Page 208 - Start Up Mathematics_8 (Non CCE)

Ê 1 ˆ 2 Ê 51ˆ 2
= 50,000 1+ 50¯ ˜ = 50,000 Á 50¯
Á
˜
Ë
Ë
51 51
= 50,000 × × = 52,020
50 50
Hence, the population of the town after 2 years will be 52,020.
Example 18: The population of a city was 4,60,000 in 2009. In 2010, it grew by 4%. In 2011, it dropped
by 5%. In 2012, it again increased by 10%. What will be the population in 2013?
Solution: Let the population in 2013 be P.
Given, Initial population (P ) = 4,60,000
1
Growth rate in 2010 (R ) = 4%
1
Growth rate in 2011 (R ) = –5%
2
Growth rate in 2012 (R ) = 10%
3

Ê R ˆ Ê R ˆ Ê R ˆ
2
3
1
P = P 1+ 100¯ Ë 1+ 100¯ Ë 1+ 100¯ ˜
˜ Á
1Á
˜ Á
Ë
Ê 4 ˆ Ê ( - 5) ˆ Ê 10 ˆ Ê 1 ˆ Ê 1 ˆ Ê 1 ˆ
fi P = 4,60,000 1+ 100¯ Ë 1+ 100 ¯ Ë 1+ 100¯ = 4,60,000 1+ 25¯ Ë 1- 20¯ Ë 1+ 10¯ ˜
˜ Á
˜ Á
Á
˜ Á
Á
˜ Á
˜
Ë
Ë
Ê 26ˆ Ê 19ˆ Ê 11ˆ
= 4,60,000 × Á 25¯ Ë 20¯ Ë 10¯ ˜ = 4,99,928
˜ Á
˜ Á
Ë
Hence, the population in 2013 will be 4,99,928.
Example 19: In a factory, the production of motorcycles increased from 10,000 units to 13,310 units in
3 years. Find the annual rate of growth.
Solution: Present production (A) = 13,310 units, Previous production (P) = 10,000 units
Time (n) = 3 years
Let the annual rate of growth be R%.
Ê R ˆ n
A = P 1+ 100¯ ˜
Á
Ë
Ê R ˆ 3 13 310, Ê R ˆ 3
fi 13,310 = 10,000 1+ 100¯ ˜ fi 10 000, = Á 1+ 100¯
˜
Á
Ë
Ë
Ê R ˆ 3 1 331, Ê 11ˆ 3 R 11
fi 1+ 100¯ ˜ = 1 000, = Á 10¯ fi 1 + 100 = 10
˜
Á
Ë
Ë
R 11 11 10- 1 100
fi = - 1 = = fi R = = 10
100 10 10 10 10
\ The annual rate of growth = 10%
Example 20: Given that carbon-14 (C ) decays at a constant rate in such a way that it reduces to 50% in
14
5,568 years, find the age of an old wooden piece in which the carbon is only 12 1 % of the
original. 2

200