NEW HERE? LEARN MORE
closeclose
Last visit was: 24 Apr 2024, 16:19 It is currently 24 Apr 2024, 16:19
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

project GMAT ps

User avatar
mand-y
Manager
Manager
Joined: 05 Oct 2005
Posts: 134
Own Kudos [?]: 33 [0]
Given Kudos: 0
Send PM
project GMAT ps [#permalink] New post   22 Nov 2005, 16:20
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 2 sessions
Hide Show timer Statistics
:)
A set consists of all integers between 600 and 1999, inclusive. If a number is selected at random from this set, what is the probability that it is divisible by both 5 and 13?

(A) 65/1399
(B) 13/280
(C) 1/50
(D) 3/200
(E) 21/1399

plz explain your work
thanks



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
User avatar
desiguy
Senior Manager
Senior Manager
Joined: 11 May 2004
Posts: 276
Own Kudos [?]: 154 [0]
Given Kudos: 0
Location: New York
Send PM
Re: project GMAT ps [#permalink] New post   22 Nov 2005, 19:57
I think its D

1999-600+1 = 1400

So eliminate A and E

I was kind stuck from there, so i started seeing how many numbers are usually divisible by 5 and 13 and i realized it would take ME more than 2 mins, so I just guessed. :wink:

Anyone have any thoughts???
User avatar
gamjatang
Director
Director
Joined: 14 Sep 2005
Posts: 525
Own Kudos [?]: 1212 [0]
Given Kudos: 0
Location: South Korea
Send PM
Re: project GMAT ps [#permalink] New post   22 Nov 2005, 20:59
GCD of 5 & 13 = 65

65 * 10 = 650 > 600
65 * 30 = 1950 < 1999

30-10+1=21
1999-600+1=1400

21/1400
= 3/200
User avatar
B
GMATT73
CEO
CEO
Joined: 29 Jan 2005
Posts: 2887
Own Kudos [?]: 1117 [0]
Given Kudos: 0
Send PM
Re: project GMAT ps [#permalink] New post   22 Nov 2005, 21:05
E is the trap answer because 1999-600= 1400 (inclusive) Nix A and E right away.

to be both a multiple of 13 and 5, the number also has to be a multiple of 13*5=65

65*10=650 (lowest common multiple)
65*30=1950 (highest common multiple)

30-10=21 (inclusive)

Probability= #desired outcomes/#possible outcomes

21/1400--->3/200 (D)
User avatar
rlevochkin
Manager
Manager
Joined: 03 Nov 2005
Posts: 149
Own Kudos [?]: 109 [0]
Given Kudos: 17
Location: Chicago, IL
Send PM
Re: project GMAT ps [#permalink] New post   22 Nov 2005, 21:23
D is the answer
I've done this way. First, I,ve found how many numbers between 600 and 1999 are perfectly divisible by both 5 and 13. Since both 5 and 13 are primes, their LCM is 65. So. there are 600/65=9 and 1999/65=30. 30-9=21 intergers are divisible by 65 between 600 and 1999. number of consecutive intergers between 600 and 1999 is 1999-600+1=1400, 21/1400=3/200. The answer is D.
User avatar
ywilfred
SVP
SVP
Joined: 07 Jul 2004
Posts: 2004
Own Kudos [?]: 1899 [0]
Given Kudos: 0
Location: Singapore
Send PM
Re: project GMAT ps [#permalink] New post   22 Nov 2005, 22:12
Numer of integers = 1400
Divisible by 5 and 13: must be divisible by 65
1999/65 = 30.75 -> highest multiple of 65 is 30
600/65 = 9.2 --> lowest multiple of 65 is 10

So Total = 30-10+1 = 21
P = 21/1400 = 3/200

Ans: D
User avatar
mand-y
Manager
Manager
Joined: 05 Oct 2005
Posts: 134
Own Kudos [?]: 33 [0]
Given Kudos: 0
Send PM
Re: project GMAT ps [#permalink] New post   23 Nov 2005, 09:28
my approach is
from 60 to 1999 is 1399 but add 1 since inclusive
both 5 AND 13 are prime number so anumber to be divisible by both 13 and 5 must be a multiple of both 5 and 13 HENCE 65

SO I found the number of multiple of 65 in the 14OO numbers
14OO/ 65 = 21 of such number

HENCE final result is 21/1400

By the way i don 't understand why some of you are doing the following :
65*10=650 (lowest common multiple)
65*30=1950 (highest common multiple)

30-10=21 (inclusive)

thanks
User avatar
nakib77
Director
Director
Joined: 28 May 2005
Posts: 986
Own Kudos [?]: 3461 [0]
Given Kudos: 0
Location: Dhaka
Send PM
Re: project GMAT ps [#permalink] New post   23 Nov 2005, 11:09
first of all in order to be divisible by 5 and 13 it needs to be multiple of 65.

65*10 = 650 so we need to start from there.

650*30 = 1950. so between 10 to 30 inclusive we have 21 of them

and from 600 to 1999 there are 1400 numbers.

so probability is 21/1400 = 3/200
so D is the answer.



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
gmatclubot
Re: project GMAT ps [#permalink]
23 Nov 2005, 11:09
Moderators:
Bunuel
Math Expert
92900 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions