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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Probability and Time Distance Question [#permalink]

Show Tags

20 Jul 2009, 00:52

For q1;

The total time is = time traveled 60 mph + time traveled 30 mph. (x-5)/60 hours + 5/30 hours = (x+5)/60 hours. if he had traveled at a constant 60 mph speed. His travel time would have been x/60 hours. Thus we get the percentage => Time difference = (x+5)/60-x/60= 5/60 hours = 1/12 hours. Time for constant travel = x/60 hours. So the percentage difference is (1/12).100 / (x/60)= 5.100/x = 500/x %

Re: Probability and Time Distance Question [#permalink]

Show Tags

20 Jul 2009, 00:56

For question 2; There are 4 cases; 1- St.A plays good 2- St.A plays bad but st.B plays good 3- St.A and St.B plays bad but st.C plays good. 4-All plays bad. Prob of 1st case is 0,3 Prob of 2nd case is 0,7.0,3 Prob of 3rd case is 0,7.0,7.0,3 Prob of 4th case is 0,7.0,7.0,7 Sum of all is 0,3 + 0,21 + 0,147 + 0,343 = 1 In only the last case the song is bad. So the prob. that a good song is played is 1-0,343=0,657 D

Re: Probability and Time Distance Question [#permalink]

Show Tags

20 Jul 2009, 02:31

Q2) Prob of Leo to hear a song he likes = Prob of Radio A playing songs which Leo likes +Prob of Radio B playing songs which Leo likes +Prob of Radio C playing songs which Leo likes = 0.3 + 0.3 * 0.7 +0.3 *0.3 *0.7 = 0.3+0.21+0.147 =0.657 Option D

Re: Probability and Time Distance Question [#permalink]

Show Tags

20 Jul 2009, 02:39

Time spent with 5 mile@30mph = X-5/60 + 5/30 = (X+5)/60 Time spent with entire trip @60mph = X/60. Time Diff = 1/12 Q: % of diff with total time @60mph: (1/12) / (X/60)*100 = (500/X)%

Re: Probability and Time Distance Question [#permalink]

Show Tags

20 Jul 2009, 18:17

Hi Guys,

Time distance was so simple... i just screwed it ... dont know y ... thanks for that one .. but for the probability one can anyone please elaborate the below mentioned logic for me as i dont know much about the proability:

1. Radio A --> 0.3 (I understand) 2. Radio B --> 0.3 (For radio B) * 0.7 (for the case where radio A is not played) 3. Radio C --> 0.3 (For radio C) * 0.3 (Why this 0.3 and not 1 - 0.3 = 0.7) * 0.7.

I dont understand the logi behind two times 0.3 in the last case as mentioned by irajeevsingh.

i know it might b simple... not clicking to me ....