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NEW SET of good PS(3)

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Re: NEW SET of good PS(3) [#permalink] New post 18 Oct 2009, 03:26
8th question

Factors of 210 = 2,3,5,7
These can form 4! numbers = 24

but u can also have 1,5,(2*3), 7 as a factor
and these can also form 4! numbers = 24 ways

So total numbers are 48

(C)
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Re: NEW SET of good PS(3) [#permalink] New post 18 Oct 2009, 03:39
Economist wrote:
yangsta8 wrote:
This question is wordy and confused me at first. If P is prime it's only factors are P and 1. So no number below it will have a common factor with it except 1. Therefore answer should just be P-1.
ANS = A

Quote:
I get B. number of factors less than n and which don't have a common factor except one should be p-2.

p = 2, then f(p) = 0
p= 3, then f(p) = 1 ( 2 is the only integer less than 3 and don't have a common factor)
p=5, then f(p) = 3 ( 2,3,4 are the integers ).
and so on...
So, basically for any p, we have to deduct 2 from the value of p ( 1 and itself ).


if p=2, then f(p) = 1 (1 is an integer which does not have a common factor with 2)

So the answer is p-1
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 05:37
rohitbhotica wrote:
Economist wrote:
yangsta8 wrote:
This question is wordy and confused me at first. If P is prime it's only factors are P and 1. So no number below it will have a common factor with it except 1. Therefore answer should just be P-1.
ANS = A

Quote:
I get B. number of factors less than n and which don't have a common factor except one should be p-2.

p = 2, then f(p) = 0
p= 3, then f(p) = 1 ( 2 is the only integer less than 3 and don't have a common factor)
p=5, then f(p) = 3 ( 2,3,4 are the integers ).
and so on...
So, basically for any p, we have to deduct 2 from the value of p ( 1 and itself ).


if p=2, then f(p) = 1 (1 is an integer which does not have a common factor with 2)

So the answer is p-1


Answer should be P-2, because 1 is also a factor of 1 and also a common factor with n. Hence we should deduct 2(1 and itself)
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 06:07
connectshilpa wrote:
Answer should be P-2, because 1 is also a factor of 1 and also a common factor with n. Hence we should deduct 2(1 and itself)


but the questions states "no positive factor in common with n other than 1". I took that to imply that 1 is allowable.
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 08:59
Ill take a shot -

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?
(C) 20
: 6 vertices, 3 to chose from so 6C3 = 20.

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) =
(B) P-2
: for a prime p, all of the numbers preceding it (except 1 will not be a factor of p). Since there p-1 #'s preceding it and we dont count 1, f(p) = p-1-1 = p-2.

3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(D) 63
: used Bunuels trick. I'll let him explain since he was the one who helped me with this.


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?
(C) 48
: Let R = mS + c. Then 6 = m*30 + c and 24 = m*60+c; substituting for c, c = 6-30*m we get 24 = 60*m + 6-30*m, so m = 18/30 = 3/5. Solving for c, c = -12. So for S = 100, R = 3/5*100 -12 = 48.

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
: No idea.

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?
(E) 1/(2r+3)
: Let I = income earned, Sa = amt saved, Sp1 = amt avail. to spend this year and Sp2 = amt avail. to spend next year.
Need to find Sa/I such that Sp2 = Sp1/2.
I = Sa + Sp1 -> [i];
Amt saved this year * (1+r) = amount avail to spend next year, so Sa(1+r) = Sp2. Given Sp2 = Sp1/2, Sp1/2 = Sa(1+r) or Sp1 = 2*Sa*(1+r) -> (ii)
Combining (i) and (ii), I = Sa + 2*Sa*(1+r) or I = Sa*(1+2+2r) so Sa/I = 1/(3+2r).


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?
(C) 50+I/40
: Not sure I understand this correctly, but I'll give it a try anyway. T = 0.02*I + (100+0.01*I)/2 = 0.025*I+50 = I/40 + 50.

