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Vehicle X cost $80,000, useful life 5 years, residual value $5,000, depreciated straight-line. After 3 years, part-exchange value was $20,000. What was the loss on disposal?
Option B ($15,000) is correct.
Annual depreciation = ($80,000 − $5,000) ÷ 5 = $15,000
Accumulated depreciation after 3 years = $45,000
Carrying value = $80,000 − $45,000 = $35,000
Loss on disposal = $35,000 − $20,000 = $15,000
What are the dimensions of the magnetic field $B$ in terms of $C$ (coulomb), $M$, $L$, and $T$?
From the Lorentz force: $F = qvB$
$[B] = \dfrac{[F]}{[q][v]} = \dfrac{MLT^{-2}}{C \cdot LT^{-1}} = MT^{-1}C^{-1} = M^1L^0T^{-1}C^{-1}$
Cause: The 4f electrons have a poor shielding effect (they cannot effectively shield each other or outer electrons from the increasing nuclear charge). As protons are added (Z increases from 58 to 71) with poor f-electron shielding, the effective nuclear charge experienced by outer electrons steadily increases, pulling them inward.
L and M are partners sharing profits 3:2. Data at year-end:
| Item | L ($) | M ($) |
|---|---|---|
| Capital | 200,000 | 150,000 |
| Interest on Capital (8%) | - | - |
| Drawings | 30,000 | 20,000 |
| Interest on Drawings (5%) | - | - |
| Salaries | 22,000 | 17,000 |
If L's share of residual profit was $24,000, what was the total profit for the year before appropriation?
1. Total Residual Profit = $24,000 / 0.6 = $40,000.
2. Total Interest on Capital = ($200k + $150k) $\times$ 8% = $28,000.
3. Total Salaries = $22k + $17k = $39,000.
4. Total Interest on Drawings = ($30k + $20k) $\times$ 5% = $2,500.
Profit = Residual ($40,000) + Interest on Cap ($28,000) + Salaries ($39,000) - Interest on Drawings ($2,500) = $104,500.
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