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If x,y,z are positive and x=3 and y=5z, what's the ratio of

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If x,y,z are positive and x=3 and y=5z, what's the ratio of [#permalink]  23 Jun 2017, 11:33
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If x,y,z are positive and x=3 and y=5z, what's the ratio of x+y: x+z?

a. $$\frac{1}{2}$$

b. $$\frac{7}{15}$$

c. $$\frac{8}{15}$$

d. $$\frac{8}{9}$$

e. $$\frac{10}{9}$$
[Reveal] Spoiler: OA

Last edited by Carcass on 24 Jun 2017, 03:07, edited 1 time in total.
Edited the question
Intern
Joined: 08 Apr 2017
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Re: Question on ratio [#permalink]  23 Jun 2017, 11:35
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Cant i be solved like this?

as y=5z therefore y/z=5/1

i.e. y = 5k and z=1k

therefore x+y +z = 3+5k/x+k

now, substitute values in k =1,2,3,... to get some values and match it with options available?
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Re: If x,y,z are positive and x=3 and y=5z, what's the ratio of [#permalink]  24 Jun 2017, 03:09
Expert's post
Please format the question in the proper way: using the right title, and using the expression. They are just above the corpus of the message.

Post the question in the right sub-forum.

Thank you
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WE: Business Development (Energy and Utilities)
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Re: Question on ratio [#permalink]  24 Jun 2017, 11:36
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3Newton wrote:
Cant i be solved like this?

as y=5z therefore y/z=5/1

i.e. y = 5k and z=1k

therefore x+y : x+z = 3+5k/x+k

now, substitute values in k =1,2,3,... to get some values and match it with options available?

the answer should match for all values of k not just for some. This question doesn't make sense.

What is the source?
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Re: Question on ratio   [#permalink] 24 Jun 2017, 11:36
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