8. How many positive integers less than 10,000 are such that the product of their digits is 210?
(B) 30
: Boy this is tough; 210 = 2*3*5*7. If we take all 4 primes as separate digits, then 4*3*2*1 = 24 different #'s. We can also make #'s from the digits 6 (2*3), 5 and 7 = 3*2*1 = 6 different #'s so total 30 #'s. Any other combination of these primes will give a digit > 9 and hence will not get the required result.

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
: Not getting the answer .. I thought it should be 7*6*5*4 since 7 letters and 4 spots.

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

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) 76
: 9 possible options for vertices, need to chose any three to make a triangle so 9C3 = 84. However, 8 (3 along the length, 3 along the height and 2 diagonals)of these 3 sets of points will not make a triangle since they are in a straight line so 84-8 = 76.
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 09:31
badgerboy wrote:
Ill take a shot -

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?
(C) 20
: 6 vertices, 3 to chose from so 6C3 = 20.

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) =
(B) P-2
: for a prime p, all of the numbers preceding it (except 1 will not be a factor of p). Since there p-1 #'s preceding it and we dont count 1, f(p) = p-1-1 = p-2.

3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(D) 63
: used Bunuels trick. I'll let him explain since he was the one who helped me with this.


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?
(C) 48
: Let R = mS + c. Then 6 = m*30 + c and 24 = m*60+c; substituting for c, c = 6-30*m we get 24 = 60*m + 6-30*m, so m = 18/30 = 3/5. Solving for c, c = -12. So for S = 100, R = 3/5*100 -12 = 48.

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
: No idea.

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?
(E) 1/(2r+3)
: Let I = income earned, Sa = amt saved, Sp1 = amt avail. to spend this year and Sp2 = amt avail. to spend next year.
Need to find Sa/I such that Sp2 = Sp1/2.
I = Sa + Sp1 -> [i];
Amt saved this year * (1+r) = amount avail to spend next year, so Sa(1+r) = Sp2. Given Sp2 = Sp1/2, Sp1/2 = Sa(1+r) or Sp1 = 2*Sa*(1+r) -> (ii)
Combining (i) and (ii), I = Sa + 2*Sa*(1+r) or I = Sa*(1+2+2r) so Sa/I = 1/(3+2r).


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?
(C) 50+I/40
: Not sure I understand this correctly, but I'll give it a try anyway. T = 0.02*I + (100+0.01*I)/2 = 0.025*I+50 = I/40 + 50.

8. How many positive integers less than 10,000 are such that the product of their digits is 210?
(B) 30
: Boy this is tough; 210 = 2*3*5*7. If we take all 4 primes as separate digits, then 4*3*2*1 = 24 different #'s. We can also make #'s from the digits 6 (2*3), 5 and 7 = 3*2*1 = 6 different #'s so total 30 #'s. Any other combination of these primes will give a digit > 9 and hence will not get the required result.

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
: Not getting the answer .. I thought it should be 7*6*5*4 since 7 letters and 4 spots.

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

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) 76
: 9 possible options for vertices, need to chose any three to make a triangle so 9C3 = 84. However, 8 (3 along the length, 3 along the height and 2 diagonals)of these 3 sets of points will not make a triangle since they are in a straight line so 84-8 = 76.


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 9th question we have to take 3 cases
1) where only one of each letter is chosen = 6C3 ways = 15 ways
2) Where either 2 E's or 2N's are chosen = 2 * 5C2 = 20 ways
3) Where 2 E's and 2 N's are chosen = 1 way

Hence answer = 36 ways

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.
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 10:17
rohitbhotica 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 9th question we have to take 3 cases
1) where only one of each letter is chosen = 6C3 ways = 15 ways
2) Where either 2 E's or 2N's are chosen = 2 * 5C2 = 20 ways
3) Where 2 E's and 2 N's are chosen = 1 way

Hence answer = 36 ways

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.


