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NEW SET of good PS(3)

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

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New post 15 Nov 2009, 11:48
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h2polo wrote:
Bunuel,

Quality problems as usual... and also as usual they have kicked my butt. :-D

Can you please post your solution to Question 9?

It is driving me nuts!

Thanks again!


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

This problem is not the one you'll see on real GMAT. As combinatorics problems on GMAT are quite straightforward. But still it could be good for practice.

We have 8 letters from which 6 are unique.

Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.

Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.

15+10+10+1=36

Answer: B.

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

This one should go relatively easy after we solved the previous. So we have:

A. 15 4 letter words with all distinct letters. # of arrangements of 4 letter word is 4!, as we have 15 such words, then = 15*4!=360;
B. 10 4 letter words with two N-s and two other distinct letters. The same here except # of arrangements would be not 4! but 4!/2! as we need factorial correction to get rid of the duplications=10*4!/2!=120;
C. The same as above: 10*4!/2!=120;
D. 2 N-s and 2 E-s, # of arrangement=4!/2!*2!=6.

Total=360+120+120+6=606.

Hope it's clear.
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Re: NEW SET of good PS(3)  [#permalink]

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New post 02 Jan 2010, 12:42
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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

210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).

So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.

{1,6,5,7} # of combinations 4!=24
{2,3,5,7} # of combinations 4!=24
{6,5,7} # of combinations 3!=6

24+24+6=54.

Answer: D.


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

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...

From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.

We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}

So the final answer would be; 9C3-8=84-8=76

Answer: B.

Hope it's clear.
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Re: NEW SET of good PS(3)  [#permalink]

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New post 03 Jan 2010, 05:44
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.


The solution provided by you is well explained.. However for the text marked in red - is there a faster way for solving the eq for k...? Since time crunch is a big issue in GMAT exam? Please let me know your views...

Thanks,
JT
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New post 03 Jan 2010, 23:53
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jeeteshsingh wrote:
The solution provided by you is well explained.. However for the text marked in red - is there a faster way for solving the eq for k...? Since time crunch is a big issue in GMAT exam? Please let me know your views...

Thanks,
JT


Do you mean from this point: (k+1)(k+2)(k+3)=3!*120=720?

I solved this in the following way: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try 5*6*7=210<720, next triplet 8*9*10=720, bingo!
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New post 03 Mar 2010, 00:24
sandeep25398 wrote:
Bunuel wrote:
.
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.

hi Bunuel,
could you please, clarify "evenly divisible" does not mean the quotient should be multiple of 2?
or evenly divisible is equivalent to completely divisible in this context.
thanks.


Evenly divisible=divisible, --> remainder=0.
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New post 06 Mar 2010, 01:43
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amod243 wrote:
Bunuel.. Can you explain the solution of Problem #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

Regular pentagon is a pentagon where all sides are equal. In such pentagon center is not collinear to any two vertices, so ANY three points (from 5 vertices and center point) WILL form the triangle.

The question basically asks how many triangles can be formed from the six points on a plane with no three points being collinear.

As any 3 points from 6 will make a triangle (since no 3 points are collinear), then:

6C3=20

Answer: C.

Hope it helps.
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New post 09 Aug 2010, 15:03
masland wrote:
ExecMBA2010 wrote:
A query on Q7:What is the meaning of this portion?

'to the average(arithmetic mean)of 100units of Country R's currency'


I'm curious as well. The language of this question really threw me off.

The language is from GMATPrep


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

Tax is the sum of the following:
2 percent of one's annual income - \(0.02I\);
The average (arithmetic mean) of 100 units of country R's currency and 1 percent of one's annual income - \(\frac{100+0.01I}{2}\).

\(Tax=0.02*I+\frac{100+0.01*I}{2}=\frac{0.04*I+100+0.01*I}{2}=50+\frac{0.05*I}{2}=50+\frac{I}{40}\).

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

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New post 25 Apr 2011, 05:07
Bunuel wrote:
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?[/b]
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...

From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.

