Question

A piece of equipment cost a certain factory Rs. 600000. If it depreciates in value, \[15\% \] the first, $ 13.5\% $ the next year, $ 12\% $ the third year and so on. What will be it’s value at the end of 10 years, all percentages applying to the original cost?

Answer 1

Hint: To solve this question, first we can see that this question belongs to Arithmetic Progression. So, first we will find the sum of all the 10 terms (i.e..10 years) and then with the help of the concluded sum, we can find the value of equipment and finally we can find the original cost after 10 years.

Complete step by step solution:
Let the cost of an equipment be Rs. 100.
Now, the percentages of depreciation at the end of 1st, 2nd, 3rd years are 15, 13.5, 12, which are in A.P, with \[a = 15\] (first term of the progression) and $ d = - 1.5 $ (difference between the two adjacent terms).
Hence, percentage of depreciation in the tenth year :
 \[{T_n} = a + (n - 1)d\]
or, $ {T_{10}} = 15 + (10 - 1)( - 1.5) = 1.5 $
Now, use the formula to get the sum of \[n\] terms:
 $ {S_n} = \dfrac{n}{2}[2a + (n - 1)d] $
So, we have to find the sum of 10 terms:
 \[\Rightarrow {S_{10}} = \dfrac{{10}}{2}[2 \times 15 + (10 - 1)( - 1.5)] = 5[30 - 13.5] = 82.5\]
So, total value depreciated in the tenth year= 82.5 (Sum of 10 terms of the AP)
Hence, the value of equipment at the end of 10 years $ = 100\% - 82.5\% = 17.5\% $
Now,
Original cost after 10 years
 $
   = 17.5\% \,of\,600000 \\
   = 17.5 \times \dfrac{{600000}}{{100}} \\
   = 17.5 \times 6000 \\
   = 105000\,Rs. \;
  $
So, the correct answer is “ $ 105000\,Rs $ ”.

Note: For any progression, the sum of \[n\] terms can be easily calculated. For an AP, the sum of the first \[n\] terms can be calculated if the first term and the total terms are known. The formula for the arithmetic progression sum is explained below: Consider an AP consisting of \[n\] terms.