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Re: TOUGH & TRICKY SET Of PROBLEMS
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25 Jan 2013, 04:55
MegW wrote: Bunuel wrote: angel2009 wrote: I've some problem to understand this ; as the two events are mutually exclusive , independent. So why not the answer is 3/9*2/9 = 2/27 Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9. Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters. Hey can you please help with the solution using combination formula P(1R1W) = 3C1 x 2C1 / (9C2) = 1/6 WHY? what am i doing wrong? Hint: notice that each ball is put back after it is drawn, so each time we choose 1 ball out of 9.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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07 Feb 2013, 03:12
Sachin9 wrote: bro bunuel, do u think we would need to knw these formulas and such a question can ever appear on gmat? you mentioned that u have never seen any gmat question that requires us to know formula. Do you need to know these formulas? No. But if you do know them and a 100 other formulas, you could save time and energy during the test. Can such a question appear on the GMAT? Sure. It can easily be solved without using any exotic formulas as done by Bunuel above and me in this post: m13q12124158.html?hilit=distance%20between%20circle%20and%20lineGenerally, there are multiple ways of arriving at the answer. Some ways use some formulas, some don't. Try to understand the methods which don't use formulas if you don't already know them.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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01 Sep 2013, 00:12
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. Hi Bunuel, I always have problem in understand the probablity problems. If this is asked without replacement then how we can solve this problem:???? Can we solve the original question with nCr method??? Please clear my doubts. Thanks in advance, Rrsnathan.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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02 Sep 2013, 02:11
rrsnathan 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. Hi Bunuel, I always have problem in understand the probablity problems. If this is asked without replacement then how we can solve this problem:???? Can we solve the original question with nCr method??? Please clear my doubts. Thanks in advance, Rrsnathan. If we had without replacement case, then the answer would be 2*3/9*2/8.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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17 Oct 2013, 06:03
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, If I shall encounter this question or one like it on the GMAT, will I be given the information as to what the ratio of regular year to leap year is? If not, is there a list of little things like this that are important to know?



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Re: TOUGH & TRICKY SET Of PROBLEMS
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17 Oct 2013, 08:08
ronr34 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, If I shall encounter this question or one like it on the GMAT, will I be given the information as to what the ratio of regular year to leap year is? If not, is there a list of little things like this that are important to know? GMAC says that we must solve the questions using your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise). Guess that the meaning of a leap year is considered as everyday facts, though this question is still not perfect so I wouldn't worry about it at all.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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16 Dec 2013, 04:02
Dear Brunel , Please explain special cases from AP 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 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



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Re: TOUGH & TRICKY SET Of PROBLEMS
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16 Dec 2013, 04:08
archit wrote: Dear Brunel , Please explain special cases from AP 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 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 Sum of first n positive integers: \(1+2+...+n=\frac{1+n}{2}*n\). For example, the sum of first 5 positive integers is \(1+2+3+4+5=\frac{1+5}{2}*5=15\) Sum of n first odd numbers: \(a_1+a_2+...+a_n=1+3+...+a_n=n^2\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n1\). Given \(n=5\) first odd integers, then their sum equals to \(1+3+5+7+9=5^2=25\). Sum of n first positive even numbers: \(a_1+a_2+...+a_n=2+4+...+a_n\)\(=n(n+1)\), where \(a_n\) is the last, \(n_{th}\) term and given by: \(a_n=2n\). Given \(n=4\) first positive even integers, then their sum equals to \(2+4+6+8=4(4+1)=20\). Hope it helps.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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18 Sep 2014, 12:21
BunuelHi Bunuel.. U r really a math expert ! Kudos for all your post. My Query : Q4.ADDITION PROBLEM: AB + CD = AAA, where AB and CD are twodigit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C? (A) 1(B) 3(C) 7(D) 9(E) Cannot be determined AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111. So, A=1. 1B+CD=111. How did you creach to the conclusion that 111 can be the only result of addition? how is A=1 Q8. " As the min temperature is 45, max would be 2504*45=70 > The range=70(max)45(min)=25 " Can you please explain the logic behind 4*45. How do we get the max value here ? Can you please explain ! Thanks in advance



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Re: TOUGH & TRICKY SET Of PROBLEMS
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19 Sep 2014, 02:22
GuptaDarsh wrote: BunuelHi Bunuel.. U r really a math expert ! Kudos for all your post. My Query : Q4.ADDITION PROBLEM: AB + CD = AAA, where AB and CD are twodigit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C? (A) 1(B) 3(C) 7(D) 9(E) Cannot be determined AB and CD are two digit integers, their sum can give us only one three digit integer of a kind of AAA it's 111. So, A=1. 1B+CD=111. How did you creach to the conclusion that 111 can be the only result of addition? how is A=1 Q8. " As the min temperature is 45, max would be 2504*45=70 > The range=70(max)45(min)=25 " Can you please explain the logic behind 4*45. How do we get the max value here ? Can you please explain ! Thanks in advance Q4: AB and CD are twodigit integers. Ask yourself what threedigit integer with the same digits can be theirs sum? Can it be 222, 333, 444, ... Q8: There are 5 days from Monday to Friday. the sum of the temperatures is 250 and the minimum temperature is 45. Now, to get max temperature possible consider 4 days out of 5 to have that minimum temperature and subtract that from 250.
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Re: TOUGH & TRICKY SET Of PROBLEMS
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10 Oct 2014, 13:22
Hi Bunnel,
In question number 3 why have you taken 3/4 for no probability of any child in leap year.



