Solutions — Depreciation and Appreciation

1. Straight-line depreciation. Fluency

   • (a) P=$8000, r=10%, n=3 (SL): V = 8000(1−0.10×3) = 8000×0.70 = $5600.
   • (b) P=$15000, D=$1500, n=4: V = 15000−1500×4 = 15000−6000 = $9000.
   • (c) P=$6000, r=8%, n=5: V = 6000(1−0.08×5) = 6000×0.60 = $3600.
   • (d) Find n when V=$12000: 20000−2000n = 12000 ⇒ 2000n = 8000 ⇒ n = 4 years.

2. Reducing balance depreciation. Fluency

   • (a) P=$10000, r=15%, n=5: V = 10000×(0.85)⁵ = 10000×0.44371 ≈ $4437.
   • (b) P=$5000, r=20%, n=3: V = 5000×(0.80)³ = 5000×0.512 = $2560.
   • (c) P=$12000, r=10%, n=6: V = 12000×(0.90)⁶ = 12000×0.53144 ≈ $6377.
   • (d) P=$8000, r=25%, n=4: V = 8000×(0.75)⁴ = 8000×0.31641 ≈ $2531.

3. Appreciation. Fluency

   • (a) P=$300000, r=5%, n=3: V = 300000×(1.05)³ = 300000×1.15763 ≈ $347288.
   • (b) P=$50000, r=8%, n=10: V = 50000×(1.08)¹⁰ = 50000×2.15892 ≈ $107946.
   • (c) P=$1000, r=3%, n=5: V = 1000×(1.03)⁵ = 1000×1.15927 ≈ $1159.
   • (d) P=$20000, r=4%, n=20: V = 20000×(1.04)²⁰ = 20000×2.19112 ≈ $43822.

4. Find the unknown. Fluency

   • (a) Find n (SL): V = P(1−rn) ⇒ 6000 = 10000(1−0.1n) ⇒ 0.6 = 1−0.1n ⇒ 0.1n = 0.4 ⇒ n = 4 years.
   • (b) Find D (SL): 4000 = 10000−5D ⇒ 5D = 6000 ⇒ D = $1200 per year.
   • (c) First year below $5000 (RB, 10%): n=5: V=10000×0.9⁵=10000×0.59049=$5905 — above. n=6: V=10000×0.9⁶=10000×0.53144=$5314 — above. n=7: V=10000×0.9⁷=10000×0.47830=$4783 — below $5000. First below $5000 in year 7.
   • (d) Find r (appreciation): 133100 = 100000×(1+r)³ ⇒ (1+r)³ = 1.331 ⇒ 1+r = 1.1 (since 1.1³=1.331) ⇒ r = 10% p.a.

5. Read from the graph: straight-line vs reducing balance. Understanding

   • (a) Value after 5 years: SL (n=5): V = 10000(1−0.1×5) = $5000. RB (n=5): V = 10000×0.9⁵ ≈ $5905. The orange (RB) curve sits above the blue (SL) line.
   • (b) From which year does RB retain more?: At n=1: SL=$9000, RB=$9000 — equal. At n=2: SL=$8000, RB=$8100 — RB is higher. From year 2 onwards, reducing balance always retains more value.
   • (c) When does SL reach $0?: V = 10000(1−0.1n) = 0 ⇒ 1−0.1n = 0 ⇒ n = 10. After 10 years.
   • (d) RB at n=15: V = 10000×(0.9)¹⁵ = 10000×0.20589 ≈ $2059. Even after 15 years the asset still has value; it never reaches zero.

6. Useful life of a machine. Understanding

   • (a) Formula: V = 50000(1 − 0.125n) = 50000 − 6250n.
   • (b) When V = $25000: 50000(1−0.125n) = 25000 ⇒ 1−0.125n = 0.5 ⇒ 0.125n = 0.5 ⇒ n = 4 years.
   • (c) When V = $0: 1−0.125n = 0 ⇒ n = 1/0.125 = 8 years.
   • (d) Reducing balance at 12.5% after 8 years: V = 50000×(0.875)⁸ = 50000×0.27285 ≈ $13643. The machine still has significant residual value under RB depreciation at the same time the SL asset has reached $0.

7. Comparing two depreciation methods. Understanding

   • (a) After 2 years: Car A (SL): V = 25000(1−0.15×2) = 25000×0.70 = $17500. Car B (RB): V = 25000×(0.85)² = 25000×0.7225 = $18063.
   • (b) After 6 years: Car A (SL): V = 25000(1−0.15×6) = 25000×0.10 = $2500. Car B (RB): V = 25000×(0.85)⁶ = 25000×0.37715 ≈ $9429.
   • (c) Which has higher resale value?: Car B ($9429 > $2500). Reducing balance loses a larger percentage early on, but slows down — so after 6 years its value is much higher than straight-line, which has almost completely depreciated the car.
   • (d) When does SL reach $0?: 1−0.15n = 0 ⇒ n = 1/0.15 = 6.67 years. The car is fully written off after 6 years and 8 months (or in year 7, the value would be negative — the model breaks down below $0).

8. Property appreciation. Understanding

   • (a) Formula: V = 450000 × (1.045)ⁿ
   • (b) After 5 years: V = 450000×(1.045)⁵ = 450000×1.24618 ≈ $560778.
   • (c) After 10 years: V = 450000×(1.045)¹⁰ = 450000×1.55297 ≈ $698836.
   • (d) Growth over 10 years: 698836 − 450000 = $248836. The house gained nearly a quarter of a million dollars in value.

9. Car depreciation. Problem Solving

   • (a) After 3 years: V = 32000×(0.82)³ = 32000×0.55136 ≈ $17644. Formula: V = 32000×(0.82)ⁿ.
   • (b) First year below $10000: n=5: V=32000×(0.82)⁵=32000×0.37008≈$11842 — above $10000. n=6: V=32000×(0.82)⁶=32000×0.30347≈$9711 — below $10000. Year 6.
   • (c) Value between $20000–$21000: n=2: V=32000×(0.82)²=32000×0.6724≈$21517 — above range. n=3: V=32000×(0.82)³≈$17644 — below range. The value passes through $20000–$21000 during year 3 (between n=2 and n=3).
   • (d) SL at 15%, when V < $20000: V = 32000(1−0.15n) < 20000 ⇒ 1−0.15n < 0.625 ⇒ 0.15n > 0.375 ⇒ n > 2.5. First complete year: n = 3 (V=32000×0.55=$17600 < $20000).

10. Investment property — variable appreciation. Problem Solving

    • (a) After first 5 years at 5%: V = 500000×(1.05)⁵ = 500000×1.27628 ≈ $638141.
    • (b) Next 5 years at 3%: V = 638141×(1.03)⁵ = 638141×1.15927 ≈ $739873.
    • (c) Equivalent single rate: 500000(1+r)¹⁰ = 739873 ⇒ (1+r)¹⁰ = 1.47975. Try r=0.04: 1.04¹⁰=1.48024 ≈ 1.480 ✓. r ≈ 4% p.a.
    • (d) Growth over 10 years: 739873 − 500000 = $239873. The property grew by about $240000 (nearly 48% total) over the decade.