Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Jun 2016, 12:55

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If two different numbers are to be selected from set {1, 2,

Author Message
Current Student
Joined: 11 May 2008
Posts: 555
Followers: 8

Kudos [?]: 126 [0], given: 0

If two different numbers are to be selected from set {1, 2, [#permalink]

### Show Tags

29 Jul 2008, 10:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

. If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number?
A. 1/2
B. 1/3
C. 1/9
D. 2/9
E. 4/9

here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once?
Intern
Joined: 26 May 2008
Posts: 40
Followers: 0

Kudos [?]: 5 [0], given: 0

### Show Tags

29 Jul 2008, 10:51
Here perfect squares can be either 4 or 9.

4 = 3+1 or 1+3 (2 ways)
9 = 3+6 or 6 +3 (2 ways)
= total 4 ways

Total ways of selecting to numbers = 6 *6 = 36 ways

Required probabilty = 4/36 = 1/9
Current Student
Joined: 11 May 2008
Posts: 555
Followers: 8

Kudos [?]: 126 [0], given: 0

### Show Tags

29 Jul 2008, 10:57
what abot 4,5 and 5,4?
also total ways of selecting two nos. from 6 is 6c2 rite?
Senior Manager
Joined: 14 Mar 2007
Posts: 317
Location: Hungary
Followers: 2

Kudos [?]: 21 [0], given: 3

### Show Tags

29 Jul 2008, 11:02
We have 15 possible option and 5 good option. Probability=5/15=1/3
Current Student
Joined: 11 May 2008
Posts: 555
Followers: 8

Kudos [?]: 126 [0], given: 0

### Show Tags

29 Jul 2008, 11:10
how 5 good option? it shud be 6 rite?
Senior Manager
Joined: 14 Mar 2007
Posts: 317
Location: Hungary
Followers: 2

Kudos [?]: 21 [0], given: 3

### Show Tags

29 Jul 2008, 12:09
I think we have 5 good options: (1,3) (2,6) (3,5) (3,6) (4,5)

I hpe these are the perfect squere numbers: 4, 8, 9, ...

I am not sure, may be I am not correct.
GMAT Tutor
Joined: 24 Jun 2008
Posts: 1181
Followers: 378

Kudos [?]: 1278 [0], given: 4

### Show Tags

29 Jul 2008, 17:39
Expert's post
arjtryarjtry wrote:
. If two different numbers are to be selected from set {1, 2, 3, 4, 5, 6}, what is the probability that the sum of two numbers is a perfect square number?
A. 1/2
B. 1/3
C. 1/9
D. 2/9
E. 4/9

here what confuses me is , in the total poss cases, do we take ,for eg 1,3 and also along with that 3,1 or should they be considered only once?

2^2 = 4 and 3^2 = 9 are the only perfect squares we can get as a sum of two numbers from the set. Note that 8 is not a perfect square (8 is a cube, however).

Now, we can either assume order matters, or assume it doesn't- we just need to be consistent. If order matters, there are 6*5 = 30 different pairs of numbers we can choose. Again, if order matters, we can get squares in the following ways:
1,3
3,1
3,6
4,5
5,4
6,3

So the probability should be 6/30 = 1/5. That isn't among the answer choices- where is the question from?
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Intern
Joined: 01 Jul 2008
Posts: 41
Followers: 1

Kudos [?]: 10 [0], given: 20

### Show Tags

30 Jul 2008, 00:14
number of ways that you can select 2 numbers out of 6 would be 6C2=15 ways
favourable outcomes are {1,3}, {4,5}, & {3,6} = 3 ways

hence the probability would be 3/15=1/5 (well it is not one of the options!!!)
Senior Manager
Joined: 14 Mar 2007
Posts: 317
Location: Hungary
Followers: 2

Kudos [?]: 21 [0], given: 3

### Show Tags

30 Jul 2008, 00:18
hi0parag,

Thank you for correcting my mistake.
Manager
Joined: 27 May 2008
Posts: 143
Followers: 1

Kudos [?]: 12 [0], given: 0

### Show Tags

30 Jul 2008, 03:26
Wrong Question.

Re: probability   [#permalink] 30 Jul 2008, 03:26
Display posts from previous: Sort by