Question

 A piece of equipment costs Rs. 600000 in a certain factory. If it depreciates in value 1% the first year, 13.5% the next year, 12% the third year and so on. What will be its value at the end of 10 years, all percentages applying to the original costs?
A. 125000.
B. 115000.
C. 100000.
D. 105000.

Answer 1

Hint: We will first observe that the decrease (in percentage) of the value of the equipment is the same in all the preceding years, hence we will solve this question by applying Arithmetic Progression.

Complete step by step answer:
We are given that piece of equipment costs Rs. 600000 in a certain factory and the value of it depreciates as 1% in the first year, 13.5% in the next year, 12% in the third year and so on.
We have to calculate the value of it at the end of 10 years.
We solve this question using Arithmetic Progression because we have the relative decrease in amount every year is a constant value.
According to the given conditions of the question we have,
First term a = 15% and common difference d = 13.5 % - 15% = -1.5 %. Since the value of the amount decreases every year so the common difference came to be negative.
The percentage of depreciation in the tenth year is given by the nth term of the A.P.
We have to formula for nth term of the A.P. as,
\[{{T}_{n}}=a+(n-1)d\], where a is the first term, d common difference and n is the year in which we have to calculate the total cost. Here we have n = 10 years.
Substituting the value of a, d and n in above we get,
\[\begin{align}
  & {{T}_{n}}=15+(10-1)(-1.5) \\
 & \Rightarrow {{T}_{n}}=15+(9)(-1.5) \\
 & \Rightarrow {{T}_{n}}=15-13.5 \\
 & \Rightarrow {{T}_{n}}=1.5 \\
\end{align}\]
Hence, the percentage of depreciation in the tenth year is given by \[{{T}_{n}}=1.5\].
Also, total value depreciated in 10 years (in percentage) is given by Sum of n=10 terms of the AP.
The formula of sum of n terms of an A.P. is given by,
\[{{S}_{n}}=\dfrac{n}{2}\left[ 2a+(n-1)d \right]\], where n, d and a are as defined above.
Then, substituting the values of all the above variables we get,
\[\begin{align}
  & {{S}_{n}}=\dfrac{10}{2}\left[ 2(15)+(10-1)(-1.5) \right] \\
 & \Rightarrow {{S}_{n}}=5\left[ 30+9(-1.5) \right] \\
 & \Rightarrow {{S}_{n}}=5\left[ 30-13.5 \right] \\
 & \Rightarrow {{S}_{n}}=5\left[ 16.5 \right] \\
 & \Rightarrow {{S}_{n}}=82.5 \\
\end{align}\]
Hence, we obtain the total value depreciated in 10 years is 82.5%.
Therefore, rest percentage = 100% - 82.5 % = 17.5 %
Now we calculate the original cost after 10 years by computing 17.5% of 600000, which is given by,
The original cost after 10 years =17.5 % of 600,000
\[\Rightarrow \] The original cost after 10 years \[=\text{ }17.5\text{ }\times \text{ }\dfrac{600,000}{100}\]
\[\Rightarrow \] The original cost after 10 years = Rs. 105000.
Therefore, we obtain the final value after 10 years as Rs. 105000.

Note: The possibility of error in the question is you can start to solve this question by using Compound or Simple Interest formula observing that the rates are given in the question, which is wrong. You should always first check if the common difference remains the same in all the years then proceed to apply Arithmetic progression to solve the question.