Find all School-related info fast with the new School-Specific MBA Forum

It is currently 22 Jul 2014, 09:49

Close

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

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

TOUGH & TRICKY SET Of PROBLEMS

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 09 Feb 2012
Posts: 72
Location: India
Concentration: Marketing, Strategy
GMAT 1: 640 Q48 V31
GPA: 3.45
WE: Marketing (Pharmaceuticals and Biotech)
Followers: 1

Kudos [?]: 21 [0], given: 41

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 27 Feb 2012, 08:14
Thankx for the set of questions ...
Senior Manager
Senior Manager
avatar
Joined: 07 Sep 2010
Posts: 340
Followers: 2

Kudos [?]: 90 [0], given: 136

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 27 Mar 2012, 06:11
Hi Bunuel,
I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong.

Probability = favorable/ total outcomes.
Total outcomes = 9C2 (i.e total ways if drawing 2 balls)
Favorable = 3C2*4C1
Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong.
Please reply.
Thanks
H


Quote:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):



We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [0], given: 2601

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 27 Mar 2012, 08:08
Expert's post
imhimanshu wrote:
Hi Bunuel,
I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong.

Probability = favorable/ total outcomes.
Total outcomes = 9C2 (i.e total ways if drawing 2 balls)
Favorable = 3C2*4C1
Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong.
Please reply.
Thanks
H


Quote:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):



We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


Since we have the case with replacement then we can not use C^2_9 for it, because it gives total # of ways to select 2 different balls out of 9 without replacement.

If you wan to solve this question with combinations approach you still should consider two scenarios: P(RW)=2*\frac{C^1_3*C^1_2}{C^1_9*C^1_9}=\frac{4}{27}, we are multiplying by 2 for the same reason: there are two possible wining scenarios RW and WR..

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Senior Manager
Senior Manager
avatar
Joined: 07 Sep 2010
Posts: 340
Followers: 2

Kudos [?]: 90 [0], given: 136

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 28 Mar 2012, 04:43
Thanks Bunuel for clearing the concept.. Appreciate that. :-)

Regards,
H
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Intern
Intern
avatar
Joined: 25 Jan 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 03 Apr 2012, 08:29
Bunuel wrote:
SOLUTION:
3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.



Hi Bunuel - Regarding this question, I believe the answer should be 2 i.e. (B).

It is because the probability of a person's birth year to be a leap yr = 1/2, this is identical to a rainy/non-rainy day situation, where a year is either a leap or normal year.


Case I - No of people = 1
Probability of atleast 1 person's birth yr falling on a leap year = 1/2 = 50%
.....INCORRECT

Case II - No of people = 2
Probability of none of guys' birth year falling on a leap year = 1/2*1/2 = 1/4
Probability of atleast one of them was born on a leap year = 1 - 1/4 = 3/4 = 75%
.....CORRECT

Hence (B).

What do you think? Thanks!
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [0], given: 2601

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 03 Apr 2012, 09:04
Expert's post
theamwan wrote:
Bunuel wrote:
SOLUTION:
3. LEAP YEAR:
How many randomly assembled people are needed to have a better than 50% probability that at least 1 of them was born in a leap year?

A. 1
B. 2
C. 3
D. 4
E. 5

Probability of a randomly selected person NOT to be born in a leap year=3/4
Among 2 people, probability that none of them was born in a leap = 3/4*3/4=9/16. The probability at least one born in leap = 1- 9/16=7/16<1/2
So, we are looking for such n (# of people), when 1-(3/4)^n>1/2
n=3 --> 1-27/64=37/64>1/2

Thus min 3 people are needed.

Answer: C.



Hi Bunuel - Regarding this question, I believe the answer should be 2 i.e. (B).

It is because the probability of a person's birth year to be a leap yr = 1/2, this is identical to a rainy/non-rainy day situation, where a year is either a leap or normal year.


Case I - No of people = 1
Probability of atleast 1 person's birth yr falling on a leap year = 1/2 = 50%
.....INCORRECT

Case II - No of people = 2
Probability of none of guys' birth year falling on a leap year = 1/2*1/2 = 1/4
Probability of atleast one of them was born on a leap year = 1 - 1/4 = 3/4 = 75%
.....CORRECT

Hence (B).

