GMAT Challenge Series (700+): Q8

Question:

Pam and Robin each roll a pair of fair, six-sided dice. What is the probability that Pam and Robin will both roll the same set of two numbers?

A. 1/216

B. 1/36

C. 5/208

D. 11/216

E. 1/18

$$\textit{36 ways that Pam or Robin can roll a pair of fair, six-sided dice}$$  $$\textit{Scenario 1: When the two numbers on the dice are the same}$$ $$\textit{Probabillity of a pair} = \frac{6}{36}$$ $$\textit{Probabillity of the same pair} = \left(\frac{6}{36}\right) \left(\frac{1}{36}\right) = \frac{1}{216}$$  $$\textit{Scenario 2: When the two numbers on the dice are different}$$ $$\textit{Probabillity numbers are different} = \frac{30}{36}$$ $$\textit{Probabillity of the same different numbers} = \left(\frac{30}{36}\right) \left(\frac{2}{36}\right) = \frac{10}{216}$$  $$\textit{Probabillity same set of two nunbers} = \frac{1}{216} + \frac{10}{216} = \frac{11}{216}$$

GMAT Challenge Series (700+): Q7

Question:

An alloy of copper and aluminum has 40% copper. An alloy of Copper and Zinc has Copper and Zinc in the ratio 2:7. These two alloys are mixed in such a way that in the overall alloy, there is more aluminum than Zinc, and copper constitutes x% of this alloy. What is the range of values x can take?

A. 30% ≤ x ≤ 40%

B. 32.5% ≤ x ≤ 40%

C. 32.5% ≤ x ≤ 42%

D. 33.33% ≤ x ≤ 40%

E. 33.33 % ≤ x ≤ 42%

$$\text{Alloy 1}$$ $$\textit{Aluminium} = \text{60%}$$ $$\textit{Copper} = \text{40%}$$ $$\textit{Zinc} = \text{0%}$$  $$\text{Alloy 1}$$ $$\textit{Aluminium} = \text{0%}$$ $$\textit{Copper} = \frac{2}{9} \approx \text{22%}$$ $$\textit{Zinc} = \frac{2}{9} \approx \text{78%}$$  $$\text{Mixture of Alloy 1 and Alloy 2}$$ $$\textit{Let y equal the % of Alloy 1 in the mixture}$$ $$\textit{Aluminum in the mixture} = \left( \frac{3}{5} \right) \textit{y}$$ $$\textit{Zinc in the mixture} = \left( \frac{7}{9} \right) \left( {1} - \textit{y} \right)$$ $$\textit{Aluminum in the mixture} > \textit{Zinc in the mixture}$$ $$\left( \frac{3}{5} \right) \textit{x} > \left( \frac{7}{9} \right) \left( {1} - \textit{y} \right)$$ $$\left( \frac{54}{90} \right) \textit{x} > \left( \frac{7}{9} \right) \textit{(1-y)}$$ $${124}\textit{y} > {70}$$ $$\textit{y} > \frac{35}{62}$$  $$\textit{Lowest value of x is when y} = \frac{35}{62}$$ $$\textit{Lowest value of x} = \left( \frac{35}{62} \right)\left( \frac{40}{100} \right) + \left( \frac{27}{62} \right)\left( \frac{2}{9} \right)$$ $$\textit{Lowest value of x} = \frac{14}{62} \frac{6}{62}$$ $$\textit{Lowest value of x} = \frac{20}{62} \approx \textit{32.26%}$$  $$\textit{Highest value of x is when y equals 100% in which case x equals 40%.}$$  $${\textit{Best range: 32.5%} \leq \textit{x}} \leq \textit{40%}$$

B. 32.5% ≤ x ≤ 40%

GMAT Challenge Series (700+): Q6

Question:

Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?

A. 12.5%

B. 25%

C. 31.25%

D. 68.75%

E. 75%

$$\text{Assume that each team has a 50% chance of winning each game.}$$ $$\text{Series will go to game 7 if each team has won 3 games from the first 6 games.}$$ $$\text{Example: A, A, A, B, B, B, ?}$$ $$\textit{Probability(7 Games)} = {{\left(\frac{1}{2}\right)^{6}}{6\,!} \over \left({3\,!}\right)\left({3\,!}\right)}$$ $$\textit{Probability(7 Games)} = {{\left(\frac{1}{64}\right)}{720} \over \left({6}\right)\left({6}\right)}$$ $$\textit{Probability(7 Games)} = \frac{20}{64} = \frac{5}{16} = \textit{31.25%}$$ $$\textit{Probability(Fewer than 7 Games)} = {1} - \textit{Probability(7 Games)}$$ $$\textit{Probability(Fewer than 7 Games)} = {1} - \textit{31.25%} = \textit{68.75%}$$

D. 68.75%

GMAT Challenge Series (700+): Q5

Question:

In racing over a given distance d at uniform speed, A can beat B by 20 yards, B can beat C by 10 yards, and A can beat C by 28 yards. Then d, in yards, equals?

