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Baker's Dozen
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13 Mar 2017, 09:01
7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive?A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000 Solution: http://gmatclub.com/forum/bakersdozen ... l#p1057509Let F represents Fritz, L for Luis, A for Alfred and W for Werner Fritz owns 2/3rd of the shares of the other three shareholders > F = 2/3 (3,600,000  F) > 3F = 7,200,000  2F > F = 1,440,000 Luis owns 3/7th of the shares of the other three shareholders > L = 3/7 (3,600,000  L) > 7L = 10,800,000  3L > L = 1,080,000 Alfred owns 4/11th of the shares of the other three shareholders > A = 4/11 (3,600,000  A) > 11A = 14,400,000  4A > A = 960,000 W= 3,600,000 (F + L + A) > 3,600,000  (1,440,000 + 1,080,000 + 960,000) = 120,000 Answer: D



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Re: Baker's Dozen
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27 Mar 2017, 18:07
Here's my solution for #2 (see visual below). It helps to use brackets!
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Baker's Dozen
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19 Dec 2017, 03:51
Bunuel niks18Quote: A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215 Quote: Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\). I solved the same problem taking hint from number theory that a consecutive odd number is represented by: 2x+1 , 2x+3 My approach had slightly complex calculations but it ensured that my first no in series is always odd.Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones)
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Re: Baker's Dozen
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19 Dec 2017, 04:03
adkikani wrote: Bunuel niks18Quote: A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215 Quote: Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\). I solved the same problem taking hint from number theory that a consecutive odd number is represented by: 2x+1 , 2x+3 My approach had slightly complex calculations but it ensured that my first no in series is always odd.Is this understanding correct? Or can I still take a random integer x as start and then add +2 (to get consecutive ones) In this question, as shown, you can take 7 consecutive odd integers to be \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), and \(x+12\). For some number properties questions, you should represent odd integers, as 2k + 1, 2k + 3, ... Here you can also take the integers, to be 2k + 1, 2k + 3, ... but it's not necessary.
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Re: Baker's Dozen
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26 Mar 2018, 04:59
nahid78 wrote: Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. I know this is how we solve this question. But can anyone please clarify why i don't get the same result if i work differently .... Suppose other two numbers are also same 6,6,6, and 1,1 or 6,6,6 and 2,2 so on... or other two are different. 6,6,6,1, and 2, or 6,6,6,2,and 4 You do get the same result in this case too. If the other 2 numbers are same, number of ways = 9C1 * 5!/3!*2! = 90 If the other 2 numbers are different, number of ways = 9C2 * 5!/3! = 720 Total = 90+720 = 810
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Re: Baker's Dozen
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27 Oct 2019, 13:24
Bunuel wrote: 7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000
Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders > Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders > Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders > Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;
Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).
Answer: D.  Bunuel: Where are you coming up with the "2", "3", and "4" in the denominators of "2/(2+3)", "3/(3+7)", "4/(4+11)", respectively? What is the rule that is enabling you to quickly use that as a ratio of Fritz, Luis, and Alfred's respective share of the whole? Best.



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Re: Baker's Dozen
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27 Oct 2019, 20:59
GMAT2645 wrote: Bunuel wrote: 7. Metropolis Corporation has 4 shareholders: Fritz, Luis, Alfred and Werner. Number of shares that Fritz owns is 2/3 rd of number of the shares of the other three shareholders, number of the shares that Luis owns is 3/7 th of number of the shares of the other three shareholders and number of the shares that Alfred owns is 4/11 th of number of the shares of the other three shareholders. If dividends of $3,600,000 were distributed among the 4 shareholders, how much of this amount did Werner receive? A. $60,000 B. $90,000 C. $100,000 D. $120,000 E. $180,000
Fritz owns is \(\frac{2}{3}\)rd of the shares of the other three shareholders > Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares; Luis owns is \(\frac{3}{7}\)th of the shares of the other three shareholders > Luis owns \(\frac{3}{3+7}=\frac{3}{10}\)th of all shares; Alfred owns is \(\frac{4}{11}\)th of the shares of the other three shareholders > Alfred owns \(\frac{4}{4+11}=\frac{4}{15}\)th of all shares;
Together those three own \(\frac{2}{5}+\frac{3}{10}+\frac{4}{15}=\frac{29}{30}\)th of all shares, which means that Werner owns \(1\frac{29}{30}=\frac{1}{30}\). Hence from $3,600,000 Werner gets \($3,600,000*\frac{1}{30}=$120,000\).
Answer: D.  Bunuel: Where are you coming up with the "2", "3", and "4" in the denominators of "2/(2+3)", "3/(3+7)", "4/(4+11)", respectively? What is the rule that is enabling you to quickly use that as a ratio of Fritz, Luis, and Alfred's respective share of the whole? Best. Check here: https://gmatclub.com/forum/bakersdoze ... l#p1059585
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Re: Baker's Dozen
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30 Oct 2019, 06:42
Bunuel wrote: 5. There are 7 red and 5 blue marbles in a jar. In how many ways 8 marbles can be selected from the jar so that at least one red marble and at least one blue marble to remain in the jar? A. 460 B. 490 C. 493 D. 455 E. 445
Total ways to select 8 marbles out of 7+5=12 is \(C^8_{12}\); Ways to select 8 marbles so that zero red marbles is left in the jar is \(C^7_7*C^1_5\); Ways to select 8 marbles so that zero blue marbles is left in the jar is \(C^5_5*C^3_7\);
Hence ways to select 8 marbles so that at least one red marble and at least one blue marble to remain the jar is \(C^8_{12}(C^7_7*C^1_5+C^5_5*C^3_7)=495(5+35)=455\).
Answer: D. Hi Bunuel, If I am correct, in the above question we are assuming that balls are numbered that is why we have taken "5 + 35"(even when nothing mentioned in the question). Otherwise, selecting 7 red marbles and 1 blue marble is only one case. How we would have solved this question if balls both red and blue are not numbered?