What do you think? Thanks!


That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 25 Jan 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 03 Apr 2012, 09:17
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.


But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [1] , given: 2601

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 03 Apr 2012, 09:32
1
This post received
KUDOS
Expert's post
theamwan wrote:
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.


But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.


Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 25 Jan 2012
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 03 Apr 2012, 09:41
Bunuel wrote:
theamwan wrote:
Bunuel wrote:

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.


But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.


Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.



Thats proves Probability is my weakest link. Thanks Bunuel.
Intern
Intern
avatar
Joined: 28 Sep 2011
Posts: 35
Location: India
WE: Consulting (Computer Software)
Followers: 1

Kudos [?]: 15 [0], given: 18

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 02 Jun 2012, 10:58
Very Nice Shortcut!!!!

Bunuel wrote:
SOLUTION:
1. THE SUM OF EVEN INTEGERS:
The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?

(A) 79
(B) 80
(C) 81
(D) 157
(E) 159

The solution given in the file was 79, which is not correct. Also the problem was solved with AP formula thus was long and used the theory rarely tested in GMAT. Here is my solution and my notes about AP. Some may be useful:

The number of terms in this set would be: n=(k-1)/2 (as k is odd)
Last term: k-1
Average would be first term+last term/2=(2+k-1)/2=(k+1)/2
Also average: sum/number of terms=79*80/((k-1)/2)=158*80/(k-1)
(k+1)/2=158*80/(k-1) --> (k-1)(k+1)=158*160 --> k=159

Answer E.

MY NOTES ABOUT AP:
ARITHMETIC PROGRESSION
Sequence a1, a2,…an, so that a(n)=a(n-1)+d (constant)
nth term an = a1 + d ( n – 1 )
Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n-1))/2
Special cases:
I. 1+2+…+n=n(1+n)/2 (Sum of n first integers)
II. 1+3+5+… (n times)=n^2 (Sum of n first odd numbers). nth term=2n-1
III. 2+4+6+… (n times)=n(n+1) (Sum of n first even numbers) nth term=2n

SOLUTION WITH THE AP FORMULA:
Sequence of even numbers
First term a=2, common difference d=2 since even number
Sum to first n numbers of AP:
Sn=n*(a1+an)/2=n(2*2+2(n-1))/2=n(n+1)=79*80
n=79 (odd)
Number of terms n=(k-1)/2=79 k=159

OR
Sum of n even numbers n(n+1)=79*80
n=79
k=2n+1=159

_________________

Kudos if you like the post!!!

Intern
Intern
avatar
Joined: 11 Apr 2012
Posts: 1
Followers: 0

Kudos [?]: 5 [0], given: 1

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 09 Jun 2012, 19:31
Hi Bunuel...
Kudos++ for this post...
I am having fun solving these questions...
_________________

Cheers,
BhatM

Intern
Intern
avatar
Joined: 25 Jun 2011
Posts: 49
Location: Sydney
Followers: 0

Kudos [?]: 1 [0], given: 7

GMAT Tests User
Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 29 Jun 2012, 19:10
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [1] , given: 2601

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 30 Jun 2012, 03:17
1
This post received
KUDOS
Expert's post
dianamao wrote:
Bunuel wrote:
SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.


The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana


No, in that case we would be asked "what is the the probability of the first ball being red and the second ball being white?"
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 25 Jun 2011
Posts: 49
Location: Sydney
Followers: 0

Kudos [?]: 1 [0], given: 7

GMAT Tests User
Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 30 Jun 2012, 03:52
Thanks again Bunuel :)
Intern
Intern
avatar
Joined: 29 Mar 2012
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 10 Oct 2012, 11:00
[quote="Bunuel"][quote="asterixmatrix"][quote="Bunuel"]SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

Let x be the speed of B.
Write the equation:

(440-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(440-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(440-48)/x-6=(440-144)/x+2
x=12

I solved this question and got -12 as an answer. I see that in the equation above 6 is subtracted and 2 is added but I think If B lost to A in the first heat by 1/10th of a minute then we would have to add those secs in B's time to equate it to A's time and similarly in the second heat B reached 2 secs before so we will have to subtract those 2 secs for equation. I solved for x in the above equation and the answer is +12. PLease help
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [0], given: 2601

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 10 Oct 2012, 11:05
Expert's post
Intern
Intern
avatar
Joined: 29 Mar 2012
Posts: 8
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink] New post 10 Oct 2012, 11:30
Bunuel wrote:
B lost by 6 seconds, so B's time is greater than A's time: B-6=A.



