DS question

IMO: (A)

On seeing first statement, we come to know that 3 units of x is proportional to 4 units of y and 5 units of z. So, clearly x is greater among x, y and z.

So, statement (1) is alone sufficient.

On rephrasing statement (2),

xyz > y^2
xz > y

The above condition satisfy at many values of x, y and z.

For example, both x and z could be negative and y may be positive but still the product of xz could be greater than y.

So, its ambiguous and can't deduce any specific answer.

Hence IMO: (A)

The answer should be C

1. x:y:z=3:4:5
This statement tell us about the magnitude of x, y and z but nothing about the sign. If all are positive numbers z is greatest, if all the negative z is smallest. INSUFFICIENT

2. xyz - y^2 is a positive integer
Do not oversimplify this! Y^2 is always positive. If we have (something) - positive = positive, then we know that something is positive and bigger in magnitude than what is subtracted. INSUFFICIENT

Combine the two, we know that z has the biggest magnitude and all the numbers are positive, thus Z is greatest.

I agree that I forget about the importance of sign. Definitely the answer is C.

simplyjat wrote:The answer should be C

If all are positive numbers z is greatest, if all the negative z is smallest. INSUFFICIENT

Though the real answer doesn't matter in the DS question, but still I would like to know that are you sure about your point mentioned above?

I think the x should be greatest if all the numbers are positive and z should be greatest if all the numbers are negative because 3 units of x are proportionate to 5 units of z.

Let me know your comment.

Yeah I am sure about what I mentioned in the previous post...

For example, lets take three numbers x=25 y=75 z=125.
x:y = 25:75 = 1:3
y:z = 75:125 = 3:5
x:z = 25:125 = 1:5
so we get x:y:z = 1:3:5 and we know that 125 (z) is the greatest and one having the highest proportion...

To remember this concept you have to look at proportions link this
x:y:z = 1:3:5 means that
x = 1*a
y = 3*a
z = 5*a
where a is a common term (25 in above example)

Thanks!!! I got it forever.

Had it been a ps problem the negativity factr could be avoided.
But not in a ds q.

So c is the answer.