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There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 00:44
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There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors? (A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42
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There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 05:37
LevanKhukhunashvili wrote: My skills were torn apart while tackling this question; very tough one. I'm not gifted to see any shortcuts or to feel myself in the skin of "overlapping sets". Greetings "Brute Force", my old friend. If g is the number of green balls and r is the number of red balls g+\(\frac{5r}{7}\)=58 (we know that \(\frac{2r}{7}\) balls have both colors, so only red will be 1\(\frac{2}{7}\)=\(\frac{5}{7}\)) r+\(\frac{4g}{7}\)=58 Two equations, two unknowns > child game, BUT I could not solve for g and r. Maybe I go wrong somewhere, maybe these two equations are not correlated, I can't say. In the first equation, I just inserted multiples of 7 for r, so I deduced to r=42 g=28 58=42+28both Both=12 Probability of picking multicolor ball: \(\frac{12}{58}\)=\(\frac{6}{29}\) IMO Ans: DSir chetan2u if you find some time, maybe you can introduce some light in this "dark" question Hi.. You have gone correct till the equations are formed... Now total are 58 balls... If 2/7 of red are painted blue too, so 5r/7 are painted only red.So total =58=g+5r/7 Now the balls painted with both colors are equal to 2r/7 and 3g/7, so both have to be equal.. 2r/7=3g/7 or r=(3/2)g.. Now substitute this value in any of the equations.. g+\(\frac{5r}{7}\)=58....g+\(\frac{5}{7}*\frac{3g}{2}\)=58.......g+\(\frac{15g}{14}\)=58..........\(\frac{29g}{14}\)=58...g=58*14/29=28.. So, both =3g/7=3*28/7=12 Probability=12/58=6/29
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There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 01:16
Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 GMATinsight sir how can we solve this question using cases or pairs ? my take below: total red with green ; 2/7 and only red ; 5/7 total green with red ; 3/7; only green ; 4/7 total balls = 58 probability that a ball randomly picked from the jar will have both red and green colors total both ; 2/7+3/7 ; 5/7 I am not able to solve this question any further ..



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There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 05:10
My skills were torn apart while tackling this question; very tough one. I'm not gifted to see any shortcuts or to feel myself in the skin of "overlapping sets". Greetings "Brute Force", my old friend. If g is the number of green balls and r is the number of red balls g+\(\frac{5r}{7}\)=58 (we know that \(\frac{2r}{7}\) balls have both colors, so only red will be 1\(\frac{2}{7}\)=\(\frac{5}{7}\)) r+\(\frac{4g}{7}\)=58 Two equations, two unknowns > child game, BUT I could not solve for g and r. Maybe I go wrong somewhere, maybe these two equations are not correlated, I can't say. In the first equation, I just inserted multiples of 7 for r, so I deduced to r=42 g=28 58=42+28both Both=12 Probability of picking multicolor ball: \(\frac{12}{58}\)=\(\frac{6}{29}\) IMO Ans: DSir chetan2u if you find some time, maybe you can introduce some light in this "dark" question



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There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 05:12
Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 Please check the solution as per attachment Answer: Option D Archit3110
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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 06:52
GMATinsight ; great solution sir, thanks.. GMATinsight wrote: Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 Please check the solution as per attachment Answer: Option D Archit3110



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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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16 Apr 2019, 17:50
Honestly, I got stuck on this one. So when I noticed that 29 is the only factor of 58 that shows up in the denominators of the answers, I just guessed that one. It worked.



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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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17 Apr 2019, 04:43
GMATinsight wrote: Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 Please check the solution as per attachment Answer: Option D Archit3110Thank you for the solution GMATinsight but why is the sum of balls not 58? (R=21 , B=14 and R+G=6) . I am confused.



