Page 209 - Start Up Mathematics_8 (Non CCE)

Solution: Let the rate of decay be R% per annum and the age of the wooden piece be n years.
Let the amount of carbon (originally) in the wooden piece be P.

P
Then, in 5,568 years, the amount left =
2

P Ê R ˆ 5568 1 Ê R ˆ 5568
\ = P 1- ˜ fi = 1- ˜ ...(1)
Á
Á
2 Ë 100¯ 2 Ë 100¯
1 25 P
After n years, the carbon left in the wooden piece = 12 % P = of P =
2 200 8
P Ê R ˆ n 1 Ê R ˆ n
\ = P 1- ˜ fi = 1- ˜
Á
Á
8 Ë 100¯ 8 Ë 100¯
R
Ê
1ˆ
fi Ê Á ˜ 3 = 1- 100¯ ˆ ˜ n ...(2)
Á
Ë
Ë
2¯
From (1) and (2), we get
Ï Ê Ô R ˆ 5568 ¸ 3 Ê R ˆ n
Ô
Ì Á 1- ˜ ˝ = Á 1- ˜
Ó Ë Ô 100¯ ˛ Ô Ë 100¯
Ê R ˆ 5568 ¥ 3 Ê R ˆ n
fi 1- 100¯ ˜ = Á Ë 1- 100¯ ˜
Á
Ë
R
R
Ê
ˆ
ˆ
fi 1- 100¯ 16704 = Ê 1- 100¯ n
˜
˜
Á
Á
Ë
Ë
fi n = 16,704 ( If the bases are same, we can equate the powers.)
Hence, the age of the wooden piece is 16,704 years.
Example 21: The population of a place is increased to 54,000 in 2012 at the rate of 5% per annum.
Find the population in 2010. (NCERT)
Solution: Population of a place in 2012 (A) = 54,000, Rate of increase = 5%
Let the population in 2010 be P .
0
Number of years (n) = 2 years
Ê R ˆ n
\ A = P 1+ 100¯ ˜
0Á
Ë
Ê 5 ˆ 2 Ê 1 ˆ 2 Ê 21ˆ 2 Ê 21 21ˆ
fi 54,000 = P 1+ 100¯ ˜ = P 1+ 20¯ = P 0Á 20¯ ˜ = P 0Á Ë 20 ¥ 20¯ ˜
˜
0Á
0Á
Ë
Ë
Ë
441
fi P = 54,000
400 0
54 000 400, ¥
fi P = = 48,979.59
0
441
fi P = 48,980 (approx.)
0
Hence, the population in 2010 = 48,980 (approx.)

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