Question

${\bf{\underbrace{{Question}}}}$
A machine which was purchased 3 year ago at rupees 6, 25, 000 depreciates at the rate of 8 p.c.p.a. What is its value today?​

Answer 1

Answer:

Present cost of machine = Rs.32000 Depreciation percentage every year = 5% So, depreciation at the end of first year = 32000×5/100 = 1600/- Then cost of machine at the end of first year = Rs.32000 - Rs.1600 = Rs.30400/- So, the depreciation amount at the end of second year = 30400×5/100 = 1520/- Therefore, total depreciation at the end of 2 years = Rs.1600 + Rs.1520 = Rs.3120/-

Step-by-step explanation:

Answer 2

★ Concept :-

Here the concept of Amount Value derived from Compound Interest has been used. We see that we are given the price of the machine which was some years ago. Also, we see the price of machine depreciates at some rate per annum. So we need to calculate the price of machine today. Firstly we can modify the given values according to the time and need. And then we can apply them in the formula and find the answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Amount\;=\;\bf{P\;\times\;\bigg(1\;+\;\dfrac{R}{100}\bigg)^{T}}}}}$

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★ Solution :-

Given,

» Price of machine 3 years ago = Rs. 6,25,000

» Time since machine is bought = 3 years

» Rate of depreciation of price of machine = 8 % per annum (p.c.a)

From this we can derive that,

• Principal Amount = P = 6,25,000

• Time period = T = 3 years

• Rate = R = - 8 %

(since the price depreciates each other so money will be subtracted according to rate from the principal amount)

• Present Price of Machine = Amount

We have got the values. Let's now apply these in the formula.

$\\\;\sf{\rightarrow\;\;Amount\;=\;\bf{P\;\times\;\bigg(1\;+\;\dfrac{R}{100}\bigg)^{T}}}$

By applying values, we get

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(1\;+\;\dfrac{(-8)}{100}\bigg)^{3}}}$

By taking the LCM of the fraction, we get

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{100}{100}\;+\;\dfrac{(-8)}{100}\bigg)^{3}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{100\;+\;(-8)}{100}\bigg)^{3}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{100\;-\;8}{100}\bigg)^{3}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{92}{100}\bigg)^{3}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{92}{100}\:\times\:\dfrac{92}{100}\:\times\:\dfrac{92}{100}\bigg)}}$

This can be written as,

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{46}{50}\:\times\:\dfrac{46}{50}\:\times\:\dfrac{46}{50}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{23}{25}\:\times\:\dfrac{23}{25}\:\times\:\dfrac{23}{25}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\bigg(\dfrac{23\:\times\:23\:\times\:23}{25\:\times\:25\:\times\:25}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\:\;\bigg(\dfrac{12167}{15625}\bigg)}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{(625000)\;\times\;\dfrac{12167}{15625}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{625000\;\times\;\dfrac{12167}{15625}}}$

$\\\;\sf{\Longrightarrow\;\;Amount\;=\;\bf{40\;\times\;12167}}$

$\\\;\bf{\Longrightarrow\;\;Amount\;=\;\bf{\red{Rs.\;\;486680}}}$

Since, we have already subtracted the rate from principal sum (as rate is taken as negative) and thus got this amount. Thus this is the required answer.

$\\\;\underline{\boxed{\tt{Value\;\:of\;\:machine\;\:today\;=\;\bf{\purple{Rs.\;\;486,680}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;S.I.\;=\;\dfrac{P\:R\:T}{100}}$

• Derivation of Formula of Amount from C. I.

We know that,

✒ Amount = Principal + CI

And we know that,

$\\\;\tt{\leadsto\;\;C.I.\;=\;\bf{P\;\times\;\bigg(1\;+\;\dfrac{R}{100}\bigg)^{T}\;-\;P}}$

Equating the both equations, we get

$\\\;\tt{\leadsto\;\;Amount\;=\;\bf{P\;+\;\bigg[P\;\times\;\bigg(1\;+\;\dfrac{R}{100}\bigg)^{T}\bigg]\;-\;P}}$

Cancelling P, we get

$\\\;\tt{\leadsto\;\;Amount\;=\;\bf{P\;\times\;\bigg(1\;+\;\dfrac{R}{100}\bigg)^{T}}}$

Thus this is the required equation.