  Question

The population of a particular area of a city is $$5000$$. It increases by $$10 \%$$ in $$1^{st}$$ year. It decreases by $$20 \%$$ in the $$2^{nd}$$ year because of some reason. In the $$3^{rd}$$ year, the population increases by $$30 \%$$. What will be the population of area at the end of $$3$$ years?

A
5120  B
5300  C
5720  D
5620  Solution

The correct option is C $$5720$$Population of a city at the starting of first year $$=5000$$It increases by $$10\%$$ in $$1^{st}$$ year.i.e., Population at the end of $$1^{st}$$ year $$=$$ Population at the starting of $$2^{nd}$$ year $$=5000+10\%\space of\space5000\\=5000+\dfrac{10}{100}\times5000\\=5500$$It decreases by $$20\%$$ in $$2^{nd}$$ year.i.e., Population at the end of $$2^{nd}$$ year $$=$$ Population at the starting of $$3^{rd}$$ year $$=5500-20\%\space of\space5500\\=5500-\dfrac{20}{100}\times5500\\=5500-1100\\=4400$$It increases by $$30\%$$ in $$3^{rd}$$ year.i.e., Population at the end of $$3^{rd}$$ year $$=4400+30\%\space of\space4400\\=4400+\dfrac{30}{100}\times4400\\=4400+1320\\=5720$$So, $$\text{C}$$ is the correct option.Mathematics

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