Question

3. 60 percent of women and 40 percent of men who took a class to quit smoking remained nonsmokers for at least 1 year after completing the class. Those nonsmokers then attended a success party at the end of the year. If 70% of the original class was male, a. What percentage of the people attending the party were women? b. What percentage of the original class attended the party?

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Rawan Badran

Answer 1

a. 39.13%

Let the original number of women and men attending the class be W and M respectively. Hence the total number oe people attending the class is W+M. We have been given that 70% of the class was men. Hence we have the below

M = 0.7*(W+M) = 0.7W+0.7M

0.3M = 0.7W

M = (0.7/0.3)W = 2.333W........................................Equation 1

Now we know that 60% of women and 40% of men remained non-smokers and attended the party. Hence number of women who attended the party are 0.6W and total number of people who attended the party are 0.6W+0.4M. We need to find out of the total people who attended the party what percentage were women. Hence we have to find below

Percentage = 0.6W 0.6W +0.4M 0.6W 0.6W +0.4 * 2.3333W

We get the above equation by substituting equation 1 for M

$Percentage = \frac{0.6W}{1.5333W} = 0.3913 = 39.13\%$

Thus 39.13% of the people attending the party were women.

b. 46%

We know that 60% of women and 40% of men remained non-smokers and attended the party. total number of people who attended the party are 0.6W+0.4M. The total number of people who attended the class were W+M. Hence we need to find the below percentage

$Percentage = \frac{0.6W+0.4M}{W+M} = \frac{0.6W+0.4*2.333W}{W+2.333W}$

$Percentage = \frac{1.5333W}{3.333W} =0.46 = 46\%$

Hence 46% of the original class attended the party