Advertisements
Advertisements
Question
A piece of equipment cost a certain factory Rs 60,000. If it depreciates in value, 15% the first, 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?
Solution
In the given problem,
Cost of the equipment = Rs 600,000
It depreciates by 15% in the first year. So,
Depreciation in 1 year
= 600000 − 495000
= 105000
= 90000
It depreciates by 13.5% of the original cost in the 2 year. So,
Depreciation in 2 year `= (13.5)/100 (600000) = 81000`
Further, it depreciates by 12% of the original cost in the 3 year. So,
Depreciation in 3 year `= 12/100 (600000)=72000`
So, the depreciation in value of the equipment forms an A.P. with first term as 90000 and common difference as −9000.
So, the total depreciation in value in 10 years can be calculated by using the formula for the sum of n terms of an A.P.
`S_n = n/2 [2a + (n-1) d]`
We get,
`S_n = 10/2 [2(90000) +(10-1)(-9000)]`
`=10/2 [180000 + (9)(-9000)]`
`=5(180000 - 81000)`
` = 5(99000)`
= 495000
So, the total depreciation in the value after 10 years is Rs 495000.
Therefore, the value of equipment = 600000 − 495000 = 105000
So, the value of the equipment after 10 years is Rs 105,000.
APPEARS IN
RELATED QUESTIONS
The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of the mth and nth terms is (2m – 1) : (2n – 1)
Which term of the AP `20, 19 1/4 , 18 1/2 , 17 3/4 ` ,..... is the first negative term?
The 7th term of the an AP is -4 and its 13th term is -16. Find the AP.
Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms.
If the numbers (2n – 1), (3n+2) and (6n -1) are in AP, find the value of n and the numbers
Write an A.P. whose first term is a and common difference is d in the following.
a = 6, d = –3
If `4/5` , a, 2 are three consecutive terms of an A.P., then find the value of a.
Suppose the angles of a triangle are (a − d), a , (a + d) such that , (a + d) >a > (a − d).
Q.3
The ratio of the 11th term to the 18th term of an AP is 2 : 3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.