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# Good set of PS 2

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Q49  V41
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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
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Re: Good set of PS 2 [#permalink]
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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)

This is hard to explain without a diagram .. I'll do my best.

Distance from one of the vertices to the surface of the sphere = Diagonal of the top 1/4th square - radius of the sphere.

Diameter of the sphere = side of the cube = 10. Hence radius = 5.

Length of the diagonal of the square = Sqrt ( 5^2 + 5^2 ) = 5 sqrt (2).

So the distance is 5 sqrt(2) - 5 . Option E.
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Q49  V41
Re: Good set of PS 2 [#permalink]
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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 = (sqrt300-10)/2 = 5(sqrt 3 - 1)

ANS = D
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Q49  V41
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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Q49  V41
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 = (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
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Re: Good set of PS 2 [#permalink]
yangsta8 wrote:
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?

This is a good question. Well, you did everything right, though it could be done easier, but conclusion is not correct. Look again at the remainders... You should see the pattern.
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Q49  V41
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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Q49  V41
Re: Good set of PS 2 [#permalink]
Bunuel wrote:
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

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
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Re: Good set of PS 2 [#permalink]
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yangsta8 wrote:
Easiest one so far Bunuel... your questions are killers (to me anyway haha)

You are right not every question is 700+... Though I try to post toughest problems from my collection.

BTW 7 is not correct, try again. You have good speed and almost every answer from you is correct. Check DS set too if you like such "killers"
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Q49  V41
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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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Q49  V41
Re: Good set of PS 2 [#permalink]
Bunuel wrote:
yangsta8 wrote:
Easiest one so far Bunuel... your questions are killers (to me anyway haha)

You are right not every question is 700+... Though I try to post toughest problems from my collection.

BTW 7 is not correct, try again. You have good speed and almost every answer from you is correct. Check DS set too if you like such "killers"

I'm still stuck on 5 and 7. I'll wait for your OA's my head hurts
+1 for posting these.
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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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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

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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Re: Good set of PS 2 [#permalink]
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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 1-7 inclusive
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Re: Good set of PS 2 [#permalink]
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 !!
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Re: Good set of PS 2 [#permalink]
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..
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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 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

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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