Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 06:52

1

This post received KUDOS

Economist wrote:

yangsta8 wrote:

Bunuel wrote:

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

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.

What does the ques ask...which of the following is divisible by EACH of the integers from 1 thru 7...so I interpret it as which is the option that is divisible by 1 and 2 and 3...It does not ask : each of the following is divisible by the product of each of the integers from 1 thru 7 ! or am I missing something badly !!

Your answer as well as Asterixmatrix's are correct according to the wording of the question. I just assumed it would be harder, since Bunuel posts tough questions haha. Let's see what the OA says.

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 08:42

1

This post received KUDOS

Heres are my answers:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (B) 32 - Total ways to seat them with one parent driving: 2 options for driver * 4 options for other seats = 2*4*3*2*1 = 48. Assuming the 2 sisters do sit together, front seat we have 2 options for driver and 2 options for passenger (the other parent + son) = 4 ways. Back seats: sisters sit together so either they sit in seats 1 and 2 or seats 2 and 3 = 2 ways. Also sisters themselves can sit in 2 ways so total = 2*2 = 4. So total ways sisters can sit together = 4*4 =16. So ways they wont sit together = 48-16 = 32 ways.

2. What is the probability that a 3-digit positive integer picked at random will have one or more "7" in its digits? (C) 7/25 - Total #'s between 100 and 999 inclusive = 9*10*10 = 900. # of numbers with no digit 7 = 8*9*9 = 648. So #'s with a 7 = 1 - 648/900 = 252/900 = 7/25.

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (D) 5( sqrt3 - 1) - Length of the diagonal of cube = ((10^2+10^2) + 10^2)^(1/2) = 10 sqrt3. Diagonal includes diameter of circle (10) + 2*length from vertex to sphere. Length of vertex to sphere = (10*sqrt3- 10)/2 = 5( sqrt3 - 1)

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (C) 6 - rains for 5 days from day 56-60. So 10 guys worked for 55 days and accomplished half of the work. If 6 more guys are added to the job then the rate is 16/1100. (since one man's rate is 1/1100). Half the job left means 550/1100 is left. Therefore 550/16 = 34.375 days of more work. Since there were 40 days between day 60 and job completion, it must've rained for 40-34.375 = 5.625 or ~6 days. (I'm not sure if this is correct)

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? (E) 45 - 64.12 = 6412/100 or 1603/25. 1603/25 gives a remainder of 3, 3206/50 gives remainder of 6 and so on ..pattern = factors of 3. so to get remainder of 45, we multiply everything by 15: 1603*15/(25*15) = 24045/375.

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? (E) 635 - 2 options: 2 men and 4 women -(i), or 3 men and 3 women -(ii). Starting with women: for (i): 5C4=5 and for (ii): 5C3 = 10. Men for (i): 2 further options if 2 men - a) neither of the problematic men chosen = 6C2 = 15 or b) one of them is chose - 2C1*6C1 = 12 so 12+15= 27 ways if (i) (ii): same 2 options - 6C3 + 2C1*6C2 = 20 + 30 = 50. Combining men and women - (i): 5*27 = 135 and (ii): 10*50 = 500 so total ways = 635.

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 (D) I and II only - if x = 1/3, then I is correct, if x = 4/5, then II is correct. I couldnt find a way to get III to work.

8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (D) -6 - Area of bxh of triangle = 24. Given base is 4, and drawing the line, easy to see that y intercept has to be -6.

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? (D) 105 - Let x be # who applied to only X and y who applied to only Y. Then 0.2(x+15) = 15 and 0.25(y+15) = 15. So x = 60 and y = 45 so x+y = 105.

10. What is the lowest positive integer that is divisible by each of the integers 1 through 7, inclusive? (A) 420 - 420 is divisible by all #'s from 1 thru 7.

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 10:55

[quote="yangsta8] 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 [/quote][/quote][/quote]

if x= 24/25 then option II x^2<1/x<2x will hold good and answer will be D

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 11:02

yangsta8 wrote:

Your answer as well as Asterixmatrix's are correct according to the wording of the question. I just assumed it would be harder, since Bunuel posts tough questions haha. Let's see what the OA says.

would agree with yangsta8 ....I have been a recent joinee to the forum... but Bunuel posts do bring the quant jitters ....(NOM- I frankly enjoy solving them)

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 12:37

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

ANSWERS (OA):

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

Answer: B.

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.

3. A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere? (A) \(10(\sqrt{3}- 1)\) (B) \(5\) (C) \(10(\sqrt{2} - 1)\) (D) \(5(\sqrt{3} - 1)\) (E) \(5(\sqrt{2} - 1)\)

Shortest distance=(diagonal of cube-diameter of sphere)/2= \(\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)\)

Answer: D.

