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If x = 10a, y = 3b, z = 7c, and x : y : z = 10 : 3 : 7, then
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14 Apr 2019, 01:40
14 Apr 2019, 01:40
Expert Reply
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Question Stats:
90% (02:02) correct
9% (02:45) wrong based on 22 sessions
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If \(x = 10a\), \(y = 3b\), \(z = 7c\), and \(x : y : z = 10 : 3 : 7\), then \(\frac{7x+2y+5x}{8a+b+3c}=\)
(A) \(\frac{111}{12}\)
(B) \(\frac{7}{6}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{108}{123}\)
(E) \(\frac{135}{202}\)
(A) \(\frac{111}{12}\)
(B) \(\frac{7}{6}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{108}{123}\)
(E) \(\frac{135}{202}\)
Re: If x = 10a, y = 3b, z = 7c, and x : y : z = 10 : 3 : 7, then
[#permalink]
13 Oct 2019, 02:45
13 Oct 2019, 02:45
1
Carcass wrote:
If \(x = 10a\), \(y = 3b\), \(z = 7c\), and \(x : y : z = 10 : 3 : 7\), then \frac{7x+2y+5x}{8a+b+3c}
(A) \(\frac{111}{12}\)
(B) \(\frac{7}{6}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{108}{123}\)
(E) \(\frac{135}{202}\)
(A) \(\frac{111}{12}\)
(B) \(\frac{7}{6}\)
(C) \(\frac{8}{15}\)
(D) \(\frac{108}{123}\)
(E) \(\frac{135}{202}\)
Sorry Carcass, there is a typo in the equation. Should be \(\frac{7x+2y+5z}{8a+b+3c}=\). Thanks
gmatclubot
Re: If x = 10a, y = 3b, z = 7c, and x : y : z = 10 : 3 : 7, then [#permalink]
13 Oct 2019, 02:45
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