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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

A plane flying north at 500 kmph passes over a city at 12 noon. A plane flying east at the same attitude passes over the same city at 12.30 pm. The plane is flying east at 400 kmph. To the nearest hundred km, how far apart are the two planes at 2 pm?

A) 600 km B) 1000 km C) 1100 km D) 1200 km E) 1300 km
_________________

Re: Tough PS: a, b, c are three distinct integers from 2 to 1 [#permalink]

Show Tags

15 Nov 2012, 20:44

2

This post received KUDOS

PraPon wrote:

a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets).

======== Choices ======== A. 4 B. 5 C. 6 D. 7 E. 8 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!

Lets decipher statements one by one. a,b,c are distinct and among {2,3,4,5,6,7,8,9 10} Exactly one of ab,bc and ca is odd => 2 among a,b,c are odd and one is even but we dont know which ones are odd and which one is even. abc is multiple of 4 => since two numbers are odd, only one is even. and that one is a multiple of 4. only possibilities for such even number in the set {4,8} arithmetic mean of a and b is an integer => a+b is even => a and b both are odd. So now we know a,b are odd and thus could be from {3,5,7,9} and c is even and could be from{4,8}

arithmetic mean of a, b and c is also integer = sum of a+b+c is a multiple of 3. if C is 4: possibilities for a and b {3,5} ,{5,9} if C is 8: possibilities for a and b {3,7}, {7,9}

Two boys begin together to write out a booklet containing 53 [#permalink]

Show Tags

15 Nov 2012, 13:34

1

This post received KUDOS

Two boys begin together to write out a booklet containing 535 lines. The first boy starts with the first line, writing at the rate of 100 lines an hour; and the second starts with the last line then writes line 534 and so on, backward proceeding at the rate of 50 lines an hour. At what line will they meet?

A) 356 B) 277 C) 357 D) 267 E) 286
_________________

How many five digit positive integers are divisible by 3 [#permalink]

Show Tags

15 Nov 2012, 14:52

1

This post received KUDOS

How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated?

=========== Choices =========== A. 15 B. 96 C. 216 D. 120 E. 625 ===========
_________________

There are 10 seats around a circular table. If 8 men and 2 [#permalink]

Show Tags

15 Nov 2012, 14:56

1

This post received KUDOS

There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P : Q equals

========== Choices ========== A. 9 : 1 B. 72 : 1 C. 10 : 1 D. 8 : 1 E. 25 : 1 ==========
_________________

Tough PS: a, b, c are three distinct integers from 2 to 1 [#permalink]

Show Tags

15 Nov 2012, 15:11

1

This post received KUDOS

a, b, c are three distinct integers from 2 to 10 (both inclusive). Exactly one of ab, bc and ca is odd. abc is a multiple of 4. The arithmetic mean of a and b is an integer and so is the arithmetic mean of a, b and c. How many such triplets are possible (unordered triplets).

======== Choices ======== A. 4 B. 5 C. 6 D. 7 E. 8 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!
_________________

TOUGH PS!! Bob is about to hang his 8 shirts in the wardrobe [#permalink]

Show Tags

15 Nov 2012, 15:14

1

This post received KUDOS

Bob is about to hang his 8 shirts in the wardrobe. He has four different styles of shirt, two identical ones of each particular style. How many different arrangements are possible if no two identical shirts are next to one another?

======== Choices ======== A. 764 B. 864 C. 876 D. 964 E. 1064 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!
_________________

Re: If F(x, n) be the number of ways of distributing "x" toys [#permalink]

Show Tags

15 Nov 2012, 20:09

1

This post received KUDOS

PraPon wrote:

If F(x, n) be the number of ways of distributing "x" toys to "n" children so that each child receives at the most 2 toys then F(4, 3) = _______?

======== Choices ======== A. 3 B. 6 C. 5 D. 4 E. 2 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!

