Question

The value of a machine depreciates at 10% per year. If the present value is 1,62,000, what
is the worth of the machine after two years.​

Answer 1

Answer:

The machine's cost after 2 years is Rs. 1,31,220.

Step-by-step explanation:

Given :

Rate of depreciation = 10%

Present value = Rs. 1,62,000

Time = 2 years

To find :

The worth of machine after 2 years

Solution :

• P = 162000
• N = 2
• R = 10%

$\boxed{\sf{Depreciation= P\left( 1 - \frac{R}{100}\right)^{N}}}$

$\sf{\longrightarrow} \: Depreciation = 162000 \times \left( 1 - \dfrac{10}{100}\right)^{2} \\ \\ \sf{\longrightarrow} \: Depreciation = 162000 \times \dfrac{90}{100} \times \dfrac{90}{100} \\ \\ \sf{\longrightarrow} \: Depreciation = 1620 \times 9 \times 9 \\ \\ \sf{\longrightarrow} \: Depreciation = 1620 \times 81 \\ \\ \sf{\longrightarrow} \: Depreciation = 131220$

Worth of machine after 2 years = Rs. 1,31,220

$\therefore$ The machine's cost after 2 years is Rs. 1,31,220.

Answer 2

AnswEr :

Depreciation = Rs.131220

$\mathfrak{Given}\begin{cases}\sf{Principal,[P]=Rs.162000}\\ \sf{Rate,[R]=10\%}\end{cases}}}$

$\bf{\large{\underline{\underline{\tt{To\:FinD\::}}}}}}$

The worth is of the machine after two years.

$\bf{\large{\underline{\underline{\tt{\green{ExplanatiOn\::}}}}}}$

Formula use :

$\bf{\large{\boxed{\rm{Depreciation=P\bigg(1-\frac{R}{100} \bigg)^{n} }}}}}$

$\dashrightarrow\tt{Worth\:of\:machine\:=\:162000\bigg(1-\dfrac{10}{100} \bigg)^{2}} \\\\\\\\\dashrightarrow\tt{Worth\:of\:machine\:=\:162000\bigg(\dfrac{100-10}{100} \bigg)^{2} }\\\\\\\\\dashrightarrow\tt{Worth\:of\:machine\:=\:162000*\bigg(\dfrac{90}{100} \bigg)^{2} }\\\\\\\\\dashrightarrow\tt{Worth\:of\:machine\:=\:162\cancel{000}*\dfrac{90}{\cancel{100}} *\dfrac{9\cancel{0}}{\cancel{10}\cancel{0}}} \\\\\\\\\dashrightarrow\tt{Worth\:of\:machine=Rs.(162*90*9)}$

$\dashrightarrow\tt{\green{Worth\:of\:machine\:=\:Rs.131220}}$

Thus,

The worth is of the machine after two years is Rs.131220.