MAGOOSH OFFICIAL SOLUTION:


Frequently Asked Questions

Q: For statement 1, why can't we multiply both sides by y to get x < 1/2y? Wouldn't that be sufficient on its own?

A: You have good intuition, and this might work for an equation, but there's a crucial difference between equations and inequalities: if you divide (or multiply) both sides by a negative number, then you have to change the direction of the inequality sign.

If I have -5x > 10

and divide both sides by -5, I need to flip the inequality so I have:

x < -2

Let's see why with actual numbers:

Equation:

8 = 8

divide both sides by - 2:

-4 = -4 <--- still true!

Inequality:

16 > 8

divide both sides by -2

-8 > -4 ....no! We changed the relationship when dividing by -2. So we need to flip the sign:

-8 < -4

Now this is correct.

In this problem, we have to ask "Is y negative?" If it isn't, then we can rework the equation as you did. If it is negative, then we have to flip the sign. But we don't know if it's negative or not! We can't multiply by y.

So, we cannot multiply across an inequality with variables unless we know for certain whether they're positive or negative.

Q: Can you please explain to me how I know that the √y² = y, is positive? Why can't it be positive and negative? For example, √4 is +2 or -2.

A: The radical sign (more commonly known as the square root sign) actually means the POSITIVE square root of a given number. This positive square root is often referred to as the principal square root. Anytime we see the radical sign, we want the principal (positive) square root.

Also, from Wikipedia

"Every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by [square root symbol], where [square root symbol] is called radical sign.

For example, the principal square root of 9 is 3, denoted √9 = 3, because 3² = 3 × 3 = 9 and 3 is non-negative."

So if x²= 4, then x = 2 or -2

But if √4 = x, then x = 2 only

√4 is the positive root of 4 only.

When you see a "√" sign with something inside, there is only one possible output.

So √(x^2) = |x|

Also check out this blog on square roots for more information.
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1) x/y < 1/2
At X=1,Y=3 above inequality holds but is x>y ? NO
At X=-1,Y=-4 above inequality holds but is x>y ? YES
NOT SUFFICIENT.

(2) \(\sqrt{x}= x\) and \(\sqrt{y} = y\)
As X and Y are integers, we can infer X=Y=1
SUFFICIENT.

ANSWER :B

Statement 1: x/y < 1/2
test numbers
If x=-3, y=2 No
if x=3 y=-2 yes
Insufficient

Statement 2:
x, y = 1
sufficient

Answer: B

Bunuel could you please explain this more briefly as why B is not the answer as the above fluff is bouncing all above the head.
if x^2=x and y^2=y this implies that x=1 and y=1. what am i missing here?????

Hi,

The way I approached is-

1) x/y< 1/2 => 2x < y

Now if x = -3 and y = -1 => 2x < y => -6 < -1. Then is x - y > 0 - NO.

if x = 1 and y = 3 => 2x < y => 2 < 3. Then is x - y > 0 - NO.

if x = -1/2 and y = -1/6 => 2x < y => -1 < -1/6. Then is x - y > 0 - NO.

if x = 1/6 and y = 1/2 => 2x < y => 1/3 < 1/2. Then is x - y > 0 - NO.

To simply put, shouldn't A be sufficient enough to say that if 2x < y then x - y will never be greater zero??

Please suggest.

Thanks it is clear now...



.

Any positive x and y satisfy the second statement, not only x = 1 and y = 1.
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\(\left( * \right)\,\,\,\,x,y\,\,\, \ne 0\,\,\,\,{\rm{ints}}\,\,\,\,\left( {xy \ne 0} \right)\,\,\,\,\)

\(x\,\,\mathop > \limits^? \,\,y\)

\(\left( 1 \right)\,\,{x \over y} < {1 \over 2}\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1, - 3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,\left\{ \matrix{
\,\left| x \right| = x\,\,\,\, \Rightarrow \,\,\,\,x \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,x > 0 \hfill \cr
\,\left| y \right| = y\,\,\,\, \Rightarrow \,\,\,\,y \ge 0\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,;\,\,\,\,\,\,\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1,2} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
\,\,{x \over y} < {1 \over 2}\,\,\,\,\,\mathop \Rightarrow \limits_{y\, > \,\,0}^{ \cdot \,\,2y} \,\,\,\,\,2x < y \hfill \cr
\,\,\,1 < 2\,\,\,\,\,\mathop \Rightarrow \limits_{\,x\, > \,\,0}^{ \cdot \,\,x} \,\,\,\,\,\,\,x < 2x \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,x < 2x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\)



This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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