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

It is currently 22 May 2013, 18:29
Customize  |  Hide

If x, y, and z are positive integers and 3x=4y=7z, then the

  Question banks Downloads My Bookmarks Reviews  
Author Message
TAGS:
Intern
Intern
Joined: 26 Jul 2006
Posts: 5
Followers: 0

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

GMAT Tests User
If x, y, and z are positive integers and 3x=4y=7z, then the [#permalink] New post 06 Sep 2006, 16:37
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!
VP
VP
User avatar
Joined: 02 Jun 2006
Posts: 1278
Followers: 2

Kudos [?]: 15 [0], given: 0

GMAT Tests User
 [#permalink] New post 06 Sep 2006, 16:49
(D) 61

Given 3x = 4y = 7z

LCM(3,4,7) = 84
=> x = 28, y = 21, z = 12

Sum = 61.
VP
VP
User avatar
Joined: 29 Dec 2005
Posts: 1356
Followers: 6

Kudos [?]: 17 [0], given: 0

GMAT Tests User
Re: very interesting #'s properties q [#permalink] New post 06 Sep 2006, 18:10
chrismeiyu wrote:
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!


since 3, 4 and 7 has no common factors, the only way to have 3x=4y=7z is 3 (4x7) = 4 (3x7) = 7 (3x4). so x = 28, y= 21, and z = 12.

therefore x + y + z = 28+21+12 = 61.
Senior Manager
Senior Manager
Joined: 11 May 2006
Posts: 262
Followers: 1

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

GMAT Tests User
Re: very interesting #'s properties q [#permalink] New post 06 Sep 2006, 22:33
chrismeiyu wrote:
If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is

(A) 33
(B) 40
(C) 49
(D) 61
(E) 84

I used the following approach in order to try and solve this problem:

First I set all variable in terms of x which resulted in

x+y+z = x+(3/4)x+(3/7)x

simplifying, x+(3/4)x+(3/7)x = (61/28)x

This is where I got stuck. Any suggestions?

Thanks!


you are almost there :-)
now x + y + z will be a +ve integer since x,y, and z each are +ve integers.

so (61/28)x will be a integer => x is a multiple of 28.
now (61/28)x will be minimum when x = 28
hence x+y+z = 61
Re: very interesting #'s properties q   [#permalink] 06 Sep 2006, 22:33
    Similar topics Author Replies Last post
Similar
Topics:
New posts If x, y, and z are positive integers and 3x=4y=7z then the guy123 6 16 Nov 2003, 18:24
New posts If x, y, and z are positive integers and 3x = 4y = 7z, then Juan8 2 10 May 2005, 00:06
New posts If x, y and z are positive integers and 3x=4y=7z, then the SimaQ 2 04 Jun 2006, 03:50
New posts If x, y, and z are positive integers and 3x = 4y = 7z, then SimaQ 7 19 Oct 2006, 10:33
New posts if x, y, and z are positive integers and 3x=4y=7z, then the young_gun 4 29 Oct 2007, 20:04
Display posts from previous: Sort by

If x, y, and z are positive integers and 3x=4y=7z, then the

  Question banks Downloads My Bookmarks Reviews  


cron

GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.