The value of a machinery is ₹10000, which is depreciating at the rate of 15% annually. will be the value the machinery after 2 years.

₹7000

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₹7205

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₹7225

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Solution

The correct option is C ₹7225
Use the concept of depreciating value:
$A = P (1 - r)^{n}$
$A = 10000 \times {[1 - \frac{15}{100}]}^{2} = 7225$

Therefore, after 2 years the value of machinery will be ₹7225.

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The value of a machinery is ₹10000, which is depreciating at the rate of 15% annually. will be the value the machinery after 2 years.

The correct option is C ₹7225Use the concept of depreciating value: A=P(1−r)n A=10000×[1−15100]2=7225 Therefore, after 2 years the value of machinery will be ₹7225.

The correct option is C ₹7225
Use the concept of depreciating value:
$A = P (1 - r)^{n}$
$A = 10000 \times {[1 - \frac{15}{100}]}^{2} = 7225$

Therefore, after 2 years the value of machinery will be ₹7225.