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AbhinavKumar
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gmatophobia
DS Question 1 - Mar 01 z is an integer such that |z| 3 (2) |z| = 2 Source: Others | Difficulty: Hard*
statement 1:
we simplify to get z>5 or z<-1
as there is constraint that -6So, Z is indeed negative thus this statement is sufficient

statement 2:
z=2 or z=-2 not sufficient


Option A
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B
how did u approach this ? can u briefly explain
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AbhinavKumar
how did u approach this ? can u briefly explain
Explanation can be found here:
standing-on-the-origin-of-an-xy-coordinate-plane-john-takes-127511.html?style=12#p1044207
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AbhinavKumar
how did u approach this ? can u briefly explain
As John wants to return to the origin he can take the following combinations

UDUD - All possible arrangements
LRLR - All possible arrangements

UDLR - All possible arrangements

Let’s calculate the probability for each set (UDUD / LRLR)

(1/4)^4 * 2 * 4!/(2!*2!) = 3/64

Let’s calculate the probability for each set (UDLR)

(1/4)^4 * 4! = 6/64

Total = 3/64 + 6/64 = 9/64
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gmatophobia
DS Question 1 - Mar 01 z is an integer such that |z| 3 (2) |z| = 2 Source: Others | Difficulty: Hard*

gmatophobia
PS Question 1 - Mar 01 Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again? (A) 7/64 (B) 9/64 (C) 11/64 (D) 13/64 (E) 15/64 Source: Manhattan | Difficulty: Hard

DS Question 1 - March 03

Is x divisible by 405 ?

(1) 27 is a factor of x
(2) x is a multiple of 15

Source: GMAT Ninja | Difficulty: Medium

PS Question 1 - March 03

If |m|/m 1

B. m > -2

C. |m| 1

Source: Others | Difficulty: Hard
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The PS question is indicated as "Hard" for a reason ;)
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gmatophobia
The PS question is indicated as "Hard" for a reason ;)
so depending on +ve/-ve nature of m :

if, m0:
we have m>1 as common region

if we combine all then option B, m > -2 will satisfy all conditions !?

then option B ?
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But if you take -1.5 then it doesn’t hold

Cause when m is -ve then we have -1-2
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note that m>-2 just covers the positions -11, so we need not have to take other value to prove it otherwise... gmatophobia to comment on this reasoning :upsidedown
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I would have picked B too though in exam :upsidedown cause I was confidently able to eliminate rest of the options
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You can have two cases: When X is a +ve integer or when it is a negative proper fraction

rickyric395
I would have picked B too though in exam :upsidedown cause I was confidently able to eliminate rest of the options
B is fine, the range of -11 is included in x>-2

Whatever value of X is chosen, it’s always greater than -2

So, x>-2
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Agree! The catch in this question is that the premise always holds true.

The question is similar to this official question, if someone wants to give it a try -

If 4 2
III. -(x+5) is positive

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III
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D
4 2. (II.)
Also, x < -5 means x + 5 < 0, which means -(x + 5) is positive. (III.)
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gmatophobia
If 4 2 III. -(x+5) is positive (A) II only (B) III only (C) I and II only (D) II and III only (E) I, II and III
on simplifying we get x<-5 , so I is out, since x<-5 , |x+3| will always be greater than 2 , as it is equal to 2 with x=-5 and and x is lesser than -5 so II is correct , since x<-5 , x+5 will always be -ve and hence -(x+5) will always be +ve. Answer is D
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gmatophobia
DS Question 1 - March 03 Is x divisible by 405 ? (1) 27 is a factor of x (2) x is a multiple of 15 Source: GMAT Ninja | Difficulty: Medium

gmatophobia
PS Question 1 - March 03 If |m|/m 1 B. m > -2 C. |m| 1 Source: Others | Difficulty: Hard

gmatophobia
If 4 2 III. -(x+5) is positive (A) II only (B) III only (C) I and II only (D) II and III only (E) I, II and III

DS Question 1 - Mar 05

At a certain animal shelter, there are nine dogs available for adoption. If Susan chooses two dogs at random and p is the probability that Susan chooses two labradors, is p < 1/2 ?

(1) The probability of picking one labrador and one non-labrador is greater than 1/2.
(2) The probability that neither dog selected is a Labrador is greater than 1/10.

Source: GMATNinja | Difficulty: Hard

PS Question 1 - Mar 05

If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at random from the set {5, 6, 7}, and z is chosen at random from the set {7, 8, 9}, what is the probability that x + y + z will be greater than 19 ?

A. 1/3
B. 2/9
C. 1/9
D. 1/18
E. 1/36

Source: GMATNinja | Difficulty: Medium
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gmatophobia
DS Question 1 - Mar 05 At a certain animal shelter, there are nine dogs available for adoption. If Susan chooses two dogs at random and p is the probability that Susan chooses two labradors, is p < 1/2 ? (1) The probability of picking one labrador and one non-labrador is greater than 1/2. (2) The probability that neither dog selected is a Labrador is greater than 1/10. Source: GMATNinja | Difficulty: Hard
is it B?

gmatophobia
PS Question 1 - Mar 05 If x is chosen at random from the set {2, 3, 4, 5}, y is chosen at random from the set {5, 6, 7}, and z is chosen at random from the set {7, 8, 9}, what is the probability that x + y + z will be greater than 19 ? A. 1/3 B. 2/9 C. 1/9 D. 1/18 E. 1/36 Source: GMATNinja | Difficulty: Medium
We can have z=9, y=7, x=5,4 ; z=9, y=6, x=5 ; z=8, y=7, x=5; so total 4 favourable outcome. And total number of outcomes are 4*3*3 (ways of selecting a number from each set). Answer = 4/(4*3*3) = 1/9
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