Question

In 2000, the circulation of a local newspaper was 3,250. In 2001, its circulation was 3,640. In 2002, the circulation was 4,100.

Find the percent of increase in the newspaper’s circulation from 2000 to 2001 and from 2001 to 2002.
Which period had a higher percent of increase, 2000 to 2001, or 2001 to 2002?

Answer 1

2000 - 2001: 3640 - 3250 = 390

2001 - 2002: 4100 - 3640 = 460

390 divided by 3250 = 0.12

460 divided by 3640 = 0.1263736 = 1.26

Answer 2

Percent increase from 2000 to 2001 is 12% and from 2001 to 2002 is 12.6%. Thus period of 2001 to 2002 has higher increase percentage.

Solution:

Given that,

Total circulation of local newspaper in 2000 = 3250

Total circulation of local newspaper in 2001 = 3640

Total circulation of local newspaper in 2002 = 4100

Finding percent of increase in the newspaper’s circulation from 2000 to 2001:

`\text { Percent increase from } 2000 \text { to } 2001=\frac{\text { change in circulation from 2001 - 2000 }}{\text { circulation in } 2000} * 100`

`\begin{array}{l}{=(3640-3250)/(3250) * 100} \\\\ {=(390)/(3250) * 100=12 \%}\end{array}`

Finding percent of increase in the newspaper’s circulation from 2001 to 2002:

`\begin{array}{l}{\text { Percent increase from } 2001 \text { to } 2002=\frac{\text { change in circulation }}{\text {circulation in } 2001} * 100} \\\\ {=(4100-3640)/(3640) * 100} \\\\ {=(460)/(3640) * 100=12.6 \%}\end{array}`

As we can see 12% < 12.6%

So period of 2001 to 2002 has higher increase percentage.