GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Sep 2018, 13:35

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

How many subordinates does Marcia have?

Author Message
TAGS:

Hide Tags

VP
Joined: 22 Nov 2007
Posts: 1058
How many subordinates does Marcia have?  [#permalink]

Show Tags

16 Dec 2007, 03:13
7
45
00:00

Difficulty:

95% (hard)

Question Stats:

25% (01:41) correct 75% (01:27) wrong based on 1318 sessions

HideShow timer Statistics

How many subordinates does Marcia have?

(1) There are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) There are 28 ways that she could decide which 2 subordinates she will recommend promoting.
Math Expert
Joined: 02 Sep 2009
Posts: 49271
How many subordinates does Marcia have?  [#permalink]

Show Tags

25 Feb 2010, 18:24
17
12
BarneyStinson wrote:
How many subordinates does Marcia Have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.

Little correction:
For (2) we have $$\{s_1,s_2,s_3,...s_n\}$$. Each subordinate $$(s_1,s_2,s_3,...s_n)$$ has TWO options: either to be included in the list or not. Hence total # of lists - $$2^n$$, correct. But this number will include $$n$$ lists with 1 subordinate as well as 1 empty list.

As the lists should contain at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.

Lists with 1 subordinate - n: $$\{s_1,0,0,0...0\}$$, $$\{0,s_2,0,0,...0\}$$, $$\{0,0,s_3,0,...0\}$$, ... $$\{0,0,0...s_n\}$$.
List with 0 subordinate - 1: $$\{0,0,0,...0\}$$

So we'll get $$200<2^n-n-1<500$$, --> $$n=8$$. Sufficient.

For (1):
$$C^2_n=28$$ --> $$\frac{n(n-1)}{2!}=28$$ --> $$n(n-1)=56$$ --> $$n=8$$. Sufficient.

_________________
Math Expert
Joined: 02 Aug 2009
Posts: 6797
Re: How many subordinates does Marcia have?  [#permalink]

Show Tags

30 Aug 2018, 08:56
2
marcodonzelli wrote:
How many subordinates does Marcia have?

(1) There are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) There are 28 ways that she could decide which 2 subordinates she will recommend promoting.

(1) There are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
So atleast 2 means, she makes list of 2 people ,3 people ... till n people..
example- say 5 people are there, so she makes a list of 2 out of 5 do 5C2, 3 out of 5 so 5C3, 4 out of 5 so 5C4 and 5 out of 5 so 5C5 people in a list
Here let the number be n, so nC2+nC3+nC4+....nCn is what we are looking for...
Now you should remember the formula nC0+nC1+....nCn=2^n
So nC2+nC3+....nCn=2^n-nC0-nC1=2^n-n-1
This is between 200 and 500....
2^7=128, so take next value 8.......
2^8=256 so 2^8-8-1=256-8-1=247 Between 200 and 500 yes
Take next value 2^9-9-1=512-9-1=502.. outside the range
Only 1 value that is 8
Sufficient

(2) There are 28 ways that she could decide which 2 subordinates she will recommend promoting ..
nC2=28......n(n-1)=28*2......n(n-1)=56=8*7 so n =8
Sufficient

D
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

GMAT Instructor
Joined: 04 Jul 2006
Posts: 1252

Show Tags

18 Dec 2007, 10:59
6
1
marcodonzelli wrote:
How many subordinates does Marcia Have?

1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

I found it very hard..

For (1) remember that ii is often easier to count the converse of what is asked. Of the 2^n subsets of a list of n objects, how many contain less than 2 objects?
General Discussion
CEO
Joined: 17 Nov 2007
Posts: 3481
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

16 Dec 2007, 03:55
B

1. Maybe I've not understood correctly but It seems to be insufficient.

2. nC2 describes this situation.

nC2=n(n-1)/2=28 ==> n(n-1)=56 ==> n=8
CEO
Joined: 17 Nov 2007
Posts: 3481
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

16 Dec 2007, 08:53
2
really D.

first condition:

7: N=7C7+7C6+7C5+7C4+7C3+7C2=119<200
8: N=8C8+8C7+8C6+8C5+8C4+8C3+8C2=238
9: N=9C9+9C8+9C7+9C6+9C5+9C4+9C3+9C2=504>500

but it is a terrible way. Any shortcuts?
SVP
Joined: 29 Aug 2007
Posts: 2413

Show Tags

16 Dec 2007, 13:08
marcodonzelli wrote:
How many subordinates does Marcia Have?

1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

I found it very hard..

