5.1 A car's engine overheated at an increasing compound rate of 26% per hour due to a leak in one of the water pipes - English General - NSC Technical Mathematics - Question 5 - 2023 - Paper 1

The initial value of the printer is R81 000. If it depreciates to ( \frac{1}{3} ) of its original value, the value is:

$\text{Value} = \frac{R81 000}{3} = R27 000$

Thus, the value of the printer will be R27 000.

Using the formula for depreciation:

$A = P(1 - r)^t$

Where:

• ( A = 27 \text{ 000} )
• ( P = 81 \text{ 000} )
• ( r = 0.2 ) (depreciation rate)

Substituting the values, we have:

$27 000 = 81 000(1 - 0.2)^t$

( 1 - 0.2 = 0.8 ):

$27 000 = 81 000(0.8)^t$

Dividing both sides by 81 000:

$\frac{27 000}{81 000} = (0.8)^t \Rightarrow 0.3333 = (0.8)^t$

Taking the logarithm of both sides:

$t = \frac{log(0.3333)}{log(0.8)} \approx 4.92 \text{ years}$

Therefore, it will take approximately 4.92 years for the printer to depreciate to ( \frac{1}{3} ) of its original value.

To find the total savings, we will calculate the future value of each investment separately:

1. First Investment:

   • Amount: R30 000

   • Time: 5 years

   • Interest: 10% for 2 years, 12% for 3 years

   $A_1 = 30 000(1 + \frac{10}{100})^2(1 + \frac{12}{100})^3$

   Calculating the future value:

   $A_1 = 30 000(1.1)^2(1.12)^3 \approx R56 611.73$

2. Second Investment:

   • Amount: R20 000
   • Time: 3 years
   • Interest: 12%

   $A_2 = 20 000(1 + \frac{12}{100})^3$

   $A_2 \approx R28 328.10$

3. Third Investment:

   • Amount: R10 000
   • Time: 2 years
   • Interest: 12%

   $A_3 = 10 000(1 + \frac{12}{100})^2$

   $A_3 \approx R12 544.00$

4. Calculate Total Savings:

   • Total = A_1 + A_2 + A_3

   $Total \approx 56 611.73 + 28 328.10 + 12 544.00 \approx R97 483.83$

The upgrade cost is R95 000. Since R97 483.83 > R95 000, the municipality will be able to save enough money for the clinic upgrade.