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Mathematics

The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :

(i) at the end of 2009.

(ii) at the beginning of 2007.

Compound Interest

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Answer

(i) For depreciation :

Value after n years = Present value ×(1r100)n\times \Big(1 - \dfrac{r}{100}\Big)^n

The cost of machine depreciates at the beginning of the next year, or we can say that at end of each year.

So, Value at end of 2009 = Value at beginning of 2010.

Value at beginning of 2010 =44000×(112100)2=44000×(88100)2=44000×(2225)2=44000×484625=34073.60\text{Value at beginning of 2010 }= 44000 \times \Big(1 - \dfrac{12}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{88}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{22}{25}\Big)^2 \\[1em] = 44000 \times \dfrac{484}{625} \\[1em] = ₹ 34073.60

Hence, value at the end of 2009 = ₹ 34073.60

(ii) Let value at beginning of 2007 be ₹ x.

After one year it becomes ₹ 44000.

For depreciation :

Value after n years = Present value ×(1r100)n\times \Big(1 - \dfrac{r}{100}\Big)^n

Substituting values we get :

44000=x×(112100)144000=x×(88100)44000=88x100x=44000×10088x=50000.\Rightarrow 44000 = x \times \Big(1 - \dfrac{12}{100}\Big)^1 \\[1em] \Rightarrow 44000 = x \times \Big(\dfrac{88}{100}\Big) \\[1em] \Rightarrow 44000 = \dfrac{88x}{100} \\[1em] \Rightarrow x = \dfrac{44000 \times 100}{88} \\[1em] \Rightarrow x = ₹ 50000.

Hence, value at beginning of 2007 = ₹ 50000.

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