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

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Manager
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Status: Current MBA Student
Joined: 19 Nov 2009
Posts: 120
Concentration: Finance, General Management
GMAT 1: 720 Q49 V40
If x, y, and z are positive integers and 3x = 4y = 7z, then  [#permalink]

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New post 16 May 2011, 15:35
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A
B
C
D
E

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  25% (medium)

Question Stats:

88% (01:07) correct 12% (02:29) wrong based on 17 sessions

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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
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Re: GMAC paper test question  [#permalink]

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New post 16 May 2011, 17:51
1
given 3x=4y=7z

x+y+z in terms of x

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

now checking with each of the answers and see which value gives a minimum integer value.

A x = 28*33/61 , not an integer

B,C can be ruled out similarly.

D is minimum value as x = 61*28/61 = 28

Answer is D.
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Re: GMAC paper test question  [#permalink]

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New post 16 May 2011, 17:59
1
tonebeeze wrote:
Can someone walk me through this one. thanks


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


My solution in another post:

When I see more than one '=', my instinct is to make them equal to k to get to a 'common ground'. Thereafter, it is easier to think in terms of just one variable.

\(3x = 4y = 7z = k\)

\(x = \frac{k}{3}; y = \frac{k}{4}; z = \frac{k}{7}\)
Now, all x, y and z are positive integers so k should be divisible by 3, 4 and 7. The minimum value of k will be 3*4*7.

\(x + y + z = \frac{k}{3} + \frac{k}{4} + \frac{k}{7} = \frac{3*4*7}{3} + \frac{3*4*7}{4} + \frac{3*4*7}{7} = 61\)[/quote]
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Re: GMAC paper test question  [#permalink]

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New post 16 May 2011, 20:30
1
3x = 4y = 7z

LCM = 3 * 4 * 7 = 84
84 / 3 = 28
84 / 4 = 21
84 / 7 = 12

28 + 21 + 12 = 61
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Re: GMAC paper test question  [#permalink]

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New post 16 May 2011, 20:44
y=3/4x and z=3/7x

x+y+z = 61/28 *x = 61 for x=1.
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Re: GMAC paper test question &nbs [#permalink] 16 May 2011, 20:44
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