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# Good set of PS 2

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Re: Good set of PS 2 [#permalink]

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16 Jan 2011, 01:41
praveenvino 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

I am getting a single equation like .2X + 0.25Y = 15. I feel like missing something. Can someone explain what it could be?

OA-s and solutions are given in my post on 2nd page. Solution for this question:

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

Question means the following:
15 people applied to both college X and Y, these 15 people represent 20% of all applicants who applied to X and 25% of all applicants who applied to Y.

$$0.2*x=15$$ --> $$x=75$$ --> # of people who applied only to X is: $$75-15=60$$;
$$0.25*y=15$$ --> $$y=60$$ --> # of people who applied only to Y is: $$60-15=45$$;

# of people who applied only to X OR Y is: $$60+45=105$$.

Hope it's clear.
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Re: Good set of PS 2 [#permalink]

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22 Feb 2011, 05:15
Fantastic set! Thanks.
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Re: Good set of PS 2 [#permalink]

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22 Feb 2011, 08:45
The answers below may not be correct. They're not matched with the OA's. Please comment should you find any discrepancy in the logic or solution.

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

Sol:
***
Total ways when father is driving: 4!=24

M*
***
Total ways when Mother is driving: 4!=24

We need to subtract the arrangements when two daughters are sitting together in the back seat;
(2!*2!) for every seating arrangement when Mother, Father and Brother(once when father is driving and again when mother is driving) are sitting on the co-driver's seat which totals to 4.
So; 4*(2!*2!) = 16 arrangements when both daughters are sitting together.

Thus total number of possible seating arrangements = 48-16=32.

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

Sol:
Total number of 3 Digit positive integers= 999-100+1=900
Numbers without any 7 = 8*9*9=648

Probability that none will be 7 = 648/900
Probability that at least one is 7 = 1-648/900 = 252/900=7/25

Ans: "C"
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)

Sol:
The distance of the vertex to the center of the sphere
$$5\sqrt{2}$$ (Hypotenuse for a isosceles right angled triangle 45-90-45 and side 5)
The shortest distance between vertex and point on the circumference of the circle
$$5\sqrt{2}-5$$
$$5(\sqrt{2}-1)$$

Ans: "E"
Bunuel wrote:
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

Sol:
It rained from 56th to 60th day. The crew of 10 men worked for 55 days prior to the rain.
If they finish 1 work in 110 days.
They must have finished 1/2 work in 55 days.
Thus; work left to be completed = 1/2

6 men added. Total workers: 16
1 work:
10 men -> 110 days
1 man -> 110*10 Days
16 men -> 110*10/16 Days

1/2 Work;
16 Men -> (110*10)/(16 *2) Days = 34.33 or 35 days

It is mentioned in the question that the job was done on the 100th day.

From 61 to 100; there are 40 days. However, the actual number of days required for the work is 35 days.

Thus; the crew didn't work for those 5 days it rained.

Ans: "B"
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

Sol:
$$\frac{s}{t}=64.12$$
$$\frac{s}{t}=64+\frac{12}{100}{$$
Remainder r=12
Divisor, d=100
$$\frac{r}{d}=\frac{12}{100}$$
$$\frac{r}{d}=\frac{3}{25}$$
$$25r=3d$$

Implies; r contains a multiple of 3 and d contains a multiple of 25.

If Remainder must be a multiple of 3; the only option is 45.

Ans: "E"
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

Sol:
Let's find the number of selections without the restriction of 2 mutually exclusive men.

3Men of 8Men and 3Women of 5Women
OR
2Men of 8Men and 4Women of 5Women

$$C^8_3*C^5_3+C^8_2*C^5_4$$
$$560+140=700$$

Find the number of possible selections with those supposedly mutually exclusive men put together.

$$C^6_1*C^5_3+C^6_0*C^5_4$$
$$60+5=65$$

Thus the number of selections where 2 particular men are not together = 700-65=635.

Ans: "E"
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

Sol:
We'll test it with 4 numbers; x=0.01,1,2,5

x=0.01
-------
x^2=0.0001
2x=0.02
1/x=1/0.01=100

x^2<2x<1/x.
I. could be correct.

x=1
----
x^2=1
2x=2
1/x=1

x^2=1/x<2x. No such inequality given. Ignore.

x=2
----
x^2=4
2x=4
1/x=0.5
1/x<x^2=2x. Ignore

x=5
---
x^2=25
2x=10
1/x=1/5=0.2

1/x<2x<x^2. Not given.

Don't want to try any more numbers. Will go with I only.

