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Good set of PS 2 [#permalink]
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Please find below new set of PS problems: 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 2. What is the probability that a 3digit 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 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( sqrt3 1) (B) 5 (C) 10( sqrt2  1) (D) 5( sqrt3  1) (E) 5( sqrt2  1) 4. A contractor estimated that his 10man 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 61st 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 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 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 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 8. In the xy plane, Line k has a positive slope and xintercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the yintercept 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 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 Please share your way of thinking, not only post the answers. OA and explanations to follow. Also you can check new set of DS problems: goodsetofds85413.html
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Re: Good set of PS 2 [#permalink]
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16 Oct 2009, 20:42
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Bunuel wrote: Please find below new set of PS problems:
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
The total possible position given that one of the parents must drive is: 2x4x3x2x1 = 48 (the first two represents the driver seat). I want to find the number of positions where the two daughters sit together and subtract this from 48. There are 4 possibilities where they are sitting together and that is: D1, D2, _ D2, D1, _ _, D1, D2 _, D2, D1 For each of these 4 possibilities the other 3 members can be seated in 2x2x1 ways. So the number of positions where the two daughters sit together = 4x2x2x1 = 16 Total positions that meet the question requirements = 4816 = 32 ANS = B Looking back on this I'm sure there's more elegant methods Bunuel wrote: 2. What is the probability that a 3digit 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
P(at least one 7) = 1  P(no 7's) Total 3 digit numbers are 9x10x10 = 900 (first digit cannot have a 0) P(no 7's) = (8x9x9)/900 = 648/900 P(at least one 7) = 1  P(no 7's) = 1  648/900 = 252/900 = 7/25 ANS = C



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Re: Good set of PS 2 [#permalink]
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17 Oct 2009, 13:37
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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 arrangementsarrangements with sisters sitting together=2*4*3!2*2(sisters together)*2*2*1(arrangement of others)=4816=32 Answer: B.2. What is the probability that a 3digit 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)=1P(no 7)=1648/900=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 cubediameter of sphere)/2= \(\frac{10*\sqrt{3}10}{2}=5(\sqrt{3}1)\) Answer: D.4. A contractor estimated that his 10man 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 61st 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 61st 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 10060=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: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\) 70065 = 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 \(\frac{2x^21}{x}<0\), the expression \(2x^21\) 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 \(\frac{2x^21}{x}>0\), the expression \(2x^21\) 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 xintercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the yintercept of line K ?(A) 3 (B) 6 (C) 3 (D) 6 (E) 4 Positive slope, positive (4) xintercept > negative yintercept. > 1/2*4*y=12 > y=6. > y=6 Answer: D9. 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=7515=60 25%Y=X&Y=15 > Y=60 > Only Y=6015=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 Answer: A.
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Bunuel wrote: 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( sqrt3 1) (B) 5 (C) 10( sqrt2  1) (D) 5( sqrt3  1) (E) 5( sqrt2  1)
The shortest distance from a vertice to the cube would follow the line that is the diagonal to the cube. Shortest distance = (Diagonal of cube  Diameter of cube) / 2 We divide by 2 otherwise we get the distance from the cube to the vertex on each side of the diagonal. (Sorry don't have graphics software to draw it out). Diagonal^2 = (diagonal of base)^2 + (height)^2 = \((10*\sqrt{2})^2 + 10^2 = 300\) Diagonal = \sqrt{300} Diameter = same as height of cube = 10 Shortest Distance = (sqrt30010)/2 = 5(sqrt 3  1) ANS = D



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Re: Good set of PS 2 [#permalink]
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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 = (566) 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



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Re: Good set of PS 2 [#permalink]
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Bunuel wrote: ANSWERS:
4. A contractor estimated that his 10man 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 61st 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 61st 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 10060=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



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Re: Good set of PS 2 [#permalink]
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Bunuel 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
s/t = 64.12 = 6412/100 => 6412 div 100 Remainder = 12 = 3206/50 => 3206 div 50 Remainder = 6 = 1603/25 => 1603 div 25 Remainder = 3 No more common factors. I don't see how the remainder could be anything but 3,6,12,24,48. What am I doing wrong here?



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Re: Good set of PS 2 [#permalink]
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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
Two situations we need to consider: 0 < x < 1, or x > 1 (since squaring a number for the first makes the number smaller for the former and larger for the latter). Situation 1: let X = 1/3 x^2 = 1/9 2x = 2/3 1/x = 3 Correct ordering: x^2 < 2x < 1/x Option I is the only possibility. You could choose B based on this. Check Situation 2: Let X = 3 x^2 = 9 2x = 6 1/x = 1/3 Correct ordering 1/x < 2x < x^2 Not in any of the possibilities. ANS = B



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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.



