GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Oct 2019, 11:43

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Two distinct numbers a and b are chosen randomly from the set of first

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 03 Feb 2015, 09:45
4
18
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

42% (03:22) correct 58% (02:55) wrong based on 128 sessions

HideShow timer Statistics

Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 09 Feb 2015, 04:50
4
4
Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Any number could be written in the form of one among the following:

(i) 3k

(ii) 3k + 1

(iii) 3k + 2

Only if the 2 numbers are of different types would |a – b| not be divisible by 3:

Numbers of type 3k --> 3, 6, 9, ......... 18 --> Total 6 numbers

Numbers of type 3k + 1 --> 1, 4, 7, .... 19 --> Total 7 numbers

Numbers of type 3k + 2 --> 2, 5, 8, 11 ...20 -->Total 7 numbers

Total possible outcomes = 20 * 19 = 380

Favorable Cases:

1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84

2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84

3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98

Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer.
_________________
Most Helpful Community Reply
Intern
Intern
avatar
Joined: 07 Nov 2013
Posts: 25
Location: Brazil
Concentration: Strategy, Finance
GMAT 1: 580 Q44 V27
WE: Corporate Finance (Retail)
GMAT ToolKit User
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 03 Feb 2015, 13:20
2
3
Here is what i thought:

From 20 first positive integers, we get that 6 of them are divisible by 3. For the equation not to be divisible by 3, we have the following possible combinations:

both of the numbers not being divisible by 3:

14/20 * 13/19 = 91/190

one of them is divisible by 3 and the other isn't:

14/20 * 6/19 = 42/190

because we need combination 1 OR combination 2 to achieve the requested, we need to sum both fractions:

91/190 + 42/190 = 133/190

For me, answer E. i guess i forgot something and made it completly wrong.
General Discussion
Manager
Manager
avatar
Joined: 27 Dec 2013
Posts: 199
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 03 Feb 2015, 13:31
Firstly,

The pattern is

1 4 7 10 13 16 19 = Pick two numbers from 7 -> 7p2= 42

2 5 8 11 14 17 20 = pick two number from 7 -> 7p2= 42

3 6 9 12 15 18= Pick two numbers from 6-> 6P2= 30.

Total permuations that are divisible by 3 are 114.

Total permutions are 20p2= 190.

1- 114/190.

This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\

Bunuel please guide me what I am doing wrong.






Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.

_________________
Kudos to you, for helping me with some KUDOS.
Manager
Manager
avatar
Joined: 15 Aug 2013
Posts: 50
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 03 Feb 2015, 17:50
3
1
I think you just made the mistake that 20P2 will be 20 * 19 = 380 and NOT 190..

If you consider this, that would make 114/380 (probability that the no is divisible by 3)
Subtract this probability from 1,
1- 114/380 = 266/380 = 133/190

Ans (E)

shriramvelamuri wrote:
Firstly,

The pattern is

1 4 7 10 13 16 19 = Pick two numbers from 7 -> 7p2= 42

2 5 8 11 14 17 20 = pick two number from 7 -> 7p2= 42

3 6 9 12 15 18= Pick two numbers from 6-> 6P2= 30.

Total permuations that are divisible by 3 are 114.

Total permutions are 20p2= 190.

1- 114/190.

This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\

Intern
Intern
avatar
Joined: 31 Dec 2014
Posts: 8
Location: United States
Concentration: Marketing
GMAT 1: 700 Q44 V41
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 03 Feb 2015, 18:03
1
I started to write the numbers out. Eventually, I thought that since a little less than a third of the numbers are divisible by three, the probability of the absolute value of any number minus another number not being divisible by three would be a little more than 2/3, so I went with answer E.
Manager
Manager
avatar
Joined: 07 Dec 2009
Posts: 86
GMAT Date: 12-03-2014
GMAT ToolKit User
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 22 Feb 2015, 12:20
Bunuel wrote:
Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Any number could be written in the form of one among the following:

(i) 3k

(ii) 3k + 1

(iii) 3k + 2

Only if the 2 numbers are of different types would |a – b| not be divisible by 3:

Numbers of type 3k --> 3, 6, 9, ......... 18 --> Total 6 numbers

Numbers of type 3k + 1 --> 1, 4, 7, .... 19 --> Total 7 numbers

Numbers of type 3k + 2 --> 2, 5, 8, 11 ...20 -->Total 7 numbers

Total possible outcomes = 20 * 19 = 380

Favorable Cases:

1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84

2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84

3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98

Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer.


Hi Bunuel,

Why exactly are you multiplying by 2 when you are calculating the favourable cases.

Many thanks
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58402
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 22 Feb 2015, 13:20
bhatiavai wrote:
Bunuel wrote:
Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.


VERITAS PREP OFFICIAL SOLUTION:

Any number could be written in the form of one among the following:

(i) 3k

(ii) 3k + 1

(iii) 3k + 2

Only if the 2 numbers are of different types would |a – b| not be divisible by 3:

Numbers of type 3k --> 3, 6, 9, ......... 18 --> Total 6 numbers

Numbers of type 3k + 1 --> 1, 4, 7, .... 19 --> Total 7 numbers

Numbers of type 3k + 2 --> 2, 5, 8, 11 ...20 -->Total 7 numbers

Total possible outcomes = 20 * 19 = 380

Favorable Cases:

1) Picking from Type 1 and Type 2: 2 * (6 * 7) = 84

2) Picking from Type 1 and Type 3: 2 * (6 * 7) = 84

3) Picking from Type 2 and Type 3: 2 * (7 * 7) = 98

Hence, the required probability = (84 + 84 + 98) / 380, and E is the correct answer.


Hi Bunuel,

Why exactly are you multiplying by 2 when you are calculating the favourable cases.

