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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
 27 Feb 2012, 09:14
Thankx for the set of questions ...
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
28 Mar 2012, 05:43
Thanks Bunuel for clearing the concept.. Appreciate that.  Regards, H
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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 --> 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!
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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 --> 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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
03 Apr 2012, 10:32
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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.
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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=(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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
09 Jun 2012, 20:31
Hi Bunuel... Kudos++ for this post... I am having fun solving these questions...
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
30 Jun 2012, 04:17
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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?"
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
30 Jun 2012, 04:52
Thanks again Bunuel
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
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:
(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
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
10 Oct 2012, 12:05
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Re: TOUGH & TRICKY SET Of PROBLMS [#permalink]
10 Oct 2012, 12:30
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
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A and B ran, at their respective constant rates [#permalink]
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: (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.
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Re: A and B ran, at their respective constant rates [#permalink]
01 Nov 2012, 03:07
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-6So, \frac{480}{a} = \frac{480-48}{b}-6Similarly \frac{480}{a} = \frac{480-144}{b}+2So, \frac{480-48}{b}-6 = \frac{480-144}{b}+2\frac{144-48}{b} = 8So, b=12 Answer is A Kudos Please... If my post helped.
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Re: A and B ran, at their respective constant rates [#permalink]
01 Nov 2012, 06:44
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Re: A and B ran, at their respective constant rates
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