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A certain bag contains red balls, green balls, and blue ball
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06 Nov 2019, 09:34
06 Nov 2019, 09:34
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Question Stats:
99% (01:14) correct
0% (00:00) wrong based on 14 sessions
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A certain bag contains red balls, green balls, and blue balls and no other balls. The ratio of the number of red balls to the number of blue balls is 2:3, and the ratio of the number of blue balls to the number of green balls is 4:3. The number of blue balls in the bag is what fraction of the total number of balls in the bag?
A. \(\frac{3}{8}\)
B. \(\frac{12}{29}\)
C. \(\frac{7}{13}\)
D. \(\frac{15}{23}\)
E. \(\frac{12}{17}\)
Kudos for the right answer and explanation
A. \(\frac{3}{8}\)
B. \(\frac{12}{29}\)
C. \(\frac{7}{13}\)
D. \(\frac{15}{23}\)
E. \(\frac{12}{17}\)
Kudos for the right answer and explanation
Re: A certain bag contains red balls, green balls, and blue ball
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05 May 2020, 05:03
05 May 2020, 05:03
1
Carcass wrote:
A certain bag contains red balls, green balls, and blue balls and no other balls. The ratio of the number of red balls to the number of blue balls is 2:3, and the ratio of the number of blue balls to the number of green balls is 4:3. The number of blue balls in the bag is what fraction of the total number of balls in the bag?
A. \(\frac{3}{8}\)
B. \(\frac{12}{29}\)
C. \(\frac{7}{13}\)
D. \(\frac{15}{23}\)
E. \(\frac{12}{17}\)
Kudos for the right answer and explanation
A. \(\frac{3}{8}\)
B. \(\frac{12}{29}\)
C. \(\frac{7}{13}\)
D. \(\frac{15}{23}\)
E. \(\frac{12}{17}\)
Kudos for the right answer and explanation
Here,
Ratio of RED : BLUE = 2 : 3
Ratio of BLUE : GREEN = 4 : 3
Now, we have to make the blue common :
R : B
2 : 3
8 : 12
B : G
4 : 3
12 : 9
or R :B : G = 8 : 12 : 9
Total nos. = 8 + 12 + 9 = 29
No. of blue balls = 12
Fraction of no. of blue balls to the total number of balls in the bag= \(\frac{12}{29}
\)
gmatclubot
Re: A certain bag contains red balls, green balls, and blue ball [#permalink]
05 May 2020, 05:03
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