Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 10 Feb 2012
Posts: 71
Location: India
Concentration: Marketing, Strategy
GPA: 3.45
WE: Marketing (Pharmaceuticals and Biotech)

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
27 Feb 2012, 09:14
Thankx for the set of questions ...



Senior Manager
Joined: 07 Sep 2010
Posts: 327

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
27 Mar 2012, 07:11
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



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
27 Mar 2012, 09:08
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 the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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? Extrahard Quant Tests with Brilliant Analytics



Senior Manager
Joined: 07 Sep 2010
Posts: 327

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
28 Mar 2012, 05:43
Thanks Bunuel for clearing the concept.. Appreciate that. Regards, H
_________________
+1 Kudos me, Help me unlocking GMAT Club Tests



Intern
Joined: 25 Jan 2012
Posts: 3

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
03 Apr 2012, 09:29
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 > 127/64=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/nonrainy 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!



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
03 Apr 2012, 10:04
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 > 127/64=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/nonrainy 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 the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Jan 2012
Posts: 3

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
03 Apr 2012, 10:17
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.



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
03 Apr 2012, 10:32
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 the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Jan 2012
Posts: 3

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
03 Apr 2012, 10:41
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
Joined: 28 Sep 2011
Posts: 33
Location: India
WE: Consulting (Computer Software)

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
02 Jun 2012, 11:58
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=(k1)/2 (as k is odd) Last term: k1 Average would be first term+last term/2=(2+k1)/2=(k+1)/2 Also average: sum/number of terms=79*80/((k1)/2)=158*80/(k1) (k+1)/2=158*80/(k1) > (k1)(k+1)=158*160 > k=159
Answer E.
MY NOTES ABOUT AP: ARITHMETIC PROGRESSION Sequence a1, a2,…an, so that a(n)=a(n1)+d (constant) nth term an = a1 + d ( n – 1 ) Sn=n*(a1+an)/2 or Sn=n*(2a1+d(n1))/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=2n1 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(n1))/2=n(n+1)=79*80 n=79 (odd) Number of terms n=(k1)/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
Joined: 11 Apr 2012
Posts: 1

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
09 Jun 2012, 20:31
Hi Bunuel... Kudos++ for this post... I am having fun solving these questions...
_________________
Cheers, BhatM



Intern
Joined: 25 Jun 2011
Posts: 47
Location: Sydney

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
29 Jun 2012, 20:10
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



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
30 Jun 2012, 04:17
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 the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
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? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 25 Jun 2011
Posts: 47
Location: Sydney

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
30 Jun 2012, 04:52
Thanks again Bunuel



Intern
Joined: 29 Mar 2012
Posts: 8

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
10 Oct 2012, 12:00
[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:
(44048)/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)=(440144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)
(44048)/x6=(440144)/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



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
10 Oct 2012, 12:05



Intern
Joined: 29 Mar 2012
Posts: 8

Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
Show Tags
10 Oct 2012, 12:30
Bunuel wrote: B lost by 6 seconds, so B's time is greater than A's time: B6=A. Oh! oops. yeah right. that was silly. Thanks



Intern
Joined: 12 Jun 2012
Posts: 41

A and B ran, at their respective constant rates [#permalink]
Show Tags
01 Nov 2012, 02:18
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: (48048)/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)=(480144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat) (48048)/x6=(480144)/x+2 x=12 Answer: A.
_________________
If you find my post helpful, please GIVE ME SOME KUDOS!



Moderator
Joined: 02 Jul 2012
Posts: 1223
Location: India
Concentration: Strategy
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: A and B ran, at their respective constant rates [#permalink]
Show Tags
01 Nov 2012, 03:07
1
This post was BOOKMARKED
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:
(48048)/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)=(480144)/x (time of B for second heat) + 2 (seconds, time B won to A second heat)
(48048)/x6=(480144)/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 (48048)m be t seconds. A travels the distance of 480m in t6 seconds \(\frac{48048}{b} = t\) \(\frac{480}{a} = t6\) So, \(\frac{480}{a} = \frac{48048}{b}6\) Similarly \(\frac{480}{a} = \frac{480144}{b}+2\) So, \(\frac{48048}{b}6 = \frac{480144}{b}+2\) \(\frac{14448}{b} = 8\) 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



Math Expert
Joined: 02 Sep 2009
Posts: 39698

Re: A and B ran, at their respective constant rates [#permalink]
Show Tags
01 Nov 2012, 06:44




Re: A and B ran, at their respective constant rates
[#permalink]
01 Nov 2012, 06:44



Go to page
Previous
1 2 3 4 5 6 7 8 9
Next
[ 166 posts ]