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

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

\(s\) divided by \(t\) yields the remainder of \(r\) can always be expressed as: \(\frac{s}{t}=q+\frac{r}{t}\) (which is the same as \(s=qt+r\)), where \(q\) is the quotient and \(r\) is the remainder.

Given that \(\frac{s}{t}=64.12=64\frac{12}{100}=64\frac{3}{25}=64+\frac{3}{25}\), so according to the above \(\frac{r}{t}=\frac{3}{25}\), which means that \(r\) must be a multiple of 3. Only option E offers answer which is a multiple of 3

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: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\)

700-65 = 635

Answer: E.

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)

Answer: D.

8. In the xy plane, Line k has a positive slope and x-intercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the y-intercept of line K ? (A) 3 (B) 6 (C) -3 (D) -6 (E) -4

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

Answer: D.

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

The integer should be divisible by: 2, 3, 4(=2^2), 5, 6(=2*3), and 7. LCM=2^2*3*5*7=420

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 19:02

Bunuel wrote:

ANSWERS:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

Answer: E.

In the second case, we have 1 P in the front seat, so the other can be occupied by the Son, no one else...so we have 2 ways to arrange people in front seats, and 2 ways to arrange backseaters ( with 2 daughters on window seats )...so total there are 2*2 ways...hence I think ans should be 28 and not 32.

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 19:28

Expert's post

Economist wrote:

Bunuel wrote:

ANSWERS:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

As most of the combination problems this one can be solved in more than 1 way:

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

Answer: E.

In the second case, we have 1 P in the front seat, so the other can be occupied by the Son, no one else...so we have 2 ways to arrange people in front seats, and 2 ways to arrange backseaters ( with 2 daughters on window seats )...so total there are 2*2 ways...hence I think ans should be 28 and not 32.

If I understood correctly: you are talking about the case when the sisters are sitting on the back seat by the window? We have two front seats (one is driver's seat) And we have three back seats. Consider this: 1. 2 sisters by the windows can be arranged 2!=2 ways; 2. Drivers seat either mother or father=2 ways; 3. Second front seat either the son or the parent which is not driving=2 ways 4. Only 1 way (option) will be left between the sisters (either son, or the parent who is not driving, but only one option)=1

So, 2*2*2=8

24+8=32.

Well again if I understood your point correctly. Please let me know. _________________

Re: Good set of PS 2 [#permalink]
17 Oct 2009, 20:11

You are right..i misunderstood the question...i thought there should be 1 P in the front seat and 1 P in the backseat(to take care of the kids in the backseat , lol )

Re: Good set of PS 2 [#permalink]
19 Oct 2009, 08:02

2

This post received KUDOS

Bunuel wrote:

ANSWERS:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

I solved in a more easier way I think:

1) 10 man 110 days --> need for 1100 man.days 2) 55 days with 10 men --> 550 man.days 3) 40 days with 16 men --> 640 man.days

--> total man.days equals 1190 vs need for 1100 --> days of rain equals 90/16 max --> 5.625 --> rounded to 5

Re: Good set of PS 2 [#permalink]
19 Oct 2009, 10:23

pleonasm wrote:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

Driver 1 can be taken in 2 ways ( M & F ) Front seat can be taken in 5 ways ( M, F , D1, D2 and S ) The last 3 seats can be taken in 6 ways :

D1 S D2 D2 S D1 F S D2 F S D1 M S D2 M S D1

Total = 5*2*6 = 60 ways.

Not very confident, I could be wrong.

Edit: Front seat can be taken in 4 ways ( M or F , D1, D2 and S ) Ans is 4*2*6 = 48

I got a slightly different answer, here was my approach - I split the scenario into two cases - Case 1 - One of the daughters takes the front passenger seat - Drivers seat can be occupied in 2 ways (M or F) AND Front passenger seat can be occupied in 2 ways (D1 or D2) AND the other three can sit in the back seat in any order ie !3 = 6 Or Case 2 - One of the daughters doesn't take the front passenger seat Drivers seat can be occupied in 2 ways (M or F) AND Front passenger seat can be occupied in 2 ways (One of the parent or the Son) AND in the back row the daughters occupy the window seats - 2 ways (the middle seat is occupied by the remaining person i.e. one of the parents or the son so only 1 way to do this).

