Last visit was: 14 Dec 2024, 15:26 It is currently 14 Dec 2024, 15:26
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
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Dec 2024
Posts: 97,877
Own Kudos:
685,892
 []
Given Kudos: 88,271
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 97,877
Kudos: 685,892
 []
Kudos
Add Kudos
6
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
askhere
Joined: 12 Nov 2014
Last visit: 13 Dec 2024
Posts: 64
Own Kudos:
204
 []
Given Kudos: 14
Posts: 64
Kudos: 204
 []
5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
avatar
shaunaklaad
Joined: 28 Feb 2015
Last visit: 13 Nov 2015
Posts: 1
Own Kudos:
1
 []
Given Kudos: 1
Posts: 1
Kudos: 1
 []
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
ManojReddy
Joined: 28 Jan 2013
Last visit: 24 Sep 2024
Posts: 31
Own Kudos:
124
 []
Given Kudos: 32
Location: United States
Concentration: General Management, International Business
GPA: 3.1
WE:Information Technology (Consulting)
Posts: 31
Kudos: 124
 []
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
1, 2, 3, 5, 8, 13, 21, 34 and 55.
In the series above, take any number after 2. The sum of previous 2 numbers will be equal to the number chosen.

lets take number 3 : sum of previous 2 numbers (1+2 = 3)
lets take number 55: sum of previous 2 numbers (21+34 = 55)

So in total we have 7 possibilities : (1,2) (2,3) (3,5) (5,8) (8,13) (13,21) (21,34)

probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box = \(7/(9c2)\) = 7/36

Ans: C
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,341
Own Kudos:
3,705
 []
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,341
Kudos: 3,705
 []
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A box contains nine slips that are each labeled with one number: 1, 2, 3, 5, 8, 13, 21, 34 and 55. Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?

A. 7/72
B. 1/6
C. 7/36
D. 15/36
E. 21/36

Kudos for a correct solution.

Total number of cases = 9C2 = 36

Total favorable scenarios =

1,2
2,3
3,5
5,8
8,13
13,21
21,34

= 7

Thus the probability = 7/36, C is the correct answer.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 14 Dec 2024
Posts: 97,877
Own Kudos:
685,892
 []
Given Kudos: 88,271
Products:
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 97,877
Kudos: 685,892
 []
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
A box contains nine slips that are each labeled with one number: 1, 2, 3, 5, 8, 13, 21, 34 and 55. Two of the slips are drawn at random from the box without replacement. What is the probability that the sum of the numbers on the two slips is equal to one of the numbers left in the box?

A. 7/72
B. 1/6
C. 7/36
D. 15/36
E. 21/36

Kudos for a correct solution.

800score Official Solution:

Let us consider the numbers printed on the slips i.e. {1, 2, 3, 5, 8, 13, 21, 34, 55}. There is a pattern in the numbers here, which if decoded makes the problem simpler. Starting from the third number (which is 3), each number is the sum of the previous two numbers. So: 3 = 2 + 1, 5 = 3 + 2, 8 = 5 + 3…and so on.

Now, if we want the sum on the two selected slips to be equal to one of the numbers from the set, then these two numbers have to be any two consecutive numbers from the set except the last two numbers. So, there are seven favorable pairs of numbers: {1, 2}, {2, 3}, {3, 5}, {5, 8}, {8, 13}, {13, 21} and {21, 34}. In total, there are 14 ways that these pairs can be chosen ((1, 2) and (2, 1) are two different solutions). If there are 14 possible favorable outcomes and 9 × 8 (there are 9 slips and after pulling one there are 8 left) possible outcomes, then the total probability of a favorable outcome is 14/(9 × 8) = (7 × 2)/(9 × 4 × 2) = 7 /36.

The correct answer is C.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 35,825
Own Kudos:
Posts: 35,825
Kudos: 930
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderator:
Math Expert
97877 posts