Many thanks


1) a is from Type 1 and b is from Type 2 + b is from Type 1 and a is from Type 2... The same for (2) and (3).
_________________
Intern
Intern
avatar
Joined: 15 Feb 2015
Posts: 13
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 27 Feb 2015, 11:50
1
1. Pick 2 from {1,2, ..... 20} 20*19 = 380
2. figure out the probability that |a–b| would be divisible by 3
1) {3, 6, 9, 12, 15, 18 } 6*5 = 30
2) {1,4,7,10,13, 16, 19} 7*6 = 42
3) {2, 5, 8, 11, 14, 17, 20} 7*6 = 42
3. the probability that |a–b| would not be divisible by 3 = (380 - 30-42-42) / 380 = 133/190

So the correct answer is E
Intern
Intern
User avatar
B
Joined: 03 Jul 2015
Posts: 10
Location: India
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 09 Aug 2015, 08:13
1
shriramvelamuri wrote:
Firstly,

The pattern is

1 4 7 10 13 16 19 = Pick two numbers from 7 -> 7p2= 42

2 5 8 11 14 17 20 = pick two number from 7 -> 7p2= 42

3 6 9 12 15 18= Pick two numbers from 6-> 6P2= 30.

Total permuations that are divisible by 3 are 114.

Total permutions are 20p2= 190.

1- 114/190.

This is exactly where I am struck. I am getting a strange answer. I guess, I have to wait for someone to crack this one. \\

Bunuel please guide me what I am doing wrong.






Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.


when you select 2 numbers from a set of numbers, you use combination and not permeutation (one of the very common mistakes), so you should do = \(1 - (7C2 + 7C2 + 6C2)/ 20C2\)
Intern
Intern
avatar
Joined: 10 Jun 2013
Posts: 17
Concentration: General Management, Technology
GMAT Date: 06-26-2015
WE: Corporate Finance (Venture Capital)
GMAT ToolKit User
Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 10 Aug 2015, 06:05
I have handled this one with no real mathematics involved.

For each number A selected you have 19 possible results for |a-b|, because when you select A you have 19 numbers remainding.

You know that only 6 results of |a-b| will give you a number divisible by 6 ( 3, 6, 9, ...18) . Then, 14 results out of 19 won't be divisible by 3.

Look, we've just said that 14 won't be divisible by 3 out of 19 possibles outcomes... 133/190 is approximatively this number.
Manager
Manager
avatar
Joined: 20 Jul 2011
Posts: 79
GMAT 1: 660 Q49 V31
GMAT ToolKit User
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 10 Aug 2015, 06:29
2
Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190
t

Kudos for a correct solution.


Let find the probability that |a-b| is divisible by 3.

Then |a-b| has to be 3, 6, 9, 12, 15, 18 (cannot be 0 since a and b are distinct)

To get |a-b| as 3, |a-b| could be |4-1|, |5-2|.....|20-17| => 17 Times.
But since this is modulus, |4-1| = |1-4|.

Hence to get 3, there are 34 ways.
Similarly to get 6 => 14*2 = 28
9 => 11*2 = 22
12 => 8*2 = 16
15 => 5*2 = 10
18 => 2*2 = 4

Hence total ways to get |a-b| divisible by 3 = 114

Total Outcomes = 20 x 19 = 380

Probability to get |a-b| divisible by 3 is 114 / 380 => 57/190

Probability that |a-b| is not divisible by 3 is = 1 - (57/190) = 133/190

Option E

------------------
Please give kudos if you like this post
Intern
Intern
avatar
Joined: 01 Jul 2015
Posts: 6
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 16 Nov 2015, 10:08
1
Bunuel wrote:
Two distinct numbers a and b are chosen randomly from the set of first 20 positive integers. What is the probability that |a–b| would not be divisible by 3?

A. 49/380
B. 21/95
C. 49/190
D. 42/95
E. 133/190


Kudos for a correct solution.

I have a better way to do this problem:
number 1-20 we have 19*20= 380 ways to choose a pair.
from 1-20 each number have 6 ways to choose a number that a-b divisible by 3: EX 20 must subtract by 17,14,11,8,5,3 to divisible by 3:
so from 1-20 we have 20*6= 120 ways to choose, but we didnt want a same number, hence 120-6= 114
so the percentage is 114/380 to choose a pair that divisible by 3,
===> hence the chance to pick a pair that not divisible by 3 is 1- (114/380)= 266/380 =133/190 and this is final answer.
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15263
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 12 Feb 2018, 13:59
Hi All,

The answer choices to this question provide a great 'logic shortcut' that you can take advantage of.

The calculation that we're asked to consider (|A-B|) and the limited possible values for the two distinct variables (1-20, inclusive) means that we're dealing with possible answers that are consecutive integers: 1 through 19, inclusive.

When dealing with consecutive integers, only 1 out of every "set" of 3 will be divisible by 3. This question asks for the probability that the calculation will NOT be divisible by 3. Thus, the answer MUST be greater than 1/2 (closer to about 2/3). There's ONLY one answer that's greater than one half....

Final Answer:

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
Image


The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Intern
Intern
avatar
B
Joined: 28 Apr 2016
Posts: 16
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

Show Tags

New post 25 Feb 2019, 11:05
|a-b| can have 19 values from 1 to 19 and out of these 19 values only 6 i.e. 3,6,9,12,15 and 18 are divisible by 3. Thus, 13 values are "not" divisible by 3. The probability can is thus 13/19 or 130/190. A close approximate to 133/190.
In any case, the probability should be more than 1/2. Only option E is >1/2
GMAT Club Bot
Re: Two distinct numbers a and b are chosen randomly from the set of first   [#permalink] 25 Feb 2019, 11:05
Display posts from previous: Sort by

Two distinct numbers a and b are chosen randomly from the set of first

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne