is it E?

this_time_i_will wrote:is it E?

on what grounds?

IMO: A

xz>y
solving inequality : x>y/z

saurabhmahajan wrote:IMO: A

xz>y
solving inequality : x>y/z

but the OA is E

Hi Sanju

Lets consider Option 1

xz > y

In the inequality problems , when we multiply or divide by any number, we need to consider whether the number is positive or negative.
Ex. In the above equation xz >y
If we divide both sides by z , it can result in x > y/z or x<y/z based on whether z>0 or z<0

So thgis statement is not sufficient to find if x> y/z

Lets consider option 2

x^2 > y/z
Taking squareroots, we get
x > + sqrt (y/z) or
x > - sqrt (y/z)

We dont have sufficient information to find if x > y/z

Combining both statements doesnt help as well

IMO E

Hope this helps
--Arati

selfmade wrote:Hi Sanju

Lets consider Option 1

xz > y

In the inequality problems , when we multiply or divide by any number, we need to consider whether the number is positive or negative.
Ex. In the above equation xz >y
If we divide both sides by z , it can result in x > y/z or x<y/z based on whether z>0 or z<0

So thgis statement is not sufficient to find if x> y/z

Lets consider option 2

x^2 > y/z
Taking squareroots, we get
x > + sqrt (y/z) or
x > - sqrt (y/z)

We dont have sufficient information to find if x > y/z

Combining both statements doesnt help as well

IMO E

Hope this helps
--Arati

Hi Arati (or Aarti?),

We are not provided with options in DS, those are rather known as statements perhaps; and if x^2 > y/z, then

taking squareroots, we get

x > + sqrt (y/z) or

x < - sqrt (y/z)

...and why combining both statements doesnt help as well? Please throw some light