Is x

Question

Is  x < y ?

(1)  \sqrt{x} < y

(2)  x^2 < y

ccooley · Answer

Interesting problem! Since there are only two variables, and the math looks like it's simplified already, I'm going to start by testing cases. Statement 1 : If \sqrt{x} < y , we could have a very small x and a very large y: for instance, x = 1 and y = 100. In that case, x < y, so the answer to the question is yes . On the other hand, we could have a situation where \sqrt{x} is smaller than y, but x itself is actually bigger than y. You could do this by picking an x that's only a little bit bigger than y. Then, when you take the square root, it gets much smaller. For instance, x = 4 and y = 3. In that case, x > y, so the answer to the question is no . We got both 'yes' and 'no' answers to the question, so the statement is insufficient. Cross off A and D. Statement 2 : Our first case will work here as well. x = 1 and y = 100. So, x < y, so the answer to the question is yes . Now we need to think of a situation where x^2 is smaller than y, but where x is actually bigger than y. That means x would have to be bigger than x^2 . That only happens when x is a fraction. Let's say that x = 1/2, so x^2 = 1/4. Then we need to choose a value of y that's bigger than 1/4, but smaller than 1/2. Let's say x = 1/2 and y = 1/3. That fits the statement, and x > y, so the answer is no . We got both 'yes' and 'no' answers to the question, so the statement is insufficient. Cross off B. Both statements together : The problem here is that you can't get a 'no' answer if both statements are true. If statement 2 is true, the only way to get a 'no' answer is if x and y are fractions. If statement 1 is true, the only way to get a 'no' answer is if x is bigger than \sqrt{x} . That doesn't happen if x is a fraction. So, there's no way to get a 'no' answer from both statements at the same time. The answer must be 'yes'. So, the right answer is C .

hellosanthosh2k2 · Answer

Tricky question.

x < y ?

before we attack statements, we must recollect this.

if y is positive and less than 1, then \(\sqrt{y} > y > y^2 > y^3\) ......

if y is positive and greater than 1, then \(\sqrt{y} < y < y^2 < y^3\) .......

Statement 1:
\(\sqrt{x} < y\)

since sq root is always positive => y is positve, we can safely square both sides,
\(x < y^2\) => insufficient, as could be the case, \(x < y < y^2\) OR \(y < x < y^2\)

Statement 2:
\(x^2 < y\)

=> Insufficient, if x is less than 1, say x = 0.1, y = 0.02, \(x^2 = 0.01\) < 0.02 (y), but x > y
=> x is positive, since it is possible to take square root.
=> Also, y is positive, since is greater than squared value,
Taking square root on both sides of statement 2, \(|x| < \sqrt{y}\)

Statement 1 + 2:
x and y are both positive,
so from statement 1, \(x < y^2\) and from statement 2 : \(x < \sqrt{y}\)

if y < 1, then we have , \(\sqrt{y} > y > y^2 > y^3\) => \(x < y^2 < y < \sqrt{y}\), so x is less than y
if y > 1, then we have, \(\sqrt{y} < y < y^2 < y^3\) => \(x < \sqrt{y} < y < y^2\), so x is less than y

Sufficient => Answer (C)

PyjamaScientist · Answer

Only Statement 2 is sufficient.

Is  x   <  y ?

Statement 1:  \sqrt{x}  <  y 
For x = 2, and y = 1.8
 \sqrt{2}  < 1.8 ( \sqrt{2}  = 1.42)
But, x IS NOT < y.
For x = 2, and y = 2,
 \sqrt{2}  < 2
But, x IS NOT < y
For x = 2, and y = 3,
 \sqrt{2}  < 3
And, x IS < y.
Since there are varying results. This statement is not sufficient.

Statement 2:  x^2 < y 
Since  x^2  will always be a positive number, and no matter what "x" is (+ve or -ve or 0) it is < y (given). 
Thus, x will always be < y for x>1. (You can confirm this by imagining numbers  x ,  x^2  and  y  on the number line.
But, for 0<x<1, x can be 0.9, and y can be 0.85. Then, 0.9^2 (0.81) is less than 0.85, but 0.9 is greater than 0.85.
This statement is not sufficient.

Together, (1) & (2),
For  \sqrt{x}  <  y  and  x^2 < y  together, x has to be less than y.
So, together sufficient.

JerryAtDreamScore · Answer

Statement 1 Alone: We are also given y > 0 since y is greater than a square root. We can square both sides to get x < y^2 since both sides are positive. However, we may have x > y (e.g., y = 3 and x = 5 < 3^2 ) so statement 1 is insufficient. Statement 2 Alone: Again, we know that y must be positive since it is greater than a square. We may have x = 0.9 and y = 0.85 > 0.81 so x > y is still possible. Thus statement 2 is insufficient. Both Statements Combined: Combined we have \sqrt{x} < y and x^2 < y . We don't know who is bigger between \sqrt{x} and x^2 , but one of them must be bigger than x. Therefore we must have y > x, and thus by combining the statements it is sufficient. Answer: C

shubhi19 · Answer

we can determine by the number line method as between 0 and 1 as the power increases the value decreases and beyond 1 as the power increases the value of x will also increase. From statement 1 We can figured that between 0 and 1 x will be less than y but beyond 1 x can be greater than y , equal to y or less than y so, clearly not sufficient. From statement 2 between 0 and 1 x can be greater , equal or less than y but beyond 1 x is less than y , so not sufficient. On combining the statements x

bumpbot · Answer

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Is x < y? (1) x^(1/2) < y (2) x^2 < y

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