1: i am not sure how st 1 is sufficient as it says 200-500 lists from at least 2 subordinates.

2: 28 ways = nc2 = n(n-1)/2 = 8x7/2 = 28
so n = 8.

so is not it B?
CEO
Joined: 17 Nov 2007
Posts: 3481
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

18 Dec 2007, 12:45
kevincan wrote:
2^n

Thanks for that! It's really simple!
VP
Joined: 22 Nov 2007
Posts: 1058

Show Tags

23 Dec 2007, 04:42
2
I actually didn't catch this expalantion..I have had another explanation as well, but even this one is not clear to me.... let's exclude from the 2^n lists, the only that have only one name and the 1 that has none, so 200<2^n - n - 1<500. so we have n=8.

Why do we have to consider 2^n lists? what's that mean: the n that have only one name and the one that has none...please help me!
Manager
Joined: 21 May 2007
Posts: 119

Show Tags

23 Dec 2007, 19:31
2
1
marcodonzelli wrote:
I actually didn't catch this expalantion..I have had another explanation as well, but even this one is not clear to me.... let's exclude from the 2^n lists, the only that have only one name and the 1 that has none, so 200<2^n - n - 1<500. so we have n=8.

Why do we have to consider 2^n lists? what's that mean: the n that have only one name and the one that has none...please help me!

nc0 + nc1 + nc2 + nc3 + ..... + ncn = 2^n

How? If you are familiar with Binomial expansion of (a + b)^n, substitute a=1 and b=1
http://www.mathwords.com/b/binomial_theorem.htm
Director
Joined: 03 Sep 2006
Posts: 833

Show Tags

24 Dec 2007, 07:34
marcodonzelli wrote:
How many subordinates does Marcia Have?

1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

I found it very hard..

2) is absolutely clear.

1)...I am not even able to comprehend the statement. I am not familiar with binomials, could someone please explain the "simple logic" behind 2^n.

I do understand that...... 2^n gives the number of combinations possible for n things.

But what does it have to do with lists? what is list implying in the question? and number of lists are not fixed in the question... it says it is varying from 200 to 500 ( in case we assume that 200 and 500 are the number of combinations)

CEO
Joined: 17 Nov 2007
Posts: 3481
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

Show Tags

24 Dec 2007, 08:20
11
7
LM wrote:
1)...I am not even able to comprehend the statement. I am not familiar with binomials, could someone please explain the "simple logic" behind 2^n.

I do understand that...... 2^n gives the number of combinations possible for n things.

But what does it have to do with lists? what is list implying in the question? and number of lists are not fixed in the question... it says it is varying from 200 to 500 ( in case we assume that 200 and 500 are the number of combinations)

1. we have n subordinates: S={1,2,3,4....,n}
2. each subordinate may be included or not be included in a list. ( two possibilities)
3. our list we can image like a={1,0,0,1,1,0,1,0,....1} - where 1 - in the list, 1 - out of the list.
4. How many lists we can compose? N=2*2*2......2=2^n
5. N is not any number and is from set: {2^1,2^2,2^3,...,2^n,...}
6. Now, we have 2^7=128,2^8=256,2^9=512
7. we should exclude n(1-subordinate lists)+1(empty list)
8. 120,247,502
Senior Manager
Joined: 29 Jan 2007
Posts: 429
Location: Earth

Show Tags

14 Apr 2008, 18:14
1
Boy. This was cool. I guess just reading answer gave me some cool missing insights.

Sounds tough for exam though. Hope, now I get it

Thanks all contributors.
Intern
Joined: 19 Oct 2009
Posts: 3

Show Tags

20 Oct 2009, 00:45
The answer has to be B and not D as stated. Since a range is mentioned in A. when there is a range multiple answers arise which is against DS. Hence B, unless there is some other explaination. Please provide source of the question, to judge the questions authencity
Intern
Joined: 29 Nov 2009
Posts: 7

