A) 9,00,000
B) 5.00,000
C) 6,00,000
D) 7,00,000
Correct Answer: D
Solution :
Let the population two years ago \[=x\] After 1 year, it remained \[=x-\frac{6}{100}x\] After 2 years, it remained \[=\left( x-\frac{6x}{100} \right)-\left( x-\frac{6x}{100} \right)\times \frac{4}{100}\] Since, present population = population after 2 years \[\Rightarrow 631680=x-\frac{6x}{100}-\frac{4x}{100}+\frac{24x}{10000}\] \[\Rightarrow 631680=\frac{10000x-600x-400x+24x}{10000}\] \[\Rightarrow 631680=\frac{9024x}{10000}\] \[\Rightarrow \frac{631680\times 10000}{9024}=x\] \[\Rightarrow x=700000\] Thus, the population two years ago was 700000
You need to login to perform this action.
You will be redirected in
3 sec
