Question

$\huge\red{Question}$


The value of a property decrease every year at the rate of 5%. If its present value Rs. 4,11,540, what was its value three years ago.



​

Answer 1

Value of property before 3 years = $\boxed{ \bf\red{4,80,000 \: RS} }$

$\underline{\underline{ \sf Given :- }}$

• Value of property decreases at rate of 5%

• Present value is 4,11,540 RS

$\underline{\underline{ \sf To \: Find:- }}$

• value before 3 years

$\huge\underline{\underline{ \sf Formula \: - }}$

$\boxed{ \sf \red{ A = P\bigg( 1 + \frac{r}{100} \bigg) ^{ n} }}$

• A = Amount

• P = principal amount

• r = rate

• n = number of years

$\huge\underline{\underline{ \sf Answer :- }}$

To Find value 3 years ago ,

$\boxed{ \sf \red{ A = P\bigg( 1 - \frac{r}{100} \bigg) ^{ n} }}$

$\bigg($ $\sf \frac{-r}{100}$ indicates value before the year $\bigg)$

• A = 4,11,540 RS

• n = 3

• r = 5

Substituting the values ,

• $\red{ \implies}$ $\sf 4 11540 = p \bigg(1 - \frac{5}{100} \bigg) {}^{3}$

• $\red{ \implies}$ $\sf411540 = p \bigg({ \frac{95}{100} \bigg) }^{3}$

• $\red{ \implies}$ $\sf411540 =p \bigg( {\frac{19}{20} } \bigg)^{3}$

• $\red{ \implies}$$\sf411540 = p \bigg( \frac{6859}{8000} \bigg)$

• $\red{ \implies}$$\sf \: p = 411540 \times \frac{8000}{6859}$

• $\red{ \implies}$$\sf \blue{ p = 480000}$

$\rule{10cm}{0.005cm}$

Value of property before 3 years = 4,80,000 RS

$\rule{10cm}{0.002cm}$

Answer 2

Answer:

A] present value = Rs 411540

rate of depreciation = 5% p.a

t = 3 years

value 3 yrs ago

=

p(1−

100

R

) n

⇒ 411540= p(1−

100

5

) 3

⇒ 411540=

p(

100

95

) 3

⇒ 411540=

p(

20

19

) 3

⇒ 411540=

p(

8000

6859

) 3

⇒

6859

411540×80000

= Rs480000