A. Cannot be determined

B. 58

C. 100

D. 116

E. 120

$$\text{Moments:}$$ $$\textit{Moment 1: A crosses the line with B 20 yards behind and C 28 yards behind.}$$ $$\textit{Moment 2: B crosses the line with C 10 yards behind.}$$  $$\text{Takeaways:}$$ $$\textit{Takeway 1: Between moment 1 and 2 B has gone 20 yards and C has gone 18 yards.}$$ $$\textit{Takeway 2: Between moment 1 and 2 B beats C by 2 yards.}$$  $$\textit{Over the distance d, B beats C by 10 yards.}$$ $$\textit{d} = \left(\frac{10}{2}\right) \left({20}\right) = {100}$$

C. 100

GMAT Challenge Series (700+): Q4

Question:

Pipe A and Pipe B fill water into a tank of capacity 1000 liters at the rate of 100 liters/minute and 25 liters/minute respectively. Pipe C drains water from the same tank at the rate of 50 liters/minute. Pipe A is kept open for a minute and then closed, then pipe B is kept open for a minute and then closed, followed by a drain pipe C for a minute and then closed. This process is repeated until the tank is full. How long will it take to fill the tank?

A. 39 min

B. 40 min

C. 39 min 20 sec

D. 37 min

E. 41 min

$$\text{The Cycle Over Time:}$$ $$\textit{0 min: 0L}$$ $$\textit{1 min: 100L (+100L Pipe A)}$$ $$\textit{2 min: 125L (+25L Pipe B)}$$ $$\textit{3 min: 75L (-50L Pipe C)}$$ $$\textit{4 min: 175L (+100L Pipe A)}$$ $$\textit{...}$$  $$\text{There is a 3 min cycle whereby the water rises and then falls.}$$ $$\text{At the end of every 3 min cycle the water has increased by 75L.}$$ $$\textit{est. cycles} = \frac{1000}{75} = {13}\frac{1}{3}$$ $$\textit{est. volume after 13 cycles/39 min} = \left({13}\right) \left({75}\right) = 975$$ $$\text{Since the cylce is not constant we need to check before and after this moment.}$$ $$\textit{...}$$ $$\textit{36 min: 900L (-50L Pipe C)}$$ $$\textit{37 min: 1000L (+100L Pipe A)}$$ $$\textit{38 min: 1000L (+25L Pipe B)}$$ $$\textit{39 min: 950L (-50L Pipe C)}$$ $$\textit{40 min: 1000L (+100L Pipe A)}$$ $$\text{First moment the tank is full is at 37 min.}$$

D. 37 min

GMAT Challenge Series (700+): Q3

Question:

On a partly cloudy day, Derek decides to walk back from work. When it is sunny, he walks at a speed of s miles/hr (s is an integer) and when it gets cloudy, he increases his speed to (s + 1) miles/hr. If his average speed for the entire distance is 2.8 miles/hr, what fraction of the total distance did he cover while the sun was shining on him?

A. 1/4

B. 4/5

C. 1/5

D. 1/6

E. 1/7

$$\text{We know that s equals 2 as s is an integer, and 2.8 must be between s and s + 1.}$$ $$\text{Let x be the percentage of time that it is sunny}$$ $$\left({3}\right) \left({1-x}\right) + \left({2}\right) \left({x}\right) = {2.8}$$ $$\left({3 - 3x}\right) + \left({2x}\right) = {2.8}$$ $${3} - {x} = {2.8}$$ $${x} = {0.2}$$  $$\text{Therefore, 20% of time it is sunny and 80% of time it is cloudy.}$$  $$\text{distance} = \text{speed} \times \text{time}$$ $$\text{total distance} = {2.8} \times \text{t} = \text{2.8t}$$ $$\text{sunny distance} = {{2} \times \text{0.2t}} = \text{0.4t}$$ $${\text{sunny distance} \over \text{total distance}} = {\text{0.4t} \over \text{2.8t}} = \frac{1}{7}$$

E. 1/7

GMAT Challenge Series (700+): Q2

Question:

A box contains three pairs of blue gloves and two pairs of green gloves. Each pair consists of a left-hand glove and a right-hand glove. Each of the gloves is separate from its mate and thoroughly mixed together with the others in the box. If three gloves are randomly selected from the box, what is the probability that a matched set (i.e., a left-hand glove and right-hand glove of the same color) will be among the three gloves selected?

