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Q.(a) A recent accounting graduate opened a new business and installed a computer system that costs ₹ 45,200. The computer system will be depreciated linearly over 3 years and will have a scrap value of ₹ 0.

(i) What is the rate of depreciation ?
(ii) Give a linear equation that describes the computer system's book value at the end of ttht^{th} year, where 0≤t≤30 \leq t \leq 3.
(iii) What will be the computer system's book value at the end of the first year and a half ?
(OR)
(b) Find the effective rate which is equivalent to normal rate of 10% p.a. compounded :
(i) semi-annually.
(ii) quarterly. [Given (1.05)2=1.1025(1.05)^2 = 1.1025, (1.025)4=1.1038(1.025)^4 = 1.1038]
CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★
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  1. Straight-line: rate =33.3%=33.3\%, v(t)=45200−452003tv(t)=45200-\tfrac{45200}{3}t, v(1.5)=₹22,600v(1.5)=₹22{,}600.
  2. re=(1+rm⋅100)m−1r_e=(1+\tfrac{r}{m\cdot100})^{m}-1 gives 10.25%10.25\% (semi-annual) and 10.38%10.38\% (quarterly).

Linear depreciation: annual depreciation =Cost−ScrapLife=\dfrac{\text{Cost}-\text{Scrap}}{\text{Life}}; rate =annual depreciationCost×100=\dfrac{\text{annual depreciation}}{\text{Cost}}\times100; book value v(t)=Cost−(annual depreciation) tv(t)=\text{Cost}-(\text{annual depreciation})\,t. Effective rate: re=(1+im)m−1r_e=\left(1+\dfrac{i}{m}\right)^{m}-1 with per-period rate i/mi/m and mm periods per year.

Part (a): Cost =₹45,200=₹45{,}200, scrap =₹0=₹0, life =3=3 years.

  1. Annual depreciation =45200−03=₹452003=\dfrac{45200-0}{3}=₹\dfrac{45200}{3}.
  2. (i) Rate of depreciation =45200/345200×100=1003=33.3%=\dfrac{45200/3}{45200}\times100=\dfrac{100}{3}=33.3\% per annum.
  3. (ii) Book value at end of year tt: v(t)=45200−452003 t,0≤t≤3v(t)=45200-\dfrac{45200}{3}\,t,\quad 0\le t\le 3. …

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