Question

A machine costs Rs.97, 000 and its effective life is estimated to be 12 years. If the scarp realises Rs.2000 only, what amount should be retained out of profits at the end of each year to accumulate at compound interest of 5% per annum in order to buy a new machine after 12 years? $(use{1.05^{12}} = 1.769)$

Answer 1

Hint: To solve this problem we need to have knowledge about annuity concepts. Annuity means a series of fixed payments over a period of time.

Complete step-by-step answer:
Let us write the given data
Cost of the machine = Rs.97, 000
Value of scrap = Rs.2000
Effective cost of machine = cost of machine – value of scrap
                                              = Rs.97, 000 - Rs.2000
                                             =Rs.95, 000
Therefore the Effective cost of machine (Required money) = Rs.95, 000
 Future annuity = $\dfrac{A}{r}\left[ {{{\left( {1 + r} \right)}^n} - 1} \right]$
Here Future annuity (required amount) =Rs.95, 000
Rate of interest = 5% =0.05, n=1
On substitute the value in the formula we get
$ \Rightarrow 95000 = \dfrac{A}{{0.05}}\left[ {{{\left( {1 + 0.05} \right)}^{12}} - 1} \right]$
\[ \Rightarrow A = \dfrac{{0.05 \times 95000}}{{{{(1.05)}^2} - 1}} = \dfrac{{0.05 \times 95000}}{{0.769}} = 6176.85\]
Therefore the Annuity amount (that is the amount that has to be retained at the end each year) =Rs.6176.85
NOTE: Annuity is the amount that we have to pay every year without any profits. Here future annuity means the required amount after removing scrap from cost price. Future annuity is also known as effective cost. To solve these kinds of problems we need to know the meaning of value.