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Re: TOUGH & TRICKY SET Of PROBLEMS
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Re: TOUGH & TRICKY SET Of PROBLEMS
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24 Oct 2014, 00:48
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. If there are 3 people required then shouldn't this too be true? case 1: when only 1 person was there, the probability would be 1/4 = 0.25 case 2: when only 2 people are there, the probability would be (1/4)*(1/4) = 0.0625 case 3: when 3 people are there, the probability would be (1/4)*(1/4)*(1/4) = 0.015625 So, shouldn't the answer be: 0.25 + 0.0625 + 0.015625 = 0.0328125? (I know my answer is incorrect, but fail to understand why my way of calculating is wrong?)



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Re: TOUGH & TRICKY SET Of PROBLEMS
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24 Oct 2014, 03:47
kamaln 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. If there are 3 people required then shouldn't this too be true? case 1: when only 1 person was there, the probability would be 1/4 = 0.25 case 2: when only 2 people are there, the probability would be (1/4)*(1/4) = 0.0625 case 3: when 3 people are there, the probability would be (1/4)*(1/4)*(1/4) = 0.015625 So, shouldn't the answer be: 0.25 + 0.0625 + 0.015625 = 0.0328125? (I know my answer is incorrect, but fail to understand why my way of calculating is wrong?) The probability that among 3 people at least 1 of them is born in a leap year is: P(at least 1) = P(exactly 1) + P(exactly 2) + P(all 3) = 3*1/4*3/4*3/4 + 3*1/4*1/4*3/4 + 1/4*1/4*1/4 = 37/64. Or P(at least 1) = 1  P(none) = 1  (3/4)^3 = 37/64. Some "at least" probability questions to practice: leilaisplayingacarnivalgameinwhichsheisgiven140018.htmlafaircoinistossed4timeswhatistheprobabilityof131592.htmlforeachplayersturninacertainboardgameacardis132074.htmlastringof10lightbulbsiswiredinsuchawaythatif131205.htmlashipmentof8tvsetscontains2blackandwhitesetsand53338.htmlonashelfthereare6hardbackbooksand2paperbackbook135122.htmlinagroupwith800people136839.htmltheprobabilityofamanhittingabullseyeinonefireis136935.htmlforeachplayersturninacertainboardgameacardis141790.htmltheprobabilitythataconveniencestorehascansoficed128689.htmltripletsadambruceandcharlieenteratriathlonif132688.htmlamanufacturerisusingglassasthesurface144642.htmltheprobabilityis12thatacertaincoinwillturnuphead144730.html (OG13) afaircoinistobetossedtwiceandanintegeristobe148779.htmlinagameoneplayerthrowstwofairsixsideddieatthe151956.htmlinadrawerofshirts8areblue6aregreenand4are24418.html
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Re: TOUGH & TRICKY SET Of PROBLEMS
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07 Apr 2018, 13:07
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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/thesumoft ... 68732.html2. THE PRICE OF BUSHEL: The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of 5x cents per day while the price of wheat is decreasing at a constant rate of 2^1/2*xx cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat? (A) $4.50 (B) $5.10 (C) $5.30 (D) $5.50 (E) $5.60 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/thepriceof ... 70101.html3. 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/howmanyran ... 89297.html4. ADDITION PROBLEM: AB + CD = AAA, where AB and CD are twodigit numbers and AAA is a three digit number; A, B, C, and D are distinct positive integers. In the addition problem above, what is the value of C? (A) 1 (B) 3 (C) 7 (D) 9 (E) Cannot be determined OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/abcdaaawh ... 36903.html5. 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/aandbran ... 06921.html6. 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 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/abagcontai ... 8350.html7. THE DISTANCE BETWEEN THE CIRCLE AND THE LINE: What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x  3? A) 1.4 B) sqrt (2) C) 1.7 D) sqrt (3) E) 2.0 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/whatisthe ... 85184.html8. THE AVERAGE TEMPERATURE: The average of temperatures at noontime from Monday to Friday is 50; the lowest one is 45, what is the possible maximum range of the temperatures? A. 20 B. 25 C. 40 D. 45 E. 75 OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/theaverage ... 16784.html9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8: If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8? A. 25% B. 50% C. 62.5% D. 72.5% E. 75% OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifanintege ... 26654.html10. SUM OF INTEGERS: If the sum of five consecutive positive integers is A, then the sum of the next five consecutive integers in terms of A is: A. A+1 inquiry B. A+5 C A+25 D 2A E. 5A OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/ifthesumo ... 02768.html
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Re: TOUGH & TRICKY SET Of PROBLEMS &nbs
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07 Apr 2018, 13:07



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