An article valued at $P when new, depreciates by 10% of it’s value at the end of every year. if at the end of the third year ,the article is valued at $7,290 . find the value of P.

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★Understanding The Question properly

It is Given that the Principal when it start and the Rate is in Depreciate by 10% at the End of every Year . Now at the end of 3 year’s the price of an article be $7,290.

From the Above Statement We get

Rate of Depreciates = 10%

Time = 3 year’s

Amount of article = $7,290

To Find

Principal

Note: If the compounding frequency per annum is 1 i.e. if the interest is compounded annually, the compound interest formula is given as:

we know that

If rate is in depreciates then Amount Formula will be

[tex] \underline {\boxed{\bf{Amount= Principal \Bigg\{ 1-\dfrac{Rate \:}{100}\Bigg\}}^{time}}}[/tex]

Putting the value of Amount, rate of Depreciate and time in formulawe get

[tex]\\\sf{~~~~~~:\implies 7290= Principal \Bigg\{ 1-\dfrac{10}{100}\Bigg\}^3}[/tex]

[tex]\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{100 – 10}{100}\Bigg\}^3}[/tex]

[tex]\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{90}{100}\Bigg\}^3}[/tex]

★Understanding The Question properlyat theRate is in Depreciate by 10%. Now at theEnd of every Yearthe price of anend of 3 year’s.article be $7,290From the Above StatementWe get## To Find

Note:

If the compounding frequency per annum is 1 i.e. if the interest is compounded annually, the compound interest formula is given as:we know that[tex] \underline {\boxed{\bf{Amount= Principal \Bigg\{ 1-\dfrac{Rate \:}{100}\Bigg\}}^{time}}}[/tex]

Putting the value of Amount, rate of Depreciate and time in formulawe get[tex]\\\sf{~~~~~~:\implies 7290= Principal \Bigg\{ 1-\dfrac{10}{100}\Bigg\}^3}[/tex]

[tex]\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{100 – 10}{100}\Bigg\}^3}[/tex]

[tex]\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{90}{100}\Bigg\}^3}[/tex]

[tex]\sf{~~~~~~:\implies 7290= Principal \times \dfrac{9 \times 9 \times 9 }{10 \times 10 \times 10}}[/tex]

[tex]\sf{~~~~~~:\implies Principal = \dfrac{ 7290\times 1000}{729}}[/tex]

[tex]\sf{~~~~~~ :\implies Principal = \dfrac{ \cancel{729}^{1} \: 0\times 1000}{ \cancel{729}}}[/tex]

[tex]\sf{~~~~~~ :\implies Principal = 10000}\\\\[/tex]