Question

The value of a machine depreciates every year at the rate of 10 % it was purchased for rupees 15000 when new and it was sold for rupees 10935 find the number of years the machine was used before selling

Answer 1

SOLUTION:


Given : The value of a machine depreciates every year at the rate of 10%. it was purchased for Rs 15000 when new and sold for Rs 10935. So

Let , the number of years the machine was used before selling = t years

$10935 =15000( 1 - \frac{10}{100} ) {}^{t} \\ \\ ( 1 - \frac{10}{100} ) {}^{t} = \frac{10935}{15000} \\ \\ (\frac{100 - 10}{100} ){}^{t} = \frac{729}{1000} \\ \\ (\frac{90}{100} ) {}^{t} = \frac{729}{1000} \\ \\( \frac{9}{10} ) {}^{t} = ( \frac{9}{10} ) {}^{3} \\ \\ \cancel{(\frac{9}{10}) }{}^{t} = \cancel{(\frac{9}{10}) }{}^{3} \\ \\ t = 3$
The number of years the machine was used before selling = $<b>3 years.$


Thanks for the question!

Answer 2

Rate(r) = -10% (It is negative because value is depreciating)

P(Initial price) = Rs.15000

Q(Final price) = RS. 10935

Time(t) = ?

We know,

Q = P (1 + r/100)^t (Using compound interest formula)

Placing the values,

Rs 10935 = Rs 15000( 1 - 10/ 100)^ t

⇒ Rs 10935 / Rs 15000 = (1 - 10/100)^ t

⇒ Rs 10935 / Rs 15000 = (90/100)^ t

⇒ Rs 10935/Rs 15000 = (9/10)^ t

⇒ Rs 729/ Rs 1000 = (9/10) ^t

⇒ (9/10)³ = (9/10)^t

⇒ 3 = t

⇒ t = 3

So, the number of years, the machine was used before selling is 3 years.

(Note: ^t is the time taken as the power)