IMO: E

BR,

ajp

MBA.Aspirant wrote: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?

(a) 2,00,000
(b) 1,05,000
(c) 4,05,000
(d) 6,50,000
(e) 6,00,000

Depreciation the first year = (.15)*600,000 = 90,000.
Each year the amount of depreciation decreases by (.015)*600,000 = 9000.
Thus, after 9 years, the amount of depreciation in the 10th year = 90,000 - 9*9000 = 9000.
Since 90,000 is a multiple of 9000, and the amount of depreciation decreases by 9000 each year, the amounts of depreciation are the 10 consecutive multiples of 9000 between 90,000 and 9000, inclusive.

Given a set of evenly spaced integers:
Average = (biggest+smallest)/2
Sum = number * average

Thus, given the 10 multiples of 9000 between 90,000 and 9000, inclusive:
Average = (90,000 + 9000)/2 = 49,500.
Sum = 10 * 49,500 = 495,000.

Thus, the amount remaining = 600,000 - 495,000 = 105,000.

The correct answer is B.

Well i did not know how to solve it...but i guessed it right(B)...now i knw...

Thanks GmatGuruNY

You could also apply the same reasoning directly to the percentages.

15% + 13.5% + ...
if you count out the 10 terms the last one is
... +1.5%

So, to sum a list of evenly spaced numbers:
sum = average * number of terms
average = (first + last)/2 = (15% + 1.5%) = 16.5%/2 = 8.25%
number of terms = 10
sum = 8.25% * 10 = 82.5% depreciation

So, 17.5% of the value remains. Multiply by 600,000.