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Master | Next Rank: 500 Posts
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by beater » Wed Oct 08, 2008 12:24 pm
Is 2y>3z?

1. 3y > 2z
2. y < 2z

Is there a way to solve this using transitive properties?
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Source: — Data Sufficiency |
Newbie | Next Rank: 10 Posts
Posts: 6
Joined: Wed Oct 08, 2008 7:34 am

looking for a solution or a specific method?

by gace780 » Wed Oct 08, 2008 1:42 pm
2y>3z same as 6y>9z

A:
===
3y>2z ===> 6y > 4z

B:
===
y<2z ====> 6y <12z

Both together 4z < 6y <12z

Therefore, 6y may or may not be greater than 9z. Answer is E!
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by gmat009 » Wed Oct 08, 2008 9:09 pm
IMO A
What's OA
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by namitrajiv » Wed Oct 08, 2008 9:17 pm
should be E
We need to find if y>1.5 z
first statement says y>0.66 Z , clearly insufficient because if y >0.66 z it can or cannot be greater than 1.5 z

2nd statement says y<2z, again insufficient because if y <2z it can or cannot be greater than 1.5 z

combined 0.66z<y<2z
again y can be greater than or less than 1.5 z , possible values of y are
1.2 z(<1.5z), 1.7z(>1.5z)
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by gmat009 » Wed Oct 08, 2008 9:21 pm
namitrajiv wrote:should be E
We need to find if y>1.5 z
first statement says y>0.66 Z , clearly insufficient because if y >0.66 z it can or cannot be greater than 1.5 z

2nd statement says y<2z, again insufficient because if y <2z it can or cannot be greater than 1.5 z

combined 0.66z<y<2z
again y can be greater than or less than 1.5 z , possible values of y are
1.2 z(<1.5z), 1.7z(>1.5z)
Yes you are right....
By mistake I understood in a wrong way.
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