II wrote:If xy is less than 3, is x less than 1

1) y is greater than 3
2) x is less than 3

thanks.

II, the Qs is saying -
xy <3> 3 let say y = 4 and xy = 2 so x = 1/2
Again, y = 3.1 and xy = 2.9 so x = 0.935
So x <1> SUFF
stmt - 2, x < 3. But x can be between 1 and 3 or x can be less than 1.
So it is INSUFF.

So IMO A.

Is there another approach to solving this, apart from plugging in numbers ?

Thanks.
II

You can break it out into equations.

Given
x*y < 3 Need to know if x < 1
Therefore... let's look at various possible y
Positive y: x < 3 / y
y = 0 : x can be anything
Negative y: x > 3 / y (remember to flip the signs)

Statement 1:
y > 3. This is the positive case above. Therefore:
x < 3 / y

For all y greater than 3, 3 / y is going to be less than 1. Therefore:
x < 1 SUFFICIENT

Statement 2:
x < 3
Obviously this practically tells you that you do not know whether it is less than 1 or not. All you have to do is a logic check to confirm that there is no other information that can block 1 < x < 3 from being a real answer.
Nothing to our knowledge. There is some y that can be multiplied by 1 < x < 3 and be less than 3.
Therefore INSUFFICIENT

So the answer is (A)

Is there another approach to solving this, apart from plugging in numbers ?

Stmt I

xy<3 (1)

y>3 i.e

3<y (2)

Add I and II

xy+3<3+y

xy-y<0
y(x-1) < 0

Either y<0 or x-1 < 0

Given y>3 so x-1<0 or x<1

SUFF


Another simple approach for Stmt I

xy<3
xy-3<0 (1)

3<y (2)

Add 1) and 2) (inqualities facing same direction)

xy-3+3<0+y
xy<y

Divide by y since we know y is positive(doesnt change the inequality )

x<1

SUFF






Stmt II

Pick numbers to prove insuff

A)

cramya wrote:
Another simple approach for Stmt I

xy<3
xy-3<0 (1)

3<y (2)

Add 1) and 2) (inqualities facing same direction)

xy-3+3<0+y
xy<y

Divide by y since we know y is positive(doesnt change the inequality )

x<1

SUFF )

Sexy..

x*y < 3

1: y is positive

so x < 3/y

so x < 1

Suff...

x*y < 3

1: y is positive

so x < 3/y

so x < 1

Suff...

Ronnie,
Good to see a nice approach for a change rather than a IMO.
Jus kidding!

I try not be blunt unlike my buddy Logitech!!!

Good luck with ur preps and keep up the good work!


Regards,
Cramya

X can be 2. x<3/y, and if y= 1.5, x=2.
So your condition doesn't work here.

man man wrote:X can be 2. x<3/y, and if y= 1.5, x=2.
So your condition doesn't work here.

In A, it is specifically mentioned that Y>3

so, Y canot be 1.5 for sure

xy < 3 Given

From 1 : y>3 , Let's say y=3 then to satisfy the above equation x<1 and since y>3 x has to be < 1. Sufficient

From 2 : x<3 doesn't mention anything about y and so x can take any value less than 3 : Insufficient

hence A. Loved the other approaches mentioned here.

- pradeep

II wrote:If xy is less than 3, is x less than 1

1) y is greater than 3
2) x is less than 3

We are given that the product of x and y is less than 3 and must determine whether x < 1.

Statement One Alone:

y is greater than 3.

We are given that y is greater than 3. If x is greater than or equal to 1, then xy will be greater than 3. Therefore, in order for xy to be less than 3, x must be less than 1. Statement one alone is sufficient to answer the question.

Statement Two Alone:

x is less than 3.

The information in statement two is not sufficient to answer the question. For instance, if x = 2, and y = 1, then xy = 2(1) = 2. However, if x = 1/2 and y = 4, then xy = (1/2)(4) = 2. In the former case, x > 1; in the latter case, x < 1. Statement two alone is not sufficient to answer the question.

Answer: A