least possible value from old GMAT Prep test

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san2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Fri Apr 16, 2010 1:02 am; Thanked: 1 times

least possible value from old GMAT Prep test

by san2009 » Sat Jul 10, 2010 1:30 am

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

OA is D

The way I approached problem was to assume the smallest possible value for x. Since x must be a positive integer, then x =1, from that we get y=3/4 and z=3/7
when we add those up we don't get anywhere close to the answer choices
if you were to take y=1 and solve with that or z=1 and solve from there - you would get a higher value for x+y+z

Can an instructor please advise what am i missing here??

thanks in advance

Quote

Source: Beat The GMAT — Problem Solving | Sat Jul 10, 2010 1:30 am

selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 am; Thanked: 135 times; Followed by:7 members

by selango » Sat Jul 10, 2010 1:46 am

Get the ratios of x:y:z

x:y=4:3=28:21

y:z=7:4=21:12

x:y:z=28:21:12

To get the least value put x=1,y=1,z=1

x+y+z=28+21+12=61

Last edited by selango on Sat Jul 10, 2010 2:01 am, edited 1 time in total.

--Anand--

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jaskaran: Senior | Next Rank: 100 Posts; Posts: 61; Joined: Mon Nov 09, 2009 3:52 am; Location: India

by jaskaran » Sat Jul 10, 2010 1:57 am

The condition mentioned in the stem is 3x=4y=7z, so you need to find the LCM among 3,4,7 and that is 84, then you divide 84 by 3,4,7 to get values of x,y,z (28,21,12) and then x+y+z(61).

I am not sure though how to clarify your method, but if you add the fractions, you do get a 61 in the numerator.

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san2009: Master | Next Rank: 500 Posts; Posts: 171; Joined: Fri Apr 16, 2010 1:02 am; Thanked: 1 times

by san2009 » Sat Jul 10, 2010 2:00 am

thanks both of you...i like your methods but still not sure what's wrong with my method
indeed, i get 61/28....but how do i get from there to just plain 61
instructors please help!

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selango: Legendary Member; Posts: 1460; Joined: Tue Dec 29, 2009 1:28 am; Thanked: 135 times; Followed by:7 members

by selango » Sat Jul 10, 2010 2:27 am

If you put only x=1 it means you are assuming least value only for 1.y and z can be of maximum values.

Similarly for y=1 or z=1

To get the least value of x+y+z,we need to substitute x,y and z values as 1(ie) x=1,y=1 and z=1.

We need to find the ratios of x,y,z so that we can get all the values.

Hope this clarify

--Anand--

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 am; Thanked: 335 times; Followed by:98 members

by Patrick_GMATFix » Sat Jul 10, 2010 10:26 am

san2009 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

OA is D

The way I approached problem was to assume the smallest possible value for x. Since x must be a positive integer, then x =1, from that we get y=3/4 and z=3/7
when we add those up we don't get anywhere close to the answer choices
if you were to take y=1 and solve with that or z=1 and solve from there - you would get a higher value for x+y+z

Can an instructor please advise what am i missing here??

thanks in advance

Hi San,

The question stem tells us that x, y and z are positive integers. As you wrote yourself, if x=1 then "we get y=3/4 and z=3/7". This contradicts the question stem because y and z must be positive integers.

My approach is to focus on 3x=4y=7z. Since the variables are integers, 3x, 4y and 7z must also be integers. They all equal some mystery number. This mystery number is a multiple of 3, 4 and 7. The smallest values of x, y, z will result in the smallest possible mystery number. Thus this number is the least common multiple (LCM) of 3, 4, and 7. This is 84. Thus 3x=4y=7z=84. x=28, y=21, z=12. x+y+z=61

Hope that helps,
-Patrick