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Is x > y?

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Is x > y?

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[Math Revolution GMAT math practice question]

Is x > y?

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

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

Is x > y?

1) x+y > 2
2) x^2 < 2y
$$x\,\,\mathop > \limits^? \,\,y$$

$$\left( 1 \right){\kern 1pt} \,\,x + y\,\, > \,\,2\,\,\,\,\,\left\{ \matrix{
\,\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {3,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

$$\left( 2 \right)\,\,{x^2} < 2y\,\,\,\left\{ \matrix{
\,\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{1 \over 2},{1 \over 4}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\,$$

$$\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{
\,\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,3} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {{3 \over 2},{4 \over 3}} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

The correct answer is (E).


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

Regards,
Fabio.

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Statement 1
$$x+y>2$$
$$x>2-y$$
$$x>2+\left(-y\right)$$
x is positive and y is negative ; x is definitely greater than y, hence Statement 1 is SUFFICIENT.

Statement 2
$$x^2<2y$$
$$square\ rooting\ both\ sides$$
$$x<\sqrt{2y}$$
$$x<\sqrt{2}\cdot\sqrt{y}$$
$$x^1=x\ and\ \ y^{\frac{1}{2}}=\sqrt{y}$$
$$x^1<\sqrt{2}\cdot y^{\frac{1}{2}}$$
Looking at their exponents 1 is greater than 1/2

For Statement 2
x is greater than y ; , hence statement 2 is SUFFICIENT
Both statements alone are SUFFICIENT..
$$answer\ is\ option\ D$$

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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)
If x = 4/3 and y = 1, then x + y > 2, x^2<2y and x > y. The answer is ‘yes’.
If x = 1 and y =3/2, then x + y > 2, x^2<2y and x < y. The answer is ‘no’.

Thus, both conditions together are not sufficient, since they do not yield a unique solution.

Therefore, the correct answer is E.
Answer: E

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.

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