The population of Hermitville increased from 142,340 to 186,480 in the last 5 years. During the same time period, Crabville incr

Question

3 years ago

6

The population of Hermitville increased from 142,340 to 186,480 in the last 5 years. During the same time period, Crabville incr

eased its population by 26%. Which town is increasing at the greatest rate and by what factor? (round to nearest hundredth)
A) Hermitville by a factor of 2.71
B) Hermitville by a factor of 1.19
C) Crabville by a factor of 2.71
D) Crabville by a factor of 1.19

Mathematics

2 answers:

goblinko [34] · Answer 1 · 2021-12-07T14:58:27+03:00

<span>B) Hermitville by a factor of 1.19 Let's see by what percentage Hermitville increased it's population: (186480 - 142340)/142340 = 44140/142340 = 0.310102571 = 31.0102571% Since 31% is greater than 26%, that means that Hermitville is growing faster. Now by what factor? For that, we can simply divide the larger value by the smaller, so: 31.0102571 / 26 = 1.192702197 And looking at the available optoins, option "B" matches.</span>

Musya8 [376] · Answer 2 · 2021-12-07T12:32:42+03:00

The rate of increase of population of Hermitville is given by:
PR=(Present Population-Past Population)/(Past population)*100
PR=(186480-142340)/142340×100
PR=31.01%
The population increase in Hermitville is 31.01;
This implies that it's population growth is faster than that of Crabville by a factor of
31.01/26=1.19.
The answer is B