F(4,3) means - "4" toys to "3" children so that each child receives at the most 2 toys. There is no restriction on minimum number of toys per child. Possibilities are: Everyone gets 1 toy, and 1 remaining toys to be given to any one of 3 children = 3 possiblities 2 children get 2 toys each and one gets 0. Or in easier words, possibilities of a child getting 0 = 3 possiblities

Re: Tough PS: There are 6 boxes numbered 1,2,...6. [#permalink]

Show Tags

15 Nov 2012, 20:29

1

This post received KUDOS

PraPon wrote:

There are 6 boxes numbered 1,2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

======== Choices ======== A. 5 B. 21 C. 33 D. 60 E. 27 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!

Not sure if we are still preparing for GMAT

Anyway, for atleast 1 box to contain green ball and green balls to be consecutive numbered. We need to calculate for each case one by one.. If only 1 green ball: We have 6 places to place the green ball = 6C1 = 6 when 2 green balls: We have 6 places, 2 green balls to be placed together = 5C2 = 5 when 3 green balls: We have 6 places, 3 green balls to be placed together = 4C3 = 4 when 4 green balls: We have 6 places, 4 green balls to be placed together = 3C2 = 3 when 5 green balls: We have 6 places, 5 green balls to be placed together = 2C1 = 2 when all 6 green balls: We have 6 places, 6 green balls (to be placed together) = 1

Hence total number of possibilities = 6+5+4+3+2+1 = 21

A plane flying north at 500 kmph passes over a city at 12 noon. A plane flying east at the same attitude passes over the same city at 12.30 pm. The plane is flying east at 400 kmph. To the nearest hundred km, how far apart are the two planes at 2 pm?

A) 600 km B) 1000 km C) 1100 km D) 1200 km E) 1300 km

Check the attached picture. Plane flying north travels 1000 km, plan flying east travels 600 km. Hence distance between two planes = \(\sqrt{(1000^2+600^2)}\) = approx 1200 km

The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is

Choices: A. 12 B. 8 C. 9 D. 10 E. 7
_________________

Tough PS: There are 6 boxes numbered 1,2,...6. [#permalink]

Show Tags

15 Nov 2012, 15:20

There are 6 boxes numbered 1,2,...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is:

======== Choices ======== A. 5 B. 21 C. 33 D. 60 E. 27 ========

A detailed Official explanation will follow after few attempts by GMAT clubbers!
_________________

Re: Tough: The number of ways of arranging n students in a row [#permalink]

Show Tags

15 Nov 2012, 21:57

PraPon wrote:

The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is

Choices: A. 12 B. 8 C. 9 D. 10 E. 7

First thing to notice is: if we are able to arrange n students such that no two boys sit together and no two girls sit together, that means boys and girls will sit in alternate positions. Second point to notice: we are also able to arrange n+1 students such that no two boys sit together and no two girls sit together.

What it means is that n is even and 'girls & boys' or 'boys & girls' can take alternate seats. And n+1 is odd and girl & boys take alternate seats starting from whoever is more in number.

(Note: if n were to be odd, we wouldnt have been able to arrange n+1 students in alternate seats.) ------ Solving the problem:

Now lets assume n =2k. (for ease in calculation and formatting, formatting fractions is $#@.. I'm sure you get it ) Number of arrangement in first case = 2 * k!*k! (boy or girl can take first seat hence 2, k boys (or girls) in k seat hence k!)

Number of arrangement in second case = (k+1)!*k! (whoever is k+1 (boy or girl) must take first seat and be arranged in alternate k+1 seat hence (k+1)!, rest k boys (or girls) in k seat hence k!)

Coming back to the problem statement: Number of arrangement in second case = 3* Number of arrangement in first case =>\((k+1)!*k! = 3*( 2*k!*k!)\) =>\((k+1)*k!*k! = 3*( 2*k!*k!)\) =>\((k+1) = 3*2\) =>\(k=5\) Hence n =2k =10

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...