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by amit2k9 » Thu Jun 16, 2011 8:59 pm
DS 3

a xr/100 = (24000-x)*7/100 two variables x and r. not sufficient.

b x*5/100 = (24000-x)*7/100 only one variable x. sufficient.

B it is.
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by amit2k9 » Thu Jun 16, 2011 9:05 pm
DS2

a z can be 1 or 5 for t = -3 or -3/5. not sufficient.

b t= -5,-6 or so z = 3/5 or 1/2 thus always a ratio.

Hence B.
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by amit2k9 » Thu Jun 16, 2011 9:08 pm
can been discussed a lot.

here it is - https://gmatclub.com/forum/gmat-prep-xy- ... 60260.html
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by abidshariff » Mon Jun 20, 2011 10:01 am
For DS2...its E...

zt <-3, is z<4?

by using a i.e z<9, we don't know anything about t..so z can be < or > 4.
now take b. say t=-5(t<-4), and z=3, so zt=-15<-3, take z=5 zt=-25<-3. So z can be < or >4

So C is clearly not the ans as z=3,5<9.

So E
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by treker » Tue Jun 21, 2011 5:38 am
amit2k9 wrote:DS 3

a xr/100 = (24000-x)*7/100 two variables x and r. not sufficient.

b x*5/100 = (24000-x)*7/100 only one variable x. sufficient.

B it is.
For DS 3, Ans is C and not B. Without stmt 1, how can you assume that the interest on both amounts are equal?
Thanks!
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