A. 3/10

B. 23/60

C. 7/12

D. 41/60

E. 5/6

$$\text{Gloves:}$$ $$\text{LB = left blue}$$ $$\text{RB = right blue}$$ $$\text{LG = left green}$$ $$\text{RG = right green}$$  $$\text{No. of Gloves:}$$ $$\text{LB: 3}$$ $$\text{RB: 3}$$ $$\text{LG: 2}$$ $$\text{RG: 2}$$  $$\text{Probability(Match)} = {1} - \text{Probability(No Match)}$$  $$\text{No match situations:}$$ $$\text{Probability(LB, LB, LB)} = \left( \frac{3}{10} \right) \left( \frac{2}{9} \right) \left( \frac{1}{8} \right) = \frac{1}{120}$$ $$\text{Probability(RB, RB, RB)} = \frac{1}{120} \textit{ (same as above)}$$ $$\text{Probability(LB, LB, and a G)}= \left( {3} \right) \left( \frac{3}{10} \right) \left( \frac{2}{9} \right) \left( \frac{4}{8} \right) = \frac{1}{10}$$ $$\text{Probability(RB, RB, and a G)} = \frac{1}{10} \textit{ (same as above)}$$ $$\text{Probability(LG, LG, and a B)} = \left( {3} \right) \left( \frac{2}{10} \right) \left( \frac{1}{9} \right) \left( \frac{6}{8} \right) = \frac{1}{20}$$ $$\text{Probability(RG, RG, and a B)} = \frac{1}{20} \textit{ (same as above)}$$

$$\text{Probability(No Match)} = \left(\frac{1}{120} + \frac{1}{120} + \frac{1}{10} + \frac{1}{10} + \frac{1}{20} + \frac{1}{20} \right) = \left(\frac{1}{60} + \frac{12}{60} + \frac{6}{60} \right) = \frac{19}{60}$$

$$\text{Probability(Match)} = {1} - \frac{19}{60} = \frac{41}{60}$$

D. 41/60

GMAT Challenge Series (700+): Q1

Question:

What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

A. 1/6

B. 1/4

C. 1/2

D. 21/216

E. 32/216

Approach 1:

$$\text{Possible options to get 8: 21 ways}$$ $$\text{(6,1,1): 3 ways}$$ $$\text{(5,2,1): 6 ways}$$ $$\text{(4,3,1): 6 ways}$$ $$\text{(4,2,2): 3 ways}$$ $$\text{(3,3,2): 3 ways}$$  $$\text{Possible options to get 14: 15 ways}$$ $$\text{(6,4,4): 3 ways}$$ $$\text{(6,5,3): 6 ways}$$ $$\text{(6,6,2): 3 ways}$$ $$\text{(5,5,4): 3 ways}$$  $$\text{Possible ways to get 8 or 14} = {15} + {21} = {36}$$ $$\text{Total combinations} = {6} \times {6} \times {6} = {216}$$  $$\text{Probability (Sum of 8 or 14 with 3 dice):} = {\text{Possible ways to get 8 or 14} \over \text{Total combinations}}$$ $$\text{Probability (Sum of 8 or 14 with 3 dice):} = \frac{36}{216} = \frac{1}{6}$$

Approach 2:

$$\text{Probabilities of sums with 1 die:}$$ $${1} = \frac{1}{6}$$ $${2} = \frac{1}{6}$$ $${3} = \frac{1}{6}$$ $${4} = \frac{1}{6}$$ $${5} = \frac{1}{6}$$ $${6} = \frac{1}{6}$$  $$\text{Probabilities of sums with 2 dice:}$$ $${2} = \frac{1}{36}$$ $${3} = \frac{2}{36}$$ $${4} = \frac{3}{36}$$ $${5} = \frac{4}{36}$$ $${6} = \frac{5}{36}$$ $${7} = \frac{6}{36}$$ $${8} = \frac{5}{36}$$ $${9} = \frac{4}{36}$$ $${10} = \frac{3}{36}$$ $${11} = \frac{2}{36}$$ $${12} = \frac{1}{36}$$  $$\text{Probability of 8 with 3 dice:}$$ $$\text{Roll a 1 and then 7 on the next two dice}$$ $$\text{Roll a 2 and then 6 on the next two dice}$$ $$\text{...}$$ $$\text{Roll a 6 and then 2 on the next two dice}$$ $$\text{Probability (Sum of 8 with 3 dice):} = \frac{1}{6} \left({{6} + {5} + {4} + {3} + {2} + {1}} \over {36}\right) = \frac{1}{6} \left({21} \over {36}\right) = \frac{21}{216}$$  $$\text{Probability of 14 with 3 dice:}$$ $$\text{Roll a 2 and then 12 on the next two dice}$$ $$\text{Roll a 3 and then 11 on the next two dice}$$ $$\text{...}$$ $$\text{Roll a 6 and then 8 on the next two dice}$$ $$\text{Probability (Sum of 14 with 3 dice):} = \frac{1}{6} \left({{5} + {4} + {3} + {2} + {1}} \over {36}\right) = \frac{1}{6} \left({15} \over {36}\right) = \frac{15}{216}$$  $$\text{Probability (Sum of 8 or 14 with 3 dice):} = \frac{21}{216} + \frac{15}{216} = \frac{36}{216} = \frac{1}{6}$$

