Applied Maths - Compound Interest & Depreciation

Grade 12 Maths — Applied Maths - Compound Interest & Depreciation: 25 practice questions with instant scoring and explanations.

1. Compound Interest formula: A = P(1 + r/100)^n where:
2. CI = A − P = P[(1 + r/100)^n − 1]. For P = Rs. 10,000, r = 10%, n = 2 years, CI is:
3. For semi-annual compounding, the effective rate for r% per annum is:
4. For quarterly compounding (n quarters), rate per quarter = :
5. Monthly compounding for 1 year (12 months): A = P(1 + r/1200)^12. Effective rate ≈ :
6. Doubling time (in years) for principal at r% per annum: t ≈ 72/r (Rule of 72). For r = 8%, t ≈ :
7. Present value (PV) given future amount A after n years at r%: PV = :
8. Depreciation means:
9. Straight-line depreciation: Depreciation per year = (Original cost − Scrap value) / Useful life. For cost Rs. 1,00,000, scrap Rs. 10,000, life 10 years:
10. Book value after n years in straight-line depreciation = :
11. After 5 years, book value = Rs. 1,00,000 − (9,000 × 5) = :
12. Declining balance depreciation (compound): V = P(1 − r/100)^n where V = final value, r = depreciation rate. More realistic for:
13. For asset worth Rs. 50,000 with 20% annual depreciation for 3 years, value = :
14. Comparison of depreciation methods: Straight-line is:
15. Rate of depreciation (r%) is typically determined by:
16. Appreciation (opposite of depreciation): V = P(1 + r/100)^n applies to:
17. Property worth Rs. 5,00,000 appreciates at 8% annually. After 3 years, value ≈ :
18. Depreciation provision (reserves) is kept to:
19. Book value and market value are:
20. Salvage value (scrap value) is:
21. Accelerated depreciation (like double declining balance) means:
22. Useful life of an asset is determined by:
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