Answer = 2*2*6+2*2*2= 32 hence B. _________________

Re: Good set of PS 2 [#permalink]
19 Oct 2009, 11:31

Expert's post

atish wrote:

pleonasm wrote:

1. A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? (A) 28 (B) 32 (C) 48 (D) 60 (E) 120

Driver 1 can be taken in 2 ways ( M & F ) Front seat can be taken in 5 ways ( M, F , D1, D2 and S ) The last 3 seats can be taken in 6 ways :

D1 S D2 D2 S D1 F S D2 F S D1 M S D2 M S D1

Total = 5*2*6 = 60 ways.

Not very confident, I could be wrong.

Edit: Front seat can be taken in 4 ways ( M or F , D1, D2 and S ) Ans is 4*2*6 = 48

I got a slightly different answer, here was my approach - I split the scenario into two cases - Case 1 - One of the daughters takes the front passenger seat - Drivers seat can be occupied in 2 ways (M or F) AND Front passenger seat can be occupied in 2 ways (D1 or D2) AND the other three can sit in the back seat in any order ie !3 = 6 Or Case 2 - One of the daughters doesn't take the front passenger seat Drivers seat can be occupied in 2 ways (M or F) AND Front passenger seat can be occupied in 2 ways (One of the parent or the Son) AND in the back row the daughters occupy the window seats - 2 ways (the middle seat is occupied by the remaining person i.e. one of the parents or the son so only 1 way to do this).

Answer = 2*2*6+2*2*2= 32 hence B.

Atish your solution is absolutely correct. You can refer to my post with OA and explanations (the one which starts with "ANSWERS..."): the same approach was taken. _________________

Re: Good set of PS 2 [#permalink]
22 Oct 2009, 12:59

Jgroeten wrote:

Bunuel wrote:

ANSWERS:

4. A contractor estimated that his 10-man crew could complete the construction in 110 days if there was no rain. (Assume the crew does not work on any rainy day and rain is the only factor that can deter the crew from working). However, on the 61-st day, after 5 days of rain, he hired 6 more people and finished the project early. If the job was done in 100 days, how many days after day 60 had rain? (A) 4 (B) 5 (C) 6 (D) 7 (E) 8

This one was solved incorrectly: Days to finish the job for 10 people 110 days. On the 61-st day, after 5 days of rain --> 5 days was rain, 55 days they worked, thus completed 1/2 of the job, 1/2 is left (55 days of work for 10 people). Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375, BUT after they were hired job was done in 100-60=40 days --> so 5 days rained. They needed MORE than 34 days to finish the job, so if it rained for 6 days they wouldn't be able to finish the job in 100(40) days.

Answer: B.

I solved in a more easier way I think:

1) 10 man 110 days --> need for 1100 man.days 2) 55 days with 10 men --> 550 man.days 3) 40 days with 16 men --> 640 man.days

--> total man.days equals 1190 vs need for 1100 --> days of rain equals 90/16 max --> 5.625 --> rounded to 5

Then 6 more people was hired --> speed of construction increased by 1.6, days needed to finish 55/1.6=34.375 This is the key part of the solution here. If this is well understood, any problem of this type can be solved. Thanks Bunuel. _________________

In the land of the night, the chariot of the sun is drawn by the grateful dead

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? 28 32 48 60 120

Re: family seating [#permalink]
19 Nov 2009, 06:23

Expert's post

kairoshan wrote:

A family consisting of one mother, one father, two daughters and a son is taking a road trip in a sedan. The sedan has two front seats and three back seats. If one of the parents must drive and the two daughters refuse to sit next to each other, how many possible seating arrangements are there? 28 32 48 60 120

Refer to the question #1 in the set.

Sisters sit separately: 1. one of them is on the front seat (2 ways). Others (including second sister) can be arranged in: 2 (drivers seat)*3! (arrangements of three on the back seat)=12 ways. Total for this case: 2*12=24 Or 2. both by the window on the back seat (2 ways). Others can be arranged in: 2 (drivers seat)*2 (front seat)*1(one left to sit between the sisters on the back seat)=4 ways. Total for this case=8. Total=24+8=32.

Another way: Total number of arrangements-arrangements with sisters sitting together=2*4*3!-2*2(sisters together)*2*2*1(arrangement of others)=48-16=32

Re: Good set of PS 2 [#permalink]
20 Nov 2009, 11:54

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.

Re: Good set of PS 2 [#permalink]
21 Dec 2009, 05:53

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.

gmatclubot

Re: Good set of PS 2
[#permalink]
21 Dec 2009, 05:53

Hey, everyone. After a hectic orientation and a weeklong course, Managing Groups and Teams, I have finally settled into the core curriculum for Fall 1, and have thus found...

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

After I was accepted to Oxford I had an amazing opportunity to visit and meet a few fellow admitted students. We sat through a mock lecture, toured the business...