rohan2345 wrote:
Of the students in a certain school, 65 percent are enrolled in an art class and 70 percent are enrolled in a music class. What percent of the students in the school are enrolled in both an art class and a music class?
(1) 25 percent of the students are enrolled in only a music class.
(2) The ratio of the number of students who are enrolled in both an art class and a music class to the number of students who are enrolled in neither an art class nor a music class is 9 to 2.
Let us suppose the following,
Total students = x,
People enrolled in only art class = a,
People enrolled in only music class = c,
People enrolled in both art and music class = b
From prompt we know,
a+b = 0.65x, --- (1)
c+b = 0.70x --- (2)
we need to find
b as a percent of xStatement 1: 25 percent of the students are enrolled in only a music class.
Okay, so we are given,
c = 0.25x --- (3)
From (2) and (3), we can easily find b.
Hence sufficientStatement 2: The ratio of the number of students who are enrolled in both an art class and a music class to the number of students who are enrolled in neither an art class nor a music class is 9 to 2.
On translating this into variables we get,
\(\frac{b}{x-(a+b+c)} = \frac{9}{2}\)
Adding and removing the variable 'b' ,
\(\frac{b}{x-((a+b)+(b+c)-b)} = \frac{9}{2}\) --- (4)
From (1), (2) and (4), we get
\(\frac{b}{x-(0.65x+0.75x - b)} = \frac{9}{2}\)
\(\frac{b}{x - 1.35x + b} = \frac{9}{2}\)
\(\frac{b}{- 0.35x + b} = \frac{9}{2}\)
From this we should be able to take out 'b' as a percent of 'x'.
Hence sufficientAnswer = D _________________
Regards,
AD
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