Question

The value of machine depreciates every year at the rate of 10% p.a. If the present value of machine is Rs 6561. What was its value 3 year ago?​

Answer 1

Given :

• Rate = - 10% p.a. ( The rate is negative since the price is depriciating.)

• Time = 3 years

• Final price = Rs. 6561

To Find :

• Initial price

Solution :

Let initial price = x

Now, We know that,

$\underline{\boxed{\sf Final \: Price = Initial \: Price \Bigg( 1 + \dfrac{Rate}{100} \Bigg) ^{time}}}$

By putting values,

$: \implies \sf Rs. \: 6561 = x \times \Bigg( 1 + \dfrac{- 10}{100} \Bigg) ^{3}$

$: \implies \sf Rs. \: 6561 = x \times \Bigg( 1 - \dfrac{ 1\cancel{0}}{10\cancel{0}} \Bigg) ^{3}$

$: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{10}{10} - \dfrac{ 1}{10} \Bigg) ^{3}$

$: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{10 - 1}{10} \Bigg) ^{3}$

$: \implies \sf Rs. \: 6561 = x \times \Bigg( \dfrac{9}{10} \Bigg) ^{3}$

$: \implies \sf Rs. \: 6561 = x \times \dfrac{9}{10} \times \dfrac{9}{10} \times \dfrac{9}{10}$

$: \implies \sf Rs. \: \cancel{6561}^{729} \times \dfrac{10}{\cancel{9}} \times \dfrac{10}{9} \times \dfrac{10}{9} = x$

$: \implies \sf Rs. \: \cancel{729}^{81} \times 10 \times \dfrac{10}{\cancel{9}} \times \dfrac{10}{9} = x$

$: \implies \sf Rs. \: \cancel{81}^{9} \times 10 \times 10 \times \dfrac{10}{\cancel{9}} = x$

$: \implies \sf Rs. \: 9 \times 10 \times 10 \times 10= x$

$: \implies \sf Rs. \: 9000= x$

$: \implies \sf x = Rs. \: 9000$

$\large \underline{\boxed{\sf x = Rs. \: 9000}}$

So, the price of the machine 3 years ago = Rs. 9000.