We'll have 8 sets of collinear points of three:
3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}...
3 vertical
2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}

So the final answer would be; 9C3-8=84-8=76

Answer: B.

Hope it's clear.


Bunuel,

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

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

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New post 27 May 2011, 05:53
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yogesh1984 wrote:

Bunuel,

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

TIA ~ Yogesh


Check out this thread:
ps-right-triangle-pqr-71597.html?hilit=how%20many%20triangles#p830694
It discusses what to do in case of a larger range.
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New post 12 Jun 2011, 10:27
if anyone can help please to clarify the methos:

let's say that the Q was:
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≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance
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New post 12 Jun 2011, 13:09
dimri10 wrote:
if anyone can help please to clarify the methos:

let's say that the Q was:
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≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance


While I seriously doubt whether one could encounter such long range question
(esp. because calculating # diagonals is going to be little tricky here) unless you are shooting for 51 in quant.

That said let me try my hands-

Think about when it will be horizontal collinear- all the y values are same for a given set of X values. so we have 6 values where Y can be same (it has to be integer coordinate)- so total # horizontal collinear points- 6
You can have similar argument for vertical (constant X and vary Y) set of collinear points- 6

For # diagonals (please refer tot the attachment, I sketched only one side of the diagonals ) - you should be able to count the numbers now. For one side it comes out that we will have 16 such pairs (of 3 points) so by symmetry you need to multiply by 2. SO a total # diagonals will be 32.

I hope that helps.

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

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New post Updated on: 19 Mar 2012, 03:40
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dimri10 wrote:
if anyone can help please to clarify the methos:

let's say that the Q was:
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≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance


I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind).
Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

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≤6 and 1≤y≤6?

Check out this post for the solution:
http://www.veritasprep.com/blog/2011/09 ... mment-2495

*Edited the post to fix the problem.
Attachments

Ques2.jpg
Ques2.jpg [ 16.67 KiB | Viewed 6746 times ]


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Originally posted by VeritasKarishma on 12 Jun 2011, 19:17.
Last edited by VeritasKarishma on 19 Mar 2012, 03:40, edited 1 time in total.
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Re: NEW SET of good PS(3)  [#permalink]

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New post 21 Aug 2011, 08:56
Economist wrote:
yangsta,
i liked your solution for 4. I didnt know we can use the definition of linear equation to solve such problems.

I used the guessing method.
we have two relationships...6--30 and 24---60.
This means when R is increased 4 times, S increases 2 times, so if R is increased 2 times S will increase 1 time.
Now, 30*3 ~ 100, so 3 times increase in S will have atleast a 6 times increase in R, i.e. R should be something greater than 36..closest is 48 :)


Another method (let me call it intuition method) :

6 on scale R corresponds to 30 on scale S and 24 on scale R corresponds to 60 on scale S. If we notice the relationship, we will see that for every 6 points on scale R, 10 points move on scale S. So, 90 points on scale S corresponds to 42 points on Scale R and another 6 points of scale S for another 10 points on scale R. Hence 100 on scale S corresponds to 42+6 = 48 on scale R.

I hope I am making sense :)
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Re: NEW SET of good PS(3)  [#permalink]

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New post 22 Jan 2012, 04:49
VeritasPrepKarishma wrote:
dimri10 wrote:
if anyone can help please to clarify the methos:

let's say that the Q was:
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≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance


I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind).
Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

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≤6 and 1≤y≤6?

Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there?
Look at the diagram:

Attachment:
Ques2.jpg

The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points.
These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points.
Total triplets of collinear points = 64 + 16 = 80
Therefore, total number of triangles you can make = 36C3 - 80


Similarly you can work with 1<=x<=5 and -9<=y<=3.
The number of red dots in this case = 11*3 = 33
So number of collinear triplets represented by red arrows will be = 33*4 = 132
Number of black arrows will be 3 + 11 + 3 + 11 = 28
Total triplets of collinear points = 132 + 28 = 160
Total triangles in this case = 65C3 - 160


Ma'am,
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).
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Re: NEW SET of good PS(3)  [#permalink]

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New post 22 Jan 2012, 05:08
akhileshankala wrote:
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).


