A committee is reviewing a total of \(20x\) black-and-white

[VJesus12]

A committee is reviewing a total of \(20x\) black-and-white films and \(6y\) color films for a festival. If the committee selects \(\dfrac{y}x\%\) of the black-and-white films and all of the color films, what fraction of the selected films are in color?

A. \(\dfrac1{130}\)
B. \(\dfrac15\)
C. \(\dfrac3{13}\)
D. \(\dfrac{10}{13}\)
E. \(\dfrac{30}{31}\)

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

[swerve]

A committee is reviewing a total of \(20x\) black-and-white films and \(6y\) color films for a festival. If the committee selects \(\dfrac{y}x\%\) of the black-and-white films and all of the color films, what fraction of the selected films are in color?

A. \(\dfrac1{130}\)
B. \(\dfrac15\)
C. \(\dfrac3{13}\)
D. \(\dfrac{10}{13}\)
E. \(\dfrac{30}{31}\)

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

Use smart numbers for this one

\(y=100\)
\(x=2\)

Then Color films 600
Black and White are 20

Color/Total \(= \frac{600}{620}=\frac{30}{31}\)

[Scott@TargetTestPrep]

A committee is reviewing a total of \(20x\) black-and-white films and \(6y\) color films for a festival. If the committee selects \(\dfrac{y}x\%\) of the black-and-white films and all of the color films, what fraction of the selected films are in color?

A. \(\dfrac1{130}\)
B. \(\dfrac15\)
C. \(\dfrac3{13}\)
D. \(\dfrac{10}{13}\)
E. \(\dfrac{30}{31}\)

[spoiler]OA=B[/spoiler]

Source: Manhattan GMAT

We can let y = 10 and x = 1. So there are a total of 20 black-and-white films and 60 color films. Thus 0.1 x 20 = 2 black-and-white films and 60 color films are selected. Therefore, the fraction of color films selected is 60/62 = 30/31.

Alternate Solution:

y/x% of 20x is 20x*[(y/x)/100] = [20x*(y/x)]/100 = 20y/100 = y/5. Thus, a total of y/5 + 6y = y/5 + 30y/5 = 31y/5 films are selected, 6y of which are color films. Then, the fraction of color films selected is 6y/(31y/5) = 30y/31y = 30/31.

Answer: E