Question

The price of a machine depreciates by 20% every year at the end of 3 years the price come down to rupees 12800 what was the price of the machine 3 years back.​

Answer 1

Answer:

25,000

Step-by-step explanation:

A = 12800

T = 3 YEARS

R = 20 p.c.p.a

To find : P

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

= 12800 = P ( 1 - 20/100 ) ^3

= 12800 = P ( 1 - 1/5 ) ^3

=12800 = P ( 4/5 ) ^3

= 12800*5*5*5/4*4*4 = P

=  200 * 125

= 25,000

Therefore the final answer is 25,000. THNX and PLZ mark me as the brainliest.

Answer 2

Answer:

25000

Step-by-step explanation:

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

12800 = P ( 1 - 20/100)^3  [Substitute formula]

12800 = P ( 100/100 - 20/100)^3  [1/1 = 100/100]

12800 = P (80/100)^3  [Cancel the zeros]

12800 = P * 8/10 * 8/10 *8/10 [Cube]

12800 * 10/8 * 10/8 * 10/8 = P  [Transposition rule]

25000 = P

P = 25000