This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6
In this question we have been given that a piece of equipment cost a certain factory Rs 600,000
If it depreciates in value, 15% the first, 13.5% the second year, 12% the third year and so on.
Niw we have to find what its value will be at the end of 10 years, all percentages applying to the original cost.
This is very conceptual and important question.
CBSE DHANPAT RAI PUBLICATIONS
Understanding CBSE Mathematics
Class :- 10th
Question no 68
Cost of equipment = Rs 600,000
Depreciation value in first year = 15% of 600,000 = 90,000
Depreciation value in second year = 13.5% of 600000 = 81,000
Depreciation value in third year = 12% of 600000 = 72,000
So, first term(a) = 90000 and common difference(d) = 81000 – 90000 = –9000.
By using the formula of the sum of n terms of an A.P.
Sn = n[2a +(n – 1)d] / 2
Total Depreciation amount in 10 years = S10
= 10[2(90000) + (10 – 1)(–9000)] / 2
= 5[180000 – 81000]
= 5[99000]
= Rs 495,000
Value of equipment = Cost – Depreciation at end of 10 years
= 600000 – 495000
= Rs 105000
Hence, value at the end of 10 years is Rs 105000.