Question

baf3m : introduction à la comptabilité financière enseignant : jean-roger cavé ty bikanh activité 4.4 exercice1 : une compagnie d’assurance achète une balayeuse au prix de $4000. la vie utile de cette machine est estimée à 9 ans et sa valeur résiduelle est estimée à 300$. construis un tableau d’amortissement en utilisant la méthode de l’amortissement linéaire. tableau d’amortissement table with columns période (n), amortissement linéaire, amortissement cumulé, solde de l’actif non amorti and rows 0 - 9 exercice2. une compagnie de location achète une voiture aux prix de 30 000$ si la voiture perd 15% de sa valeur par année et que la méthode de l’amortissement est linéaire. quelle est la valeur de l’actif non amorti après : i. 5 ans ii. après 7ans

Explanation:

Exercice 1:

Step 1: Calculate annual linear depreciation

The formula for linear (straight - line) depreciation is $D=\frac{C - R}{n}$, where $C$ is the cost of the asset, $R$ is the residual value, and $n$ is the useful life.
Given $C = 4000$, $R=300$, and $n = 9$.
So, $D=\frac{4000 - 300}{9}=\frac{3700}{9}\approx411.11$ (rounded to two decimal places)

Step 2: Determine the initial non - amortized balance (Period 0)

At period 0, the non - amortized balance (Solde de l'actif non amorti) is equal to the cost of the asset, so it is $4000$. The cumulative amortization (Amortissement cumulé) and the annual amortization (Amortissement linéaire) are $0$.

Step 3: Calculate values for each period

• For period $n$:
• Amortissement linéaire: Constant at $\frac{3700}{9}\approx411.11$
• Amortissement cumulé: For period $n$, it is the sum of the annual amortizations up to period $n$. So, for period $1$, it is $411.11$; for period $2$, it is $411.11\times2$, and so on.
• Solde de l'actif non amorti: For period $n$, it is $C- \text{Amortissement cumulé}$. So, for period $1$, it is $4000 - 411.11 = 3588.89$; for period $2$, it is $4000-411.11\times2 = 3177.78$, and so on until period 9, where the solde de l'actif non amorti should be equal to the residual value ($300$).

The completed table is as follows:

Période (n)	Amortissement linéaire	Amortissement cumulé	Solde de l'actif non amorti
1	$\approx411.11$	$\approx411.11$	$4000 - 411.11=3588.89$
2	$\approx411.11$	$\approx822.22$	$4000 - 822.22 = 3177.78$
3	$\approx411.11$	$\approx1233.33$	$4000 - 1233.33 = 2766.67$
4	$\approx411.11$	$\approx1644.44$	$4000 - 1644.44 = 2355.56$
5	$\approx411.11$	$\approx2055.55$	$4000 - 2055.55 = 1944.45$
6	$\approx411.11$	$\approx2466.66$	$4000 - 2466.66 = 1533.34$
7	$\approx411.11$	$\approx2877.77$	$4000 - 2877.77 = 1122.23$
8	$\approx411.11$	$\approx3288.88$	$4000 - 3288.88 = 711.12$
9	$\approx411.11$	$\approx3700$	$300$

Exercice 2:

i. After 5 years

Step 1: Calculate annual linear depreciation

The formula for linear depreciation is $D=\frac{C - R}{n}$. But in this case, since the depreciation is linear and the asset loses 15% of its value per year (but we are using linear depreciation, so we first find the annual depreciation amount). Wait, actually, if the depreciation is linear and the asset has an initial value of $C = 30000$ and we assume that at the end of its useful life (we can find the useful life or just calculate the annual depreciation as follows). Wait, the problem says "la méthode de l'amortissement est linéaire". If the asset loses 15% of its value per year, but for linear depreciation, the annual depreciation is constant. Wait, maybe there is a mis - understanding. If we consider that the linear depreciation is such that the annual depreciation is 15% of the initial value? Wait, no, linear depreciation is based on the formula $D=\frac{C - R}{n}$. But the problem says "la voiture perd 15% de sa valeur par année et que la méthode de l'amortissement est linéaire". Maybe it is a straight - line depreciation where the annual depreciation is 15% of the initial cost? Wait, if $C = 30000$ and the annual depreciation rate is 15% (linear, so constant depreciation per year), then the annual depreciation $D=0.15\times30000 = 4500$.

Step 2: Calculate the cumulative amortization after 5 years

The cumulative amortization after 5 years is $D\times5=4500\times5 = 22500$

Step 3: Calculate the non - amortized balance after 5 years

The non - amortized balance (Solde de l'actif non amorti) is $C-\text{Amortissement cumulé}=30000 - 22500=7500$

ii. After 7 years

Step 1: Calculate cumulative amortization after 7 years

Using the annual depreciation $D = 4500$ (from above), the cumulative amortization after 7 years is $D\times7=4500\times7 = 31500$

Step 2: But wait, the cost of the asset is $30000$, so we can't depreciate more than the cost minus the residual value (assuming residual value is 0 here). Wait, maybe the residual value is 0. So, the non - amortized balance after 7 years is $30000-31500$. But this is negative, which means that after 6.67 years ($\frac{30000}{4500}=6.67$), the asset is fully depreciated. So, after 7 years, the non - amortized balance is $0$.

Answer:

s:

Exercice 1 Table:

Période (n)	Amortissement linéaire	Amortissement cumulé	Solde de l'actif non amorti
1	$\approx411.11$	$\approx411.11$	3588.89
2	$\approx411.11$	$\approx822.22$	3177.78
3	$\approx411.11$	$\approx1233.33$	2766.67
4	$\approx411.11$	$\approx1644.44$	2355.56
5	$\approx411.11$	$\approx2055.55$	1944.45
6	$\approx411.11$	$\approx2466.66$	1533.34
7	$\approx411.11$	$\approx2877.77$	1122.23
8	$\approx411.11$	$\approx3288.88$	711.12
9	$\approx411.11$	$\approx3700$	300

Exercice 2:

i. After 5 years, the value of the non - amortized asset is $\boldsymbol{7500}$ (assuming residual value is 0 and annual depreciation is 15% of 30000)
ii. After 7 years, the value of the non - amortized asset is $\boldsymbol{0}$ (since the asset is fully depreciated after approximately 6.67 years)