Question

An auto repair company is advertising on television and on the internet. It hires a statistician to poll 500050005000 people across the country at the beginning of every month for six months. Eight hundred people saw the advertisement on television at the beginning of the first month with an increase of 200200200 people each month. Also, 100010001000 people saw the advertisement on the internet at the beginning of the first month with an increase of xxx percent each month. At the beginning of the third month, there were 240240240 more people who saw the advertisement on the internet than on television. To the nearest percent, what is the value of xxx

Answer 1

Answer:

xxx = 20%

Step-by-step explanation:

No. of people who saw the advert on television:

At the beginning of the first month = 800

Monthly increase = 200

At the beginning of the third month = 1,200 (800 + 200 * 2)

If 240 more people saw the advert on the internet than on television at the beginning of the third month,

This implies that, at the beginning of the third month, 1,440 (1,200 + 240) saw the advert on the internet, increasing from 1,000 the previous two months.

Therefore, the rate increase in two months = 1,440/1,000 = 1.44

and the square root of 1.44 = 1.2

xxx = 0.2 (1.2 - 1) = 20%