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Mathematics

The cost of a machine depreciates every year by 10%; the percentage decrease during two years will be:

  1. 20%

  2. 18%

  3. 19%

  4. 21%

Compound Interest

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Answer

Let initial value of machine be ₹ x.

For first year :

P = ₹ x

Rate of depreciation = 10%

n = 2 years

Value of machine after 2 years =x(110100)2=x(10010100)2=x×(90100)2=x×(910)2=81x100.\text{Value of machine after 2 years } = x\Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] = x\Big(\dfrac{100 - 10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{90}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] = \dfrac{81x}{100}.

Depreciation = Initial value - Final value = x81x100=19x100x - \dfrac{81x}{100} = \dfrac{19x}{100}.

By formula,

Percentage depreciated =Total depreciationInitial value×100=19x100x×100=19100×100=19%\text{Percentage depreciated }= \dfrac{\text{Total depreciation}}{\text{Initial value}} \times 100\\[1em] = \dfrac{\dfrac{19x}{100}}{x} \times 100\\[1em] = \dfrac{19}{100} \times 100\\[1em] = 19\%

Hence, option 3 is the correct option.

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