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PS Question 1 - Dec 06

A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?

A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5

Source: GMAT Club Tests | Difficulty: Hard
X is length of wire cut which forms permiter
X=4s
s=x/4
Area needs to be greater than one so
(x/4)^2>1
x>4
Therefore we need the longer cut to be more than 4 meters.
We can aproximate that the cut must be between 0cm and 99cm or 401cm and 500cm.
99*2/500
Must be E, as it is the only one remotley close.
I’m assuming this is an aproximate probability since it will be slightly less than 2/5 (since you cant cut at 100cm or 400cm)
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DS Question 1 - Dec 6

Given that x is a real number, is −1 < x < 3?

(1) ||x − 1| − 1| < 1

(2) (x + 1)(x − 3) < 0

Source: Others | Difficulty: Medium
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[quote="gmatophobia"] DS Question 1 - Dec 6

Given that x is a real number, is −1 this is we want

B). |x-1| > 0
x can be anything but 1

now here’s the confusion - how do I treat this information from B ?? we combine inferences from A and B, saying that -1if so, then answer is "option D"
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PS Question 1 - Dec 06

A 5 meter long wire is cut into two pieces. If the longer piece is then used to form a perimeter of a square, what is the probability that the area of the square will be more than 1 if the original wire was cut at an arbitrary point?

A) 1/6
B) 1/5
C) 3/10
D) 1/3
E) 2/5

Source: GMAT Club Tests | Difficulty: Hard
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PS Question 1 - Dec 7

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers?

A. –8
B. 12
C. 22
D. 52
E. 252

Source : Official Guide | Difficulty: Medium
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gmatophobia
PS Question 1 - Dec 7

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers?

A. –8
B. 12
C. 22
D. 52
E. 252

Source : Official Guide | Difficulty: Medium
C. 22
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gmatophobia
DS Question 1 - Dec 6

Given that x is a real number, is −1 this is we want

B). |x-1| > 0
x can be anything but 1

now here’s the confusion - how do I treat this information from B ?? we combine inferences from A and B, saying that -1<x<3 but x !=1
if so, then answer is "option D"

AbhinavKumar - The output of the modulus operation will always yield a non negative value. Hence once we couple that information that value of x will always lie between -1 and 3.

We can solve Statement 1

| |x-1| - 1 | < 1

0 <= | |x-1| - 1 | < 1

0 <= |x-1| - 1 < 1 or -1 < |x-1| - 1 <= 0
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DS Question 1 - Dec 7

Let S be a set of outcomes and let A and B be events with outcomes in S. Let ∼B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A) ?

(1) P(A ⋃ B) = 0.7
(2) P(A ⋃∼B) = 0.9

Source : Official Guide | Difficulty: Hard
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gmatophobia
PS Question 1 - Dec 7

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers?

A. –8
B. 12
C. 22
D. 52
E. 252

Source : Official Guide | Difficulty: Medium
A

Shikhar29
gmatophobia
PS Question 1 - Dec 7

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers?

A. –8
B. 12
C. 22
D. 52
E. 252

Source : Official Guide | Difficulty: Medium
A
Sorry C :-P
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DS Question 1 - Dec 7

Let S be a set of outcomes and let A and B be events with outcomes in S. Let ∼B denote the set of all outcomes in S that are not in B and let P(A) denote the probability that event A occurs. What is the value of P(A) ?

(1) P(A ⋃ B) = 0.7
(2) P(A ⋃∼B) = 0.9

Source : Official Guide | Difficulty: Hard
should be C

recall that 1=P(A)+P(B)-P(A and B)+P(not A or B)

You get each piece from 1 and 2
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You get each piece from 1 and 2
What’s the value of P(A) you got ?
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gmatophobia
PS Question 1 - Dec 7

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers?

A. –8
B. 12
C. 22
D. 52
E. 252

Source : Official Guide | Difficulty: Medium
A is possible: -5+-3
B is possible: 5+7
D:25+27 possible
E: 125 plus 127 possible
C: 13 plus 11 is 24, 7 plus 11 is 19 - not a possible sum given the conditions
C

AbhinavKumar
What’s the value of P(A) you got ?
I didnt calculate since its DS
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AbhinavKumar
What’s the value of P(A) you got ?
but if I am correct: from 1 you find everything that is not A or B which is 0.3, and from 2 you find B (since 2 is everything that is A or not B) which is .1
then you do 1=P(A)+P(B)-Both plus Neither:
P(A and B) can be found from 1:
0.7=P(A)+0.1-Both

1=P(A)+0.1+0.3-0.6+P(A)
1.2=2P(A)
p(A)=0.6
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mysterymanrog
but if I am correct: from 1 you find everything that is not A or B which is 0.3, and from 2 you find B (since 2 is everything that is A or not B) which is .1
then you do 1=P(A)+P(B)-Both plus Neither:
P(A and B) can be found from 1:
0.7=P(A)+0.1-Both

1=P(A)+0.1+0.3-0.6+P(A)
1.2=2P(A)
p(A)=0.6
thanks for the explanation, p(A) is indeed coming out as 0.6
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mysterymanrog
but if I am correct: from 1 you find everything that is not A or B which is 0.3, and from 2 you find B (since 2 is everything that is A or not B) which is .1
then you do 1=P(A)+P(B)-Both plus Neither:
P(A and B) can be found from 1:
0.7=P(A)+0.1-Both

1=P(A)+0.1+0.3-0.6+P(A)
1.2=2P(A)
p(A)=0.6
thanks for the explanation, p(A) is indeed coming out as 0.6
calculating takes too long imo I wouldnt have done it in real GMAT but since its practice its okay
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even I would refrain myself from doing long unnecessary calculations in DS. I was just checking on the value just for practice. Thanks again !
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