Quant Question of the Day Chat

DS Question 1 - Feb 27

If set T is obtained by multiplying every elements of set S by 2, is the median of set T greater than that of set S ?

(1) All elements of set S are positive.

(2) The median of set S is positive.

Source: GMAT Club Tests | Difficulty: Medium

PS Question 1 - Feb 27

The mean of four integers will not change if all the integers are multiplied by any constant. What is always true about this set of numbers?

I. The mean of the set is 0
II. The sum of the largest member and the smallest member of the set is 0
III. The set contains both positive and negative integers

A. I only
B. II only
C. III only
D. I and II only
E. I, II, and III

Source: GMAT Club Tests | Difficulty: Hard
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DS Question 1 - Mar 01

z is an integer such that |z| 3
(2) |z| = 2

Source: Others | Difficulty: Hard*

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
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I’d go for A instead


B

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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how did u approach this ? can u briefly explain
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Explanation can be found here:
standing-on-the-origin-of-an-xy-coordinate-plane-john-takes-127511.html?style=12#p1044207

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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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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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

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
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I would have picked B too though in exam cause I was confidently able to eliminate rest of the options

You can have two cases: When X is a +ve integer or when it is a negative proper fraction


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

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.)

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

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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is it B?


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