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Re: Good set of PS 2 [#permalink]
21 Dec 2009, 06:40

Expert's post

whuannou wrote:

1A Combination with one or the other of the daughters in the non-driver front seat 2 * 2 * 3! = 24 With no daughter in front seat 2 * 2 * 1(they cannot sit together so only one possibility) = 4 Total gives 28

2C = 1 - Proba it has no digit 7 at all Total nber of 3 digits number = 900 nbers with no 7 at all = 8*9*9= 648 yields (900-648)/900 = 252/ 900 = 7/25

3E Draw a square with side 2 and inscribe in a circle of radius 1 Then using pythagore theorem twice and a substraction, compute the distance that goes tangently from one vertex to the circle

4E 5 I did not really understand the question wording 6E 7B 8D 9B 10A

For the rest I will type in my explanations later. Thanks for the question sir.

The OA's are given on page 2. Some of your answers are not correct. Please refer to the solutions given along with OA's. Please ask if any question remains.

bidishabarpujari wrote:

for Q10 i am also confused! so far as i can understand, its asking for the lowest possible integer divided by each integer from 1 through 7.( i think it not asking for an integer divided by 7!). Bunuel, please clarify this.

You are right it's not asking about the integer divisible by 7!. Refer to the solution on the page 2. The answer is 420, which is LCM of the integers from 1 to 7. _________________

Re: Good set of PS 2 [#permalink]
21 Dec 2009, 08:11

hi.... OA given for q7 is D..... however i feel it should be E.... III. 2x<x^2<1/x ......take x=1.1.....subst in equality 2.2<1.21<0.9... so III is also correct .... it is just given x is +ive and not +ive int so x can be taken 1.1.... any one finds any objn ,pl let me know...

Re: Good set of PS 2 [#permalink]
21 Dec 2009, 08:30

Expert's post

chetan2u wrote:

hi.... OA given for q7 is D..... however i feel it should be E.... III. 2x<x^2<1/x ......take x=1.1.....subst in equality 2.2<1.21<0.9... so III is also correct .... it is just given x is +ive and not +ive int so x can be taken 1.1.... any one finds any objn ,pl let me know...

How is the red part correct?

Anyway solution for this question:

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).

\(x>2\)

\(1<x<2\)

\(0<x<1\)

When \(x>2\) --> \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) --> \(2x\) is greatest then comes \(x^2\) and no option is offering this.

So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:

I. \(x^2<2x<\frac{1}{x}\) --> can \(2x<\frac{1}{x}\)? Can \(\frac{2x^2-1}{x}<0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

II. \(x^2<\frac{1}{x}<2x\) --> can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^2-1}{x}>0\), the expression \(2x^2-1\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)

Re: Good set of PS 2 [#permalink]
31 Dec 2009, 05:37

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

Dear Bunuel!

Could you please explain hod did you get: 3 digit number with no 7=8*9*9- where it coms from?

Re: Good set of PS 2 [#permalink]
31 Dec 2009, 07:04

1

This post received KUDOS

Expert's post

sher1978 wrote:

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

Dear Bunuel!

Could you please explain hod did you get: 3 digit number with no 7=8*9*9- where it coms from?

Thank you in advance

3 digit number with no 7, I mean without 7 = 8*9*9 = 648:

First digit can take 8 values from 1 to 9 excluding 7 (1xx, 2xx, ... 9xx, but not 7xx); Second and third digits can take 9 values from 0 to 9 excluding 7 (eg. for second digit: x0x, x1x, ... x9x but not x7x).

Re: Good set of PS 2 [#permalink]
01 Jan 2010, 09:31

1

This post received KUDOS

Bunuel wrote:

sher1978 wrote:

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (A) 271/900 (B) 27/100 (C) 7/25 (D) 1/9 (E) 1/10

Total 3 digit numbers 900, 3 digit number with no 7 =8*9*9=648, P(at least one 7)=1-P(no 7)=1-648/900=252/900=7/25

Answer: C.

Dear Bunuel!

Could you please explain hod did you get: 3 digit number with no 7=8*9*9- where it coms from?

Thank you in advance

3 digit number with no 7, I mean without 7 = 8*9*9 = 648:

First digit can take 8 values from 1 to 9 excluding 7 (1xx, 2xx, ... 9xx, but not 7xx); Second and third digits can take 9 values from 0 to 9 excluding 7 (eg. for second digit: x0x, x1x, ... x9x but not x7x).

Hope it's clear.

Thank you very much.This will help me iin solving such problems.I used to do it manually

Re: Good set of PS 2 [#permalink]
04 Jan 2010, 04:50

Folks,

I have seen many Business Schools vouchers related to Business School admission process (Eg. Brandeis International Business School), i found out they need TOEFL score including GMAT. Before starts my study to get into Business School i tot of GMAT only. can anyone please tell me is that compulsory for all the Business Schools.

