Question

The value of a machine worth 5,00.000 depreciates at the rate of 10% every
year. In how many years will its value be? 3,64,500?
​

Answer 1

$\rm\blue{CORRECT~QUESTION}$

The value of a machine worth 5,00,000 depreciates at the rate of 10% every

year. In how many years will its value be? 3,64,500? (If the value (decreases) depreciates then the amount cannot be more than the principal.)

$\huge\rm\orange{Given}$

Principal = Rs5,00,000

Rate = 10%

Amount = Rs364500

$\huge\rm\orange{Find:-}$

Time = ?

$\huge\rm\red{Solution:-}$

$\rm\orange{As~we~know,}$

$Amount = P[1+ \dfrac{r}{100}]^n$

But here the value has depreciated (decreased) so we will (-) it.

$Amount = P[1- \dfrac{r}{100}]^n$

Put the given values.

$364500 = 500000[1- \dfrac{10}{100}]^n$

$\dfrac{364500}{500000} = [1- \dfrac{1}{10}]^n$

The zeros will be cancelled.

(Divide both the digits. Both are divisible by 5...)

(729×5=3645)

(1000×5= 5000)

$\dfrac{729}{1000} = (\dfrac{10-1}{10})^n$

$\dfrac{729}{1000} = (\dfrac{9}{10})^n$

$(\dfrac{9}{10})^3 = (\dfrac{9}{10})^n$

Both the bases are same. Hence, they will be cancelled. and the powers will be left...

3 = n

Time = 3 years

In 3 years the value will depreciate to Rs364500

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Additional Information

Formula to find C.I. if amount and principal are given.

➣ C.I. = A - P

▪ Formula to find C.I. if principal and time are given.

➣$C.I. = P[({\dfrac{r}{100}})^n-1]$

▪ Formula to find Amount if principal and Compound Interest are given.

➣ A = C.I. + P

▪ Formula to find Interest.

➣ $\sf\:S\:I\:=\:\dfrac{P\:\times\:R\:\times\:T}{100}$

▪Formula to find amount when principal, time and rate is given.

.➣$Amount = P[1+ \dfrac{r}{100}]^n$

NOTE..

The positive sign changes to negative sign when the value depreciates.

Answer 2

$<font color = blue>$

AnSWeR

.

.

.

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➡It is given that

Present value = 500000

Reduced value = 364500

Rate of depreciation = 10% p.a.

Consider n years as the period

We know that

A/P = (1−r/100) n

➡Substituting the values

364500/500000=(1−10/100) n

➡By further calculation

(9/10) n

=729/1000=(9/10) ³

So we get

n=3

➡Therefore, the period in which its value be reduced to 364500 is 3 years.