Question 4 the day : March 25, 2003
The question for the day is from the topic Progressions. It is an Arithmetic Progression question where the common difference is negative.

Question
A piece of equipment cost a certain factory Rs. 600,000. If it depreciates in value, 15% 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 cost?

(1) 2,00,000
(2) 1,05,000
(3) 4,05,000
(4) 6,50,000

Explanatory Answer

Let the cost of an equipment is 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 and d = - 1.5.

Hence, percentage of depreciation in the tenth year = a + (10-1) d = 15 + 9 (-1.5) = 1.5
Also total value depreciated in 10 years = 15 + 13.5 + 12 + ... + 1.5 = 82.5

Hence, the value of equipment at the end of 10 years=100 - 82.5 = 17.5.
The total cost being Rs. 6,00,000/100 * 17.5 = Rs. 1,05,000.

Question on finding sum of terms of a GP, Problem on finding a particular term of an AP, given sum of two other terms, Summation of AP having negative common difference, Finding terms of an AP, given sum and product of some of its terms, Sum of series of numbers in arithmetic progression, A CAT 2003 question to find sum of terms of an AP

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