It is currently 18 Nov 2018, 23:54

### 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

# GRE Math Challenge #41-From the even numbers between 1 and 9

Author Message
TAGS:
GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4711
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 91

Kudos [?]: 1615 [1] , given: 376

GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]  09 Sep 2014, 03:15
1
KUDOS
Expert's post
00:00

Question Stats:

63% (00:56) correct 36% (00:41) wrong based on 30 sessions
From the even numbers between 1 and 9, two different numbers are to be chosen at random. What is the probability that their sum will be 8 ?

A. 1/6
B. 3/16
C. 1/4
D. 1/3
E. 1/2

[Reveal] Spoiler: Explanation
here are four even numbers between 1 and 9 (2, 4, 6,8).

From this set two distinct numbers can be selected in 4C2 ways :

Hence, there are 6 ways to select 2 numbers from the given set.

The number of pairs which sum 8 is 1 ( 2+6 ).
[Reveal] Spoiler: OA

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 26 Mar 2015
Posts: 3
Followers: 0

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

Re: GRE Math Challenge #41 [#permalink]  23 Apr 2015, 15:44
Intern
Joined: 12 Apr 2018
Posts: 2
Followers: 0

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

Re: GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]  12 Apr 2018, 22:27
I would say B
Manager
Joined: 26 Jan 2018
Posts: 172
Followers: 0

Kudos [?]: 104 [0], given: 3

Re: GRE Math Challenge #41 [#permalink]  12 Apr 2018, 23:23
sagar wrote:

The person answering this is expert.

@sandy can you please elaborate how 4C2 is further resolved and in what conditions we use it?
GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4711
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 91

Kudos [?]: 1615 [1] , given: 376

Re: GRE Math Challenge #41 [#permalink]  13 Apr 2018, 01:22
1
KUDOS
Expert's post
The way we can select 2 objects from a list of 4 is $$C^{4}_{2}=\frac{4!}{2! \times 2!}=6$$. Hence there are 6 ways of choosing a pair of number from 4 numbers.

Now when we eyeball the data we see only one number pair whose sum is 8 (2+4).

Hence the probability is $$\frac{1}{6}$$
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 28 Nov 2017
Posts: 44
Followers: 1

Kudos [?]: 30 [2] , given: 22

Re: GRE Math Challenge #41-From the even numbers between 1 and 9 [#permalink]  22 Apr 2018, 00:43
2
KUDOS
Option A :

Different Approach

Sum of two Numbers to be 8 which leaves us with only two possible numbers (2 & 6)

So for the 1st Number we have 2 options either to select 2 or 6 from total of 4 numbers : 2/4

For the 2nd Number since we have already selected 1 out of 2 or 6 , the # of ways to select the other number : 1/3 (Since 1 Number is already selected we only have 3 Numbers Left)

Solutions = 2/4*1/3 = 2/12 = 1/6
Re: GRE Math Challenge #41-From the even numbers between 1 and 9   [#permalink] 22 Apr 2018, 00:43
Display posts from previous: Sort by