Question

The value of a car depricates 15% every year. If the present value of the car is Rs 3,61,250 after two years What was the original price of the car.​

Answer 1

Answer:

Original Price:3,08,102.35

Answer 2

Answer:

Given :

• ➻ Depreciation of car every year = 15%
• ➻ Present value of car after two years = Rs.3,61,250.
• ➻ Time period = 2 years

$\begin{gathered}\end{gathered}$

To Find :

• ➻ Original price of car

$\begin{gathered}\end{gathered}$

Using Formula :

$\longrightarrow{\small{\underline{\boxed{\pmb{\sf{A = P {\bigg(1 - \dfrac{R}{100} \bigg)}^{n}}}}}}}$

☼ Where :-

• ➤ A = Amount of depreciation
• ➤ P = Principle
• ➤ R = Rate of depreciation
• ➤ n = Time period

$\begin{gathered}\end{gathered}$

Solution :

☼ Let the :-

• Original price of car be Rs.x.

$\begin{gathered}\end{gathered}$

☼ Here :-

• ➛ A = Rs.3,61,250
• ➛ P = x
• ➛ R = 15%
• ➛ n = 2 years

$\begin{gathered}\end{gathered}$

☼ Now, finding the original price of car by substituting the values in formula :-

${\longrightarrow{\small{\sf{A = P {\bigg(1 - \dfrac{R}{100} \bigg)}^{n}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg(1 - \dfrac{15}{100} \bigg)}^{2}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg( \dfrac{(1 \times 100) - (15 \times 1)}{100} \bigg)}^{2}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg( \dfrac{100 - 15}{100} \bigg)}^{2}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg( \dfrac{85}{100} \bigg)}^{2}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg( \dfrac{85}{100} \times \dfrac{85}{100} \bigg)}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg(\dfrac{7225}{10000} \bigg)}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg( \cancel{\dfrac{7225}{10000}} \bigg)}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x {\bigg({\dfrac{289}{400}} \bigg)}}}}}$

${\longrightarrow{\small{\sf{Rs.361250 = x \times {\dfrac{289}{400}}}}}}$

${\longrightarrow{\small{\sf{ x = 361250 \times \dfrac{400}{289} }}}}$

${\longrightarrow{\small{\sf{ x = \cancel{361250}\times \dfrac{400}{\cancel{289}}}}}}$

${\longrightarrow{\small{\sf{ x = {1250 \times 400}}}}}$

${\longrightarrow{\small{\underline{\underline{\sf{ x = {Rs.500000}}}}}}}$

${\longrightarrow{\small{\underline{\boxed{\pmb{\sf{Original \: price = {Rs.500000}}}}}}}}$

∴ The original price of car is Rs.500000.

$\begin{gathered}\end{gathered}$

Learn More :

$\longrightarrow{\small{\underline{\boxed{\sf{\purple{ Simple \: Interest = \dfrac{P \times R \times T}{100}}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{Amount={P{\bigg(1 + \dfrac{R}{100}{\bigg)}^{T}}}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{Amount = Principle + Interest}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{ Principle=Amount - Interest }}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{Principle = \dfrac{Amount\times 100 }{100 + (Time \times Rate)}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{Principle = \dfrac{Interest \times 100 }{Time \times Rate}}}}}}$

$\longrightarrow\small{\underline{\boxed{\sf{\purple{Rate = \dfrac{Simple \: Interest \times 100}{Principle \times Time}}}}}}$

$\begin{gathered}\tiny{\overline{\underline{\star\rule{5mmpt}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}\star\rule{5mm}{1mm}}}}\end{gathered}$