Question

Machine is depreciated at the rate of 20% on reducing balance the original cost of the machine was 100000 and its ultimate scrap value was 30000 the effective life of the machine is without taking log

Answer 1

in this question,

p=100000, A=30000, i=20%

then, formula is

A=p(1-i)^n

30000=100000(1-20/100)^n

30000/100000=(1-20/100)^n

0.3=(0.8)^n

then, n=5.4

so the option b is correct

Answer 2

Answer:

n=5.4 years

Explanation:

P = 1,00,000

A = 30,000

I = 20%

A = P ( 1 - i )^n

30,000 = 1,00,000 ( 1 - 20/100 )^n

30,000/1,00,000 = ( 0.8 )^n

0.3 = ( 0.8 )^n

n = 5.4 years

hope it helps you

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