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# NEW SET of good PS(3)

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NEW SET of good PS(3) [#permalink]

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17 Oct 2009, 17:33
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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

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]

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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]

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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

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: $$x=k+8=7+8=15$$.

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]

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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

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]

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15 Nov 2009, 10:48
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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

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

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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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]

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02 Jan 2010, 11:42
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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.

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

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

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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]

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12 Jun 2011, 18:17
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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.

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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by VeritasPrepKarishma on 19 Mar 2012, 02:40, edited 1 time in total. Kudos [?]: 18140 [2], given: 236 Intern Joined: 04 Oct 2009 Posts: 19 Kudos [?]: 10 [1], given: 2 GMAT 1: 730 Q49 V40 GPA: 3.22 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 25 Oct 2009, 19:39 1 This post received KUDOS andershv wrote: Will anyone please explain what the C means in the notation? Thanks in advance Combinations: 6 choose 3 nCk = n!/(k!(n-k)!) Kudos [?]: 10 [1], given: 2 Math Expert Joined: 02 Sep 2009 Posts: 42629 Kudos [?]: 135833 [1], given: 12715 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 02 Nov 2009, 14:14 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED Talinhuu wrote: Hi. I think the answer of No. 6 is E instead of C. Can anyone confirm it? thx The answer is indeed E. Here is the solution: 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) SOLUTION: $$x$$ fraction of saving,$$I$$ income. $$(1-x)*I=2*x*I*(1+r)$$, $$I$$ cancels out. $$x=\frac{1}{3+2r}$$ Answe: E. _________________ Kudos [?]: 135833 [1], given: 12715 Intern Joined: 01 Oct 2009 Posts: 10 Kudos [?]: 4 [1], given: 16 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 02 Nov 2009, 18:32 1 This post received KUDOS Bunuel wrote: Answe: E. Hi buddy, I would appreciate if you can confirm the answer of question no.8 in the post below. new-set-of-good-ds-85441-20.html Cheers! Kudos [?]: 4 [1], given: 16 Manager Joined: 13 Aug 2009 Posts: 199 Kudos [?]: 130 [1], given: 16 Schools: Sloan '14 (S) Re: NEW SET of good PS(3) [#permalink] ### Show Tags 14 Nov 2009, 23:31 1 This post received KUDOS Bunuel, Quality problems as usual... and also as usual they have kicked my butt. Can you please post your solution to Question 9? It is driving me nuts! Thanks again! Kudos [?]: 130 [1], given: 16 Math Expert Joined: 02 Sep 2009 Posts: 42629 Kudos [?]: 135833 [1], given: 12715 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 03 Jan 2010, 22:53 1 This post received KUDOS Expert's post 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! _________________ Kudos [?]: 135833 [1], given: 12715 Math Expert Joined: 02 Sep 2009 Posts: 42629 Kudos [?]: 135833 [1], given: 12715 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 06 Mar 2010, 00:43 1 This post received KUDOS Expert's post 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. _________________ Kudos [?]: 135833 [1], given: 12715 Intern Joined: 10 Feb 2010 Posts: 34 Kudos [?]: 2 [1], given: 12 Re: NEW SET of good PS(3) [#permalink] ### Show Tags 09 Aug 2010, 14:21 1 This post received KUDOS Ahhh, like every other GMAT question I get wrong -- the solution now seems quite obvious. Thank you Bunuel Kudos [?]: 2 [1], given: 12 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7799 Kudos [?]: 18140 [1], given: 236 Location: Pune, India Re: NEW SET of good PS(3) [#permalink] ### Show Tags 27 May 2011, 04:53 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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19 Mar 2012, 02:35
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 18140 [1], given: 236 Manager Joined: 14 Nov 2011 Posts: 140 Kudos [?]: 21 [1], given: 103 Location: United States Concentration: General Management, Entrepreneurship GPA: 3.61 WE: Consulting (Manufacturing) Re: NEW SET of good PS(3) [#permalink] ### Show Tags 30 Nov 2013, 22:17 1 This post received KUDOS VeritasPrepKarishma wrote: 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. Hi Karishma, To find out the possible number of right triangles I tried as below: No. of rectangles *4 (for each orientation) For example: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 2 ≤ x ≤ 4 and 5 ≤ y ≤ 7? No. of rectangles = 9 (this by actual counting of rectangles) No. of right triangles = 4*9 = 36, Is this correct? Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2? To me it should be = 3c2*(3-1)=6, but this is not correct. Kudos [?]: 21 [1], given: 103 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7799 Kudos [?]: 18140 [1], given: 236 Location: Pune, India Re: NEW SET of good PS(3) [#permalink] ### Show Tags 01 Dec 2013, 20:45 1 This post received KUDOS Expert's post cumulonimbus wrote: VeritasPrepKarishma wrote: 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. Secondly, I tried to calculate no. of rectangles via combinations but i count understand why no. of rectangles would be 3c2*3c2? To me it should be = 3c2*(3-1)=6, but this is not correct. Make the coordinate axis, mark the coordinates and make all vertical and horizontal lines, To make a rectangle, you need 2 horizontal and 2 vertical lines. You have 3 vertical lines and 3 horizontal lines. You select 2 of each in 3C2 * 3C2 ways. Similarly, say 2 <= x <= 5 and 5<= y <= 7. Calculate manually to get 18 rectangles. Use 4C2*3C2 to get 18 rectangles too. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: NEW SET of good PS(3)   [#permalink] 01 Dec 2013, 20:45

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# NEW SET of good PS(3)

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