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

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15 Nov 2012, 13:34

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

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15 Nov 2012, 14:52

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

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15 Nov 2012, 14:56

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

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15 Nov 2012, 15:11

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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]

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15 Nov 2012, 15:14

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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!
_________________

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

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

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15 Nov 2012, 20:09

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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]

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15 Nov 2012, 20:29

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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

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

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15 Nov 2012, 20:44

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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}

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

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

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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

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