Yes, I did miss out on the non-adjacent collinear points! And on the face of it, your calculation looks correct. I will put some more time on this variation tomorrow (since today is Sunday!) and get back if needed.
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Re: NEW SET of good PS(3)  [#permalink]

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New post 14 Mar 2012, 06:37
akhileshankala wrote:
VeritasPrepKarishma wrote:
dimri10 wrote:
if anyone can help please to clarify the methos:

let's say that the Q was:
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≤6 and 1≤y≤6?
how will it be solved:
will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance


I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind).
Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

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≤6 and 1≤y≤6?

Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there?
Look at the diagram:

Attachment:
Ques2.jpg

The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points.
These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points.
Total triplets of collinear points = 64 + 16 = 80
Therefore, total number of triangles you can make = 36C3 - 80


Similarly you can work with 1<=x<=5 and -9<=y<=3.
The number of red dots in this case = 11*3 = 33
So number of collinear triplets represented by red arrows will be = 33*4 = 132
Number of black arrows will be 3 + 11 + 3 + 11 = 28
Total triplets of collinear points = 132 + 28 = 160
Total triangles in this case = 65C3 - 160


Ma'am,
It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw.
my way:
no: of collinear points=?
horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs.
hence horz and vert. lines give a total of 2*6*6C3.
next 2 diagonals give same no: of such possibs.
consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).



Can you please elaborate on the bolded part in details...
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Re: NEW SET of good PS(3)  [#permalink]

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New post 14 Mar 2012, 08:48
This is a 6x6 square. For each diagonal of this square, you have 8 parallel lines, you can draw within the square by joining the vertices that lies on the edges of the square.
eg: Join (1,2) & (2,1); (1,3) & (3,1); (1,4) & (4,1); (1,5) & (5,1); to get 4 parallel lines along the diagonal (1,6)-(6,1)
Similarly you can get 4 lines on the other side of the diagonal.

Of these, (line joining (1,2) to (2,1) is of no use to us since it contains only 2 points within the square)
the line joining point (1,3) & (3,1) contains total of 3 integer co-ordinates,
the line joining point (1,4) & (4,1) contains total of 4 integer co-ordinates, and so on.....

Any 3 points that you select from these lines will be collinear and not form a traingle.
Thus, you have 3,4,5,6,5,4,3 points collinear along the lines parallel to the diagonal.
Rest as akhilesh has mentioned.

You may draw a figure by plotting these points.

My 1st post on this forum, so Apologies for the weird explanation.
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New post 19 Mar 2012, 03:35
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yogesh1984 wrote:

Can you please elaborate on the bolded part in details...


Check out this post. I have explained this question in detail in this post. It fixes the problem my above given solution had.

http://www.veritasprep.com/blog/2011/09 ... o-succeed/
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New post 06 Jul 2012, 02:21
shreya717 wrote:
Shouldn't the answer to Question 2 be B?


Answers are given in the following post: new-set-of-good-ps-85440.html#p642321 OA for this question is A, not B.

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 a prime number then f(p)=
A. p-1
B. p-2
C. (p+1)/2
D. (p-1)/2
E. 2

If not the wording the question wouldn't be as tough as it is now. The GMAT often hides some simple concept in complicated way of delivering it.

This question for instance basically asks: how many positive integers are less than given prime number p which have no common factor with p except 1.

Well as p is a prime, all positive numbers less than p have no common factors with p (except common factor 1). So there would be p-1 such numbers (as we are looking number of integers less than p).

For example: if p=7 how many numbers are less than 7 having no common factors with 7: 1, 2, 3, 4, 5, 6 --> 7-1=6.

Answer: A.

Hope it's clear.
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New post 18 Mar 2013, 14:15
Hi, still not clear why answer to the 2nd ques should be P-1 and not P-2. Shouldn't both P and 1 be deducted from the set?
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Re: NEW SET of good PS(3) &nbs [#permalink] 18 Mar 2013, 14:15

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