GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 18 Jan 2020, 12:55 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Baker's Dozen

Author Message
TAGS:

### Hide Tags

Manager  G
Joined: 02 Jun 2015
Posts: 169
Location: Ghana

### Show Tags

3
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

Director  P
Status: Professional GMAT Tutor
Affiliations: AB, cum laude, Harvard University (Class of '02)
Joined: 10 Jul 2015
Posts: 717
Location: United States (CA)
Age: 40
GMAT 1: 770 Q47 V48
GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: Q168 V169 WE: Education (Education)

### Show Tags

Top Contributor
1
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 2197 times ]

IIMA, IIMC School Moderator V
Joined: 04 Sep 2016
Posts: 1380
Location: India
WE: Engineering (Other)

### Show Tags

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)
_________________
It's the journey that brings us happiness not the destination.

Feeling stressed, you are not alone!!
Math Expert V
Joined: 02 Sep 2009
Posts: 60480

### Show Tags

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.
_________________
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9983
Location: Pune, India

### Show Tags

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
_________________
Karishma
Veritas Prep GMAT Instructor

Intern  B
Joined: 01 Jan 2016
Posts: 22

### Show Tags

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. Math Expert V Joined: 02 Sep 2009 Posts: 60480 Re: Baker's Dozen [#permalink] ### Show Tags 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 $$3,600,000*\frac{1}{30}=120,000$$.

------------------------------------------

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/baker-s-doze ... l#p1059585
_________________
Intern  B
Joined: 20 Feb 2019
Posts: 1

### Show Tags

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 $$C^8_{12}-(C^7_7*C^1_5+C^5_5*C^3_7)=495-(5+35)=455$$.

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?
Intern  B
Joined: 28 May 2019
Posts: 10

### Show Tags

@
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.

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

@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
Intern  B
Joined: 26 Sep 2018
Posts: 1

### Show Tags

1
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. 1-y
C. -1-y
D. y-1
E. x-y

Note that $$\sqrt{a^2}=|a|$$. Next, since $$x<0$$ and $$y<0$$ then $$|x|=-x$$ and $$|y|=-y$$.

So, $$\frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}=\frac{|x|}{x}-\sqrt{(-y)*(-y)}=\frac{-x}{x}-\sqrt{y^2}=-1-|y|=-1+y$$

Just Excellent Question Bunuel...
Intern  B
Joined: 12 Oct 2019
Posts: 41
Location: India
Concentration: Marketing, General Management
GMAT 1: 720 Q48 V41
GPA: 4
WE: Information Technology (Computer Software)

### Show Tags

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: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

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 -49-51-53-55-57 = --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

Math Expert V
Joined: 02 Sep 2009
Posts: 60480

### Show Tags

1
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: $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=-185$$ --> $$(x+4)+(x+6)+(x+8)+(x+10)+(x+12)=5x+40=-185$$ --> $$(5x+20)+20=-185$$ --> $$5x+20=-205$$

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 -49-51-53-55-57 = --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

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.
_________________ Re: Baker's Dozen   [#permalink] 22 Dec 2019, 08:57

Go to page   Previous    1   2   3   4   5   6   [ 112 posts ]

Display posts from previous: Sort by

# Baker's Dozen  