Re: Good set of PS 2 [#permalink]
08 Jan 2010, 16:00

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

Re: Good set of PS 2 [#permalink]
08 Jan 2010, 16:14

Expert's post

reply2spg wrote:

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

is it 45?

Yes it is. OA-s and solutions are given in my post on second page. _________________

Re: Good set of PS 2 [#permalink]
23 Feb 2010, 01:39

Bunuel wrote:

7. If x is positive, which of the following could be the correct ordering of 1/x,2x and x^2 ? I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x

(A) None (B) I only (C) III only (D) I and II only (E) I II and III

First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between x, 1/x and x^2 in three areas: 0<1<2<. x>2 1<x<2 0<x<1

When x>2 --> x^2 is the greatest and no option is offering this, so we know that x<2. If 1<x<2 --> 2x is greatest than comes x^2 and no option is offering this.

So, we are left with 0<x<1: In this case x^2 is least value, so we are left with: I. x^2<2x<1/x --> can 2x<1/x? Can (2x^2-1)/x<0, the expression 2x^2-1 can be negative or positive for 0<x<1. (You can check it either algebraically or by picking numbers) II. x^2<1/x<2x --> can 1/x<2x? The same here (2x^2-1)/x>0, the expression 2x^2-1 can be negative or positive for 0<x<1. (You can check it either algebraically or by picking numbers)

Answer: D.

@Bunuel

I tried number plugging but couldn't find values that satisfy second condition. Can you help identifying such numbers.

Re: Good set of PS 2 [#permalink]
23 Feb 2010, 08:33

Expert's post

honeyrai wrote:

@Bunuel

I tried number plugging but couldn't find values that satisfy second condition. Can you help identifying such numbers.

Also you can show how second is correct.

Second condition: \(x^2<\frac{1}{x}<2x\)

Put \(0.9\) --> \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) --> \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.

Basically as you can see this in the solution that second condition boils down to: can \(\frac{2x^2-1}{x}>0\) be true for the range \(0<x<1\)? As x is positive: can \(2x^2-1>0\) be true for the range \(0<x<1\)? For \(x>\frac{1}{\sqrt{2}}=0.71\) it will be true and for \(0<x<\frac{1}{\sqrt{2}}\) it won't. And vise-versa for the first condition.

Re: Good set of PS 2 [#permalink]
04 Mar 2010, 21:54

yangsta8 wrote:

Bunuel wrote:

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

The committee can be formed in two ways: 1) 2 men and 4 women 2) 3 men and 3 women The answer is the sum of these.

1) 2 men and 4 women = (8C2 - 1) x 5C4 = 27 x 5 = 135 Subtract 1 since there is one combo of men that are not allowed. 2) 3 men and 3 women = (8C3 - 6) x 5C3 = (56-6) x 10 = 500 Subtract 6 since there are 6 groups of men that can include those the two that refuse to work together. Adding these together we get 135+500 = 635 ANS = E

How did you get the number 6 for the groups of men that refuse to work together?

Re: Good set of PS 2 [#permalink]
04 Mar 2010, 22:27

aisha14 wrote:

yangsta8 wrote:

Bunuel wrote:

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

The committee can be formed in two ways: 1) 2 men and 4 women 2) 3 men and 3 women The answer is the sum of these.

1) 2 men and 4 women = (8C2 - 1) x 5C4 = 27 x 5 = 135 Subtract 1 since there is one combo of men that are not allowed. 2) 3 men and 3 women = (8C3 - 6) x 5C3 = (56-6) x 10 = 500 Subtract 6 since there are 6 groups of men that can include those the two that refuse to work together. Adding these together we get 135+500 = 635 ANS = E

How did you get the number 6 for the groups of men that refuse to work together?

8 men and 2 refuse to work together find out all possibilities when 2 work together -> take these two men and the 3rd person can be filled with 6 other persons -> so in total there are 6 possibilities that THOSE TWO men work together.

Re: Good set of PS 2 [#permalink]
28 Apr 2010, 20:16

Nice set of Qs bunuel!

A suggestion for those who tried picking numbers on Q7: Try graphing the lines instead and using that to determine x values where certain orders will apply. Remember - the goal on the test is to get the right answer as quickly as possible, since there is no partial credit.

Key intercept points on the graph would be at (1,1) and at (2,4). There is another intercept that occurs between the 1/x line and the 2x line - that is the part that most are missing when they answer incorrectly.

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

I came across this problem somewhere in this club before but can't exactly remember where. ...but there was some sort of formula(trick). Does anyone have any idea????

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Please explain as to why we are subtracting one and six.