Ans: "B"
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

Sol:
Positive slope means that the line is making an acute angle with x-axis. From the given information; the line should intersect y axis below x axis; meaning thereby the y-intercept is negative.
The line is touching Ist, IIIrd and IVth quadrant. Say the line intersect y-axis at Y and x-axis at X. The triangle hence formed \
will be XOY where O is the origin(0,0)

$$\triangle{XOY}=\frac{1}{2}*\bar{OY}*4=12$$
$$\bar{OY}=6$$

Since y is minus; the intercept is -6.

Ans: "D"
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

Sol:
Say x applied for college X
And y applied for college y

$$x \cap y = 15$$
$$15=0.2x$$
$$x=75$$

$$15=0.25y$$
$$y=60$$

The candidates who applied only for x and y should be: 75+60-2(15) = 105

Ans: "D"
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

Sol:
Find LCM(1,2,3,4,5,6,7)

2=2
3= 3
4=2^2
5= 5^1
6=2^1 * 3^1
7= 7^1
Choose Highest Powers of prime factors= 2^2*3^1*5^1*7^1 = 420

Ans: "A"

Thanks for the questions, Bunuel.
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Re: Good set of PS 2 [#permalink]

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29 Mar 2011, 08:08
so good set, thank you very much...
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Re: Good set of PS 2 [#permalink]

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04 Apr 2011, 06:47
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
**************************************
there are (999-100+1=900) 3-digits numbers.
we should find 3-digits numbers that have1 digit "7" & 2 digits "7"& 3 digits "7" and them sum them.
1 digit "7" :
the form of these number = **7 or *7* or 7** (* is digit that cannot be 7 )
the numbers of **7 format=8*9=72;(first digit cannot to be 0)
the numbers of *7* format=8*9=72;
the numbers of 7** format=9*9=81;
total1=72+72+81=225
2 digits "7" :
the form of these number = *77 or 7*7 or 77* (* is digit that cannot be 7 )
the numbers of *77 format=8;(first digit cannot to be 0)
the numbers of 7*7 format=9;
the numbers of 77* format=9;
total2=8+9+9=26
3 digits "7" :
only on number :777;
total3=1

now total=total1+total2+total3=225+26+1=252;
252/900====>7/25
therefor correct answer is C;
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Re: Good set of PS 2 [#permalink]

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24 Apr 2011, 03:47
@Bunuel : can you please list down your solutions to these problems one problem per reply.
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Re: Good set of PS 2 [#permalink]

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24 Apr 2011, 03:56
gprasant wrote:
@Bunuel : can you please list down your solutions to these problems one problem per reply.

Bunuel's solutions:
good-set-of-ps-85414-20.html#p640101

My solutions:
good-set-of-ps-85414-80.html#p876910

Hope it doesn't matter whether the solutions are in one post or multiple so far they are not a hodgepodge!!
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Re: Good set of PS 2 [#permalink]

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24 Apr 2011, 22:27
yangsta8 wrote:
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 = 48-16 = 32
ANS = B

Looking back on this I'm sure there's more elegant methods

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

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

While I am not sure about 1st- whether a more smart approach can be there or not but not doubt about second one
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Re: Good set of PS 2 [#permalink]

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05 Sep 2011, 06:24
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

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

Can anyone help me by pointing out where my understanding is wrong.
Given below is my approach for the problem.

to meet the condition of "at least 2 men and 3 women" , i will first choose 2 men and 3 women and the 6th position can be any of the remaining.

C(8,2) * C(5,3) * C(8,1) [ remaining after selection of minimum is 8 people]
= 2240

Now,
To the condition of "Two men do not wish to work together" , i will find the negation of this as " two men always work together"

C(2,2) [two men always working together] * C(5,3) * C (8,1) [ remaining after selection of minimum is 8 people]
= 80

So the answer as per my logic is 2160 [ 2240 - 80 ]
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Re: Good set of PS 2 [#permalink]

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05 Sep 2011, 08:25
can you tell me the correct answers for all? or did u post the corect answer if so plz paste the link.
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Re: Good set of PS 2 [#permalink]

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05 Sep 2011, 08:30
vinodmallapu wrote:
Bunuel wrote:

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

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

Can anyone help me by pointing out where my understanding is wrong.
Given below is my approach for the problem.

to meet the condition of "at least 2 men and 3 women" , i will first choose 2 men and 3 women and the 6th position can be any of the remaining.