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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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17 Apr 2019, 04:50
neha283 wrote: GMATinsight wrote: Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 Please check the solution as per attachment Answer: Option D Archit3110Thank you for the solution GMATinsight but why is the sum of balls not 58? (R=21 , B=14 and R+G=6) . I am confused. Hi neha283That number 58 is actually a redundant information for this question. Because we only needed a ratio of two numbers in same units. However to avoid your doubts. What I assumed R=21 is actually R=42 G=14 is actually G = 28 And Common balls instead of 6 is 12 So total balls = 42+2812 = 58 I hope this helps now...
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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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17 Apr 2019, 18:01
Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 We and let r, g, and b be the number of balls that have red color only, green color only, and both colors, respectively. We can create the equations: r + g + b = 58, b = (2/7)(r + b) And b = (3/7)(g + b) Let’s simplify the second equation by multiplying it by 7: 7b = 2r + 2b 5b = 2r r = (5/2)b Likewise, let’s simplify the third equation by multiplying it by 7: 7b = 3g + 3b 4b = 3g g = (4/3)b So plug (5/2)b for r and (4/3)b for g into the first equation, we have: (5/2)b + (4/3)b + b = 58 15b + 8b + 6b = 58 x 6 29b = 58 x 6 b = 2 x 6 = 12 Therefore, the probability of picking a ball with both colors is 12/58 = 6/29. Answer: D
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There are 58 balls in a jar. Each ball is painted with at least one of
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20 Jun 2020, 06:11
ScottTargetTestPrep wrote: Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 We and let r, g, and b be the number of balls that have red color only, green color only, and both colors, respectively. We can create the equations: r + g + b = 58, b = (2/7)(r + b) And b = (3/7)(g + b) Let’s simplify the second equation by multiplying it by 7: 7b = 2r + 2b 5b = 2r r = (5/2)b Likewise, let’s simplify the third equation by multiplying it by 7: 7b = 3g + 3b 4b = 3g g = (4/3)b So plug (5/2)b for r and (4/3)b for g into the first equation, we have: (5/2)b + (4/3)b + b = 58 15b + 8b + 6b = 58 x 6 29b = 58 x 6 b = 2 x 6 = 12 Therefore, the probability of picking a ball with both colors is 12/58 = 6/29. Answer: D How did you arrive at the two equations Sir ? Can we solve this using an easier alternate method ?



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20 Jun 2020, 09:55
Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 I would like to offer a solution WHICH DOES NOT INVOLVE ANY SKILL RELATED TO PROBABILITY.The set of answer choices provided makes the problem very easy. I noticed one of the posts which has this sentence: "That number 58 is actually a redundant information for this question". However, with the given set of answer choices, one only needs to know that “There are 58 balls in a jar.”
Whatever probability is requested, it is sure that the value can be represented as a fraction  which has 58 in the denominator and any number between 1 and 58 in the numerator.
The simplified form of the fraction can only have 2 values: 58 IF NUMERATOR IS ODD 29 IF NUMERATOR IS EVEN (you can pick any number at random and verify this.) Therefore, the answer will have either 29 or 58 in the denominator.
If there be an answer choice such as 15/58 then the approach could be different.
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Re: There are 58 balls in a jar. Each ball is painted with at least one of
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17 Jul 2020, 03:25
ishitadas1994@gmail.com wrote: ScottTargetTestPrep wrote: Bunuel wrote: There are 58 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What is the probability that a ball randomly picked from the jar will have both red and green colors?
(A) 6/14 (B) 2/7 (C) 6/35 (D) 6/29 (E) 6/42 We and let r, g, and b be the number of balls that have red color only, green color only, and both colors, respectively. We can create the equations: r + g + b = 58, b = (2/7)(r + b) And b = (3/7)(g + b) Let’s simplify the second equation by multiplying it by 7: 7b = 2r + 2b 5b = 2r r = (5/2)b Likewise, let’s simplify the third equation by multiplying it by 7: 7b = 3g + 3b 4b = 3g g = (4/3)b So plug (5/2)b for r and (4/3)b for g into the first equation, we have: (5/2)b + (4/3)b + b = 58 15b + 8b + 6b = 58 x 6 29b = 58 x 6 b = 2 x 6 = 12 Therefore, the probability of picking a ball with both colors is 12/58 = 6/29. Answer: D How did you arrive at the two equations Sir ? Can we solve this using an easier alternate method ? Response:First, notice that r denotes the number of balls that have only red color, g denotes the number of balls that have only green color and b denotes the number of balls that have both red and green colors. The total number of balls is red only + green only + both; therefore, the total number of balls is r + g + b. The first equation is obtained by simply equating this quantity to 58. For the equation b = (2/7)(r + b), we use the sentence “2/7 of the balls that have red color also have green color.” Notice that the balls that have red color is r + b (i.e. only red + both) and if a ball that contains the red color also contains the green color, then it contains both colors. Thus, 2/7 of r + b should equal b and that’s how we obtain the equation (2/7)(r + b) = b. The equation b = (3/7)(r + b) is obtained similarly. Alternate Solution:As an alternate solution, let the number of balls that have both the colors be 6n. We choose the coefficient 6 because LCM(2, 3) = 6. Since 2/7 of the balls that have red color also have green color, there are 6n / (2/7) = 7 * 3n = 21n balls with red color (including the ones which also have green color). Similarly, since 3/7 of the balls that have green color also have red color, there are 6n / (3/7) = 7 * 2n = 14n balls with green color (including the ones which also have red color). We will use the following formula: Total = #(red) + #(green)  #(both) 58 = 21n + 14n  6n 58 = 29n n = 2 Thus, there are 6n = 6*2 = 12 balls containing both colors. Since the total number of balls is 58, the probability that a randomly selected ball contains both colors is 12/58 = 6/29. Answer: D
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