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# NEW SET of good PS(3)

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Joined: 02 Aug 2009
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Re: NEW SET of good PS(3) [#permalink]

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10 Feb 2016, 03:00
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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?

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

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05 Jun 2016, 01:03
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Expert's post
nishi999 wrote:
Bunuel wrote:
h2polo wrote:
Bunuel,

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

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

Hi,

I fail to understand as to why have you considered multiple scenario's for the 4 letter selection.
I believe that the formula of nCr i.e. 8C4 would consider all of the same. Are you trying to eradicate the possibility of some duplicates in the scenario?
Could you kindly enumerate the same principle in case we have to select 3 out of 5 letters from AABCD by mentioning all the possible selections, so we can understand what is not to be considered?

Thanks

Hi,
8C4 will be correct if you have to choose 4 out of 8 different items...
But if you have few items same, then few groups become SAME and thus extra..
take this example in hand-
4 letters from the word "ENTRANCE...
In 8C4 ENTRANCE will be one and ENTRANCE will be another..
But are they different NO-- they consist of same 4 letters E, N, T, R....

There fore we require to work out individually..

select 3 out of 5 letters from AABCD..
there are four different letters - ABCD so you can choose 3 out of them in 4C3 = 4 ways..
Now let the three consist of AA, so the remaining 3rd can be any of BCD - 3 ways
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: NEW SET of good PS(3) [#permalink]

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05 Jun 2016, 03:54
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Expert's post
nishi999 wrote:
chetan2u

Hi,

Thanks for the explanation.

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;;

In B - Can you confirm why only 5C2 is considered i.e. five distinct letters. Is it because if all 6 letters were considered, then each combination formed would have a duplicate like TE, TE or AE, AE because of 2 E's which would in turn lead to duplication.

P.S.
Its taken me over 20 mins to understand what i wanted to ask. Shoooo

Hi nishi999,

Why we have taken only 5C2 i.e. five distinct letters..
Hi we are looking for cases where 2 Ns are there, but if we choose 2 out of remaining 6, we will get those cases too where two Es are there..
But we are already working on BOTH 2Es and 2Ns together ...
..
so we choose first 2-Ns and other 2 different ......
then we choose where 2-Es are there and other 2 are different..
Finally we choose cases where both N and E are used twice...

If we choose 2-Ns and choose 2 from remaining 6, it will contain cases where both 2Es and 2Ns are there..
Same thing will happen when we work on 2Es, thus will result in repetitions..
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Re: NEW SET of good PS(3) [#permalink]

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18 Oct 2009, 04:17
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yangsta8 wrote:
Bunuel wrote:
Bunuel 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

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

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

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18 Oct 2009, 04:22
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
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Re: NEW SET of good PS(3) [#permalink]

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18 Oct 2009, 04:26
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]

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18 Oct 2009, 04:39
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)

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

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21 Oct 2009, 06:37
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)

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]

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21 Oct 2009, 07:07
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]

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21 Oct 2009, 09:59
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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]

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21 Oct 2009, 10:31
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

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]

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21 Oct 2009, 11:17
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

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]

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21 Oct 2009, 11:21
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

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]

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21 Oct 2009, 11:26
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]

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23 Oct 2009, 18:40
Expert's post
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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

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

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

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

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

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)

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

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

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

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?
(A) 72
(B) 76
(C) 78
(D) 80
(E) 84

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Last edited by Bunuel on 02 Nov 2009, 15:12, edited 1 time in total.
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Re: NEW SET of good PS(3) [#permalink]

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23 Oct 2009, 22:39
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]

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24 Oct 2009, 13:06
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?

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

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16 Nov 2009, 04:36
Thanks Bunuel... your explainations are always the best.
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Posts: 359
Re: NEW SET of good PS(3) [#permalink]

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02 Jan 2010, 11:08
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

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

thanks,
JT
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JT...........
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|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

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Posts: 359
Re: NEW SET of good PS(3) [#permalink]

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03 Jan 2010, 03:19
Thanks Bunuel..... ...

Cheers!
JT
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Re: NEW SET of good PS(3)   [#permalink] 03 Jan 2010, 03:19

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