Question

A machine worth rs 490740 is depreciated at 15% of its opening value each year. When its value would reduce by 90%

Answer 1

NOTE: Before solving this problem, we must get aware that in this type of problem, we use both the rules of compound and simple, decreasing or increasing formulas accordingly.

Solution:

Worth of the machine = Rs. 490740

Let, after n years, the machine's value would get reduced by 90% with 15% depreciation at the beginning of each year.

Therefore, after n years, the value would become

= Rs. 490740 * (1 - 15/100)ⁿ

= Rs. 490740 * (85/100)ⁿ

= Rs. 490740 * (0.85)ⁿ

Again, after n years, the value would get reduced by 90%,

i.e., the value of the machine after n years be

= Rs. 490740 * (1 - 90/100)

= Rs. 490740 * 10/100

= Rs. 49074

By the given condition,

490740 * (0.85)ⁿ = 49074

or, (0.85)ⁿ = 0.1

or, n * log(0.85) = log(0.1)

or, n = log(0.1) / log(0.85)

or, n ≈ 14.2

i.e., n = 14.2 years

= 14 years 2 months

∴ after 14 years 2 months, the machine's value would get reduced by 90%.

Answer 2

Answer:

14 years and 2 months

Step-by-step explanation:

In this question,

We have been that

A machine worth Rs.490740 is depreciated @ 15% of its opening value each year.

We need to find when its value will reduce by 90%.

   After n years, the value would become

= $490740\times (1\ -\ \frac{15}{100})^{n}$

= $490740\times (\frac{85}{100})^{n}$

= $490740\times (0.85)^{n}$

Again, after n years, the value would get reduced by 90%,

i.e., the value of the machine after n years be

= $490740\times  (1\ -\ \frac{90}{100})$

= $490740\times \frac{10}{100}$

= Rs. 49074

By the given condition,

$490740\times (0.85)^{n}\ =\ 49074$

or, $(0.85)^{n}\ =\ 0.1$

or, $n\times log(0.85)\ =\ log(0.1)$

or, n = $\frac{log(0.1)}{log(0.85)}$

or, n = 14.2 (approximately)-

i.e., n = 14.2 years

= 14 years 2 months

Therefore, after 14 years 2 months, the machine's value would get reduced by 90%.