Is xy

Question

Is  xy<1 ?

(1)  x^2+y^2 < 1 
(2)  x + y < 1

MayankSingh · Answer

a) x^2 & y^2 are always positive
So for (x^2 + y^2 )<1 both x & y should be less than 1 and greater than -1 (-1<x,y<1)
Therefore, xy<1 (x<1 & y<1)
Sufficient.

b) x+y<1 
( Possible x=0.2,y=0.3: therefore xy<1) but (if x=-2 & y=-4 : xy>1) 
Not Sufficient.

Ans. A

chetan2u · Answer

Is  xy<1 ?

1) x^2+y^2 < 1 
Now  x^2+y^2-2xy=(x-y)^2............x^2+y^2=(x-y)^2+2xy ..
So  x^2+y^2<1  means  (x-y)^2+2xy<1 
Now here  (x-y)^2\geq{0}  so 2xy<1 or xy<1
Sufficient

2) x + y < 1 
Say x=-1 ans y=1.5, ans is yes
x=y=-2, ans is no
Insufficient

A

fskilnik · Answer

xy\,\,\mathop < \limits^? \,\,\,1 (1) Points (x,y) that satisfy statement (1) are precisely the points inside the circle with center at the origin and radius 1, therefore we have |x| < 1 AND |y| < 1. Conclusion: xy\,\, \leqslant \,\,\left| {xy} ight|\,\, = \,\,\left| x ight| \cdot \left| y ight|\,\, < \,\,1\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{ ext{YES}}} ight angle (2) Insufficient: > Take (x,y) = (0,0) to answer in the affirmative > Take (x,y) = (-1,-1) to answer in the negative The above follows the notations and rationale taught in the GMATH method.

MathRevolution · Answer

=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2):

\(x^2 + y^2 < 1\)
\(=> x^2< 1 - y^2 ≤ 1\) since \(y^2 ≥ 0\)
\(=> x^2< 1\)
\(=> -1 < x < 1\)
and
\(x^2 + y^2 < 1\)
\(=> y^2< 1 - x^2 ≤ 1\) since \(x^2 ≥ 0\)
\(=> y^2< 1\)
\(=> -1 < y < 1\)

Combining these two inequalities yields \(-1 < xy < 1\), so \(xy < 1\). Both conditions are sufficient, when taken together.

Since this question is an inequality question (one of the key question areas), CMT (Common Mistake Type) 4(A) of the VA (Variable Approach) method tells us that we should also check answers A and B.

Condition 1)

The argument showing that conditions 1 and 2 are sufficient, when taken together, only used condition 1. Therefore, condition 1 is sufficient.

Using the same argument as above,

\(x^2 + y^2 < 1\)
\(=> x^2< 1 - y^2 ≤ 1\) since \(y^2 ≥ 0\)
\(=> x^2< 1\)
\(=> -1 < x < 1\)
and
\(x^2 + y^2 < 1\)
\(=> y^2< 1 - x^2 ≤ 1\) since \(x^2 ≥ 0\)
\(=> y^2< 1\)
\(=> -1 < y < 1\).

So, \(-1 < xy < 1\), and condition 1) is sufficient.

Condition 2)

If \(x = \frac{1}{3}\) and \(y = \frac{1}{3,}\) then \(xy = \frac{1}{9} < 1\) and the answer is ‘yes’
If \(x = -2\) and \(y = -2\), then \(xy = 4 > 1\) and the answer is ‘no’
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.

Answer: A

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.

EgmatQuantExpert · Answer

Solution

Given  
 We are given no information in the question stem.

To find  
We need to determine
  Is xy < 1

Approach and Working out

Statement 1
   x^2 + y^2  < 1
  x^2, y^2  are both ≥ 0.
 So, we can say that if  x^2 + y^2  < 1, then both x2 and y2 are < 1.
 Hence, 0 ≤  x^2, y^2  < 1
  0 ≤  x^2  < 1
  -1 < x < 1 
 0 ≤  y^2  < 1
  -1 < y < 1
  
 Thus, xy < 1.
 
 Statement 1 is sufficient alone.

Statement 2
  x + y < 1
  We have no idea about the signs of x and y. There are multiple possibilities:
  x = ½, y = 1/3
  xy = 1/6 < 1
 Yes 
 x = -2, y = -3
  xy = 6 > 1
 No  
  
 Statement 2 is insufficient alone.

Thus, option A is the correct answer.

GMATGuruNY · Answer

Statement 1:
  x^2+y^2 = 1  is the equation for a circle centered at the origin with a radius of 1.
  x^2+y^2 < 1  is the equation for every point (x, y) INSIDE this circle.
Since every x value and every y value inside this circle is between -1 and 1, exclusive, the product of x and y will always be less than 1.
Thus, the answer to the question stem is YES.

Statement 2:
If x=0 and y=0. then xy = 0, so the answer to the question stem is YES.
If x=-1 and y=-1, then xy = 1, so the answer to the question stem is NO.
Since the answer is YES in the first case but NO in the second case, INSUFFICIENT.

A

Is xy < 1 (1) x^2 + y^2 < 1 (2) x + y < 1

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