Page 211 - Start Up Mathematics_8 (Non CCE)

Example 23: A new computer costs ` 80,000. Due to advancement in technology, its value depreciates
every year. If the rates of depreciation are 4%, 5% and 10% in the three successive years,
find its value after 3 years.
Solution: Original value of the computer (P) = ` 80,000
Rate of depreciation in the first year = 4%
Rate of depreciation in the second year = 5%
Rate of depreciation in the third year = 10%
Let the depreciated value be A.

Ê 4 ˆ Ê 5 ˆ Ê 10 ˆ Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ
\ A = 80,000 1- 100¯ Ë 1- 100¯ Ë 1- 100¯ ˜ = 80,000 1- 25¯ Ë 1- 20¯ Ë 1- 10¯
Á
˜ Á
˜ Á
˜ Á
Á
˜ Á
˜
Ë
Ë
Ê 24ˆ Ê 19ˆ Ê 9 ˆ
= 80,000 Á Ë 25¯ Ë 20¯ Ë 10¯ ˜ = 65,664
˜ Á
˜ Á
Hence, the depreciated value of the computer after 3 years is ` 65,664.
EXERCISE 13.5
1. The cost of a generator is ` 1,60,000. If its value depreciates at the rate of 5% per annum, what will
be its value after: (a) 2 years (b) 3 years
2. Sonika bought land for ` 5,50,000. The value of the land depreciates at the rate of 5% every six
months. What is the value of the land after 2 years?
3. The price of silver at present is ` 8,000 per kg. Its price increases at the rate of 5% in the first year,
decreases at the rate of 10% in the second year and again decreases further at the rate of 4% in the
third year. Find its price per kg at the end of the third year.

4. A car costs ` 2,40,000. Its value depreciates at the rate of 5% every year. If after some time period,
its value is ` 2,05,770, find the time period.
5. The cost of a new washing machine is ` 22,000. Its value depreciates every year. If the depreciated
value of the washing machine after 2 years is ` 14,080, find the rate of depreciation.

AT a Glance

1. When P Æ Principal, R% Æ Rate of interest per annum, A Æ Amount, n Æ number of years,
C.I. Æ Compound Interest, m Æ number of times interest is compounded and R %, R %, R %, ..., R %
n
2
3
1
rate of interest for 1st, 2nd, 3rd, ..., nth year then,
Ô
Ê R ˆ n Ï Ê Ô R ˆ n ¸
(a) if the interest is compounded annually, A = P 1+ 100¯ ˜ and C.I. = A – P = P Á Ì Ë 1+ 100¯ ˜ - 1 ˝
Á
Ë
Ó Ô
˛ Ô
(b) if the interest is compounded m times in a year,
Ï
¸
R
ˆ
Ô
R
ˆ
Ê
A = P 1+ 100 ¯ mn and C.I. = A – P = Ì Ê Ô Á Ë 1+ 100 ¯ mn - 1 ˝
˜
˜
Á
Ë
m
m
˛ Ô
Ó Ô
(c) if different years have different rates of interest,
Ê R ˆ Ê R ˆ Ê R ˆ Ê R ˆ
n
1
3
2
A = P 1+ 100¯ Ë 1+ 100¯ Ë 1+ 100¯ ˜ ... 1+ 100¯ ˜ and C.I. = A – P
˜ Á
Á
˜ Á
Á
Ë
Ë
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