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

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1. ★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 formula we 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]

   • [tex]\pink{\text{The Value of Principal is \bf{\$10,000.}}}[/tex]

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