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Re: GMAC paper test question [#permalink]
16 May 2011, 16:59
1
This post received KUDOS
Expert's post
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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