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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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20 Mar 2019, 07:13
C812−(C77∗C15+C55∗C37)=495−(5+35)=455C128−(C77∗C51+C55∗C73)=495−(5+35)=455. How do you get (5+35) (I know that it's probably very simple...)? BunuelThank you in advance!!



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Re: Baker's Dozen
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25 Aug 2019, 02:28
Bunuel wrote: mofasser08 wrote: Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)? Why not : 9*9*5P3 ?? That's a tough 700+ problem. We are interested 666XX numbers. Now, two this can be arranged in 10 ways; 666XX; 66X6X; 6X66X; X666X; 66XX6; 6X6X6; X66X6; 6XX66; XX666; X6X66. So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's. Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5digit password unlike 5digit number can start with 0. Total \(9*9*C^3_5=810\). Hope it's clear. Bunuel1.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 I think if we have three 6 and two 9 then we get 666XX; 66X6X; 6X66X; X666X; 66XX6; 6X6X6; X66X6; 6XX66; XX666; X6X66. 10 cases But if we have three 6 and two different numbers such as 1,2 then we get 20 cases .







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