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Q.(a) A person invested ₹ 5,000 in a fund for 5 years. The value of the investment was ₹ 4,800 at the end of the second year, ₹ 6,000 at the end of the third year, ₹ 6,700 at the end of the fourth year and on maturity, the final investment sold at ₹ 8,000. Find the CAGR. [Use (1⋅6)15=1⋅098(1\cdot6)^{\frac{1}{5}} = 1\cdot098]

(OR)
(b) The annual depreciation of an asset is ₹ 50,000 and its scrap value after useful life of 10 years is ₹ 60,000. Find the original cost of the asset, using linear depreciation method.
CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★
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  1. CAGR =(80005000)1/5−1=(1.6)1/5−1=0.098=9.8%=\left(\tfrac{8000}{5000}\right)^{1/5}-1=(1.6)^{1/5}-1=0.098=9.8\%.
  2. From D=C−SnD=\tfrac{C-S}{n}, 50000=C−6000010⇒C=50000=\tfrac{C-60000}{10}\Rightarrow C= ₹ 5,60,000.

CAGR =(VfV0)1/n−1=\left(\dfrac{V_{f}}{V_{0}}\right)^{1/n}-1, where V0V_0 = beginning value, VfV_f = ending value, nn = number of years.

Linear depreciation: D=C−SnD=\dfrac{C-S}{n}, where CC = original cost, SS = scrap value, nn = useful life.

(a) CAGR of the investment

  1. Only the initial and maturity values matter: V0=5000V_0=5000, Vf=8000V_f=8000, n=5n=5 years. (The intermediate year-end values ₹4,800, ₹6,000, ₹6,700 are not used in CAGR.)
  2. CAGR=(80005000)1/5−1=(1.6)1/5−1\text{CAGR}=\left(\dfrac{8000}{5000}\right)^{1/5}-1=(1.6)^{1/5}-1.
  3. Using the given (1.6)1/5=1.098(1.6)^{1/5}=1.098: CAGR=1.098−1=0.098\text{CAGR}=1.098-1=0.098. …

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