Respuesta :

lpa19ezch

Answer:

To determine the cost of the lawnmower 10 years ago, we can use the formula for straight-line depreciation:

Depreciation per year= Initial cost−Trade-in value   ÷ ​Number of years

In this case, we are given the trade-in value (R236), the number of years (10 years), and the depreciation rate (9.2% per year). We need to find the initial cost (cost of the lawnmower 10 years ago).

Let's denote:

  • Initial cost as C
  • Trade-in value as T
  • Number of years as n
  • Depreciation rate per year as r

The formula becomes:

Depreciation per year= C−T÷ n  × r

We can rearrange this formula to solve for the initial cost (C):

C = T + (C−T÷ n  × r )×n

Now, we can substitute the given values into the formula and solve for C:

C = 236 + ( C−236 ÷ 10 × 0.092) ×10

Let's solve for C:

         C = 236 + ( C−236 ÷ 10 × 0.092) ×10

         C = 236 + 0.092 × ( C − 236)

         C = 236 + 0.092 C − 21.712

          0.908C = 214.288

          C ≈ 214.288 ÷ 0.908

          C ≈ 235.84

So, the cost of the lawnmower 10 years ago was approximately R235.84.

Step-by-step explanation:

msm555

Answer:

R236.92

Step-by-step explanation:

We can solve this problem by employing the formula for straight-line depreciation:

[tex] \boxed{\boxed{\textsf{Depreciation Rate }= \dfrac{\textsf{(Original Cost - Trade-in Value)}}{\textsf{ Useful Life}}}}[/tex]

Given:

  • Depreciation Rate = 9.2% p.a. (0.092 per year)
  • Trade-in Value = R236
  • Useful Life = 10 years

Now

Substitute the given value in the formula:

[tex] 0.092 = \dfrac{ \textsf{ Original cost} - 236 }{10}[/tex]

Multiply both sides by 10.

[tex] 0.092 \cdot 10 = \dfrac{ \textsf{ Original cost} - 236 }{\cancel{10}}\cdot \cancel{10} [/tex]

[tex] 0.92 = \textsf{ Original cost} - 236 [/tex]

Add 236 on both sides:

[tex] 0.92 + 236= \textsf{ Original cost} - 236+236 [/tex]

[tex]\textsf{ Original cost} = 236.92 [/tex]

Therefore, the cost of the lawnmower 10 years ago was approximately R236.92.

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