Question

# If the population of a town is $$64000$$ and its annual increase is $$10\%$$, then its correct population at the end of $$3$$ years will be

A
8000
B
85000
C
85100
D
85184

Solution

## The correct option is D $$85184$$Population at the end of $$3$$ years.$$= 64000{ \left( 1+\cfrac { 10 }{ 100 } \right) }^{ 3 }$$ $$=64000\times \cfrac { 11 }{ 10 } \times \cfrac { 11 }{ 10 } \times \cfrac { 11 }{ 10 }$$$$=85184$$Maths

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