A MACHINE IS PURCHASED FOR RUPEES 625000.ITS VALUE DECREASES AT THE RATE OF 8% PER ANNUM.WHAT WILL BE ITS VALUE AFTER 2 YEARS?...................................ANSWER FASTLY URGENT!
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Answered by
15
Value of the machine now: P = 625,000.
Rate of decrease in its value : 8% per annum. (depreciation)
Depreciation 1st year = P * 8/100 * 1 = 0.08 P
Value after 1 st year : P - P * 8/100 * 1
= 0.92 P
Value after 2nd year : 0.92 P - 0.92 P * 8/100
= 0.92 P (1 - 0.08)
= 0.92² P
= 0.92² * 625, 000 units
= 529, 000 units.
We could also directly use the formula for compound interest : P (1 + r/100)² , with = -8% (negative interest rate)
Rate of decrease in its value : 8% per annum. (depreciation)
Depreciation 1st year = P * 8/100 * 1 = 0.08 P
Value after 1 st year : P - P * 8/100 * 1
= 0.92 P
Value after 2nd year : 0.92 P - 0.92 P * 8/100
= 0.92 P (1 - 0.08)
= 0.92² P
= 0.92² * 625, 000 units
= 529, 000 units.
We could also directly use the formula for compound interest : P (1 + r/100)² , with = -8% (negative interest rate)
kvnmurty:
:-)
Cant understand your steps
But your answer is absolutely.correct because done with equation of compound
Interest
Answered by
2
Answer:
Price of machine = ₹625000
Depreciation rate = 8% p a
Value after 2 years = ?
Value after 2 years = 625000 × (1 -8/100)²
625000 × (1 - 2/25)²
625000 × 23/25 × 23/25
Answer = ₹ 529000
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