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Re: Baker's Dozen
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15 Nov 2019, 06:40
@ Bunuel wrote: SOLUTIONS:
1. A password on Mr. Wallace's briefcase consists of 5 digits. What is the probability that the password contains exactly three digit 6?
A. 860/90,000 B. 810/100,000 C. 858/100,000 D. 860/100,000 E. 1530/100,000
Total # of 5 digit codes is 10^5, notice that it's not 9*10^4, since in a code we can have zero as the first digit.
# of passwords with three digit 6 is \(9*9*C^3_5=810\): each out of two other digits (not 6) has 9 choices, thus we have 9*9 and \(C^3_5\) is ways to choose which 3 digits will be 6's out of 5 digits we have.
\(P=\frac{favorable}{total}=\frac{810}{10^5}\)
Answer: B. @Bunuel As per question the code could be XXXYY and XXXYZ where X=6, and Y,Z are any other digit 1.For XXXYY arrangement – 1*1*1*9*1*5!/(3!2!) = 90 2.For XXXYZ arrangement – 1*1*1*9*8*5!/3! = 1440 Hence total arrangement 90+1440=1530 Please explain where I am making a mistake



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Re: Baker's Dozen
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21 Dec 2019, 08:27
Bunuel wrote: 9. If x and y are negative numbers, what is the value of \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}\)? A. 1+y B. 1y C. 1y D. y1 E. xy
Note that \(\sqrt{a^2}=a\). Next, since \(x<0\) and \(y<0\) then \(x=x\) and \(y=y\).
So, \(\frac{\sqrt{x^2}}{x}\sqrt{y*y}=\frac{x}{x}\sqrt{(y)*(y)}=\frac{x}{x}\sqrt{y^2}=1y=1+y\)
Answer: D. Just Excellent Question Bunuel...



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Re: Baker's Dozen
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22 Dec 2019, 04:35
Bunuel wrote: 8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D. Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong? Let the smallest no. be 2n+1, so the numbers of the set will be : 2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13 given that sum of largest 5 no.s in the set is 185. so, 2n+5+2n+7+2n+9+2n+11+2n+13 = 185 solving, n=23 so the no.s are 45,47,49,51,53,55,57 sum of smallest 5 are 4951535557 = 265 Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = 185 solving n= 21 so the no.s are 41, 43, 45, 47, 49, 51, 53 sum of 5 smallest = 245 Really confused.. Please help.



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Re: Baker's Dozen
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22 Dec 2019, 08:57
dips1122 wrote: Bunuel wrote: 8. A set A consists of 7 consecutive odd integers. If the sum of 5 largest integers of set A is 185 what is the sum of the 5 smallest integers of set A? A. 165 B. 175 C. 195 D. 205 E. 215
Say 7 consecutive odd integers are: \(x\), \(x+2\), \(x+4\), \(x+6\), \(x+8\), \(x+10\), \(x+12\).
Question: \(x+(x+2)+(x+4)+(x+6)+(x+8)=5x+20=?\)
Given: \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=185\) > \((x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=185\) > \((5x+20)+20=185\) > \(5x+20=205\)
Answer: D. Hello Bunuel, I tried to solve it like this.. Can u please tell me where did I go wrong? Let the smallest no. be 2n+1, so the numbers of the set will be : 2n+1, 2n+3, 2n+5, 2n+7, 2n+9, 2n+11, 2n+13 given that sum of largest 5 no.s in the set is 185. so, 2n+5+2n+7+2n+9+2n+11+2n+13 = 185 solving, n=23 so the no.s are 45,47,49,51,53,55,57 sum of smallest 5 are 4951535557 = 265 Also, if we take into consideration that these are negative no.s, from the start we can write that 2n+1+2n+3+2n+5+2n+7+2n+9 = 185 solving n= 21 so the no.s are 41, 43, 45, 47, 49, 51, 53 sum of 5 smallest = 245 Really confused.. Please help. Since n = 21, then the seven consecutive numbers are: 45, 43, 41, 39, 37, 35, 33. The sum of the five largest is 41 + (39) + (37) + (35) + (33) = 185. The sum of the five smallest is 45 +(43) +(41) + (39) + (37) = 205.
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