Re: confusionnnn ~ [#permalink]
07 May 2010, 02:36

Expert's post

bibha wrote:

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

I came across this problem somewhere in this club before but can't exactly remember where. ...but there was some sort of formula(trick). Does anyone have any idea????

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Please explain as to why we are subtracting one and six.

OA's and solutions to every problem are given in my post on the second page.

5. If s and t are positive integer such that s/t=64.12, which of the following could be the remainder when s is divided by t? (A) 2 (B) 4 (C) 8 (D) 20 (E) 45

We know that when we are told that "s divided by t gives remainder r" can be expressed by the following formula: \(s=qt+r\), in or case \(q=64\), --> \(s=64t+r\), divide both parts by \(t\) --> \(\frac{s}{t}=64.12=64+\frac{r}{t}\) --> \(0.12=\frac{r}{t}\)--> \(\frac{3}{25}=\frac{r}{t}\) so \(r\) must be the multiple of 3. Only answer multiple of 3 is 45.

Or: \(\frac{s}{t}=64\frac{12}{100}=64\frac{3}{25}\), so if the divisor=t=25 then the remainder=r=3. Basically we get that divisor is a multiple of 25 and the remainder is a multiple of 3. Only answer multiple of 3 is 45.

Answer: E.

6. A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together? (A) 3510 (B) 2620 (C) 1404 (D) 700 (E) 635

Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): \(C^2_8*C^4_5+C^3_8*C^3_5 = 700\)

Ways to chose 6 members committee with two particular men serve together: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

Re: Good set of PS 2 [#permalink]
09 Jun 2010, 11:08

With regards to the combinatorial question with the parents and kids I thought this was kind of an intuitive way to go about it.

1) We know that one seat in the car, regardless, can have only two combos so the key is to evaluate the remaining seats without restrictions, namely ---> (N-1)! in this case 4! = 24. 2) Given the restriction of the 2 daughters not sitting together we know two things. a) The daughters can sit in two ways namely D1D2 or D2D1 therefore 2! b) We also know that given the four remaining seats there are 3 seat pairs that can be formed of which each has 2! combination.

1&2 2&3 3&4 2! x 2! x 2! = 8 3) Given all this information we can piece everything together

(24 - 8)*2! = 32 So far on most of the combo problems I've seen on the GMAT at least with restrictions some variation of this technique seems to work pretty well. Please evaluate to see if this is the kill all for combinatorics

Re: Good set of PS 2 [#permalink]
10 Jun 2010, 07:12

for q 9... 20%of those who applied X = 15 (because they applied for y also) x/5 =15 x = 75 25%of those who applied Y = 15 (because they applied for X also) y/4 = 15 y = 60

total who applied for x and y should be x+y = 135

asterixmatrix wrote:

yangsta8 wrote:

Bunuel wrote:

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 (B) 840 (C) 1260 (D) 2520 (E) 5040

Q9) 20% of total applied at X = 15 100% of total applied at X = 75 Only applied at X = 60

25% of total applied at Y = 15 100% of total applied at Y = 60 Only applied at Y = 45

Only applied at X + Only applied at Y = 60 + 45 = 105 ANS = D

Q10) 1 x 2 x 3 x 4 x 5 x 6 x 7 = 2^4 x 3^2 x 5 x 7 Option A = 420 = 2^2 x 3 x 5 x 7 (factored this out) Option B = 2 x 420 = 2^3 x 3 x 5 x 7 (just add a 2 to A) Option C = 3 x 420 = 2^2 x 3^2 x 5 x 7 (just add a 3 to A) Option D = 2 x 1260 = 2^3 x 3^2 x 5 x 7 (just add a 2 to C) Option E = 2 x 2520 = 2^4 x 3^2 x 5 x 7 (just add a 2 to D)

Actually after doing up to A you can quickly just figure out how many more 2's and 3's and determine it's E.

for q10 we need to find the lowest number - so should be 420 its divisible by all the integers from 1-7 inclusive

Re: Good set of PS 2 [#permalink]
10 Jun 2010, 07:58

Expert's post

shalu wrote:

for q 9... 20%of those who applied X = 15 (because they applied for y also) x/5 =15 x = 75 25%of those who applied Y = 15 (because they applied for X also) y/4 = 15 y = 60

total who applied for x and y should be x+y = 135

Answer 135 (A) is not correct. OA's and solutions for this set of questions are given in my post on page 2.

9. Of the applicants passes a certain test, 15 applied to both college X and Y. If 20 % of the applicants who applied college X and 25% of the applicants who applied college Y applied both college X and Y, how many applicants applied only college X or college Y? (A) 135 (B) 120 (C) 115 (D) 105 (E) 90

20%X=X&Y=15 --> X=75 --> Only X=75-15=60 25%Y=X&Y=15 --> Y=60 --> Only Y=60-15=45 Only X or Y=60+45=105

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