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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» 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. HOPE IT HELPS U!MARK AS BRAINLIEST!(: STAY HOME, BE SAFE :) |
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