The following table shows the population (in lakhs) of four cities (A, B, C and D)

over the past four years. Study the table and answer the question.

Which city has registered the highest percentage of population growth from 2018-2019 to 2020-2021?

Correct option is B

Given:

The populationof four cities (A, B, C and D)over the past four years

Formula Used:

Increased Percent = $$\frac{Increased\ Value}{Original \ Value} \times 100$$​​

Solution:

For city A population increase from 2018-19 to 2020-21 = 92-78 = 14lakh

Increase percent =$$\frac{14}{78} \times 100$$​ = 17.94%

For city B population increase from 2018-19 to 2020-21 = 84-61 = 23lakh

Increase percent =$$\frac{23}{61} \times 100$$​= 37.70%

For city C population increase from 2018-19 to 2020-21 = 92-71 = 21lakh

Increase percent = $$\frac{21}{71} \times 100$$​= 29.57%

For city D population increase from 2018-19 to 2020-21 = 95-72 = 23lakh

Increase percent =$$\frac{23}{72} \times 100$$= 31.94%

So the highest percent increase of population from 2018-19 to 2020-21 is for city B at 37.70%