C(8,2) * C(5,3) * C(8,1) [ remaining after selection of minimum is 8 people]
= 2240

Now,
To the condition of "Two men do not wish to work together" , i will find the negation of this as " two men always work together"

C(2,2) [two men always working together] * C(5,3) * C (8,1) [ remaining after selection of minimum is 8 people]
= 80

So the answer as per my logic is 2160 [ 2240 - 80 ]

I suspect you have done worng here:Ways to chose 6 members committee with two particular men serve together: $$C^2_2*C^4_5+C2^_2*C^1_6*C^3_5=5+60=65$$

it should be C^2_2*C^4_5+C2^_2*C^1_4(4 b'coz we have already selected 2 members right?so 6-2 4 left)*C^3_5.
Sry correct me if iam worng:(
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Re: Good set of PS 2 [#permalink]

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31 Mar 2012, 16:45
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

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

Bunuel,

Can you try to solve this question with the slot method? Combination was very straight forward but when I tried to use slot I got really confused. Thanks.
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Re: Good set of PS 2 [#permalink]

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20 Jul 2012, 07:39
[quote="Bunuel"]Please find below new set of PS problems:

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

R= 0,12, then:
$$\frac{R}{S}= \frac{12}{100}$$
$$\frac{R}{S}= \frac{3}{25}$$
$$25R = 3S$$
Since 45 is the only number which has 3 and 5 as his primes (E) should be the answer.
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Re: Good set of PS 2 [#permalink]

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20 Jul 2012, 07:54
Stiv wrote:
Bunuel wrote:
Please find below new set of PS problems:

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

R= 0,12, then:
$$\frac{R}{S}= \frac{12}{100}$$
$$\frac{R}{S}= \frac{3}{25}$$
$$25R = 3S$$
Since 45 is the only number which has 3 and 5 as his primes (E) should be the answer.

OA-s and solutions are given in my post on 2nd 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

$$s$$ 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

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Re: Good set of PS 2 [#permalink]

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26 Jul 2012, 23:10
For those who like visualizing, here is a solution for Question 7 using graphs.

Please, refer to the attached drawing, in which the three graphs $$y=1/x,$$ $$y=2x,$$ and $$y=x^2$$ are depicted for $$x>0$$.
The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions.

So, the correct orderings are:
If $$x$$ between 0 and A: $$x^2<2x<1/x$$
If $$x$$ between A and B: $$x^2<1/x<2x$$
If $$x$$ between B and C: $$1/x<x^2<2x$$
If $$x$$ greater than C: $$1/x<2x<x^2$$

We can see that only the first two of the above options are listed as answers (I and II).

Attachments

3Graphs.jpg [ 15.15 KiB | Viewed 1627 times ]

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Re: Good set of PS 2 [#permalink]

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24 Sep 2012, 12:02
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

Good challenging question.

Driving spot - only two variations. 1) Mother drives 2) Father drives

The permutations come up in the spots for the other 4 seats.

So, 2 default values * 4P4 = total 2*4! = 48 permutations

Total # ways - # ways together = # ways can't sit together

The two daughters can only sit together when they are together in the back seat.
So that's:
12B
21B
B12
B21

...where 1 and 2 represent daughter #1 and #2. "B" represents the boy child.

so that's 4 ways we counted.

4 ways * variations on other 3 seats
4 ways * (default mother/father drives and variations on other 2 seats)
4 ways * (2 * 2P2)
4 ways * 2 * 2! = 16

Total ways = 48
Variations together = 16
Variations not together = 48 - 16 = 32
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Re: Good set of PS 2 [#permalink]

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12 Oct 2012, 01:08
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

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

Hi Bunuel - If the question had been " how many applicants applied EITHER college to X or to college Y?
It would be 75 + 60 - 15 = 120 right?

Can you please confirm?

cheers
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Re: Good set of PS 2 [#permalink]

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12 Oct 2012, 02:43
Jp27 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

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

Hi Bunuel - If the question had been " how many applicants applied EITHER college to X or to college Y?
It would be 75 + 60 - 15 = 120 right?

Can you please confirm?

cheers

Yes, that's correct.
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Re: Good set of PS 2 [#permalink]

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16 Mar 2013, 20:41
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 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

I Couldnt get the triangle being in 4th quadrant part!Bunuel,pls help!????
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Re: Good set of PS 2 [#permalink]

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17 Mar 2013, 01:29
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Expert's post
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 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

I Couldnt get the triangle being in 4th quadrant part!Bunuel,pls help!????

Check here: in-the-xy-plane-line-k-has-a-positive-slope-and-and-100739.html

Hope it helps.
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Re: Good set of PS 2   [#permalink] 17 Mar 2013, 01:29

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