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# Baker's Dozen

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Joined: 02 Jun 2015
Posts: 190
Location: Ghana

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13 Mar 2017, 09:01
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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/baker-s-dozen ... l#p1057509

Let 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

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27 Mar 2017, 18:07
Top Contributor
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Here's my solution for #2 (see visual below). It helps to use brackets!
Attachments

Screen Shot 2017-03-27 at 6.07.16 PM.png [ 182.8 KiB | Viewed 1670 times ]

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19 Dec 2017, 03:51
Bunuel niks18

Quote:
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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19 Dec 2017, 04:03
Bunuel niks18

Quote:
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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26 Mar 2018, 04:59
1
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.

$$P=\frac{favorable}{total}=\frac{810}{10^5}$$

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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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...)? Bunuel

Re: Baker's Dozen   [#permalink] 20 Mar 2019, 07:13

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