Answer:

Given :-

• The value of a machine depreciates at the rate of 10% per annum.
• It was purchased 3 years ago, if the present value is $ 291600.

To Find :-

• What is the present value of a machine ?

Solution :-

Given :

• Depreciated Value (A) = $ 291600
• Rate of Depreciation (r%) = 10% per annum
• Time Period (n) = 3 years

According to the question by using the formula we get,

[tex]\implies \sf\boxed{\bold{A =\: P\bigg(1 - \dfrac{r}{100}\bigg)^n}}\\[/tex]

where,

• A = Depreciated Value
• P = Present Value
• r = Rate of Depreciation
• n = Time Period

So, by putting those values we get,

[tex]\implies \sf 291600 =\: P\bigg(1 - \dfrac{10}{100}\bigg)^3\\[/tex]

[tex]\implies \sf 291600 =\: P\bigg(\dfrac{100 - 10}{100}\bigg)^3\\[/tex]

[tex]\implies \sf 291600 =\: P\bigg(\dfrac{90}{100}\bigg)^3\\[/tex]

[tex]\implies \sf 291600 =\: P \times \dfrac{90}{100} \times \dfrac{90}{100} \times \dfrac{90}{100}\\[/tex]

[tex]\implies \sf 291600 =\: P \times \dfrac{729\cancel{000}}{1000\cancel{000}}\\[/tex]

[tex]\implies \sf 291600 =\: \dfrac{729P}{1000}\\[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 729P =\: 291600 \times 1000\\[/tex]

[tex]\implies \sf 729P =\: 291600000\\[/tex]

[tex]\implies \sf P =\: \dfrac{291600000}{729}\\[/tex]

[tex]\implies \sf\bold{\underline{P =\: 400000}}\\[/tex]

[tex]\therefore[/tex] The present value of a machine is $ 400000 .