Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
16 Sep 2014, 00:39
Kudos
Bookmarks
A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
(2) There are 5 more green chips in the bowl than blue chips.
(1) The ratio of blue chips to green chips in the bowl is 3:4.
(2) There are 5 more green chips in the bowl than blue chips.
16 Sep 2014, 00:39
Kudos
Bookmarks
Official Solution:
A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
If \(B=3\) and \(G=4\), then \(P=\frac{3}{7}*\frac{2}{6}\), but if \(B=6\) and \(G=8\), then \(P=\frac{6}{14}*\frac{5}{13}\). We get two different answers, so this information is not sufficient.
(2) There are 5 more green chips in the bowl than blue chips.
So, \(G=B+5\). This information is clearly insufficient by itself.
(1)+(2) Combining the information, we have \(B:G=3:4\) and \(G=B+5\). Solving this system of equations gives us: \(B=15\) and \(G=20\). With this information, we can find the probability: \(P=\frac{15}{35}*\frac{14}{34}\). Sufficient.
Answer: C
A bowl contains only green and blue chips. If two chips are drawn from the bowl at random and without replacement, what is the probability that both chips will be blue?
(1) The ratio of blue chips to green chips in the bowl is 3:4.
If \(B=3\) and \(G=4\), then \(P=\frac{3}{7}*\frac{2}{6}\), but if \(B=6\) and \(G=8\), then \(P=\frac{6}{14}*\frac{5}{13}\). We get two different answers, so this information is not sufficient.
(2) There are 5 more green chips in the bowl than blue chips.
So, \(G=B+5\). This information is clearly insufficient by itself.
(1)+(2) Combining the information, we have \(B:G=3:4\) and \(G=B+5\). Solving this system of equations gives us: \(B=15\) and \(G=20\). With this information, we can find the probability: \(P=\frac{15}{35}*\frac{14}{34}\). Sufficient.
Answer: C
10 Oct 2017, 23:04
Kudos
Bookmarks
This is how I approached this.
Given: Bowl contains only Blue and Green Chips. Replacement is not allowed. This is Value question and not yes/no
S1: Blue Chips : Green Chip = 3x : 4x
=> Probability of picking two blue =\(\frac{3x}{7x} * \frac{(3x-1)}{(7x-1)}\). This cannot be solved without knowing the value of x. Insufficient.
S2: Clearly insufficient. Blue and Green could be any number. The probability will be different for different cases.
S1+S2 : We have \(4x - 3x = 5 => x = 5\). Knowing x and using S1, we can now find the probablity. Hence sufficient.
Final answer: C
Given: Bowl contains only Blue and Green Chips. Replacement is not allowed. This is Value question and not yes/no
S1: Blue Chips : Green Chip = 3x : 4x
=> Probability of picking two blue =\(\frac{3x}{7x} * \frac{(3x-1)}{(7x-1)}\). This cannot be solved without knowing the value of x. Insufficient.
S2: Clearly insufficient. Blue and Green could be any number. The probability will be different for different cases.
S1+S2 : We have \(4x - 3x = 5 => x = 5\). Knowing x and using S1, we can now find the probablity. Hence sufficient.
Final answer: C
