Question
A marathon is a race over a distance of 42.195 km. Two runners, Eefje and Shumay, are training to run a marathon.
The two runners train in different ways:
- Eefje runs 5 km on the first day of training and then increases the distance she runs by 2 km on each subsequent day.
- Shumay runs 5 km on the first day of training and then increases the distance she runs by 13% on each subsequent day.
Determine which runner will be the first to run the distance of a marathon on a particular day of their training, and state on which day of their training this will occur.
▶️ Answer/Explanation
Detailed Solution
Step 1: Identifying the Progressions
Eefje’s running pattern follows an Arithmetic Progression (AP):
\[ a = 5, \quad d = 2 \]
The general formula for an arithmetic sequence:
\[ S_n = a + (n – 1)d \]
Setting \( S_n = 42.195 \), we solve for \( n_E \):
\[ 42.195 = 5 + (n – 1) \cdot 2 \]
Rearranging:
\[ n – 1 = \frac{42.195 – 5}{2} = 18.5975 \]
Approximating:
\[ n_E = 19.6 \quad \text{(accept 20)} \]
Shumay’s running pattern follows a Geometric Progression (GP):
\[ a = 5, \quad r = 1.13 \]
The general formula for a geometric sequence:
\[ S_n = a \cdot r^{n-1} \]
Setting \( S_n = 42.195 \), we solve for \( n_S \):
\[ 42.195 = 5(1.13)^{n-1} \]
Dividing both sides by 5:
\[ \frac{42.195}{5} = 1.13^{n-1} \]
Taking the natural logarithm:
\[ \ln\left(\frac{42.195}{5}\right) = (n – 1) \ln(1.13) \]
Solving for \( n_S \):
\[ n – 1 = \frac{\ln(8.439)}{\ln(1.13)} \]
Approximating:
\[ n – 1 = 17.4513 \]
\[ n_S = 18.5 \quad \text{(accept 19)} \]
Step 3: Conclusion
- Eefje reaches 42.195 km on day 20.
- Shumay reaches 42.195 km on day 19.
Final Answer: Shumay reaches the marathon distance first, on day 19.
…………………………..Markscheme…………………………..
- Correct recognition of AP and GP.
- Correct equations and calculations.
- Final correct answer: Shumay on day 19.
Question
On 1 January 2025, the Faber Car Company will release a new car to global markets. The company expects to sell 40 cars in January 2025. The number of cars sold each month can be modelled by a geometric sequence where \( r = 1.1 \).
(a) Use this model to find the number of cars that will be sold in December 2025.
(b) Use this model to find the total number of cars that will be sold in the year
(i) 2025,
(ii) 2026.
▶️Answer/Explanation
Detailed solution
(a) Finding the Number of Cars Sold in December 2025
The number of cars sold in each month follows a geometric sequence, where:
– First term \( u_1 = 40 \)
– Common ratio \( r = 1.1 \)
The general formula for the \( n \)th term of a geometric sequence is:
\[ u_n = u_1 \cdot r^{n-1} \]
For December 2025 (\( n = 12 \)):
\[ u_{12} = 40 \times 1.1^{12-1} \]
\[ u_{12} = 40 \times 1.1^{11} \]
\[ u_{12} \approx 114.12 \]
Thus, the number of cars sold in December 2025 is approximately **114**.
(b) Finding the Total Number of Cars Sold
(i) Total Cars Sold in 2025
The total number of cars sold in 2025 is the sum of the first 12 terms of the geometric sequence. The sum of the first \( n \) terms of a geometric sequence is given by:
\[ S_n = \frac{u_1 (r^n – 1)}{r – 1} \]
For 2025 (\( n = 12 \)):
\[ S_{12} = \frac{40 (1.1^{12} – 1)}{1.1 – 1} \]
\[ S_{12} \approx \frac{40 (3.138 – 1)}{0.1} \]
\[ S_{12} \approx \frac{40 \times 2.138}{0.1} \]
\[ S_{12} \approx 855.37 \]
So, the total number of cars sold in 2025 is approximately **855**.
(ii) Total Cars Sold in 2026
The total number of cars sold in 2026 is the sum of the next 12 terms (terms 13 to 24). The sum of the first 24 terms is:
\[ S_{24} = \frac{40 (1.1^{24} – 1)}{1.1 – 1} \]
\[ S_{24} \approx 3539.89 \]
To find the number of cars sold in 2026, we subtract the total from 2025:
\[ S_{2026} = S_{24} – S_{12} \]
\[ S_{2026} \approx 3539.89 – 855.37 \]
\[ S_{2026} \approx 2684.52 \]
So, the total number of cars sold in 2026 is approximately **2685**.
……………………………Markscheme……………………………….
(a)
$u_{12}=40\times1.1^{12-1}$
114 (114.124…)
(b) (i)
$S_{12}=\frac{40(1.1^{12}-1)}{1.1-1}$
855 (855.371…)
(b) (ii)
$S_{24}=3539.89…$
$3539.89… – 855.371…$ OR $\sum_{r=13}^{24} 40 \times 1.1^{r-1}$ OR $125.537… + 358.172…$
$\left(\frac{125.537…(1.1^{12}-1)}{1.1-1}\right)$ OR $125.537… + … + 358.172…$
Accept a calculation using $u_{13} = 125$ or $126$.
2680 (2684.52…, 2685)