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NEW SET of good PS(3)

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NEW SET of good PS(3) [#permalink] New post 17 Oct 2009, 17:33
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Question Stats:

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New questions:

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

As always please share your way of thinking.

Also you can check new set of DS problems: new-set-of-good-ds-85441.html
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Re: NEW SET of good PS(3) [#permalink] New post 17 Oct 2009, 18:21
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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
AND = C

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
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Re: NEW SET of good PS(3) [#permalink] New post 17 Oct 2009, 18:35
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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


This one is tricky, am not sure if I solved it correctly:
264600 = 2^3 * 3 * 5^2 * 431
Numbers that will divide 264600 will be made up of multiples of factors of 264600 that do not divide by 6.
Expanding out the factors that are made up of the prime factors I get:
2,4,8,5,25,431
The numbers of multiples of these are: 6C1+6C2+6C3+6C4+6C5+6C6 = 63.
But in addition 1 is also a factor and so is 3 so I would have thought the ans to = 65 but my best guess is:
ANS = D


Bunuel wrote:
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


Question says they are both related linearly so they the relationships can be represented by standard linear definition y=mx+b.
Let S scale = y. and R scale = x.
30=6m+b EQN1
60=24m+b EQN2
EQN2-EQN1 => 30=18m => m=5/3
Solving for b using either equation gives us b=20
linear relationship is represented by y=(5/3)x+20
Solve for 100:
100=5/3x+20
80=5/3x
x=48
ANS = C
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Re: NEW SET of good PS(3) [#permalink] New post 18 Oct 2009, 03:17
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] New post 18 Oct 2009, 03:22
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] 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, 08:35
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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.
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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
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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 23 Oct 2009, 22:00
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Economist wrote:
Hi Bunuel, would appreciate if you can explain the solutions for 3,5 and 9.


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.

First of all you should know the formula counting the number of distinc factors of an integer:

You have to write the number as the product of primes as a^p*b^q*c^r, where a, b, and c are prime factors and p,q, and r are their powers.

The number of factors the number contains will be expressed by the formula (p+1)(q+1)(r+1).
Let's take an example for clear understanding:Find the number of all (distinct) factors of 1435:
1. 1435 can be expressed as 5^1*17^1*19^1
2. total number of factors of 1435 including 1 and 1435 itself is (1+1)*(1+1)*(1+1)=2*2*2=8 factors.

OR
Distinct factors of 18=2*3^2 --> (1+1)*(2+1)=6. Lets check: factors of 18 are: 1, 2, 3, 6, 9 an 18 itself. Total 6.

Back to our question:
How many numbers that are not divisible by 6 divide evenly into 264,600?

264,600=2^3*3^3*5^2*7^2

We should find the factor which contain no 2 and 3 together, so not to be divisible by 6.

Clearly, the factors which contain only 2,5,7 and 3,5,7 won't be divisible by 6. So how many such factors are there?
2^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36 (the product of powers of 2, 5,and 7 added 1)

3^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36 (the product of powers of 2, 5, and 7 added 1)

So 36+36=72. BUT this number contains duplicates:

For example: 2^3*5^2*7^2--> (3+1)*(2+1)*(2+1)=36 This 36 contains the factors when the power of 2 is 0 (2^0=1)--> 2^0*5^2*7^2 giving us only the factors which contain 5-s and/or 7-s. (5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7....) number of such factors are (2+1)*(2+1)=9 (the product of powers of 5 and 7 added 1).

And the same factors are counted in formula 3^3*5^2*7^2 --> (3+1)*(2+1)*(2+1)=36: when power of 3 is 0 (3^0=1). --> 5*7=35, 5*7^2=245, 5^2*7=175, 5*7^0=5, 5^0*7=7.... such factors are (2+1)*(2+1)=9. (the product of powers of 5 and 7 added 1).

So we should subtract this 9 duplicated factors from 72 --> 72-9=63. Is the correct answer.


The problem can be solved from another side:
264,600=2^3*3^3*5^2*7^2 # of factors= (3+1)(3+1)(2+1)(2+1)=144. So our number contains 144 distinct factors. # of factors which contain 2 and 3 is 3*3=9 (2*3, 2^2*3, 2^3*3, 2*3^2, 2^2*3^2, 2^3*3^2, 2*3^3, 2^2*3^3, 2^3*3^3 total 9) multiplied by (2+1)*(2+1)=9 (powers of 5 and 7 plus 1) --> 9*9=81 ---> 144-81=63.

Hope now it's clear.
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Re: NEW SET of good PS(3) [#permalink] New post 23 Oct 2009, 22:42
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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.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute x-8 vouchers, so that each can get from zero to x-8 as at "least 2", or 2*4=8, we already booked. Let x-8 be k.

In how many ways we can distribute k identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let k=5. And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

ttttt|||
We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

ttttt|||
Means that first nephew will get all the tickets,

|t|ttt|t
Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 t's and 3 |'s are identical, so \frac{8!}{5!3!}=56. Basically it's the number of ways we can pick 3 separators out of 5+3=8: 8C3.

So, # of ways to distribute 5 tickets among 4 people is (5+4-1)C(4-1)=8C3.

For k it will be the same: # of ways to distribute k tickets among 4 persons (so that each can get from zero to k) would be (K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120.

(k+1)(k+2)(k+3)=3!*120=720. --> k=7. Plus the 8 tickets we booked earlier: x=k+8=7+8=15.

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is n+r-1C_{r-1}.

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is n-1C_{r-1}.


Hope it helps.
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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
Re: NEW SET of good PS(3)   [#permalink] 24 Oct 2009, 12:06
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