Question

The value of a machine is ₹50,000. If each year, its value depreciates by 5%, what will its value be at the end of 2 years?​

Answer 1

Answer:

5000 is the write answer.

Answer 2

Given :-

• Current value of the machine = ₹50,000
• Depreciate % = 5 %

Aim :-

• To find the depreciated value after 2 years

Formula to use :-

$\to \sf{Depreciated \: value} = \sf{Current \:value}\bigg(1 - \dfrac{\sf{rate\%}}{100} \bigg)^{\sf{time}}$

By substituting the values,

• Let the Depreciated value be denoted as DV

$\implies \sf{DV} = 50,000\bigg(1 - \dfrac{5}{100} \bigg)^{2}$

Taking LCM as 100,

$\implies \sf{DV} = 50,000\bigg(\dfrac{100-5}{100} \bigg)^{2}$

$\implies \sf{DV} = 50,000\bigg(\dfrac{95}{100} \bigg)^{2}$

Reducing to the lowest terms,

$\implies \sf{DV} = 50,000\bigg(\dfrac{19}{20} \bigg)^{2}$

$\implies \sf{DV} = 50,000 \times \dfrac{19}{20} \times \dfrac{19}{20}$

Cancelling,

$\implies \sf{DV} = 125 \times 19 \times 19$

$\implies \sf{DV} = 45125$

Therefore after 2 years, the value of the machine will be ₹45125.

Some more formulas :-

• When the population (or) the value of a machine increases,

$\longrightarrow \sf{Appreciated \: value} = Initial \: value \bigg( 1 + \dfrac{rate\%}{100} \bigg)^{time}$

• When interest is compounded annually,

$\longrightarrow \sf{Amount} = Principal\bigg(1+ \dfrac{rate\%}{100} \bigg)^{time}$

• When interest is compounded half yearly,

$\longrightarrow \sf{Amount} = Principal\bigg(1 + \dfrac{rate\%}{200} \bigg)^{2\times time}$

• When interest is compounded quarterly,

$\longrightarrow \sf{Amount} = Principal\bigg(1 + \dfrac{rate\%}{400} \bigg)^{4\times time}$

• Simple interest

$\longrightarrow\sf{Simple \: Interest} = \dfrac{Principal \times Rate \times Time}{100}$