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Mathematics

A machine depreciates each year at 8% of its value in the beginning of the year. If its value be ₹ 57,500 at the end of the year 2015, find :

(i) its value at the end of the year 2014,

(ii) its value at the end of the year 2016.

Compound Interest

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Answer

(i) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine n years ago = ₹ V(1r100)n\dfrac{V}{\Big(1 - \dfrac{r}{100}\Big)^n}

Substituting the values in formula,

Value of machine at the end of the year 2014 =57500(18100)1=575001008100=5750092100=575002325=57500×2523=62,500\text{Value of machine at the end of the year 2014 }=\dfrac{57500}{\Big(1 - \dfrac{8}{100}\Big)^1} \\[1em] =\dfrac{57500}{\dfrac{100 - 8}{100}} \\[1em] =\dfrac{57500}{\dfrac{92}{100}} \\[1em] =\dfrac{57500}{\dfrac{23}{25}} \\[1em] =\dfrac{57500 \times 25}{23} \\[1em] =₹ 62,500

Hence, value of machine at the end of the year 2014 = ₹ 62,500.

(ii) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine after n years = [V×(1r100)n]\Big[V \times \Big(1 - \dfrac{r}{100}\Big)^n \Big]

Substituting the values in formula,

Value of machine at the end of 2016=57500×(18100)1=57500×(1008100)=57500×(92100)=57500×2325=57500×2325=52,900.\text{Value of machine at the end of 2016}=57500 \times \Big(1 - \dfrac{8}{100}\Big)^1 \\[1em] =57500 \times \Big(\dfrac{100 - 8}{100}\Big) \\[1em] =57500 \times \Big(\dfrac{92}{100}\Big) \\[1em] =57500 \times \dfrac{23}{25} \\[1em] =\dfrac{57500 \times 23}{25} \\[1em] =₹ 52,900.

Hence, value of machine at the end of the year 2016 = ₹ 52,900.

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