Oh! oops. yeah right. that was silly. Thanks :)
Intern
Intern
avatar
Joined: 12 Jun 2012
Posts: 42
Followers: 1

Kudos [?]: 18 [0], given: 28

A and B ran, at their respective constant rates [#permalink] New post 01 Nov 2012, 01:18
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


I really don't understand Bunuel's explanation, can someone explain this answer as they would to a child. Thanks!









Let x be the speed of B.
Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2
x=12

Answer: A.
_________________

If you find my post helpful, please GIVE ME SOME KUDOS!

Moderator
Moderator
User avatar
Joined: 02 Jul 2012
Posts: 1226
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 61

Kudos [?]: 623 [0], given: 116

GMAT Tests User Premium Member
Re: A and B ran, at their respective constant rates [#permalink] New post 01 Nov 2012, 02:07
jordanshl wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


I really don't understand Bunuel's explanation, can someone explain this answer as they would to a child. Thanks!









Let x be the speed of B.
Write the equation:

(480-48)/x (time of B for first heat) - 6 (seconds, time B lost to A first heat) = TIME OF A (in both heats A runs with constant rate, so the time for first and second heats are the same)=(480-144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)

(480-48)/x-6=(480-144)/x+2
x=12

Answer: A.


Let A's speed = a, B's speed = b

First Heat:

Let time taken by B to run the distance (480-48)m be t seconds.
A travels the distance of 480m in t-6 seconds

\frac{480-48}{b} = t

\frac{480}{a} = t-6

So,

\frac{480}{a} = \frac{480-48}{b}-6

Similarly

\frac{480}{a} = \frac{480-144}{b}+2

So,

\frac{480-48}{b}-6 = \frac{480-144}{b}+2

\frac{144-48}{b} = 8

So,
b=12

Answer is A

Kudos Please... If my post helped.
_________________

Did you find this post helpful?... Please let me know through the Kudos button.

Thanks To The Almighty - My GMAT Debrief

GMAT Reading Comprehension: 7 Most Common Passage Types

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 18682
Followers: 3231

Kudos [?]: 22177 [0], given: 2601

Re: A and B ran, at their respective constant rates [#permalink] New post 01 Nov 2012, 05:44
Expert's post
jordanshl wrote:
SOLUTION:
5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20


I really don't understand Bunuel's explanation, can someone explain this answer as they would to a child. Thanks!


Merging topics.

Check alternate solutions given by Karishma:

tough-tricky-set-of-problms-85211-80.html#p970679
tough-tricky-set-of-problms-85211-80.html#p970681

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Re: A and B ran, at their respective constant rates   [#permalink] 01 Nov 2012, 05:44
    Similar topics Author Replies Last post
Similar
Topics:
30 Tricky problem sets fozzzy 48 04 Oct 2013, 21:47
Experts publish their posts in the topic Sets tough Problem Try it...... mydreammba 5 03 Jan 2012, 04:48
Experts publish their posts in the topic Tough and tricky 5: Race Bunuel 5 11 Oct 2009, 17:14
2 Experts publish their posts in the topic Tough and tricky 4: addition problem Bunuel 9 11 Oct 2009, 16:42
Experts publish their posts in the topic Tough and tricky 2: Bunuel 4 11 Oct 2009, 16:17
Display posts from previous: Sort by

TOUGH & TRICKY SET Of PROBLEMS

  Question banks Downloads My Bookmarks Reviews Important topics  

Go to page   Previous    1   2   3   4   5   6   7   8    Next  [ 155 posts ] 



GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.