# An article valued at $P when new, depreciates by 10% of it’s value at the end of every year. if at the end of the third year ,the An article valued at$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 $7,290 . find the value of P.​ About the author ###### Everleigh ### there are 2 250 Marbles in a table and another table have 290 Marbles how many total total marbles are there​ ### 8. The factors of m2-256 areA. (m + 4)²C. (m – 4) (m+4)B. (m – 4)²D. None of the above​ ### 1 thought on “An article valued at$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”

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

$$\underline {\boxed{\bf{Amount= Principal \Bigg\{ 1-\dfrac{Rate \:}{100}\Bigg\}}^{time}}}$$

• Putting the value of Amount, rate of Depreciate and time in formula we get

$$\\\sf{~~~~~~:\implies 7290= Principal \Bigg\{ 1-\dfrac{10}{100}\Bigg\}^3}$$

$$\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{100 – 10}{100}\Bigg\}^3}$$

$$\sf{~~~~~~:\implies 7290= Principal \Bigg\{ \dfrac{90}{100}\Bigg\}^3}$$

$$\sf{~~~~~~:\implies 7290= Principal \times \dfrac{9 \times 9 \times 9 }{10 \times 10 \times 10}}$$

$$\sf{~~~~~~:\implies Principal = \dfrac{ 7290\times 1000}{729}}$$

$$\sf{~~~~~~ :\implies Principal = \dfrac{ \cancel{729}^{1} \: 0\times 1000}{ \cancel{729}}}$$

$$\sf{~~~~~~ :\implies Principal = 10000}\\\\$$

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