Question

hii mates plz answer it's urgent











Krishna purchased an old Scooter for $ 8000 if its cost after 2 years is $ 6480 what is the rate of depreciation ??

Answer 1

Given, Principal = 8000, A = 6480, Time n = 2 years.

It was asked for rate of depreciation.

We know that A = P(1 - r/100)^n

⇒ 6480 = 8000(1 - r/100)^2

⇒ 6480/8000 = (1 - r/100)^2

Take the common factor '80', we get

⇒ 81/100 = (1 - r/100)^2

⇒ (9/10)^2 = (1 - r/100)^2

⇒ 9/10 = (1 - r/100)

⇒ (9/10) - 1 = -r/100

⇒ -1/10 = -r/100

⇒ 1/10 = r/100

⇒ 100 = 10r

⇒ r = 10.

Therefore, rate of depreciation = 10%.

Hope it helps!    ------ > Good luck!

Answer 2

Hi there!
Here's the answer:

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¶¶¶ Points to remember :

When Interest is compounded Annually,

$Amount = P (1 + \frac{R}{100})^{n}$

where, P = Principal
Rate = R% per annulment
Time = n years

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SOLUTION :

Given:
Compound Interest on a sum of 8000₹ at a certain rate of interest compounded annually for 2 years is 6480

Amount = Compound Interest = 6480₹
Principal P = 8000₹
Time n = 2 years
let rate per annum = r%

Substitute in Formula

$6480 = 8000 (1 + \frac{R}{100})^{2}$

=> $\frac{6480}{8000} = (1 + \frac{R}{100})^{2}$

=> $(1 + \frac{R}{100})^{2} = \frac{81}{100}$

=> $(1 + \frac{R}{100})^{2} = \frac{9^{2}}{10^{2}}$

=> $(1 + \frac{R}{100})^{2} = (\frac{9}{10})^{2}$

=> $1 + \frac{R}{100} = \frac{9}{10}$

=> $\frac{R}{100} = \frac{9}{10} - 1$

=> $\frac{R}{100} = \frac{9-10}{10}$

=> $\frac{R}{100} = - \frac{1}{10}$

=> $R = - \frac{1}{10} × 100$

=> $R = - 10$

^^^ Here, the -Ve sign shows that it is a rate of depreciation.

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$\underline{\underline{Rate\: of\: Depreciation = 10\%}}$

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:)

Hope it helps