QUESTION 1

Question

1.1 Avary invested R85 000 at 11% p.a. simple interest. Calculate the total
amount of money available after 3 years. (3)
1.2 Ali wanted to calculate the number of years she needed to invest
𝑅45 000 to accumulate 𝑅90 000. She has been offered a simple
interest rate of 12% p.a. How many years will it take for the
investment to grow to 𝑅90 000. (3)
1.3 Noluthando bought a new car 5 years ago, for triple what it's worth
today. At what yearly simple interest rate did her car depreciate? (4)
[10]
QUESTION 2
2.1 Right-angled triangles 𝐾𝐿𝑀 and 𝐿𝑀𝑁 are shown in the diagram below, with
𝐿𝑁 = 11𝑐𝑚, 𝐿𝑀 = 37𝑐𝑚 and 𝐾𝑀𝐿 = 42°.
Figure 2: Diagram for Question 2.1.
Calculate the following by rounding off your answer to 1 decimal place:
2.1.1 KL (3)
2.1.2 𝑁𝐿𝑀 (4)
4

Answer 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°