The population of vultures in a particular locality decreases by a certain rate of interest (compounded annually). If the current population of vultures be 29160 and the ratio of decrease in population for the second year and 3rd year be 10 : 9. What was the population of vultures 3 years ago?

30000

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35000

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40000

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50000

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Solution

The correct option is C 40000
$\frac{D e c r e a s e s i n s e c o n d y e a r}{D e c r e a s e s i n t h i r d y e a r} = \frac{100}{100 - r} = \frac{10}{9} \Rightarrow r = 10 %$
Let the population of vultures 3 years ago be P, then
$P {(1 - \frac{10}{100})}^{3} = 29160$
$\Rightarrow$ P = 40000

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The population of vultures in a particular locality decreases by a certain rate of interest (compounded annually). If the current population of vultures be 29160 and the ratio of decrease in population for the second year and 3rd year be 10 : 9. What was the population of vultures 3 years ago?

The correct option is C 40000Decreases in second yearDecreases in third year=100100−r=109⇒ r=10% Let the population of vultures 3 years ago be P, then P(1−10100)3=29160 ⇒ P = 40000

The correct option is C 40000
$\frac{D e c r e a s e s i n s e c o n d y e a r}{D e c r e a s e s i n t h i r d y e a r} = \frac{100}{100 - r} = \frac{10}{9} \Rightarrow r = 10 %$
Let the population of vultures 3 years ago be P, then
$P {(1 - \frac{10}{100})}^{3} = 29160$
$\Rightarrow$ P = 40000