Shoot ... good catch on Q8. I forgot about the 1.
Can you explain why the (x+3)C3 for Q5?
Thanks for the explanations .. they were very helpful.
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 10:21
badgerboy wrote:
rohitbhotica 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 9th question we have to take 3 cases
1) where only one of each letter is chosen = 6C3 ways = 15 ways
2) Where either 2 E's or 2N's are chosen = 2 * 5C2 = 20 ways
3) Where 2 E's and 2 N's are chosen = 1 way

Hence answer = 36 ways

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.


Shoot ... good catch on Q8. I forgot about the 1.
Can you explain why the (x+3)C3 for Q5?
Thanks for the explanations .. they were very helpful.



no i think 1 has to be counted. What is the OA?
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Re: NEW SET of good PS(3) [#permalink] New post 21 Oct 2009, 10:26
rohit,
agreed 1 has to be counted. i read the question a second time and it made sense. I shall defer to Bunuel for OA's.
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Re: NEW SET of good PS(3) [#permalink] New post 23 Oct 2009, 17:40
Expert's post
1
This post was
BOOKMARKED
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.
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Last edited by Bunuel on 02 Nov 2009, 14:12, edited 1 time in total.
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Re: NEW SET of good PS(3) [#permalink] New post 23 Oct 2009, 21:39
Hi Bunuel, would appreciate if you can explain the solutions for 3,5 and 9.
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Re: NEW SET of good PS(3) [#permalink] New post 24 Oct 2009, 12:06
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
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Re: NEW SET of good PS(3) [#permalink] New post 16 Nov 2009, 03:36
Thanks Bunuel... your explainations are always the best.
:-D :-D :-D
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Re: NEW SET of good PS(3) [#permalink] New post 02 Jan 2010, 10:08
Bunuel wrote:
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.


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.


Hi Bunuel.. could you please explain the above answers in detail...

thanks,
JT
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Re: NEW SET of good PS(3) [#permalink] New post 03 Jan 2010, 02:19
Thanks Bunuel..... :good...

Cheers!
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Re: NEW SET of good PS(3) [#permalink] New post 03 Jan 2010, 04:39
atish wrote:
Answer to the 3rd question -
We start of by factorizing 264,600

=2^3 * 3^3 * 5^2 * 7^2

To create numbers from these factors we basically separate multiples of 2 & 3, since any combination of these will be divisible by 6.

Hence we find the number of factors for
2^3 * 5^2 * 7^2

and add it to the factors of

3^3 * 5^2 * 7^2

In case someone doesn't know how to calculate the number of factors of a given number - add the powers of it's prime factors by 1 and multiply them.

In our case it is (3+1)*(2+1)*(2+1) = 36
similarly for 3^3 * 5^2 * 7^2 it is (3+1)*(2+1)*(2+1) = 36

Now if we add the two numbers above we end up double counting the factors of 5^2*7^2 = (2+1)*(2+1) = 9

Hence the answer is 36+36-9 = 63.


Cool explanation :good +1 for this mate...

Cheers!
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Re: NEW SET of good PS(3) [#permalink] New post 03 Jan 2010, 04:44
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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Re: NEW SET of good PS(3) [#permalink] New post 02 Mar 2010, 23:24
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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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Re: NEW SET of good PS(3) [#permalink] New post 06 Mar 2010, 00:43
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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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Re: NEW SET of good PS(3) [#permalink] New post 06 Mar 2010, 00:50
Expert's post
concept wrote:
Hi Buneul,
I have a little trouble in understanding the question No.3. Will it mean the same if I rephrase it as 'How many numbers that are not divisible by 6 divide 264,600 evenly?'

Quote:
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


Yes, your rephrasing is correct. This problem is a hard one, the solution for it is given on page 1.
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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.


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Re: NEW SET of good PS(3)   [#permalink] 06 Mar 2010, 00:50
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