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# NEW SET of good PS(3)

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

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25 Oct 2009, 19:39
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andershv wrote:

Will anyone please explain what the C means in the notation?

Combinations: 6 choose 3

nCk = n!/(k!(n-k)!)
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Re: NEW SET of good PS(3) [#permalink]

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02 Nov 2009, 14:14
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Talinhuu wrote:
Hi.

I think the answer of No. 6 is E instead of C.

Can anyone confirm it?

thx

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

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02 Nov 2009, 18:32
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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!
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Re: NEW SET of good PS(3) [#permalink]

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14 Nov 2009, 23:31
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Bunuel,

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

It is driving me nuts!

Thanks again!
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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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16 Nov 2009, 03:36
Thanks Bunuel... your explainations are always the best.
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Re: NEW SET of good PS(3) [#permalink]

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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

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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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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03 Jan 2010, 02:19
Thanks Bunuel..... ...

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

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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 +1 for this mate...

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

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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

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]

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03 Jan 2010, 22:53
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jeeteshsingh wrote:
The solution provided by you is well explained.. However for the text marked in red - is there a faster way for solving the eq for k...? Since time crunch is a big issue in GMAT exam? Please let me know your views...

Thanks,
JT

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

I solved this in the following way: 720, three digit integer ending with 0, is the product of three consecutive integers. Obviously one of them must be multiple of 5: try 5*6*7=210<720, next triplet 8*9*10=720, bingo!
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Re: NEW SET of good PS(3) [#permalink]

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02 Mar 2010, 23:24
sandeep25398 wrote:
Bunuel wrote:
.
3. How many numbers that are not divisible by 6 divide evenly into 264,600?
(A) 9
(B) 36
(C) 51
(D) 63
(E) 72

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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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

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

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

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06 Mar 2010, 01:14
Bunuel wrote:
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.

Thanks Bunuel for the explanation. I just restarted my GMAT prep after a 6 month break and this set was quite a refresher. Keep them coming. Thanks
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Re: NEW SET of good PS(3) [#permalink]

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12 Mar 2010, 20:25
A query on Q7:What is the meaning of this portion?

'to the average(arithmetic mean)of 100units of Country R's currency'
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Re: NEW SET of good PS(3) [#permalink]

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22 Apr 2010, 13:47
Bunnel ... simply you rock...
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Re: NEW SET of good PS(3) [#permalink]

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09 Aug 2010, 14:03
masland wrote:
ExecMBA2010 wrote:
A query on Q7:What is the meaning of this portion?

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

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

The language is from GMATPrep

7. Before being simplified, the instructions for computing income tax in Country Rwere to add 2 percent of one's annual income to the average(arithmetic mean)of 100units of Country R's currency and 1 percent of one's annual income. Which of the following represents the simplified formula for computing the income tax in Country R's currency, for a person in that country whose annual income is I?
(A) 50+I/200
(B) 50+3I/100
(C) 50+I/40
(D) 100+I/50
(E) 100+3I/100

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

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

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

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09 Aug 2010, 14:21
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Ahhh, like every other GMAT question I get wrong -- the solution now seems quite obvious.

Thank you Bunuel
Re: NEW SET of good PS(3)   [#permalink] 09 Aug 2010, 14:21

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# NEW SET of good PS(3)

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