A. 1/6

Fake Accounts and Lies… Why I am Parting Ways with Beat the GMAT

I recently joined Beat the GMAT as a Featured Expert (a membership that costs US\$250 per month). My goal was to help the community through being an active member of the MBA social network and in turn promote my skills/business as a GMAT tutor.

My initial feelings were of pride. I was excited about joining an “open and professional community… built on trust, transparency, and professionalism”. With bundles of enthusiasm I began answering questions and giving insightful approaches to problems posted on the forum. My responses even caught the attention of the Beat the GMAT team who reached out to me on numerous occasions to thank me for the quality and volume of my posts.

Being such an eager and active member, however, revealed an alarming trend: the majority (and in fact, very close to all) of the initial questions posted were from fake accounts. The fake accounts had traits that differed greatly from real accounts (see Appendix 1 for evidence):

1. they only posted questions
2. they posted a high volume of questions
3. they posted the same number of questions per day, per forum, at the same time each day
4. they posted easy/repetitive questions, with generic one-line comments at the end, never indicating where they got stuck
5. they did not follow any Featured Experts
6. they did not answer private messages
7. they did not respond to questions regarding their posts

As I started to investigate this further it became apparent that all the most active users were fake (see Appendix 2 for evidence). At this stage the pride I had initially felt dissipated. I honestly felt sick. I reached out to Beat the GMAT with a bold accusation -  that they were artificially manipulating forum activity and profiting as a result. Beat the GMAT’s initial response was to skirt around the issue claiming they had a “small number of volunteers” who may have acted outside their “guidance”. They even offered me expert status “on the house” if I continued to participate in the forums.

Not satisfied, I set up a call with Justin Doff, Director of BTGI. This was going to be interesting! Justin explained that given Beat the GMAT’s low level of forum activity and poor SEO versus its peers the team had decided to pay people to post questions on the forum. Through discussion it was teased out that all the most active users were paid and that this had been going on for over 6 months (most likely much longer).

What was most disturbing was that Justin could not come to terms with why I found this practice unethical. I had signed up to, and agreed to pay to be a part of, a community of users who I thought genuinely wanted help solving GMAT problems. I felt lied to. I no longer trusted Beat the GMAT. If they were doing this, what other unethical practices were they doing to artificially manipulate the number of users, posts, and page views?

It is for these reasons that I have decided to part ways with Beat the GMAT. To the real/genuine users on the site, I will find an alternative way to help you with your GMAT prep. To the Beat the GMAT management team, don’t create fake accounts and lie to your community – it erodes trust, plays with people’s emotions and ultimately ruins reputations.

Appendix 1: Signs of Fake Accounts

1)      Fake accounts only post questions such that "Posts in Last 30 Days" equal "New Topics in Last 30 Days":

2)      Fake accounts post a high volume of questions (see above)

3)      Fake accounts post the same number of questions per day, per forum, at the same time each day. We can see below that user lheiannie07 will generally post 4 questions per day, one in each of the four forums: data sufficiency, problem solving, critical reasoning and sentence correction (in the same order and within the same time frame):

6th March 2018

7th March 2018

8th March 2018

4)      Fake accounts post easy/repetitive questions, with generic one-line comments at the end, never indicating where they got stuck

5)      Fake accounts do not follow any Featured Experts

6)      Fake accounts do not answer private messages

7)      Fake accounts do not respond to questions regarding their posts.  No response to this message or any similar messages:

Appendix 2: All top accounts are fake

The most active users have the same amount of posts to new conversations (this applies to all top users except Roland2rule who has only 3 more posts):