A small business has just invested in a new piece of equipment. The business's accountant estimates that the value of...

Phase 1: Owning the Dataset

Visualization Selection

This is a time-based depreciation problem, so we'll use a timeline showing the equipment value over 8 years.

Timeline Visualization:

Year:    0-------3-------6-------8
Value:   V₀      X       Y       0
         |<--3d-->|<--3d-->|<--2d-->|

Where:

• \(\mathrm{V_0}\) = initial value
• d = depreciation per year (constant)

Understanding the Depreciation

Since depreciation is constant:

• From year 0 to 3: value drops by 3d
• From year 3 to 6: value drops by 3d
• From year 6 to 8: value drops by 2d
• Total drop from \(\mathrm{V_0}\) to 0 = 8d

Phase 2: Understanding the Question

Key Relationships

From our timeline:

• After 3 years: \(\mathrm{V_0 - 3d = X}\)
• After 6 years: \(\mathrm{V_0 - 6d = Y}\)
• After 8 years: \(\mathrm{V_0 - 8d = 0}\)

From the third equation: \(\mathrm{V_0 = 8d}\)

Substituting back:

• \(\mathrm{X = 8d - 3d = 5d}\)
• \(\mathrm{Y = 8d - 6d = 2d}\)

Therefore: \(\mathrm{X/Y = 5d/2d = 2.5}\)

This means \(\mathrm{X = 2.5Y}\)

Phase 3: Finding the Answer

Systematic Check

We need Y such that 2.5Y is also in our answer choices.

If \(\mathrm{Y = 200 \rightarrow X = 2.5 \times 200 = 500}\)
Is 500 in our choices? Yes! ✓
Stop here - we found our answer.

Verification

With X = 500 and Y = 200:

• \(\mathrm{Y = 200 = 2d}\), so \(\mathrm{d = 100}\)
• Initial value \(\mathrm{V_0 = 8d = 800}\)
• After 3 years: \(\mathrm{800 - 300 = 500}\) ✓
• After 6 years: \(\mathrm{800 - 600 = 200}\) ✓
• After 8 years: \(\mathrm{800 - 800 = 0}\) ✓

Phase 4: Solution

Final Answer:

• Statement 1 (X): 500
• Statement 2 (Y): 200

These values satisfy the constant depreciation requirement where the equipment depreciates $100 per year from an initial value of $800.