Show Tags

29 Nov 2009, 01:54
sapbi wrote:
The answer has to be B and not D as stated. Since a range is mentioned in A. when there is a range multiple answers arise which is against DS. Hence B, unless there is some other explaination. Please provide source of the question, to judge the questions authencity

You are right for most cases, but here only n = 8 fits in the range of 200 to 500, so answer is D.
Intern
Joined: 30 Oct 2009
Posts: 4

Show Tags

25 Feb 2010, 15:45
the easiest way i cld think of was to draw a pascals triangle first
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1 ..... 256
1 9 36 84 126 126 84 36 9 1 ..... 512

(i) adding all the elements in the individual row, only the last before row fall in the 200 to 500 category, so n=8

(ii) match nC2 = 28 in the above triangle and the only row that satisfies is last before row. so n=8

Shashi
Senior Manager
Joined: 21 Jul 2009
Posts: 345
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

Show Tags

25 Feb 2010, 17:49
This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Senior Manager
Joined: 21 Jul 2009
Posts: 345
Schools: LBS, INSEAD, IMD, ISB - Anything with just 1 yr program.

Show Tags

25 Feb 2010, 19:19
[quote="Bunuel"][/quote]

I feel so humbled that you care for me, Math God!!!
_________________

I am AWESOME and it's gonna be LEGENDARY!!!

Manager
Status: Essaying
Joined: 27 May 2010
Posts: 119
Location: Ghana
Concentration: Finance, Finance
Schools: Cambridge
GMAT 1: 690 Q47 V37
GPA: 3.9
WE: Accounting (Education)
Re: How many subordinates does Marcia Have? 1) there are between  [#permalink]

Show Tags

28 Nov 2011, 09:35
Intern
Joined: 25 May 2013
Posts: 26

Show Tags

09 Jul 2013, 01:02
Bunuel wrote:
BarneyStinson wrote:
How many subordinates does Marcia Have?
If
(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her subordinates.
(2) there are 28 ways that she could decide which 2 subordinates she will recommend promoting.

This is a very very interesting problem and the best explanation I could come with is this -

Imagine you have nine open slots and nine digits from 1 to 9, repetition is allowed and you can fill those slots with the digits, it gives you 9^9 entire combinations.

Similarly, you have n open slots and n candidates, each candidate can either fill the slot or not. So that gives 2 possibilities for each slot. For n slots, 2^n. This is practically, all the possible combinations of filling in those slots.

Now given in stmt 1 is that - all such possible combinations range between 200 and 500. So 200 < 2^n < 500. Therefore, n = 8.

Little correction:
For (2) we have $$\{s_1,s_2,s_3,...s_n\}$$. Each subordinate $$(s_1,s_2,s_3,...s_n)$$ has TWO options: either to be included in the list or not. Hence total # of lists - $$2^n$$, correct. But this number will include $$n$$ lists with 1 subordinate as well $$1$$ empty list.

As the lists should contain at least 2 subordinates, then you should subtract all the lists containing only 1 subordinate and all the lists containing 0 subordinate.

Lists with 1 subordinate - n: $$\{s_1,0,0,0...0\}$$, $$\{0,s_2,0,0,...0\}$$, $$\{0,0,s_3,0,...0\}$$, ... $$\{0,0,0...s_n\}$$.
List with 0 subordinate - 1: $$\{0,0,0,...0\}$$

So we'll get $$200<2^n-n-1<500$$, --> $$n=8$$. Sufficient.

For (1):
$$C^2_n=28$$ --> $$\frac{n(n-1)}{2!}=28$$ --> $$n(n-1)=56$$ --> $$n=8$$. Sufficient.

Hi
I wanted to know how can the list be composed of zero subordinate and if there are other candidates as well who are not subordinate then answer will change.

Posted from my mobile device
Re: perms &nbs [#permalink] 09 Jul 2013, 01:02

Go to page    1   2    Next  [ 31 posts ]

Display posts from previous: Sort by

Events & Promotions

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.