BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

I disagree with the beatthegmat answer to this DS question

This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sat Sep 18, 2010 10:53 am
Thanked: 1 times

I disagree with the beatthegmat answer to this DS question

by peco » Tue Nov 30, 2010 1:52 am
Please help clarify the solution to this question. Thanks.

One number, k , is selected at random from a set of 11 consecutive even integers. What is the probability that
k = 10?

(1) The average (arithmetic mean) of the set is zero.

(2) The probability that k = 10 is the same as the probability that k = -10.


The answer given by BTG is A

Thanks.
Top
Source: — Data Sufficiency |
User avatar
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 am
Location: Bangalore, India
Thanked: 146 times
Followed by:24 members

by shovan85 » Tue Nov 30, 2010 2:53 am
peco wrote: One number, k , is selected at random from a set of 11 consecutive even integers. What is the probability that
k = 10?


(1) The average (arithmetic mean) of the set is zero.
When 11 consecutive even integers AM will be 0, If the sum of all the integers equals to zero.

This means the sequence is
-10 -8 -6 -4 -2 0 2 4 6 8 10

Thus the probability of k = 10 is 1/11. Sufficient.
peco wrote: (2) The probability that k = 10 is the same as the probability that k = -10.
So if the sequence is -10 -8 -6 -4 -2 0 2 4 6 8 10
then P(k = 10) = 1/11 and P(k = -10) = 1/11

But say if the sequence is something else (say 12 14 16....30 32).
n this sequence P(k = 10) = 0 and P(k = -10) = 0.

Thus Not Sufficient.

IMO A
Last edited by shovan85 on Tue Nov 30, 2010 3:06 am, edited 1 time in total.
If the problem is Easy Respect it, if the problem is tough Attack it
Top

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 1179
Joined: Sun Apr 11, 2010 9:07 pm
Location: Milpitas, CA
Thanked: 447 times
Followed by:88 members

by Rahul@gurome » Tue Nov 30, 2010 2:58 am
peco wrote:Please help clarify the solution to this question. Thanks.

One number, k , is selected at random from a set of 11 consecutive even integers. What is the probability that
k = 10?

(1) The average (arithmetic mean) of the set is zero.

(2) The probability that k = 10 is the same as the probability that k = -10.


The answer given by BTG is A

Thanks.
The answer is A indeed.
May I know what was your approach?
Here is the solution...

Given: One number, k is to be selected at random from a set of 11 consecutive even integers. Therefore the probability of k being a particular even integer is either 1/11 (in case the integer is included in the set) or 0 (in case the integer is not included).

Statement 1: The average (arithmetic mean) of the set is zero.
Only possible set of 11 consecutive even integers such that their average is zero is {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}. Thus, 10 is included in the set.

Sufficient.


Statement 2: The probability that k = 10 is the same as the probability that k = -10.
The probability may be 1/11 in case 10 and -10 both are present in the set. For example, {-10 , -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}. Or the probability may be zero in case 10 and -10 both are not present in the set. For example, {20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40}.

Not sufficient.

The corrcet answer is A.
Rahul Lakhani
Quant Expert
Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
Top
User avatar
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Sat Sep 18, 2010 10:53 am
Thanked: 1 times

by peco » Tue Nov 30, 2010 3:09 am
okay, now i think i understand the logic. i was initially of the opinion that another consecutive set of integers could give an average of zero using a set (-20,-16,-12,-8,-4,0,4,8,12,16,20) but i have just realised that this defeats the purpose of a consecutive set.

The explanation is very helpful.

Thanks a lot for clarifying. Now i feel very bad as i didnt understand the question stem before concluding.
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 332
Joined: Tue Feb 09, 2010 3:50 pm
Thanked: 41 times
Followed by:7 members
GMAT Score:720

by rishab1988 » Tue Nov 30, 2010 3:26 am
The answer should be A.

Here is my different approach.

Let the first term in the sequence =x
We know it is sequence of even integers so the next term (2) = x+2
Similarly the last term = x+20

In an A.P average = average of first and last term = (x+x+20)/2=x+10

mean(avg)=0
x=-10

last term = x+20=10

since there are 11 terms in the sequence and only 1 of them is 10.the probability =1/11.

sufficient

2) probability of -10 is same as +10.

The first case scenario is that both have probability is =1/11 [when mean =0]

But what if both have 0 probability?

the sequence could start from 12 and go on till 32... a valid sequence of 3 numbers.

Hence,insufficient

Hence the answer should be A
Top
Post new topic Post Reply
• Page 1 of 1