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Least possible value of x

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by Anurag@Gurome » Sat Dec 01, 2012 4:53 am
neeti2711 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
Let 3x=4y=7z = n. Then x = n/3, y = n/4 and z = n/7
x+y+z = n/3 + n/4 + n/ 7 = 61n/84
Since x , y and z are positive integers so x + y + z should also be a positive integer, this implies minimum value of n = 84
Hence, minimum value of x + y + z = 61

The correct answer is D.
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by GMATGuruNY » Sat Dec 01, 2012 6:05 am
neeti2711 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
3x = 4y = 7z.
Minimum possible solution:
3(4*7) = 4(3*7) = 7(3*4).
Thus, x+y+z = (4*7) + (3*7) + (3*4) = 28+21+12 = 61.

The correct answer is D.
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