If xy < 3, is x < 1? (1) y > 3 (2) x < 3

SOLUTION

If xy < 3, is x < 1?

(1) y > 3. If x were \(\geq{1}\), then xy would be more than or equal to 3, which would contradcit the condition that xy < 3. Thus x must be less than 1. Sufficient.

(2) x < 3. Not sufficient: consider x=y=0 and x=2 and y=0. Not sufficient.

Answer: A.

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Statement 1) y > 3 and if xy < 3, then X has to be < 1 (One of the variable has to compensate for the increase in other variable).
Sufficient.

Statement 2) x < 3
As there is no information of y, we cannot determine if x < 1
So insufficient.

Hence Option A)
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From S1:y>3=>x=fraction/negative integer=>x<1.Sufficient
From S2:x<3=>x=2,1,0...depending on y.It cannot be said whether x<1 or >1.Hence insufficient.
Ans.A

Statement (1). By number picking strategy, if y>3 and xy<3, then X is always less than 1. Sufficient

Statement (2). X can be greater or less than 1 and still xy<3 holds. Insufficient.

Answer A

If xy < 3, is x < 1?

(1) y > 3. If x were \(\geq{1}\), then xy would be more than or equal to 3, which would contradict the condition that xy < 3. Thus x must be less than 1. Sufficient.

(2) x < 3. Not sufficient: consider x=y=0 and x=2 and y=0. Not sufficient.

Answer: A.

I'm a bit confused after analyzing another similar question with different solution

In st1, why negative values of x are not being considered ? I came up with answer "E", considering negative values of x

However, just when I was convinced that, values of x and y need to be assumed positive in this type of prompt, I met the following question

Is xy < 6?

(1) x < 3 and y < 2.
(2) 1/2 < x < 2/3 and y^2 < 64.

Data Sufficiency, Question: 68, Page: 157, OG quant review 2nd edition

SOLUTION

Is xy < 6?

(1) x < 3 and y < 2 --> now, if both x and y are equal to zero then xy=0<6 and the answer will be YES but if both x and y are small enough negative numbers, for example -10 and -10 then xy=100>6 and the answer will be NO. Not sufficient.

(2) \frac{1}{2}<x<\frac{2}{3} and y^2<64, which is equivalent to -8<y<8 --> even if we take the boundary values of x and y to maximize their product we'll get: xy=\frac{2}{3}*8\approx{5.3}<6, so the answer to the question "is xy<6?" will always be YES. Sufficient.

Answer: B.

Here, both the positive and negative values of x and y have been considered.

Bunuel can you please explain the contradiction ?

If xy < 3, is x < 1?

(1) y > 3

Let y = 4.......4x <3.......x <3/4< 1....Yes

Let y = 100...100x <3....x <0.03 <1...Yes

Always Yes

Sufficient

(2) x < 3

Let x =1/3........1/3 y < 3 ......y < 8/3.......Answer is yes

Let x = 2...........2 y < 3........y < 2/3........Answer is NO

Insufficient

Answer: A

Hi,

here are my two cents for this questions

We should know two rules for this for one of the method to solve algebrically

1. if a>b & c<d then From one inequality it is possible to subtract term wise another inequality of the opposite sense retaining the sense of inequality from which the other was subtracted ..

Here is what i mean Say a>b & c<d then a-c > b-d.

2. 1. if a>b & c>d then Two inequality having same sense may be added term wise Here is what it will be a+c>b+d

So
From Stamt 1:
xy < 3 ---- (i)
and
y >3 -----(ii)

(i)- (ii)
Using rule 1 we have
xy-y<0
y(x-1)<0
so wither y<0 and or x-1<0
since y>3 it must be that x-1<0 or x<1
So sufficient.

From Stmt 2:
X<3-------(iii)

adding (i) and (iii)
xy+x<6
x(y+1)<6
so we have
x<6 and or y+1<6
or x<6 and or y<5
or we have
x<3 and or y<5
we can't say if x<1
Hence insufficient.

Probus
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