Search found 25 matches
Hi Mitch, I'm unable to understand how 'wants' is a Noun. Can you elaborate? Hi Mitch, While solving this problem i was searching for a ing form of verb to make the it parallel with seeking . In a state of pure commercial competition, there would be a large number of producing firms, all unfettered ...
- by Kaustubhk
Thu Aug 03, 2017 10:48 pm- Forum: Sentence Correction
- Topic: one another vs. others?
- Replies: 17
- Views: 10109
Hi,
Regarding the usage of ONE, does it really matter if it is a same entity or a different entity as long as we know what we are referring to.
Correct me if my understanding is wrong.
- by Kaustubhk
Thu Jul 27, 2017 7:07 pm- Forum: Sentence Correction
- Topic: Any medical test will sometimes fail to detect a condition
- Replies: 20
- Views: 8347
- by Kaustubhk
Fri Jul 07, 2017 5:27 am- Forum: Critical Reasoning
- Topic: Question Pack 1 Medium CR 1
- Replies: 6
- Views: 2331
Hi Experts, I solved this in a different way. Kindly correct if I'm wrong. Y^3 < |y| .. |y| > y^3 .. Reverse Y > y^3 .. For positive -y > y^3 Y < -y^3.. For Negative y^3<y < -Y^3., combining the two A) y< 1 For y= 0, 0< 0 < -0 .. Not satisfied B) y< 0 For y= -1 -1< -1< 1 For Y= -2 -8 < -2 < 8 The an...
- by Kaustubhk
Wed Jun 28, 2017 4:56 am- Forum: Data Sufficiency
- Topic: Is y^3 < |y|
- Replies: 8
- Views: 2337
Manhattan Inequality and Absolute problem
Is |x| < 1 ? (1) |x + 1| = 2|x – 1| (2) |x – 3| > 0 The OA is C Hi Mates, My approach Is |x| < 1? We need to find -1<x<1, whether x falls in this range. 1) |x+1| = 2|x-1| First i considered positive sign, x+1 = 2x-2 x=3.. This doesn't fall in the range -1<x<1, Now consider negative sign, -(x+1) ...
- by Kaustubhk
Mon Jun 12, 2017 5:55 am- Forum: Data Sufficiency
- Topic: Manhattan Inequality and Absolute problem
- Replies: 2
- Views: 1101
Hi Matt, Thanks for recommending the videos. I know the experts have explained but need to type it out here so that i'm clear with my concept. Explanation: The question is to find the probability that atleast three of them have their birthdays in the same month. I have taken the complement route. Le...
- by Kaustubhk
Sat May 27, 2017 12:25 am- Forum: Problem Solving
- Topic: Probability question
- Replies: 12
- Views: 5522
MGMAT CR or Powerscore CR..Confused
Hi Guys, Apologies if i'm asking a already discussed topic again. I'm confused which book should i refer for CR. I have MGMAT CR hardcopy with me but i have read in many forums that powerscore CR is good. Till now i haven't really focussed much on CR strategies. The practice is only the one that i h...
- by Kaustubhk
Fri May 26, 2017 11:21 pm- Forum: GMAT Strategy
- Topic: MGMAT CR or Powerscore CR..Confused
- Replies: 1
- Views: 1308
Hi Mitch, The doubt which dhonu121 had asked of ordering of the numbers. Does the catch lie in the question itself? The last part clearly states that the 'what is the probability that the sum of the three numbers on the balls selected from the box will be odd? Since the probability of sum of three n...
- by Kaustubhk
Thu May 25, 2017 9:49 am- Forum: Problem Solving
- Topic: Probability GMAT Prep
- Replies: 12
- Views: 3785
- by Kaustubhk
Thu May 25, 2017 8:58 am- Forum: Problem Solving
- Topic: Probability question
- Replies: 12
- Views: 5522
HI brent,
1st member is selected in 8ways
2nd member in 6 ways (we can't select the spouse)
3rd Member in 4 ways
8*6*4
The order in which the is not important so we must divide by the factorial of the number of choices to eliminate over-counting:
(8*6*4)/3! = 8*4 = 32
- by Kaustubhk
Wed May 24, 2017 8:51 am- Forum: Problem Solving
- Topic: permutation
- Replies: 7
- Views: 5124
Hi Brent, 144 234 126 333 522 711 531 144 can arranged in 3!/2! = 3 ways 234 can arranged in 3! = 6 ways 126 can arranged in 3! = 6 ways 333 can arranged in 1! = 1 way 522 can arranged in 3!/2! = 3 ways 711 can arranged in 3!/2! = 3 ways 531 can arranged in 3! = 6 ways Total number of ways = 28 ways.
- by Kaustubhk
Wed May 24, 2017 8:13 am- Forum: Problem Solving
- Topic: Arabian Horses - Good One!
- Replies: 17
- Views: 7568
Hi Brent,
If the question is rephrased as below
In how many different ways can a group of 9 people be divided into 3 groups?
How we will solve this?
- by Kaustubhk
Tue May 23, 2017 8:27 am- Forum: Problem Solving
- Topic: Arabian Horses - Good One!
- Replies: 17
- Views: 7568
Probability-Coins
Harvey flipped a fair coin until he got tails, at which point he stopped. If Harvey stopped after no more than three flips, what is the probability that he didn’t get tails until his third flip? A. 1/2 B. 1/3 C. 1/4 D. 1/7 E. 1/8 Here the OA is D My Approach: The question says he didn't get tails ...
- by Kaustubhk
Sun May 21, 2017 1:10 am- Forum: Problem Solving
- Topic: Probability-Coins
- Replies: 4
- Views: 1797
Hi Mitch,
This is a bit confusing why should't we consider the probability of an earlier event? Is it because it is guareented?
- by Kaustubhk
Sat May 20, 2017 9:29 am- Forum: Problem Solving
- Topic: Pleas help--Probability problem
- Replies: 12
- Views: 3432
HI Mitch, Let me ask you one thing, if the question was framed like There are 10 women and 3 men in room A. One women is picked at random from room A and moved to room B, where there are already 3 women and 5 men. If a single person is then to be picked from room B, what is the probability that a wo...
- by Kaustubhk
Sat May 20, 2017 12:34 am- Forum: Problem Solving
- Topic: Pleas help--Probability problem
- Replies: 12
- Views: 3432