A bag kept some green balls, some blue balls and seven red balls. Probability…
2020
A bag kept some green balls, some blue balls and seven red balls. Probability of picking one green ball is 1/7 more than that of picking one red ball from the bag and probability of picking one blue ball is 9/35 more than that of picking one red ball from the bag. Find total number of blue balls in the bag?
Answer: E. 16 — Let the total number of balls be T. Red balls = 7. P(green) = P(red) + 1/7 = 7/T + 1/7. P(blue) = P(red) + 9/35 = 7/T + 9/35. So green balls = T(7/T + 1/7) =…
- A.
12
- B.
14
- C.
18
- D.
20
- E.
16
Attempted by 8 students.
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Correct answer: E
Let the total number of balls be T. Red balls = 7.
P(green) = P(red) + 1/7 = 7/T + 1/7.
P(blue) = P(red) + 9/35 = 7/T + 9/35.
So green balls = T(7/T + 1/7) = 7 + T/7, and blue balls = T(7/T + 9/35) = 7 + 9T/35.
Using green + blue + red = T gives (7 + T/7) + (7 + 9T/35) + 7 = T.
Thus 21 + 2T/5 = T, so T = 35. Blue balls = 7 + 9 = 16.