Question

A a machine whose cost price is rupees 10000 ,depreciates at the rate of r% every year and its value reduces to 8464 at the end of 2 years find the value of R percent

Answer 1

Answer:

The value of r = 1058/32% or 42.32%

Step-by-step explanation:

CP of the machine = 10,000

Depreciate every year = r%

Value at the end of 2 years = 8464

r/100 × 10,000 × 2 = 8464

200r = 8464

r = 8464/200

r = 1058/32% or 42.32%

Check ⇒

42.32/100 × 10,000 = 4232

4232 × 2 = 8464

8464 = 8464

L.H.S = R.H.S

Hence, the answer is correct.

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