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Re: Good set of PS 2 [#permalink]
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Bunuel 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
s/t = 64.12 we know > s,t are +ve integers, remainder r is a +ve integer (from options). s/t = 64.12 so from this we can infer : r= 12% of t r = (12/100) x t t = (r x 100)/12 t needs to be an integer, which is only satisfied by option (E) 45.
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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
we can try and get the limit values by equating 2 side. i.e. if 2x = 1/x , then x = 1/ sqroot 2 if x^2= 1/x , then x = 1 if 2x = x^2 , then x = 2 so we need to try for x between each of these values , i.e. 0 to 1/(sqroot 2), 1/(sqroot 2) to 1, 1 to 2 , 2 to infi for x in 0 to 1/(sqroot 2) (e.g. 0.1 > x ^2 = .01 ,2x = .2 ,1/x = 10) > x^2 < 2x < 1/x for x in 1/(sqroot 2) to 1 (e.g. 0.9 > x ^2 = .81 ,2x = 1.8 ,1/x = 1.11) > x^2 < 1/x < 2x for x in 1 to 2 (e.g. 1.1 > x ^2 = 1.21 ,2x = 2.2 ,1/x = .9) > 1/x < x^2 < 2x for x in 2 to infi (e.g. 10 > x ^2 = 100 ,2x = 20 ,1/x = .1) > 1/x < 2x < x^2 options : I. x^2<2x<1/x II. x^2<1/x<2x III. 2x<x^2<1/x so I and II can be true. So D.
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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 17 inclusive



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Re: Good set of PS 2 [#permalink]
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Bunuel wrote: Please find below new set of PS problems:
4. A contractor estimated that his 10man 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 61st 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
Nobody attempted 4, let me give a try. total work = 110*10 = 1100 man days now, from day 1 to day 55, 10 men worked = 550 man days of work was done. from day 61 to day 100, 16 men worked = 640 man days of work was done. so total work done should be 1190, the 90 days offset is due to rain on few days between 61 to 100th day...so the number of rainy days should be 90/16=5.625 ~ 6. C.



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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.



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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 = 4816 = 32 ways.
2. What is the probability that a 3digit 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 10man 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 61st 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 5660. 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 4034.375 = 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 xintercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the yintercept 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.



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Re: Good set of PS 2 [#permalink]
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31 Dec 2009, 08:04
sher1978 wrote: 2. What is the probability that a 3digit 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)=1P(no 7)=1648/900=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.
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Re: Good set of PS 2 [#permalink]
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01 Jan 2010, 10:31
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Bunuel wrote: sher1978 wrote: 2. What is the probability that a 3digit 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)=1P(no 7)=1648/900=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



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Re: Good set of PS 2 [#permalink]
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13 Aug 2010, 13:50
gmat1011 wrote: Hi Bunuel
for the time and work question in this problem set can you pls explain this step in a bit more detail:
"Then 6 more people was hired > speed of construction increased by 1.6, days needed to finish 55/1.6=34.375"
How did you figure out the speed increased by 1.6?
For such problems I tend to reduce the question to how much work did the workers do in 1 day and then to how much work did each worker do in one day and then multiply that by 6 to get what 6 workers would have done in a day; add that to what 10 workers would have done in a day; take reciprocal of the fraction to see how much time 16 workers would take for that work
your method seems much better.... can you explain that step of figuring out the increased speed of 1.6.... thanks. Time, rate and job in work problems are in the same relationship as time, speed (rate) and distance. \(Time*Rate=Distance\) <> \(Time*Rate=Job \ done\). We know that 10 people need 55 days to complete the job > let the combined rate of 10 people be x > \(Time*Rate=x*55=Job \ done\); Now, if combined rate of 10 people is \(x\), then combined rate of 16 man will be \(1.6x\) (1.6 times more than x as 16 is 1.6 times more than 10), so \(new \ time*new \ rate=same \ job\) > \(t_2*1.6x=x*55\) > \(t_2=\frac{55}{1.6}\approx{34.8}\) (as rate increased 1.6 times then time needed to do the same job will decrease 1.6 times). Some work problems with solutions: timenworkproblem82718.html?hilit=reciprocal%20ratefacingproblemwiththisquestion91187.html?highlight=rate+reciprocalwhatamidoingwrongtobunuel91124.html?highlight=rate+reciprocalgmatprepps93365.html?hilit=reciprocal%20ratequestionsfromgmatpreppracticeexampleasehelp93632.html?hilit=reciprocal%20rateagoodone98479.html?hilit=rateHope it helps.
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Re: Good set of PS 2 [#permalink]
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17 Mar 2013, 02:29
mansi123 wrote: yangsta8 wrote: Bunuel wrote:
Why is it that the triangle will be in 4th quadrant? 8. In the xy plane, Line k has a positive slope and xintercept 4. If the area of the triangle formed by line k and the two axes is 12, What is the yintercept of line K ? (A) 3 (B) 6 (C) 3 (D) 6 (E) 4
Easiest one so far Bunuel... your questions are killers (to me anyway haha) If K has a positive slope it will create a triangle in the bottom right quadrant (4th) of the xy plane. This means that the Y intercept will be negative. Area = 1/2 * base * height 12 = (1/2) * 4 * height height = 6 Y Intercept = 6 AND = D I Couldnt get the triangle being in 4th quadrant part!Bunuel,pls help!???? Check here: inthexyplanelinekhasapositiveslopeandand100739.htmlHope it helps.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: Good set of PS 2
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