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

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Math Expert
Joined: 02 Sep 2009
Posts: 53800
Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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03 Feb 2015, 09:45
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Difficulty:

95% (hard)

Question Stats:

41% (03:21) correct 59% (02:55) wrong based on 125 sessions

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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.

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Posts: 53800
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.
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Joined: 27 Dec 2013
Posts: 226
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.

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Kudos to you, for helping me with some KUDOS.
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Joined: 15 Aug 2013
Posts: 53
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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. \\

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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.
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GMAT Date: 12-03-2014
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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
Joined: 02 Sep 2009
Posts: 53800
Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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).
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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 = $$1 - (7C2 + 7C2 + 6C2)/ 20C2$$
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Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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

------------------
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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.
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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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....

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Re: Two distinct numbers a and b are chosen randomly from the set of first  [#permalink]

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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
Re: Two distinct numbers a and b are chosen randomly from the set of first   [#permalink] 25 Feb 2019, 11:05
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