Find all School-related info fast with the new School-Specific MBA Forum

It is currently 28 May 2016, 00:03
GMAT Club Tests

Close

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
Your Progress

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

NEW SET of good PS(3)

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Intern
Intern
avatar
Joined: 10 Sep 2013
Posts: 3
Concentration: General Management, Technology
GMAT Date: 08-20-2014
Followers: 0

Kudos [?]: 0 [0], given: 113

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 28 May 2014, 12:48
Bunuel wrote:
5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\)
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\)
Means that first nephew will get all the tickets,

\(|t|ttt|t\)
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).


Hope it helps.



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

Hi Bunuel, Can you please help solve this problem directly using the formulas provided?

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is n-1C_{r-1}.

Regards
Harish
Intern
Intern
User avatar
Joined: 03 Feb 2014
Posts: 3
Location: Russian Federation
Followers: 0

Kudos [?]: 6 [0], given: 39

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 11 Jul 2014, 23:29
Augustus wrote:
yangsta8 wrote:
Bunuel wrote:
1. ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30


6 points in total to make triangles. I think a combination of any 3 will make a unique triangle so:
6C3 = 20


Will anyone please explain what the C means in the notation?

Thanks in advance



I guess, 6C3 stands for a Combination formula, i.e. we have 3 choices out of 6 without restrictions. This means 6!/3!*3!=20
Intern
Intern
User avatar
Joined: 21 Jul 2014
Posts: 23
Followers: 0

Kudos [?]: 1 [0], given: 13

GMAT ToolKit User
Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 09 Aug 2014, 07:06
Bunuel wrote:
ANSWERS (OAs):

As most of the problems was solved correctly, I'm posting only OAs. Please let me know if anyone needs any clarification.

1. ABCDE is a regular pentagon with F at its center. How many different triangles can be formed by joining 3 of the points A,B,C,D,E and F?
(A) 10
(B) 15
(C) 20
(D) 25
(E) 30

Answer: C.


2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2

Answer: A.


3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

Answer: D.


4.A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale?
(A) 20
(B) 36
(C) 48
(D) 60
(E) 84

Answer: C.


5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers?
(A) 13
(B) 14
(C) 15
(D) 16
(E) more than 16

Answer: C.


6. This year Henry will save a certain amount of his income, and he will spend the rest. Next year Henry will have no income, but for each dollar that he saves this year, he will have 1 + r dollars available to spend. In terms of r, what fraction of his income should Henry save this year so that next year the amount he was available to spend will be equal to half the amount that he spends this year?
(A) 1/(r+2)
(B) 1/(2r+2)
(C) 1/(3r+2)
(D) 1/(r+3)
(E) 1/(2r+3)

Answer: E.


7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100

Answer: C.


8. How many positive integers less than 10,000 are such that the product of their digits is 210?
(A) 24
(B) 30
(C) 48
(D) 54
(E) 72

Answer: D.


9. Find the number of selections that can be made taking 4 letters from the word"ENTRANCE".
(A) 70
(B) 36
(C) 35
(D) 72
(E) 32

Answer:B.

Find in the above word, the number of arrangements using the 4 letters.

Answer:606.


10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

Answer: B.


Hello Bunuel, Could you please explain the approaches to solve Q9 and Q10.
GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 9648
Followers: 465

Kudos [?]: 120 [0], given: 0

Premium Member
Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 02 Sep 2015, 00:59
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Intern
Intern
avatar
Joined: 03 Jul 2015
Posts: 33
Followers: 0

Kudos [?]: 2 [0], given: 4

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 05 Oct 2015, 17:33
2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2


Regarding the question 2, I think it should be choice B and not A
I agree to the way of solving here . BUT question says f(n) includes all integers LESS than N and does not say including N . Thus if F is a prime, shouldnt it be n-1(as n is not included) and another -1 (as 1 is a common factor, Thus n-2.

Apologies if I am wrong. PLease correct me .
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 33052
Followers: 5767

Kudos [?]: 70671 [0], given: 9856

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 06 Oct 2015, 04:52
Expert's post
shreyashid wrote:
2. The function f is defined for all positive integers n by the following rule: f(n) is the number of positive integers each of which is less than n and has no positive factor in common with n other than 1. If p is prime, then f(p) =
(A) P-1
(B) P-2
(C) (P+1)/2
(D) (P-1)/2
(E) 2


Regarding the question 2, I think it should be choice B and not A
I agree to the way of solving here . BUT question says f(n) includes all integers LESS than N and does not say including N . Thus if F is a prime, shouldnt it be n-1(as n is not included) and another -1 (as 1 is a common factor, Thus n-2.

Apologies if I am wrong. PLease correct me .


That's not correct. There is no reason whatsoever to exclude 1: 1 also does not have have a common factor with say 7 other than 1. So, you should include 1 as well.
_________________

New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 19 Jan 2016
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 1

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 10 Feb 2016, 02:39
I think your answer for the 8th question is wrong coz ur missing the cases when it is 5, 6 ,7 and 1 which are 24 more cases. So the answer should be 54 and not 30

For question 5 assume first that Mrs. Smith has given 8 tickets to her grandsons by giving 2 to each and has "x" tickets left. So she can now distribute these x tickets to her 4 grandsons in (x+3)C3 ways. This is selection without arrangement so we use this formula.
We thus get this to be equal to 120. Thus we get x+3 = 10 and x = 7
hence total tickets is 15


for the second question I think we should count 1, as there is no reason for not counting it and hence the answer should be p-1 only.[/quote]



Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?

Thanks in advance.
Expert Post
1 KUDOS received
Verbal Forum Moderator
Verbal Forum Moderator
User avatar
Joined: 02 Aug 2009
Posts: 3644
Followers: 197

Kudos [?]: 1974 [1] , given: 84

Re: NEW SET of good PS(3) [#permalink]

Show Tags

New post 10 Feb 2016, 03:00
1
This post received
KUDOS
Expert's post
alishandhanani17 wrote:
I think your answer for the 8th question is wrong coz ur missing the cases when it is 5, 6 ,7 and 1 which are 24 more cases. So the answer should be 54 and not 30

For question 5 assume first that Mrs. Smith has given 8 tickets to her grandsons by giving 2 to each and has "x" tickets left. So she can now distribute these x tickets to her 4 grandsons in (x+3)C3 ways. This is selection without arrangement so we use this formula.
We thus get this to be equal to 120. Thus we get x+3 = 10 and x = 7
hence total tickets is 15


for the second question I think we should count 1, as there is no reason for not counting it and hence the answer should be p-1 only.




Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?

Thanks in advance.[/quote]

Hi,
for your Q:-
Can anyone please explain from where this (X+3) came? And how (x+3)C3=120 is equal to x+3 = 10?
(x+3)C3= (x+3)!/3!(x+3-3)!= (x+3)!/3!(x)!
= (x+3)(x+2)(x+1)x!/3!x!= (x+3)(x+2)(x+1)/3!
now this is equal to 120..
(x+3)(x+2)(x+1)/3!=120
(x+3)(x+2)(x+1)=720..
now 720=8*9*10=(x+3)(x+2)(x+1)..
thus x=7 and x+3=10
hope it helps
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Re: NEW SET of good PS(3)   [#permalink] 10 Feb 2016, 03:00

Go to page   Previous    1   2   3   4   5   [ 88 posts ] 

    Similar topics Author Replies Last post
Similar
Topics:
266 Experts publish their posts in the topic New Set of Mixed Questions!!! Bunuel 180 01 Apr 2013, 07:48
274 Experts publish their posts in the topic New Algebra Set!!! Bunuel 168 18 Mar 2013, 07:56
Very good PS set with answers! miguelmick 1 22 Feb 2011, 16:28
PS3 Gmat prep jitendra 2 10 Apr 2010, 06:06
63 Experts publish their posts in the topic Good set of PS 2 Bunuel 108 16 Oct 2009, 20:00
Display posts from previous: Sort by

NEW SET of good PS(3)

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

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®.