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Re: Good set of PS 2
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17 Oct 2009, 07:50
Economist wrote: yangsta8 wrote: 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 I missed the word 'could' ..agree with B.. Hmmm apparrently B is still not correct, guess we can wait for OA. I couldn't find any other possibilities. I tested fractions, integers and then numbers which turn out unique properties such as 1 and 2 and still can't find one.





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Re: Good set of PS 2
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17 Oct 2009, 07:52
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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Re: Good set of PS 2
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17 Oct 2009, 09:42
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
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17 Oct 2009, 20:02
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.



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Re: Good set of PS 2
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17 Oct 2009, 20:28
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.
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Re: Good set of PS 2
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19 Oct 2009, 09:02
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
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19 Oct 2009, 11:23
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.
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Re: Good set of PS 2
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19 Nov 2009, 07:08
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



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Re: Good set of PS 2
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19 Nov 2009, 07:23
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 arrangementsarrangements with sisters sitting together=2*4*3!2*2(sisters together)*2*2*1(arrangement of others)=4816=32 Answer: B.
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Re: Good set of PS 2
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21 Dec 2009, 07:40
whuannou wrote: 1A Combination with one or the other of the daughters in the nondriver 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 (900648)/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.
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Re: Good set of PS 2
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21 Dec 2009, 09:11
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...
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Re: Good set of PS 2
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21 Dec 2009, 09:30
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 \(\frac{2x^21}{x}<0\), 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.
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Re: Good set of PS 2
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31 Dec 2009, 06:37
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



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Re: Good set of PS 2
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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
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23 Feb 2010, 02:39
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^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<1/x<2x > can 1/x<2x? The same here (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.
@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.



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Re: Good set of PS 2
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23 Feb 2010, 09:33
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 \(\frac{2x^21}{x}>0\) be true for the range \(0<x<1\)? As x is positive: can \(2x^21>0\) 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 viseversa for the first condition. Hope it's clear.
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Re: Good set of PS 2
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07 May 2010, 01:49
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.



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Re: Good set of PS 2
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07 May 2010, 03:36
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: \(C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65\) 70065 = 635 Answer: E.
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Re: Good set of PS 2
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10 Jun 2010, 08:12
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 17 inclusive



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Re: Good set of PS 2
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10 Jun 2010, 08:58
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=7515=60 25%Y=X&Y=15 > Y=60 > Only Y=6015=45 Only X or Y=60+45=105 Answer: D.
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Re: Good set of PS 2 &nbs
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