( \angle KML = 42° \), we can find \(KL\) using the sine function:

\[
\sin(42°) = \frac{KL}{LM}
\]

We know \( LM = 37 \, \text{cm} \):

\[
KL = LM \cdot \sin(42°)
\]
\[
KL = 37 \cdot \sin(42°) \approx 37 \cdot 0.6691 \approx 24.8 \, \text{cm}
\]

#### 2.1.2 \( NLM \)
To find angle \( NLM \), we can recognize it as the complementary angle to \( KML \) since in right triangles, the angles sum to \( 90° \):

\[
NLM + KML = 90°
\]
\[
NLM + 42° = 90°
\]
\[
NLM = 90° - 42° = 48°
\]

Thus, after rounding off:
1. **KL** = **24.8 cm**
2. **NLM** = **48.0°**

### **Summary of Answers:**

- **1.1** Total Amount after 3 years: R113,050
- **1.2** Years to grow to R90,000: 8.33 years
- **1.3** Yearly simple interest rate: 13.3%
- **2.1.1** Length KL: 24.8 cm
- **2.1.2** Angle NLM: 48.0°