KnowledgeBoat Logo
|

Mathematics

A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for ₹ 10,000 and is sold for ₹ 7,744. The number of years, that the machine was used is:

  1. 2

  2. 4

  3. 6

  4. 8

Compound Interest

1 Like

Answer

Given,

Initial value (P) = ₹ 10,000

Depreciated value (A) = ₹ 7,744

R = 12%

Value of machine after n years = [V×(1r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

7744=10000(112100)n7744=10000(10012100)n7744=10000(88100)n774410000=(88100)n(88100)2=(88100)nn=2.\Rightarrow 7744 = 10000 \Big(1 - \dfrac{12}{100}\Big)^{n} \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{100 - 12}{100}\Big)^n \\[1em] \Rightarrow 7744 = 10000 \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \dfrac{7744}{10000} = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{88}{100}\Big)^2 = \Big(\dfrac{88}{100}\Big)^n \\[1em] \Rightarrow n = 2.

Hence, option 1 is correct option.

Answered By

3 Likes


Related Questions