Math, asked by aroopmukherjee57, 3 days ago

the value of a machine depreciates at the rate of 20% per annum it was purchased two years ago if its present value is 40000 rupees for how much was it purchased​

Answers

Answered by ay6712124
0

Answer:

A=P(1-r/100)^T

A=40000(1-20/100)^2

=40000*80/100*80/100

=40000*4/5*4/5

=4000*4*4

=Rupees 64000.

Step-by-step explanation:

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